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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions

Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions

Electromagnetic Waves Class 12 Important Questions Very Short Answer Type

Question 1.
Name the part of the electromagnetic spectrum of wavelength 10-2 m and mention its one application. (Delhi 2008)
Answer:
Name of the part: Microwave
Applications :

  1. It is used in radar communication.
  2. It is used in microwave ovens.
  3. It is also used in analysis of fine details of molecular and atomic structure.

Question 2.
Write the following radiations in ascending order in respect of their frequencies ;
X-rays, Microwaves, UV rays and radiowaves. (Delhi 2009)
Answer:
Radiowaves, microwaves, UV-rays and X-rays.

Question 3.
Name the electromagnetic radiation to which waves of wavelength in the range of 10-2 m belong. Give one use of this part of EM spectrum. (Delhi 2009)
Answer:
Name : Microwave, Range 0.1 to 1 mm
Uses : Microwaves are used in aircraft navigation.

Question 4.
Name the part of electromagnetic spectrum which is suitable for

  1. radar systems used in aircraft navigation
  2. treatment of cancer tumours. (Delhi 2009)

Answer:

  1. Micro-waves
  2. Gamma-rays.

Question 5.
Name the EM waves used for studying crystal structure of solids. What is its frequency range? (All India 2009)
Answer:
X-rays frequency range : 1017 Hz to 1020 Hz

Question 6.
Which part of electromagnetic spectrum has largest penetrating power? (Delhi 2010)
Answer:
γ-rays are the electromagnetic waves of frequency range 3 × 1018 Hz to 5 × 1022 Hz and have the highest penetrating power.

Question 7.
Which part of electromagnetic spectrum is absorbed from sunlight by ozone layer? (Delhi 2010)
Answer:
Ultraviolet rays are absorbed from sunlight by ozone layers.

Question 8.
Which part of electromagnetic spectrum is used in radar systems? (Delhi 2010)
Answer:
Microwave region of electromagnetic spectrum is used in radar systems.

Question 9.
Name the part of electromagnetic spectrum whose wavelength lies in the range of 10-10 m. Give its one use. (All India 2010)
Answer:
Name : X-rays
Use : In medical diagnosis to look for broken bones; treatment study of crystal structure.

Question 10.
Which of the following has the shortest wavelength :
Microwaves, Ultraviolet rays, X-rays. (All India 2010)
Answer:
X-rays have the shortest wavelength.

Question 11.
Arrange the following in descending order of wavelength :
X-rays, Radio waves, Blue light, Infrared light. (All India 2010)
Answer:
Decreasing order ➝ Radio waves, Infrared light, Blue light, X-rays.

Question 12.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)
Answer:
The direction of electric field vector is along X-axis. Magnetic field vector is along Y-axis.

Question 13.
A plane electromagnetic wave travels in vacuum along x-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)
Answer:
The electric field and magnetic field vectors are in YZ-plane in the Y-direction and Z-direction respectively.

Question 14.
A plane electromagnetic wave travels in vacuum along y-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)
Answer:
The electric field and magnetic field vector are in ZX-plane in the X-direction and Z-direction respectively.

Question 15.
How are radio waves produced? (All India 2011)
Answer:
Radio waves are produced by the accelerated motion of charges in conducting wires.

Question 16.
How are X-rays produced? (All India 2011)
Answer:
X-rays are produced by sudden deceleration or acceleration of electrons in an X-ray tube.

Question 17.
How are microwaves produced? (All India 2011)
Answer:
Microwaves are produced by Klystron valve or magnetron valve.

Question 18.
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum. (Delhi 2012)
Answer:
Speed/Velocity of light remains the same.

Question 19.
What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves? (All India 2012)
Answer:
The oscillations of Image may be NSFW.
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\overrightarrow{\mathrm{E}}
and Image may be NSFW.
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\overrightarrow{\mathrm{B}}
fields are perpendicular to each other as well as to the direction of propagation of the wave.

Question 20.
The speed of an electromagnetic wave in a material medium is given by Image may be NSFW.
Clik here to view.
v=\frac{1}{\sqrt{\mu \varepsilon}}, \mu
the permeability of the medium and ε its permittivity. How does its frequency change? (All India 2012)
Answer:
Frequency remains unchanged.

Question 21.
A capacitor has been charged by a dc source. What are the magnitudes of conduction and displacement currents, when it is fully charged? (Delhi 2013)
Answer:
On full charging, the source will maintain the potential across the plates. The magnitudes of displacement current and conduction current will be zero.

Question 22.
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency. (All India 2013)
Answer:
The name of radiations is ultraviolet radiation. Its frequency range is 1015 to 1017 Hz.

Question 23.
To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019 Hz belong? (All India 2013)
Answer:
A wave of frequency 5 × 1019 Hz belongs to γ-rays region of electromagnetic spectrum.

Question 24.
To which part of the electromagnetic spectrum does a wave of frequency 3 × 1013 Hz belong? (All India 2014)
Answer:
Infra-red region of electromagnetic spectrum.

Question 25.
Why are microwaves considered suitable for radar systems used in aircraft navigation? (Delhi 2016)
Answer:
Due to their short wavelengths, microwaves are considered suitable for radar systems in aircraft navigation.

Question 26.
How is the speed of em-waves in vacuum determined by the electric and magnetic fields? (Delhi 2017)
Answer:
Speed of em-waves in vacuum is determined by the ratio of the peak values of electric and magnetic field vectors.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 1

Question 27.
Do electromagnetic waves carry energy and momentum? (All India 2017)
Answer:
Yes, they do, because of change of magnetic flux associated with circular loop.

Question 28.
Write the relation for the speed for electromagnetic waves in terms of the amplitudes of electric and magnetic fields. (All India 2017)
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 2

Question 29.
In which directions do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the x-axis? (All India 2017)
Answer:
Electric field (Image may be NSFW.
Clik here to view.
\overrightarrow{\mathrm{E}}
) oscillates along y-axis and magnetic field ( Image may be NSFW.
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\overrightarrow{\mathrm{B}}
) oscillates along z-axis;in an electromagnetic wave propagating along the x-axis.

Electromagnetic Waves Class 12 Important Questions Short Answer Type SA-I

Question 30.
The oscillating magnetic field in a plane electromagnetic wave is given by
By = (8 × 10-6) sin [2 × 10-11 t + 300 π x] T
(i) Calculate the wavelength of the electo-magnetic wave.
(ii) Write down the expression for the oscillating electric field. (Delhi 2008)
Answer:
Given: By = 8 × 10-6 sin [2 × 1011 t + 300 π x] T
(i) Standard equation is,
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 3

The oscillations of Image may be NSFW.
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\overrightarrow{\mathrm{E}}
and Image may be NSFW.
Clik here to view.
\overrightarrow{\mathrm{B}}
fields are perpendicular to each other as well as to the direction of propagation of the wave. So we take electric field in z-direction because oscillating magnetic field is in y-di recti on and propagation of the wave is in x-direction.

Question 31.
The oscillating electric field of an electromagnetic wave is given by :
E = 30 sin [2 × 1011 t + 300 π x] Vm-1
(a) Obtain the value of the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field. (Delhi 2008)
Answer:
(a) We compare the given expression with
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 4

Question 32.
How does a charge q oscillating at certain frequency produce electromagnetic waves? Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave propagating along the Z-direction. (Delhi 2009)
Answer:
As the charge q moves accelerating, the electric field and magnetic field produced will change the space and time E and B varying with time produced the other field B and E respectively and sustain the E.M. pattern.

This is from the interpretation of Maxwell supported by
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 5

Question 33.
Arrange the following electromagnetic radiations in ascending order of their frequencies:
(i) Microwave
(ii) Radiowave
(iii) X-rays
(iv) Gamma rays
Write two uses of any one of these. (Delhi 2009)
Answer:
In ascending order of their frequencies :
Radiowave < Microwave < X-rays < Gamma rays.
Two uses of microwaves are :
1. In microwave ovens.
2. In aircraft navigation.

Question 34.
Draw a sketch of a plane electromagnetic wave propagating along the z-direction. Depict clearly the directions of electric and magnetic fields varying sinusoidally with z. (All India 2009)
Answer:
Sketch of a plane electromagnetic wave propagating along the z-direction with oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction.

Question 35.
How are infrared waves produced? Why are these referred to as ‘heat waves’? Write their one important use. (Delhi 2009)
Answer:
Infrared rays are produced by hot bodies and molecules. This may involve vibration and bending of molecules. Infrared band lies adjacent to low-frequency or long-wavelength end of the visible spectrum. Infrared waves are sometimes referred to as heat waves.

Use: Infrared rays are used to take photographs in darkness. These are also used to study secret writing. They are also used in physical therapy.

Question 36.
A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. (All India 2011)
Answer:
Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field B is related to steady current I as
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 6

Maxwell showed that this relation is logically in-consistent. He accounted this inconsistency as follows :
Ampere’s circuital law for loop C1 gives
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 7

which is logically inconsistent. So, Maxwell gave idea of displacement current.

Thus displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 8

It is now called Ampere-Maxwell law. This is the generalization of Ampere’s circuital law.

Question 37.
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current? (Delhi 2012)
Answer:
A dc battery connected to an ideal capacitor ‘ provides only a momentarily charge, whereas an ac battery allows a continuous flow of current
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 9

As charge on capacitor plates changes, electric field associated with that also changes and hence giving rise to a displacement current according to
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 10

Question 38.
A capacitor of capacitance ‘C is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor. (All India 2012)
Answer:
Ammeter will definitely show a momentary deflection, which is due to the flow of electron produced in the charging process. As the capacitor plates get charging, the displacement current start flowing in the gap and thus shows a continuity of current.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 11

Question 39.
(a) An em wave is travelling in a medium with a velocity Image may be NSFW.
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\overrightarrow{\mathbf{v}}=\mathbf{v} \hat{i}
. Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.
(b) How are the magnitudes of the electric and magnetic fields related to the velocity of the em wave?
Answer:
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 12

Question 40.
A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor. (All India 2012)
Answer:
The displacement current arises due to varying electric field
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 13

If q be instantaneous charge, then E is electric field between the plates of capacitor at that time and A is area of plate; then
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 14

Question 41.
(a) How are electromagnetic waves produced?
(b) How do you convince yourself that electromagnetic waves carry energy and momentum? (Comptt. Delhi 2012)
Answer:
(a) Electromagnetic Waves : Accelerating electric charge produces electromagnetic waves.

(b) Einstein’s explanation of photoelectric effect led de Broglie to the wave-particle duality, i.e., matter exhibits wave as well as particle properties. Electromagnetic waves are characterised by wave properties, such as periodicity in space-time, wavelength, amplitude, frequency, wave velocity etc. It transports energy but no matter.

The term wave-particle duality refers to the behaviour where both wave-like and particle-like properties are exhibited under different conditions by the same entity. Hence electromagnetic waves show particle properties such as definite position, size, mass, velocity, momentum, energy etc.
For a photon of momentum (p), an associated wavelength is given by Image may be NSFW.
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\lambda=\frac{h}{p}
.

Question 42.
(a) Arrange the following electromagnetic waves in the descending order of their wavelengths :

  1. Microwaves
  2. Infra-red rays
  3. Ultra-violet radiation
  4. Gamma rays

(b) Write one use each of any two of them. (Comptt. Delhi 2013)
Answer:
(a) Arrangement:

  1. Microwaves
  2. Infra-red rays
  3. Ultra-violet radiation
  4. Gamma rays

(b) Uses :

  1. Microwaves are used in radar system.
  2. Infra-red rays are used for protecting dehydrated fruits.
  3. Ultra-violet rays are used in the study of molecular structure.
  4. Gamma rays are used to kill micro-organisms in food industry.

Question 43.
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s circuital law to include the term due to displacement current. (All India 2014)
Answer:

Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field B is related to steady current I as
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 6

Maxwell showed that this relation is logically in-consistent. He accounted this inconsistency as follows :
Ampere’s circuital law for loop C1 gives
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 7

which is logically inconsistent. So, Maxwell gave idea of displacement current.

Thus displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 8

It is now called Ampere-Maxwell law. This is the generalization of Ampere’s circuital law.

Question 44.
A capacitor is connected in series to an ammeter across a d.c. source. Why does the ammeter show a momentary deflection during the charging of the capacitor? What would be the deflection when it is fully charged? (Comptt. All India 2014)
Answer:
The momentary deflection is due to the transient current flowing through the circuit when the capacitor is getting charged.
The deflection would be zero when the capacitor gets fully charged.

Question 45.
Name the types of e.m. radiations which

  1. are used in destroying cancer cells,
  2. cause tanning of the skin and
  3. maintain the earth’s warmth.

Write briefly a method of producing any one of these waves. (Delhi 2015)
Answer:

  1. γ-rays
  2. Ultraviolet rays
  3. Infrared rays

Mode of production

  1. γ-rays are produced by radioactive decay of nucleus.
  2. Ultraviolet rays are produced when inner shell electrons in atoms move from one energy level to an other energy level.
  3. Infrared rays are produced due to vibration of atoms and molecules.

Question 46.
For a plane electromagnetic wave, propagating along the Z-axis, write the two (possible) pairs of expression for its oscillating electric and magnetic fields. How are the peak values of these (oscillating) fields related to each other? (Comptt. All India 2016)
Answer:
For the e.m. wave, propagating along the z-axis, we have
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 15

The two possible forms for electric and magnetic fields are :
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 16

The peak values of these two fields are related by
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 17

Question 47.
An e.m. wave, Y1, has a wavelength of 1 cm while another e.m. wave, Y2, has a frequency of 1015 Hz. Name these two types of waves and write one useful application for each. (Comptt. All India 2014)
Answer:
(i) Y1 ➝ Microwaves
Applications : Microwaves are used in Microwave ovens, Aircraft Navigators etc.

(ii) Y2 ➝ Ultraviolet waves
Applications : Ultraviolet rays are used in sterilizing surgical instruments, food preservation etc.

Question 48.
How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. (Delhi 2017)
Answer:
During charging, the electric flux between the plates of a capacitor keeps on changing; this results in the production of a displacement current between the plates.

Question 49.
Identify the electromagnetic waves whose wavelengths vary as
(a) 10-12 < λ < 10-8 m
(b) 10-3 m < X < 10-1 m
Write one use for each. (All India 2017)
Answer:
(a) X-rays—Used in medical science for the purpose of detection of fractures, stones in gall bladder, stones in kidney etc.

(b) Microwaves—Used in radar systems for aircraft navigation.

Question 50.
Identify the electromagnetic waves whose wavelengths lie in the range
(a) 10-11 m < λ < 10-8 m
(b) 10-4 m < λ < 10-6 m Write one use of each. (All India 2017)
Answer:
(a) Uses of X-Rays and Gamma rays :
X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Gamma rays are used in medicine to destroy cancer cells.

(b) Uses of Infrared, visible and microwaves :

  • Infrared waves are widely used in remote switches of household electronic systems such as remotes for TVs, video recorders etc.
  • Visible rays provide us information about the world.
  • Microwaves are used in the radar systems in aircraft navigation.

Question 51.
How is electromagnetic wave produced? Draw a sketch of a plane e.m. wave propagating along X-axis depicting the directions of the oscillating electric and magnetic fields. (Comptt. Delhi 2017)
Answer:
Electromagnetic waves are produced due to oscillating/accelerating charged particles.
Sketch of e.m. wave : Refer to Q. 39 (a), Page 171

Electromagnetic Waves Class 12 Important Questions Short Answer Type SA-II

Question 52.
Identify the following electromagnetic radiations as per the wavelengths given below. Write one application of each. (All India 2008)
(a) 10-3 nm
(b) 10-3 m
(c) 1 nm
Answer:
(a) 10-3 nm : γ-rays
Application :

  1. γ-rays are used in the treatment of cancer and tumour.
  2. γ-rays are used in radiation therapy. (any one)

(b) 10-3m : Microwave
Application : Microwaves are used in Radar systems for aircraft navigation.

(c) 1 nm : X-rays Application :

  1. Infra-red waves are used for taking photographs during the conditions of fog, smoke etc.
  2. These are also used as a diagonostic tool for the detection of fractures, (any one)

Question 53.
Identify the following electromagnetic radiations as per the wavelengths given below. Write one application of each.
(a) 1 mm
(b) 10-12 m
(c) 10-8 m (All India 2008)
Answer:
(a) 1 mm : Microwaves
Application : In aircraft navigation for the radar system. Also used in microwave ovens.

(b) 10-12 m : Gamma rays
Application : Gamma rays are used as medicine to destroy cancer cells

(c) 10-8 m : Ultraviolet rays
Application : Ultraviolet rays are used in LASIK eye surgery.

Question 54.
(a) How does an oscillating charge produce electromagnetic wave? Explain.
(b) Draw a sketch showing the propagation of a plane em wave along the Z-direction, clearly depicting the directions of oscillating electric and magnetic field vectors. (Comptt. Delhi 2012)
Answer:
(a) Consider a charge oscillating with same frequency. This produces an oscillating electric field in space, which produces an oscillating magnetic field which in turn is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other, as the waves propagate through the space. The frequency of the electromagnetic wave naturally equals the frequency of the oscillation of the charge.

(b)
Sketch of a plane electromagnetic wave propagating along the z-direction with oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction.

Question 55.
When an ac source is connected across a capacitor, current starts flowing through the circuit. Show how Ampere’s circuital law is generalized to explain the flow of current through the capacitor. Hence obtain the expression for the displacement current inside the capacitor. (Comptt. All India 2012)
Answer:
When an ‘ac’ source is connected across a capacitor, the charge on the capacitor also becomes time dependent. It gives rise to a time dependent electric field between the plates of capacitor. As a result the electric flux changes.

It was suggested that we need to regard this changing electric flux, between the plates of capacitor, as equivalent to a current which is called the displacement current.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 18

which is generalised form of an Ampere’s circuital law. .

Question 56.
(a) When the oscillating electric and magnetic fields are along the x- and indirection respectively

  1. point out the direction of propagation . of electromagnetic wave.
  2. express the velocity of propagation in terms of the amplitudes of the oscillating electric and magnetic fields.

(b) How do you show that the em wave carries energy and momentum?(Comptt. All India)
Answer:
(a)

  1. Along z-direction.
  2. Velocity of propogation will be, Image may be NSFW.
    Clik here to view.
    \mathbf{C}=\frac{\mathbf{E}_{0}}{\mathbf{B}_{0}}

(b) Photoelectric effect shows the particle nature of electromagnetic waves. As such the photons carry energy and momentum. The energy is given by
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 19

Question 57.
Answer the following :
(a) Name the em waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.
(b) If the. earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(c) An em wave exerts pressure on the surface on which it is incident. Justify. (Delhi 2014)
Answer:
(a) Microwaves are used in radar systems. Its frequency range : 1010 to 1012 Hz

(b) In the absence of earth’s atmosphere, there would have no ozone layer to prevent ultraviolet radiations reaching the earth, the temperature on earth’s surface would have been lower due to green house effect, making it difficult for human survival.

(c) Since em wave carries both energy and momentum, hence exerts pressure on the surface on which it is incident.

An em wave exerts negligibly very small pressure on the surface on which it is incident.

It is due to the fact that momentum of the photon is extremely small, which can be
calculated by de-Broglie relation Image may be NSFW.
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\left(\lambda=\frac{h}{p}\right)

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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 20

Question 58.
Answer the following :
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
(b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?
(c) Why is the amount of the momentum transferred by the em waves incident on the surfrace so small? (Delhi 2014)
Answer:
(a) Gamma (γ) rays are used for the treatment of certain forms of cancer. Their frequency range is 1018 Hz to 1022 Hz.

(b) The thin ozone layer on top of stratosphere absorbs most of the harmful ultraviolet rays coming from the Sun towards the Earth. They include UVA, UVB and UVC radiations, which can destroy the life system on the Earth.
Hence, this layer is crucial for human survival.

(c) Thus, the amount of the momentum transferred by the em waves incident on the surface is very small, because of small value of planks constant. For example, an electromagnetic wave of wavelength 1.00 nm will provide momentum (p) according to de-Broglie’s relation,
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 21

It is extremely small value of the momentum.

Question 59.
Answer the following questions :
(a) Name the em waves which are produced during radioactive decay of a nucleus. Write their frequency range.
(b) Welders wear special glass goggles while working. Why? Explain.
(c) Why are infrared waves often called as heat waves? Give their one application. (Delhi 2014)
Answer:
(a) γ-rays; Frequency range : 1018 Hz to 1022 Hz

(b) Because to protect eyes from intense ultra-violet radiations produced during welding; and also to protect from glare and flying sparks.

(c) Because infrared waves are em waves of higher wavelength (less frequency) and are produced by highly vibrating molecules of hot bodies.
Applications :

  1. used in the remote switches of household electronic systems.
  2. used for protecting dehydrated fruits.
  3. used in solar water heaters and cookers. (Any one)

Question 60.
Answer the following questions:
(i) Show, by giving a simple example, how em waves carry energy and momentum.
(i) How are microwaves produced? Why is it necessary in microwave ovens to select the frequency of microwaves to match the resonant frequency of water molecules?
(iii) Write two important uses of infrared waves. (Comptt. Delhi 2014)
Answer:
(i) Consider a plane perpendicular to the direction of propagation of the wave. An electric charge, on the plane, will be set in motion by the electric and magnetic fields of em wave, incident on this plane. This illustrates that em waves carry energy and momentum.

(ii) Microwaves are produced by special vacuum tubes like the Klystron/Magnetron/Gunn diode.
In microwave ovens, the frequency of microwaves is selected to match the resonant frequency of water molecules, so that energy is transferred efficiently to the kinetic energy of the molecules.

(iii) Important uses of infra-red waves :
1. These are associated with the green house effect.
2. These are used in remote switches of household electrical appliances.

Question 61.
(a) A capacitor is connected in series to an ammeter across a d.c. source. Why does the ammeter show a momentary deflection during the charging of the capacitor? What would be the deflection when it is fully charged?
(b) How is the generalized form of Ampere’s circuital law obtained to include the term due to displacement current? (Comptt. All India 2014)
Answer:
(a) The momentary deflection is due to the transient current flowing through the circuit when the capacitor is getting charged.
The deflection would be zero when the capacitor gets fully charged.

(b)
Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field B is related to steady current I as
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 6

Maxwell showed that this relation is logically in-consistent. He accounted this inconsistency as follows :
Ampere’s circuital law for loop C1 gives
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 7

which is logically inconsistent. So, Maxwell gave idea of displacement current.

Thus displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 8

It is now called Ampere-Maxwell law. This is the generalization of Ampere’s circuital law

Question 62.
Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation.
(b) used to treat muscular strain.
(c) used as a diagnostic tool in medicine.
Write in brief, how these waves can be produced. (Delhi 2014)
Answer:
(a) Microwaves
Production : Klystron/magnetron

(b) Infrared Radiations
Production ; Hot bodies/vibrations of atoms and molecules.

(c) X-Rays
Production : Bombarding high energy electrons on a metal target.

Question 63.
Write the expression for the generalized form of Ampere’s circuital law. Discuss its significance and describe briefly how the concept of displacement current is explained through charging/discharging of a capacitor in an electric circuit. (All India 2014)
Answer:
Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field B is related to steady current I as
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 22

Maxwell showed that this relation is logically inconsistent. He accounted this inconsistency as follows :
Ampere’s circuital law for loop C1 gives
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 23

which is logically inconsistent. So, Maxwell gave idea of displacement current.
Thus displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 24

It is now called Ampere-Maxwell law. This is the generalization of Ampere’s Circuital law.

Question 64.
How are em waves produced by oscillating charges?
Draw a sketch of linearly polarized em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields. (Delhi 2016)
Answer:
em waves by oscillating charges.

(a) Consider a charge oscillating with same frequency. This produces an oscillating electric field in space, which produces an oscillating magnetic field which in turn is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other, as the waves propagate through the space. The frequency of the electromagnetic wave naturally equals the frequency of the oscillation of the charge.

(b)
Sketch of a plane electromagnetic wave propagating along the z-direction with oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction.

Sketch of em waves.
Sketch of a plane electromagnetic wave propagating along the z-direction with oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction.

Question 65.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 25

where ϕ E is electric flux produced during charging of the capacitor plates. (Delhi 2016)
Answer:

Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field B is related to steady current I as
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 6

Maxwell showed that this relation is logically in-consistent. He accounted this inconsistency as follows :
Ampere’s circuital law for loop C1 gives
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 7

which is logically inconsistent. So, Maxwell gave idea of displacement current.

Thus displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Image may be NSFW.
Clik here to view.
Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 8

It is now called Ampere-Maxwell law. This is the generalization of Ampere’s circuital law.

Question 66.
(i) Identify the part of the electromagnetic spectrum which is :
(a) suitable for radar system used in aircraft navigation,
(b) produced by bombarding a metal target by high speed electrons.
(ii) Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor? Write the necessary expression to explain this observation. (All India 2016)
Answer:
(i)
(a) Microwaves
(b) X-rays

(ii) The total current
(i) is the sum of conduction current
(ic) and displacement current (id), so we have
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Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions 26

This means that outside the capacitor plates in connecting wires, we have only conduction current ic = i and no displacement current (id = 0). On the other hand, inside the capacitor, there is no conduction current (ic = 0) and there is only displacement current hence i = id.
It is why there is momentary deflection in the galvanometer at the time of charging or discharging a capacitor.

Question 67.
Name the e.m. waves in the wavelength range 10 nm to 10-3 nm. How are these waves generated? Write their two uses. (Comptt. All India 2017)
Answer:

  • e.m. waves in the wavelength range 10 nm to 10-3 nm are X-rays.
  • X-rays are generated by bombarding a metal target with high energy electrons.

• Uses :

  1. Diagnosis of bone fractures.
  2. Treatment of some forms of cancer.

Question 68.
Name the type of e.m. waves having a wavelength range of 0.1 m to 1 mm. How are these waves generated? Write their two uses. (Comptt. All India 2017)
Answer:

  1. e.m. waves having a wavelength range 0.1 m to 1 mm are MICROWAVES.
  2. Microwaves are generated by special vacuum tubes such as klystron, magnetron and gunn diodes.
  3. Microwaves are used in :
    • Radar system in aircraft navigation
    • Ovens for heating and cooking.

Question 69.
Name the type of e.m. waves having a wavelength range 10-7 m to 10-9 m. How are these waves generated? Write their two uses. (Comptt. All India 2017)
Answer:

  1. e.m. waves having a wavelength range 10-7 m to 10-9 m are ultra violet rays.
  2. Sun is an important source of UV rays. Some special lamps and very hot bodies also produce UV rays.

• Uses :

  • UV rays are used in lasik eye surgery.
  • UV lamps are being used to kill germs in water purifiers.

Important Questions for Class 12 Physics

The post Important Questions for Class 12 Physics Chapter 8 Electromagnetic Waves Class 12 Important Questions appeared first on Learn CBSE.


NTSE Odisha 2019-20 for Class X | Exam Dates, Application, Pattern and Answer Key

NTSE Odisha 2019-20: State Council of Educational Research and Training (SCERT) will conduct NTSE Odisha Stage 1 on 3rd November. The application form will be released on 1st August 2019. Students studying in class X can apply for NTSE Odisha till 5th September 2019. Candidates who want to appear for the NTSE Odisha can download their admit card on 20th October 2019. Only shortlisted candidates in the NTSE Odisha stage1 are invited to appear for the stage 2 exam to be conducted on May 2020. The final merit list will be prepared and published on the official website depends on the candidate’s entire performance.

NTSE is conducted every year to award scholarships to deserving candidates till the doctorate level of their studies. Candidates should read this article till the end to know more information about NTSE Odisha 2019-20. Information such as exam dates, application, admit card, eligibility and exam pattern, etc.

NTSE Odisha Exam Dates

EventsDates
Releasing of Application formAugust 1, 2019
Closing of ApplicationSeptember 5, 2019
Validation of Application by District Education OfficerSeptember 6 to September 25, 2019
Admit Card for Stage 1October 20, 2019
NTSE Odisha Stage 1November 3, 2019
NTSE Odisha Answer KeyNovember 2019
NTSE Odisha Stage 1 ResultLast Week of January 2020
NTSE Admit Card for Stage 2April 2020
NTSE Stage 2May 10, 2020
NTSE Stage 2 ResultSeptember 2020

NTSE Odisha Eligibility Criteria

The following are the eligibility criteria to appear for NTSE Odisha 2019-20.

  • Candidates who are studying in class X in the academic year of 2019-20.
  • Age limit for NTSE Odisha Stage 1 is less than 18 years as on 1st July 2019.
  • Candidates belonging to the Reserved category must secure at least 55% marks in Class IX.
  • Candidates belonging to the general category must secure at least 60% marks in class IX.
  • Candidates from any recognized schools of Odisha are eligible.
  • Candidates of open schooling or distance learning are also eligible.

NTSE Odisha Application Form

NTSE Odisha application form will be released online on 1st August 2019. The closing date of NTSE Odisha online application form is 5th September 2019. Candidates can check the below-mentioned steps to apply for the NTSE Odisha exam.

  • Candidates should visit the official website scertodisha.nic.in.
  • Click on the online application link to enter the necessary details.
  • Download and take a print of your application form.
  • All the necessary details should be filled properly. It should also be rechecked thoroughly to avoid rejection of the application.
  • Submit the filled application form at the Principal’s office of the respective school.
  • After that, it will be submitted to the state liaison’s officer for NTSE Odisha Stage 1.
  • Reserved candidates should submit the Caste Certificate issued by the Tahsildar if required.
  • Physically Challenged candidates should submit the Disability Certificate issued by the District Medical Board if required.
  • Candidates of open schooling and distance learning are advised to submit their applications directly to the state liaison officer.
  • Candidates should save their application number for downloading the admit card.

NTSE Odisha Application Fee

  • Applicants no need to pay any application fee for NTSE Odisha Stage 1 exam.

NTSE Odisha Admit Card

Candidates can download their admit card from the official website of SCERT, Odisha on October 20, 2019. The following are the important points to download the admit card.

  • Go to the NTSE Odisha official website scertodisha.nic.in.
  • Click on the print hall ticket link.
  • Candidate should fill the application number and date of birth and click enter.
  • After that download and take a hard copy of the hall ticket.

The NTSE Odisha admit card contains candidate’s name, roll number, exam center details. Candidates must carry their admit card in the examination hall, otherwise, they will not be allowed to sit in the exam.

NTSE Odisha Question Paper Pattern

The NTSE Odisha stage 1 exam contains two parts i.e, Scholastic Ability Test (SAT) and Mental Ability Test (MAT). MAT and SAT contains 100 questions each. Overall 200 questions will be asked in the exam. The below-given table shows the details of exam pattern for NTSE Odisha.

PapersNumber of questionsTime Allotted
Scholastic Ability Test (SAT)100120 minutes
Mental Ability Test (MAT)100120 minutes
  • Each right answer carries 1 mark. There is no negative marking for erroneous answers.
  • SAT questions will be asked from math, science, social science, physics, chemistry, biology, history, political science, economics, etc.
  •  SAT contains a total of 100 questions with the time duration of 120 minutes.
  • The questions will be asked in the exam is equivalent to the syllabus of class IX and X.
  • MAT questions are based on logical reasoning, coding-decoding, series, analogies, hidden figures, problem-solving, and pattern perception.
  • MAT contains a total of 100 questions with the time duration of 120 minutes.

NTSE Odisha Answer Key

The answer key for the NTSE Odisha will be released on the official website of SCERT by the exam conducting authority. The official answer key will be published after one month of the exam. The answer key contains the correct answers of each question asked in MAT and SAT. Candidates from Odisha can download the answer keys to estimate their scores and guess the chances of qualifying for NTSE Stage 2. The following points are given to check the NTSE Odisha Answer key:

  1. Go to the official website of SCERT, Odisha.
  2. Click on the answer key, download and save it.
  3. The expected marks can be compared with the previous year’s cut-off.
  4. The answer keys help the candidates to determine their performance in Stage 1 and chances of qualifying for NTSE stage 2.
  5. Candidates should follow the right approach for marking. The answer key can be amended based on particular proof.
  6. Finally, the amended answer key will be published on the official website.

NTSE Odisha Exam Result

NTSE Odisha exam result will be released by SCERT on the official website in the last week of January 2020. SCERT will announce the merit list of shortlisted candidates for NTSE stage 2 exam which will be held on May 10, 2020. The following are the steps to check the NTSE Odisha exam results:

  • Go to the official website of SCERT, Odisha to check the exam result.
  • Click on the result link which will be shown on the homepage of the portal.
  • Fill the roll number and date of birth to download the result list.
  • Keep the printed copies of the result list for future reference.

Details Mentioned on NTSE Odisha Exam Result

Candidates can check the NTSE Odisha exam result on the official website scertodisha.nic.in. The NTSE Odisha merit list contains the following information of the selected candidates.

  • Candidates School Address, Name, and Roll number
  • Candidates gender, caste, date of birth and disability status
  • Candidates Rank and Total marks
  • Candidates scores in MAT and SAT

NTSE Odisha Stage 1 Qualifying Scores

The minimum scores obtained by a candidate to qualify for the NTSE stage 2 exam is the qualifying scores. NTSE Odisha qualifying scores will be declared along with the result by SCERT in the last week of January 2020. The qualifying scores depends on various factors as given below:

  • Maximum marks obtained in the exam
  • Minimum marks obtained in the exam
  • Number of candidates belongs to each category

The following table shows the cut-off marks for NTSE Odisha Stage 1 category candidates

CategoryCut-Off
SC113
ST108
OBC118
General152

The following table shows the selection process based on the qualifying percentage for NTSE Odisha category candidates

PapersGeneral CategoryReserved Categories (ST, SC, and PH)
MAT40%32%
SAT40%32%

NTSE Odisha Stage 1 Reservation Criteria

Total 259 scholarships are reserved for Odisha state which will be awarded as per the reservation percentages are given below:

CategoryReservation Criteria
SC15%
ST7.5%
PH4%
OBC27%
Economically Weaker Section (EWS)10%

NTSE Odisha Stage 1 Syllabus

There is no particular syllabus for the NTSE Odisha Stage 1. So, candidates have to prepare from class IX and X syllabus with the aim of getting qualified in this exam. The below-given table shows some important topics for NTSE Odisha 2019-20.

PapersSubjectsTopics
MATReasoningCoding-Decoding, Distance and Direction, Word Problems, Calendar, Time and Clock, Mirror and Water Images, Blood Relations, Ranking, and Arrangements.
SATMathsQuadratic Equation, Surface Area and Volume, Algebraic Expression, Arithmetic Progression, Square Roots and Cube Roots, Percentage, Simple and Compound interest.
ScienceMetals and non-metals, Acid, Bases and Salt, Human Body, Food Production and Management, Motion and Force, Source of Energy, Fibers and Plastics, Structure of an Atom and Measurements.
Social SciencesIndustrial Revolution, Early Medieval Period, Resources and Development, New Empires and Kingdoms, The Mughal Empire, Delhi Sultanate, British Raj, Indian Constitution, Democracy and Elections, Atmosphere, The Judiciary, Map and Globe, Natural Vegetation and Solar System.

NTSE Scholarship Amount 2019-20

Candidates who qualified in both the stages of NTSE with minimum cut-off marks will get the amount of the scholarships every month.

Education LevelScholarship Amount
Class XI to XIIRs. 1250/-
UndergraduateRs. 2000/-
PostgraduateRs. 2000/-
PhDAccording to the UGC norms

Hope this article will help you to get more information about NTSE Odisha 2019-20. Don’t forget to add your comment for any information/queries.

The post NTSE Odisha 2019-20 for Class X | Exam Dates, Application, Pattern and Answer Key appeared first on Learn CBSE.

CBSE Class 12 Sanskrit व्याकरणम् अन्विति-प्रकरण

CBSE Class 12 Sanskrit व्याकरणम् अन्विति-प्रकरण

अन्विति-प्रकरण
(कर्ता-क्रिया-अन्वितिः/विशेषण-विशेष्य-अन्वितिः)

(क) कर्ता-क्रिया-अन्वितिः
क्रिया में वही वचन, पुरुष होता है जो कर्त्ता में होता है।

(क) यदि कर्ता एकवचन में हो तो क्रिया भी एकवचन में होगी; यथा

1. देवः याति।
2. फलं पतति।
3. रमा हसति।
तीनों उदाहरणों में यद्यपि कर्ता का लिङ्ग भिन्न है किन्तु कर्ता एकवचन में है अतः क्रियापद भी एकवचन का है।
4. अहं लिखामि।
5. त्वं पठसि।
यहाँ कर्त्ता यद्यपि भिन्न पुरुषों में है किन्तु वचन की दृष्टि से वह एकवचन में है अतः क्रिया भी एकवचन की ही प्रयुक्त हुई है।

(ख) कर्ता यदि द्विवचन है तो क्रिया भी द्विवचन में होगी; जैसे

  1. पितरौ वदतः।
  2. दम्पती गच्छतः।
  3. भवन्तौ आगच्छतः।
  4. युवां नृत्यथः।
  5. आवां कथयावः।

(ग) कर्ता यदि बहुवचन का है तो क्रियापद भी बहुवचन में होगा; जैसे –

  1. रश्मयः जले पतन्ति।
  2. सर्वे आकाशं पश्यन्ति।
  3. मित्राणि उच्चैः हसन्ति।
  4. यूयं फलानि खादथ।
  5. वयं गर्जनां करिष्यामः।

यहाँ सारे कर्तृपद बहुवचन में हैं तथा सारे क्रियापद भी बहुवचन में हैं।

(घ) कर्त्ता में जो पुरुष होता है क्रियापद में भी वही पुरुष होता है; जैसे –

  1. ताः रामायणं शृण्वन्ति।
  2. आवां रामायणीं कथां कुर्वः।
  3. यूयं देवान् पूजयत।

इनमें प्रथम वाक्य में कर्तृपद व क्रियापद में प्रथमपुरुष, द्वितीय वाक्य में कर्तृपद व क्रियापद में उत्तम पुरुष तथा अन्तिम वाक्य में कर्तृपद व क्रियापद में मध्यम पुरुष का प्रयोग हुआ है। कर्मवाच्य में इस बात का ध्यान रखा जाएगा कि क्रियापद कर्म के वचन और पुरुष के अनुसार हो। कर्मवाच्य में कर्म प्रथमा विभक्ति में होता है।

  1. तेः सिंहाः दृष्टाः।
  2. तैः नर्तकी दृष्टा।
  3. तैः मित्रं दृष्टम्।

यहाँ सिंहाः, नर्तकी, मित्रम् के अनुसार क्रियापद में वचन का ध्यान रखा गया है। क्तान्त क्रियापदों में तथा क्तवतु-अन्त क्रियापदों में लिङ्ग में भी भिन्नता आ जाती है; यथा

  1. ते पाठं पठितवत्यौ।
  2. सख्यौ पाठं पठितवत्यौ।
  3. मित्रे पाठं पठितवती।
  4. ताभ्यां पाठः पठितः।
  5. ताभ्यां फलानि खादितानि।
  6. ताभ्यां देवी दृष्टा।

निर्देश –  छात्र कर्तृपद में लिङ्ग की पहचान करें तथा देखें कि क्रियापद में किस लिङ्ग का प्रयोग हुआ है। प्रथम तीन वाक्य क्तवतु प्रत्ययान्त क्रियापद के हैं तथा कर्तृ वाच्य हैं। अन्तिम तीन क्त प्रत्ययान्त क्रियापद के हैं और कर्मवाच्य हैं।

(ख) विशेषण-विशेष्य-अन्वितिः
वाक्य में सभी कारकों में संज्ञापदों के साथ विशेषणों का प्रयोग हो सकता है। हमें यह ध्यान रखना है कि जो विभक्ति विशेष्य पद में है वही विभक्ति विशेषण पद में होनी चाहिए। जो लिङ्ग तथा वचन विशेष्य पद का होगा वही लिङ्ग तथा वचन विशेषण पद में होना चाहिए। सभी विभक्तियों के उदाहरण यहाँ प्रस्तुत हैं

(क) प्रथमा विभक्ति के उदाहरण –

  1. अश्वात् पतन सैनिकः शीघ्रम् उत्तिष्ठति।
  2. गच्छन्ती माता पुत्रम् अपि नयति।
  3. कम्पमानाः शत्रवः अधावन्।

(ख) द्वितीया विभक्ति के उदाहरण

  1. अहं प्रिय मित्रम् अपश्यम्।
  2. अहं प्रिये मित्रे अपश्यम्।
  3. अहं प्रियाणि मित्राणि अपश्यम्।

(ग) तृतीया विभक्ति के उदाहरण

  1. नीलेन पुष्पेण सह श्वेतं पुष्पं मेलयत।
  2. सः योग्यैः शिक्षकैः सह उपविष्टः आसीत्।
  3. अहं पीडिताभ्यां सखीभ्यां सह सान्त्वनावचनं कथयिष्यामि।

(घ) चतुर्थी विभक्ति के साथ

  1. सस्वराय गीताय ताम् आह्वय।
  2. शोभनायै कन्यायै शोभनं वरं पश्य।
  3. ललिताभ्यः लवङ्गलताभ्यः तस्य प्रशंसा कुरु।

(ङ) पञ्चमी विभक्ति के साथ

  1. एकस्मात् वृक्षात् पक्षी उड्डयते।
  2. द्वाभ्यां गृहाभ्याम् अन्नम् आनय।
  3. त्रिभ्यः पुस्तकेभ्यः युग्म-शब्दान् अवचिनुत।

(च) षष्ठी विभक्ति के साथ

  1. उन्नतस्य भवनस्य शिखरं मनोरमम् अस्ति।
  2. निम्नयोः वाक्ययोः अन्तरं कथय।
  3. पठितानां पुस्तकानां कः उपयोगः अस्ति?

(छ) सप्तमी विभक्ति के साथ

  1. रक्ते पुष्ये मम अभिलाषः वर्तते।
  2. तस्यां कन्यायां सा स्निह्यति।
  3. निजयोः पित्रोः सः साधुः अस्ति।

अभ्यासार्थ

निम्नवाक्यानि कोष्ठकस्य शब्दानां सहायतया कर्ता-क्रिययोः अन्विति-माध्यमेन पूरयित्वा लिखत  (1 x 5 = 5)
निम्नवाक्यों को कोष्ठक के शब्दों की सहायता से कर्ता-क्रिया की अन्विति द्वारा पूर्ण करके लिखें –

1. (i) तृणानि भूमिरुदकं वाचतुर्थी च सुनृता एतानि सतां गेहे कदाचन न ………………..। (उच्छिद्यते, उच्छिद्यन्ते)
(ii) अद्भिः गात्राणि ………………. | (शुध्यन्ति, शुध्यति)
(iii) मनः सत्येन ……………….. । (शुध्यति, शुध्यन्ति)
(iv) सत्यमेव ……………… नानृतम्। (जयति, जयते)
येन हि ऋषयः आप्तकामाः ……………….. । (आक्रमन्ति, आक्रमति)
(v) यः तु सर्वाणि भूतानि आत्मनि एव ……………….. । (अनुपश्यन्ति, अनुपश्यति)
उत्तर:
(i) उच्छिद्यन्ते
(ii) शुध्यन्ति
(iii) शुध्यति
(iv) जयते, आक्रमन्ति
(v) अनुपश्यति

2. (i) तत् पथः दुर्ग कवयो ……………….. । (वदति, वदन्ति)
(ii) यूयं सर्वे वरान् प्राप्य ……………….. । (निबोधत, निबोधत्)
(iii) अस्माकं सृष्टि-आधारम् किं भवन्तः ……………….. । (जानाति, जानन्ति)
(iv) ऋतूणां दिनरात्र्योः च कथं परिवर्तनं ……………….. । (भवेत, भवेत्)
(v) सर्वत्र एव सूर्यस्य महिमा ……………….. । (वर्णिता, वर्णितम्)
उत्तर:
(i) वदन्ति
(ii) निबोधत
(iii) जानन्ति
(iv) भवेत्
(v) वर्णिता

3. (i) अयमेव वत्सरं द्वादशसु भागेषु ……………….. । (विभनक्ति, विभक्ति)
(ii) एष एव उत्तरं दक्षिणं चायनम् ……………….. । (अङ्गीकरोषि, अङ्गीकरोति)
(iii) अनेन एव कल्पभेदाः ……………….. । (कृताः, कृतः,)
(iv) एनमेव आश्रित्य परमेष्ठिनः परार्द्धसंख्या ……………….. । (भवन्ति, भवति)
(v) गायत्री अमुम् एव ……………….. । (गायति, गीयते)
उत्तर:
(i) विभनक्ति
(ii) अङ्गीकरोति
(iii) कृताः
(iv) भवति
(v) गायति

4. (i) देशस्य कृते एव प्रजाधनस्य सदुपयोगः ……………….. । (स्यात, स्यात्)
(ii) अहं कौमुदीमहोत्सव-कारणतः कुसुमपुरम् अवलोकयितुम् ……………….. । (इच्छति, इच्छामि)
(iii) अतः सुगाङ्गप्रसादस्य उपरि स्थिताः प्रदेशाः ……………….. । (संस्क्रियन्ताम्, संस्क्रियताम्)
(iv) चन्दन वारिणा भूमि शीघ्रं ……………….। (सिञ्चन्तु, सिञ्चतु)
(v) अयम् एव देवः चन्द्रगुप्तः ……………….. । (आगताः, आगतः)
उत्तर:
(i) स्यात्
(ii) इच्छामि
(iii) संस्क्रियन्ताम्
(iv) सिञ्चन्तु
(v) आगतः

5. (i) आर्य वैहीनरे! सुगाङ्गमार्गम् ……………….. । (आदेशयतु, आदेशय)
(ii) तत्कथं न कौमुदीमहोत्सवः ……………….. । (प्रारब्धः, प्रारब्धाः)
(iii) देव! कः अन्यः जीवितुकामो देवस्य शासनम् ………………… । (अतिवर्तेत, अतिवर्तेत्)
(iv) आर्य! आचार्यचाणक्यं द्रष्टुम् ……………….. । (इच्छामः, इच्छामि)
(v) कथं मया सह दुरात्मा राक्षसः ……………….. । (स्पर्धते, स्पर्धसे)
उत्तर:
(i) आदेशय
(ii) प्रारब्धः
(iii) अतिवर्तेत
(iv) इच्छामि
(v) स्पर्धते

6. (i) एतादृशैः जनैः राजा तृणवद् ……………….. । (गण्यसे, गण्यते)
(i) आर्य! देवः चन्द्रगुप्तः आर्य शिरसा प्रणम्य ……………….. । (विज्ञापयति, ज्ञापयति)
(iii) वैहीनरे! किं तेन कौमुदीमहोत्सवस्य प्रतिषेधः ……………….. । (ज्ञातः, ज्ञातम्)
(iv) भवद्भिः एव प्रोत्साह्य वृषल: ……………….. । (कोपिताः, कोपितः)
(v) आर्य, देवेन एव अहम् आर्यस्य चरणयोः ……………….. । (प्रेषिताः, प्रेषितः)
उत्तर:
(i) गण्यते
(ii) विज्ञापयति
(iii) ज्ञातः
(iv) कोपितः
(v) प्रेषितः

7. (i) अये! सिंहासनम् वृषलः ……………….. । (अध्यास्ते, अध्यासते)
(ii) न निष्प्रयोजनं प्रभुभिः अधिकारिणः ……………….. । (आहूयते, आहूयन्ते)
(iii) कौमुदी महोत्सवस्य प्रतिषेधे किं फलम् आर्यः ……………….. । (पश्यति, पश्यन्ति)
(iv) यदि एवं तर्हि शिष्येण गुरोः आज्ञा ……………….. । (पालनीया, पालनीयाः)
(v) स्वयम् अनभियुक्तानां राज्ञाम् एते दोषाः ……………….. । (सम्भवति, सम्भवन्ति)
उत्तर:
(i) अध्यास्ते
(ii) आहूयन्ते
(iii) पश्यति
(iv) पालनीया
(v) सम्भवन्ति

8. (i) अतः इदानीं दुर्गसंस्कारः ……………….. । (प्रारब्धव्यः, प्रारब्धव्यम्)
(ii) पर्वतक-पुत्रः मलयकेतुः अस्मान् अभियोक्तुम् ……………….. । (उद्यताः, उद्यतः)
(iii) न निष्प्रयोजनं प्रभुभिः अधिकारिणः ……………….. । (आहूयते, आहूयन्ते)
(iv) आर्य! चन्द्रगुप्तः ……………….. । (प्रणमामि, प्रणमति)
(v) न प्रयोजनम् अन्तरा चाणक्यः स्वप्नेऽपि ……………….. । (चेष्टते, चेष्टन्ते)
उत्तर:
(i) प्रारब्धव्यः
(ii) उद्यतः
(iii) आहूयन्ते
(iv) प्रणमति
(v) चेष्टते

9. (i) जीवने दूरदृष्टिं विना सफलता न ……………….। (लभ्यन्ते, लभ्यते)
(ii) दूरदर्शी जनः सर्वान् पक्षान् विपक्षान् च गणयित्वा निर्णय ……………….। (करोषि, करोति)
(iii) एकस्मिन् जलाशये त्रयो मत्स्याः ……………….। (प्रतिवसन्ति स्म, प्रतिवसति स्म)
(iv) अहो, बहुमत्स्योऽयं हृदः अस्माभिः कदापि न ……………….। (अन्वेष्टः, अन्वेषितः)
(v) तद् रात्रौ एव किञ्चिद् निकटं सरः ……………….। (गम्यताम्, गम्यन्ताम्)
उत्तर:
(i) लभ्यते
(ii) करोति
(iii) प्रतिवसन्ति स्म
(iv) अन्वेषितः
(v) गम्यताम्

10. (i) अशक्तैः बलिनः रात्रोः प्रपालायनं ……………….। (कर्तव्यम्, कर्तव्यः)
(ii) प्रातः काले मत्स्यजीविनः अत्र समागत्य मत्स्यक्षयं ……………….। (करिष्यति, करिष्यन्ति)
(iii) ते विद्वांसः देहभङ्ग कुलक्षयम् च न ……………….। (पश्यन्ति, पश्यति)
(iv) भवद्भ्यां सम्यक् न ……………….। (मन्त्रितम्, मन्त्रितः)
(v) किं वाङ्मात्रेण एव पितृपैतामहादिकम् एतत् सरः त्यक्तुं ……………….। (युज्यन्ते, युज्यते)
उत्तर:
(i) कर्तव्यम्
(ii) करिष्यन्ति
(iii) पश्यन्ति
(iv) मन्त्रितम्
(v) युज्यते

11. (i) वने विसर्जितः अनाथोऽपि ……………….। (जीवति, जीवन्ति)
(ii) कृत प्रयत्नः जनः गहेऽपि ……………….। (नश्यन्ति, नश्यति)
(iii) मत्स्यजीविभिः जालैः तं जलाशयं निर्मत्स्यता ……………….। (नीतम्, नीतवान्)
(iv) अनागतविधाता प्रत्युत्पन्नमतिश्च परिजनेन सह ततः ……………….। (निष्क्रान्तः, निष्क्रान्तौ)
(v) सुरक्षितं दैवहतम् अपि ……………….। (विनश्यति, विनश्यन्ति)
उत्तर:
(i) जीवति
(ii) नश्यति
(iii) नीतम्
(iv) निष्क्रान्तौ
(v) विनश्यति

12. (i) अहो! इयं हिमानी कीदृशी ……………….। (राज्यते, राजते)
(ii) सर्वे छात्राः यात्रायाः वृत्तान्तं ज्ञातुम् अनुरोध ……………….। (कुर्वन्ति, कुर्वति)
(iii) नूनम् एतादृशानि वर्णनानि त्यागस्य भावनाः ……………….। (जनयति, जनयन्ति)
(iv) छात्राः विद्यालयस्य पत्रिकायां चित्राणि ……………….। (पश्यति, पश्यन्ति)
(v) गतवर्षे द्वादशकक्षायाः छात्राः लेह-लद्दाखनगरं ……………….। (प्रयाताः, प्रयाता)
उत्तर:
(i) राजते
(ii) कुर्वन्ति
(iii) जनयन्ति
(iv) पश्यन्ति
(v) प्रयाताः

13. (i) इदम् अभियानम् अतीव रोचकं साहसिकं च ……………….। (आसन्, आसीत्)
(ii) अस्माकं कक्षायाः सर्वे छात्राः यात्रावृत्तं द्रष्टुम् उत्सुकाः ……………….। (अस्ति, सन्ति)
(iii) सर्वे छात्राः यथास्थानम् ……………….। (उपविशन्ति, उपविशति)
(iv) एते लद्दाखप्रदेशीयाः गिरयः अतीव ……………….। (शोभते, शोभन्ते)
(v) उत्तुङ्गपर्वतानाम् उपत्यका भूमिं (जनाः) लद्दाख इति ……………….। (वदन्ति, वदति)
उत्तर:
(i) आसीत्
(ii) सन्ति
(iii) उपविशन्ति
(iv) शोभन्ते
(v) वदन्ति

14. (i) लद्दाखे आस्तृतः नीलाकाशः छत्रवत् ……………….। (प्रतीयेते, प्रतीयते)
(ii) सा उपत्यका विभजन्ती सिन्धुनदी ……………….। (भवति, अस्ति)
(iii) अयं स्तूपः रात्रौ दीपेषु प्रज्वलितेषु भव्यं आलोकं ……………….। (वितरति, वितरन्ति)
(iv) दीर्घ-दीर्घाणि एतानि स्थानानि किं मठाः……………….? (अस्ति, सन्ति)
(v) मठानां विशालता भव्यता च प्रेक्षकान् प्रसभम् …………….। (आकर्षति, आकर्षतः)
उत्तर:
(i) प्रतीयते
(ii) अस्ति
(iii) वितरति
(iv) सन्ति
(v) आकर्षति

15. (i) ‘स्टाक पैलेस’ संग्रहालये सप्तसप्ततिः कक्षाः ……………….। (अस्ति, सन्ति)
(ii) ‘लामायारु’ आदीनि ग्रीष्मपर्वाणि भगवन्तं बुद्ध प्रति भक्तिभावं ……………….। (दर्शयन्ति, पश्यन्ति)
(iii) उपत्यकाभूमिषु शीते ऋतौ महान् हिमराशिः ……………….। (निपतन्ति, निपतति)
(iv) ग्रीष्मे सः द्रवीभूय कृषकाणां भूमि सेचने भूयिष्ठम् ……………….। (उपकुर्वन्ति, उपकरोति)
(v) ग्रीष्मे ऋतौ पर्वतारोहिणोऽत्र प्रायः ……………….। (दृश्यन्ते, दृश्यते)
उत्तर:
(i) सन्ति
(ii) दर्शयन्ति
(iii) निपतति
(iv) उपकरोति
(v) दृश्यन्ते

16. (i) विपदि निपतिताः मानवाः सुभाषितैः आश्वासनं ……………….। (प्राप्नोति, प्राप्नुवन्ति)
(i) संस्कृतवाङ्मयं सहस्रशः सुमधुरवचनैः सम्यग् अलङ्कृतं ……………….। (वर्तन्ते, वर्तते)
(iii) एतानि सुभाषितानि साहित्ये सर्वत्र सुलभानि ……………….। (अस्ति, सन्ति)
(iv) येषां करणं परोपकरणं ते केषाम् न ……………….। (वन्द्यः, वन्द्याः)
(v) हुत्तम् च दत्तं च सदैव ………………..। (तिष्ठन्ति, तिष्ठति)
उत्तर:
(i) प्राप्नुवन्ति
(ii) वर्तते
(iii) सन्ति
(iv) वन्द्याः
(v) तिष्ठति

17. (i) त्यागेन, शीलेन, गुणेन, कर्मणा इति चतुर्भिः पुरुषः ……………….। (परीक्षते, परीक्ष्यते)
(ii) ये मायाविषु मायिनः न ……………….। (भवन्ति, भवति)
(iii) शठाः निशिताः इषवः इन असंवृत्ताङ्गान् प्रविश्य ……………….। (हन्ति, घ्नन्ति)
(iv) किं प्रभुः हितान् न ……………….। (संशृणुते, संश्रूयते)
(v) अपयशः मृत्योरपि भयावहम् ……………। (अस्ति, सन्ति)
उत्तर:
(i) परीक्ष्यते
(ii) भवन्ति
(iii) घ्नन्ति
(iv) संश्रूयते
(v) अस्ति

18. (i) धनवान् चारुदत्तः उदारतावश दानकरणात् शीघ्रं दरिद्रो ……………….। (जातः, भविष्यति)
(ii) दैन्येऽपि तस्य मनः कदापि भ्रष्टं न ……………..। (भवति, भवन्ति)
(iii) ततः नान्द्यन्ते सूत्रधारः तत्र ……………….। (प्रविशन्ति, प्रविशति)
(iv) अद्य बुभुक्षया पुष्करपत्रपतित जलबिन्दू इव मे अक्षिणी ……………….। (चञ्चलायेते, चञ्चलायते)
(v) एवं शोभनानां भोजनानां दात्री त्वं ……………….। (भवतु, भव)
उत्तर:
(i) जातः
(ii) भवति
(iii) प्रविशति
(iv) चञ्चलायेते
(v) भव

19. (i) अहम् अस्मादृशयोग्यं कञ्चिद् जनं निमन्त्रयितुम् ……………….। (इच्छामः, इच्छामि)
(ii) आर्य, दिष्ट्या खलु आगतः ……………….। (अस्ति, असि)
(iii) चिरं जीव, एवं शोभनानां भोजनानां दात्री ……………….। (भव, भवतु)
(iv) अहं पर्वताद् दूरमारोप्य पातितः ……………….। (अस्ति, अस्मि)
(v) अहम् अस्मादृशयोग्यं कञ्चिद् जनं निमन्त्रयितुम् ……………….। (इच्छति, इच्छामि)
उत्तर:
(i) इच्छामि
(ii) असि
(iii) भव
(iv) अस्मि
(v) इच्छामि

20. (i) एषः आर्य चारुदत्तस्य वयस्यः आर्यमैत्रेयः इतः एव ……………….। (आगच्छसि, आगच्छति)
(ii) मयापि मैत्रेयेण परस्य आमन्त्रणकानि ……………….। (अभिलषणीयानि, अभिलषानि)
(iii) इदानीम् अहं अन्यत्र भुक्त्वा तस्यावासमेव ………………..। (गच्छामि, आगच्छामि)
(iv) एष तत्रभवान् चारुदत्तः गृहदैवतानि अर्चयन् इतः एव ……………….। (आगच्छन्ति, आगच्छति)
(v) न खल्वहं नष्टां श्रियम् ……………….। (अनुशुचामि, अनुशोचामि)
उत्तर:
(i) आगच्छति
(ii) अभिलषणीयानि
(iii) गच्छामि
(iv) आगच्छति
(v) अनुशोचामि

21. (i) सखे! ‘दानं श्रेयस्करम्’ इति प्रत्ययादेव ममार्थाः क्षीणाः……………….। (जातः, जाताः)
(ii) यथा वसन्ते शरस्तम्बस्य अङ्कुराद् अङ्कुराः ……………….। (निस्सरन्ति, निस्सरति)
(iii) तथैव धनविनाश दुःखस्य पुनः पुनः चिन्त्यमानस्य नाना चिन्ताङ्कराः ……………….। (प्रादुर्भवति, प्रादुर्भवन्ति)
(iv) निर्वैराः सुहृदः अपि विमुखी ……………….। (भवति, भवन्ति)
(v) गुणरसज्ञस्य तु पुरुषस्य व्यसनं दारुणतरं मां ……………….। (प्रतिभाति, प्रतिभान्ति)
उत्तर:
(i) जाताः
(ii) निस्सरन्ति
(iii) प्रादुर्भवन्ति
(iv) भवन्ति
(v) प्रतिभाति

22. (i) सर्वे छात्राः यथासमयं कृपया तत्र ……………….। (आगच्छतु, आगच्छन्तु)
(ii) अद्य अस्माकं मध्ये प्रतिभाशालिनः छात्राः ……………….। (समुपस्थिताः, समुपस्थितः)
(iii) कृपया करतल ध्वनिना एतेषां स्वागतम् अभिनन्दनं च ……………….। (कुर्वन्तु, करोतु)
(iv) विमानशास्त्रे त्रिपुरविमानस्य एवं वर्णनं ……………….। (समुपलभते, समुपलभ्यते)
(v) तृतीयः भागः तु स्वतः एव आकाशे ……………….। (सञ्चरति, सञ्चरन्ति)
उत्तर:
(i) आगच्छन्तु
(ii) समुपस्थिताः
(iii) कुर्वन्तु
(iv) समुपलभ्यते
(v) सञ्चरति

23. (i) शालिनी अधुना सुश्रुतस्य शल्य क्रियामधिकृत्य अस्माकं ज्ञानवृद्धि ……………….। (करिष्यति, करिष्यन्ति)
(ii) उत्तम शल्य चिकित्सकाः भवितुं अस्माभिः सुश्रुत संहिता अवश्यमेव ……………….। (पठनीया, पठनीयम्)
(iii) सुश्रुतसंहितायां नासिका प्रत्यारोपणं विस्तरश: ……………….। (वर्णितम्, वर्णिताः)
(iv) यदि जम्बूवृक्षस्य पूर्वदिशि वल्मीकः ……………….। (भवेत, भवेत्)
(v) मिहिरेण वृक्षायुर्वेदपशुविज्ञानयोः उपरि बहुमूल्या सामग्री ……………….। (सङ्कलिता, सङ्कलितम्)
उत्तर:
(i) करिष्यति
(ii) पठनीया
(iii) वर्णितम्
(iv) भवेत्
(v) सङ्कलिता

24.(i) यदि आर्यभटेन शून्यस्य अविष्कारः न कृतः ……………….। (स्यात, स्यात्)
(ii) नौकायां स्थितः मनुष्यः वृक्षादीन् पृष्ठं प्रति गच्छतः……………….। (पश्यति, पश्यन्ति)
(iii) भगवतः आर्यभटस्य महिमानं वयं ……………….। (जानीमः, जानामि)
(iv) कौटिल्यस्य अर्थशास्त्रे बहवः विषयाः ……………….। (वर्णितः, वर्णिताः)
(v) अशुद्धं स्वर्णं चतुर्गुणेन सीसेन …………। (शोधयेत्, शोधयसि)
उत्तर:
(i) स्यात्
(ii) पश्यति
(iii) जानीमः
(iv) वर्णिताः
(v) शोधयेत्

25. (i) सुषमा भारते उपलब्धानां विचित्र शिल्पकलाकृतीनां परिचय ……………….। (प्रदास्यति, प्रदास्यन्ति)
(ii) भारतस्य विविधाः स्तम्भाः अद्य यावत् भारतीय वैज्ञानिकानां गौरवगाथां ……………….। (वर्णयन्ति, वर्णयति)
(iii) इदानीम् एषा गोष्ठी समाप्तिं ……………….। (यान्ति, याति)
(iv) अस्माकं प्राचार्यमहोदयाः श्वः सम्मान पत्राणि ……………….। (प्रदास्यति, प्रदास्यन्ति)
(v) सम्प्रति वयं सर्वे मिलित्वा शान्ति पाठं ……………….। (करिष्यति, करिष्यामः)
उत्तर:
(i) प्रदास्यति
(ii) वर्णयन्ति
(iii) याति
(iv) प्रदास्यन्ति
(v) करिष्यामः

NCERT Solutions for Class 12 Sanskrit

The post CBSE Class 12 Sanskrit व्याकरणम् अन्विति-प्रकरण appeared first on Learn CBSE.

Vidyasaarathi Scholarship 2019 for Undergraduates | Dates, Eligibility, Application Form

Vidyasaarathi Scholarship 2019: ACC has declared the vidyasaarathi scholarship for students for the year 2019-20. This scholarship is only available for students that are residing near the ACC plant across the PAN India. ACC vidyasaarathi scholarship is available for the candidates that are pursuing ITI, undergrad courses, BE/BTECH, or Diploma courses. This scholarship program is designed for students that are pursuing any graduation courses like bachelors in commerce, science, arts, or any subsequent branches. This scholarship program is only available for students that are residing near the ACC plant across India. It is not available for the ACC employees or their children.

Vidyasaarathi is an initiative by the NSDL e-governance infrastructure limited in order to reduce the gap in education finance in the country through online means. Students can search through this portal for various education finance schemes that they are qualified for. Industries, fund providers, corporates can promote skill development by organizing education schemes ib vidyasaarathi and these managed schemes. This solution would manage the online education finance lifecycle that is reviewing and submission of an application, disbursement of funds, scholarship awards, and renewal of education finance.

Vidyasaarathi Scholarship 2019

The online application for the vidyasaarathi scholarship begins in the first week of August 2019. While the last date for submission of application form is in the last week of September 2019.

  • The scholarship amount differs for students based on their current program.
  • The basic scholarship amount is up to Rs. 5000.
  • For diploma students, the scholarship amount is up to Rs. 20,000.
  • While for ITI students, the scholarship amount is up to Rs. 20,000.
  • For BE/BTECH students, the scholarship amount is up to 30,000.
  • For undergrads, financial assistance is up to Rs. 5000.

Vidyasaarathi Scholarship Eligibility Criteria

  • The eligibility criteria differ based on the program taken by students.
  • For undergrad, the basic minimum marks in HSC should be 50%.
  • For BE or BTech students, the minimum marks are 50% in HSC and 50% in diploma.
  • For ITI, the minimum is 35% in SSC.
  • For diploma, the minimum is 50% in SSC.
  • This scheme is available for the students who have a family income between Rs. 0 to Rs. 5,00,000 per annum.
  • This scholarship is not available for ACC employees and their children.

Documents Required for Vidyasaarathi Scholarship

  • Domicile certificate
  • Proof of address
  • Proof of identity
  • Class 10th and 12th marksheet
  • Passing marksheet of 1st year for 2nd-year students
  • Passing marksheet of 2nd year for 3rd-year students
  • Admission confirmation letter
  • Student’s bank passbook
  • College fee receipts
  • Latest income certificate
  • PAN card, voter ID, passport are optional

All these documents uploaded should be clear and must be uploaded in .png or .jpeg format only.

Vidyasaarathi Scholarship Application Form

Below are the steps to apply for ACC vidyasaarathi scholarship 2019:

  1. Go to the official portal of the vidyasaarathi scholarship which is vidyasaarathi.co.in.
  2. Sign up your profile by providing various details through the link given on the right-hand side of the homepage.
  3. Fill all the necessary details in the form and complete your profile
  4. Search for the relevant schemes available on the portal for you and apply for the scholarship.
  5. Upload the required documents and click on the submit button.

FAQ’s

Question 1.
What are some of the features of the vidyasaarathi scholarship portal?

Answer:
The vidyasaarathi portal provides a single window to access all the necessary information and make the application for the educational scholarship. Below are some of the features of the portal:

  • You can apply to multiple educational schemes and scholarship
  • There is relevant information for various educational scholarships.
  • The facility is available for corporates to download the scholarship applications and also review them.
  • Facilities for corporates or educational institutes to register online.

Question 2.
How to apply for the educational scholarship using vidyasaarathi?

Answer:
In order to apply for the scholarship, students need to login on the official portal of vidyasaarathi. Then there is an option for filling the basic details and scheme related details for the candidate.

Question 3.
What to do if I have not received the activation link on an email address?

Answer:
If you have not received any activation link on your email ID then you can check the junk folder or spam mail in your mailbox. You can also contact the customer care for more details.

The post Vidyasaarathi Scholarship 2019 for Undergraduates | Dates, Eligibility, Application Form appeared first on Learn CBSE.

CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग

CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग

धातु या शब्द के पीछे जुड़ने वाले अंश को प्रत्यय कहते हैं। धातु के बाद जुड़ने वाले प्रत्यय को कृत् तथा संज्ञा, विशेषण, क्रिया, अव्यय के पीछे जुड़नेवाले अंश को तद्धित प्रत्यय कहते हैं।

(i) कृत्-प्रत्यय –
इसमें क्त, क्तवतु, क्त्वा, ल्यप् , तुमुन्, शतृ, शानच्, क्तिन्, तव्यत्, अनीयर् आदि प्रत्यय आते हैं।
1. क्त प्रत्यय – यह भूतकालिक कृत् प्रत्यय है। अधिकतर कर्मवाच्य में प्रयुक्त होता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 1

2. क्तवतु- यह भी भूतकालिक कृत् प्रत्यय है। अधिकतर कर्तृवाच्य में प्रयुक्त होता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 2

3. क्त्वा- वाक्य में दो क्रियाओं के होने पर जो पहले समाप्त होती है उस क्रिया को बतानेवाली धातु से ‘क्त्वा’ प्रत्यय (त्वा) लगता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 3

4. ल्यप् – धातु से पूर्व उपसर्ग होने पर धातु के बाद ‘ल्यप् ‘ (य) का प्रयोग होता है।
उदाहरण –
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5. अनीयर् – ‘विधिलिङ् लकार’ के अर्थ में विधि कृदन्त अर्थात् तव्यत्, अनीयर् का प्रयोग होता है। ‘अनीयर् का ‘अनीय’ शेष रहता है।
उदाहरण –
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6. तव्यत् – ‘विधिलिङ् लकार’ के अर्थ में विधि कृदन्त अर्थात् तव्यत्, अनीयर् का प्रयोग होता है। ‘तव्यत्’ का ‘तव्य’ शेष रहता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 6

7. तुमुन्- जिस क्रिया के लिए कोई अन्य क्रिया की जाती है उसकी धातु से भविष्यत् काल के अर्थ को प्रकट करने के लिए ‘तुमुन्’ का प्रयोग होता है। ‘तुम’ का ‘तुम्’ शेष रह जाता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 7

8. क्तिन् प्रत्यय – स्त्रीलिङ्ग में भाववाचक संज्ञा बनाने के लिए धातु के साथ ‘क्तिन्’ प्रत्यय लगता है। ‘क्तिन्’ का ‘तिः’ शेष रहता है।
उदाहरण –
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9. शतृ प्रत्यय – परस्मैपदी धातुओं से ‘शतृ’ का अपूर्णकाल में प्रयोग होता है। ‘शतृ का’ ‘अत्’ शेष रह जाता है। पुं० में ‘त्’ को ‘न्’ हो जाता है।
उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 9

10. शानच् प्रत्यय – आत्मनेपद की धातुओं के साथ ‘शतृ’ (अत्/अन्) के स्थान पर ‘शान’ (आन, मान) प्रत्यय लगता है।
उदाहरण –
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(ii) तद्धित प्रत्यय
(मतुप्, इन्, ठक्, ठञ्, त्व, तल्)
संज्ञा, सर्वनाम, विशेषण, क्रियाविशेषण तथा अव्यय में ( = क्रियाओं से भिन्न शब्दों में) प्रत्यय लगाकर जो नए शब्द बनते हैं वे ‘तद्धित प्रत्ययान्त’ शब्द कहलाते हैं तथा उन प्रत्ययों को ‘तद्धित’ प्रत्यय कहते हैं। तद्धित प्रत्यय लगने पर नए शब्दों के अर्थ भी मूल शब्दों से भिन्न हो जाते हैं। जैसे- ‘दशरथ’ में इञ् (इ) तद्धित प्रत्यय लगाकर ‘दाशरथिः’ शब्द बना तथा ‘दाशरथिः’ शब्द का अर्थ ‘दशरथ की सन्तान’ होता है। इञ्’ (इ) प्रत्यय के लगने से उसकी सन्तान’ ऐसा अर्थ निकलता है अतः वह अपत्यार्थक तद्धित प्रत्यय है। इस तरह सभी तद्धित प्रत्ययों को कुल मिलाकर अपत्यार्थक, देवार्थक, समूहार्थक, अध्ययनार्थक, शैषिक, विकारार्थक, मतुबर्थक, स्वार्थक तथा अनेकार्थक- इनको, नौ भागों में बाँटा गया है। प्रत्ययों की संख्या भी बहुत अधिक है, किन्तु पाठ्यक्रम में मतुप्, इन्, ठक्, ठ, त्व, तल् इन सात तद्धित प्रत्ययों को ही रखा गया है। अत: निम्न पंक्तियों में केवल इन प्रत्ययों का ही सोदाहरण परिचय दिया जा रहा है।

(क) मतुप् प्रत्यय
‘वह इसका है’ (तत् अस्य अस्ति) या ‘वह इसमें है’ (तत् अस्मिन् अस्ति) इन अर्थों में प्रथमान्त शब्द से मतुप् (मत्) प्रत्यय हो जाता है। इस प्रत्यय का प्रयोग किसी एक वस्तु का दूसरी वस्तु में होना सूचित करने के लिए होता है। ‘मतुप्’ प्रत्यय का केवल ‘मत्’ ही शेष रह जाता है।

नियम –
1. जिन शब्दों के अन्त में अ, आ के अतिरिक्त कोई स्वर होता है। उनसे यह जैसे को तैसा जुड़ जाता है। जैसे – गो + मत् = गोमत् (= जिसकी गौएँ हैं- गावः अस्य सन्ति, इति)। यहाँ ‘गो’ ओकारान्त शब्द है अतः इसके साथ मतुप् प्रत्यय का ‘मत्’ जुड़ जाने से ‘गोमत्’ रूप बना।

इकारान्त शब्दों के उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 11

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इकारान्त शब्दों के उदाहरण –
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उकारान्त शब्दों के उदाहरण –
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CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग 14

ऊकारान्त शब्द –
वधू + मतुप् = वधूमत् (वधू युक्त), वधूमान् (पुं॰)

ओकारान्त शब्द –
गो , + मतुप् = गोमत्। (गौओं वाला), गोमान् (पुं॰)

हलन्त शब्द –
कुछ हलन्त शब्दों के बाद भी मतुप् के ‘मत्’ का प्रयोग होता है; जैसे –
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2. जिन शब्दों के अन्त में तथा उपधा में ‘म्’ हो अथवा ‘अ’ या ‘आ’ हो, तो उनके पश्चात् ‘मतुप्’ के ‘म्’ को ‘व्’ होकर ‘मत्’ के स्थान पर “वत्’ का प्रयोग होता है; जैसे –

अकारान्त शब्द –
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आकारान्त शब्द –
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हलन्त शब्द –
निम्न हलन्त शब्दों में भी मतुपु के ‘म’ को ‘व’ होकर ‘वत्’ का प्रयोग होता है, क्योंकि इनकी उपधा (अन्तिम वर्ण से पूर्व) में ‘अ’ विद्यमान है; जैसे –
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अपवाद –

विशेष – ‘मतुप्’ लगने से बने हुए सभी शब्द विशेषण रूप में प्रयुक्त होते हैं। अतः इनके रूप विशेष्य के अनुसार तीनों लिंगों में अलग-अलग बनते हैं; जैसे –
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(ख) इनि (इन्) प्रत्यय
अकारान्त शब्दों के अनन्तर ‘मतुप्’ के अर्थ में अर्थात् किसी एक वस्तु का दूसरी वस्तु में होना सूचित करने के लिए (‘तद् अस्य अस्ति’ अथवा ‘तद् अस्मिन् अस्ति’) इनि (इन्) का प्रयोग किया जाता है; जैसे –

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‘ज्ञानिन्’ या ‘ज्ञानी’ की व्याख्या होगी ‘ज्ञानम् अस्य अस्ति इति’ अर्थात् ज्ञानवान्। इसी प्रकार सभी ‘इन्’ प्रत्ययान्त शब्दों की व्याख्या समझनी चाहिए।

(ग) ठक् (इक) प्रत्यय
ठक् प्रत्यय का प्रयोग विभिन्न शब्दों के साथ निम्नलिखित पृथक्-पृथक् अर्थों में होता है (ठक्, ठन्, ठञ् आदि प्रत्ययों के स्थान पर इक हो जाता है) – ठक् प्रत्यय परे होने पर शब्द के प्रथम स्वर की वृद्धि (अ को आ, इ, ई, ए, को ऐ, उ, ऊ, ओ को औ) हो जाती है; जैसे- दिन + ठक् (इक) = दैनिकः; दैनिकम्। वर्ष + ठक् = वार्षिक:; वार्षिकम्। मूल + ठक् = मौलिकम्। देव + ठक् = दैविकः। भूत + ठक् = भौतिकः। देह + ठक् = दैहिकः। अध्यात्मक + ठक् = अध्यात्मिकः। अधिभूत + ठक् = आधिभौतिकः। अधिदेच + ठक् = आधिदैविकः। लक्ष + ठक् = लाक्षिकः इत्यादि।

1. अपत्य (सन्तान) अर्थ में –
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2. संस्कृतम् अर्थ में –
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3. चरति (जाता है, खाता है) अर्थ में –
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4. रक्षति (रक्षा करता है) के अर्थ में –

समाज + ठक् (इक) = । सामाजिकः (समाज की रक्षा करने वाला)

5. करोति (करता है) के अर्थ में –
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6. हन्ति (मारता है) के अर्थ में –
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7. ‘मतिः यस्य’ (जिसकी मान्यता है) के अर्थ में –
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8. नियुक्त अर्थ में –
आकर + ठक् (इक) = आकरिकः (आकर में नियुक्त)

9. गच्छति (गमन करना) अर्थ में –
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10. ‘अधीते’ या ‘वेद’ (पढ़ता है या जानता है) के अर्थ में –
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(घ) ठञ् ( इक) प्रत्यय

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(ङ) “त्व’ प्रत्यय
भाववाचक संज्ञा बनाने के लिए ‘त्व’ प्रत्यय का प्रयोग किया जाता है। ‘त्व’ प्रत्ययान्त पद नपुंसकलिंग में होता है।
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(च) तल् (ता) प्रत्यय
भाववाचक संज्ञा (स्त्रीलिंग) बनाने के लिए तल् प्रत्यय भी लगाया जाता है। तल् के स्थान पर ‘ता’ हो जाता है; जैसे –
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तल् प्रत्यय ग्राम, जन, बन्धु, गज तथा सहाय शब्दों के साथ समूह के अर्थ में भी किया जाता है; जैसे –
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(iii) स्त्री प्रत्यय

(क) टाप् प्रत्ययः
अजादिगण में परिगणित पुंल्लिग शब्दों को तथा अकारान्त आदि पुंल्लिग शब्दों को स्त्रीलिंग बनाने के लिए प्रायः टाप् प्रत्यय का प्रयोग होता है। टाप् प्रत्यय के ट् और प् का लोप होकर केवल ‘आ’ रूप शेष बच जाता है।

1. अजादिगण शब्द –
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2. अकारान्त शब्द –
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विशेष – कुछ शब्द अजादिगण में न होने पर भी उनके साथ टाप् प्रत्यय लगता है; जैसे –

आचार्य + टाप् (आ) = आचार्या उपाध्याय + टाप् (आ) = उपाध्याया

3. जिन शब्दों के अन्त में ‘अक’ होता है उनको स्त्रीलिंग बनाते समय भी ‘आ’ प्रत्यय लगता है, किन्तु शब्द के अन्तिम ‘क’ से पूर्व ‘इ’ लगाई जाती है; जैसे –
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(ख) ङीप् प्रत्ययः
ङीप् प्रत्यय में व प् का लोप होकर केवल ‘ई’ शेष रहता है। ङीप् प्रत्यय का प्रयोग निम्नलिखित स्थानों पर किया जाता है

1. ऋकारान्त शब्दों को स्त्रीलिंग बनाने के लिए डीप् प्रत्यय का प्रयोग होता है; जैसे –
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2. नकारात्मक शब्दों को स्त्रीलिंग बनाने के लिए ङीप् प्रत्यय का प्रयोग होता है; जैसे –
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3. प्रथम वय के वाचक अकारान्त शब्दों से स्त्रीलिंग में ङीप् (ई) प्रत्यय होता है; जैसे –
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4. ईयसुन् प्रत्यय वाले शब्दों से स्त्रीलिंग में ङीप् (ई) प्रत्यय होता है; जैसे –
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5. मतुप् प्रत्यय वाले शब्दों में स्त्रीलिंग में ङीपू (ई) प्रत्यय होता है; जैसे –
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6. वतुषु (वत्) प्रत्यय वाले शब्दों से स्त्रीलिंग में ङीप् (ई) प्रत्यय लगता है; जैसे –
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7. क्वस् (वस्) प्रत्यय वाले शब्दों से स्त्रीलिंग में ङीप् (ई) प्रत्यय होता है; जैसे –
विद्वस् + ङीप् (ई) = विदुषी  जग्मिवस् + ङीप् (ई) = जामुषी

8. शतृ (अत्) प्रत्यय वाले शब्दों से स्त्रीलिंग में ङीप् (ई) प्रत्यय लगता है और न् का आगम होता है; जैसे –
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9. गुणवाचक उकारान्त शब्दों से स्त्रीलिंग में विकल्प से ङीप् (ई) लगती है; जैसे –
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10. पतिबोधक अकारान्त शब्दों से पत्नीबोधक शब्द बनाने के लिए ङीप् (ई) प्रत्यय लगता है; जैसे –
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11. निम्नलिखित अकारान्त शब्दों से स्त्रीलिंग बनाने के लिए ङीप् (ई) प्रत्यय लगता है; जैसे –
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12. निम्नलिखित अकारान्त शब्दों से स्त्रीलिंग बनाने के लिए ङीप् (ई) प्रत्यय से पूर्व आन् का आगम होता है; जैसे –
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कुछ विशेष प्रत्ययः
ऊपर लिखे गए प्रत्ययों के अलावा कुछ विशेष महत्त्वपूर्ण प्रत्यय भी दिए जा रहे हैं जो कि पाठ्यक्रम में निर्धारित नहीं है।
1. ल्युट् (अनम्) 2. क्तिन् (ति:) 3. तृच् (तृ) 4. तसिल् (त:)

1. ल्युट् प्रत्यय- यह प्रत्यय धातु के पीछे लगाकर उन्हें भाववाचक नपुंसकलिंग एकवचन शब्द रूप में परिवर्तित कर देता है। यथा –
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2. क्तिन् प्रत्यय- यह प्रत्यय से धातु को स्त्रीलिंग (भाववाचक) एकवचन इकारांत वाले शब्द बनाए जाते है; यथा –
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3. तृच् प्रत्यय- इस प्रत्यय का प्रयोग धातु से ऋकारान्त शब्दों (करने वाला) का निर्माण करने के लिए किया जाता है; जैसे –
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4. तसिल् प्रत्यय- ये प्रत्यय शब्दों से पञ्चमी विभक्ति के रूप में उसी अर्थ को प्रकट करने के लिए लगाए जाते हैं। ये शब्द अव्यय के रूप में बने हैं; जैसे –
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प्रत्यय-प्रयोग

(क) कृत्प्रत्ययान्त पदों का प्रयोग

1. क्तान्त पदों का प्रयोग
(क) श्रीकृष्णस्य मधुराधिपते: ………………. (हस् + क्त) मधुरम् अस्ति।
(ख) तेन जलं ……………… (पा + क्त)।

2. क्तवतु-अन्त पदों का प्रयोग
(क) सः दीनान् …………… (पीड् + क्तवतु)।
(ख) सा याचकेभ्य: वस्त्राणि ……………… (दा + क्तवतु)।

3. क्त्वा-अन्त पदों का प्रयोग
(क) पुष्पं भूमि …………… (पत् + क्त्वा) अनश्यत्।
(ख) वयं नदी ……………. (गम् + क्त्वा) शीघ्रम् आयास्यामः।

4. तुमुन्-अन्त पदों का प्रयोग
(क) अहं …………… (जीव् + तुमुन्) खादामि।
(ख) स …………. (खाद् + तुमुन्) जीवति।

5. ल्यप् प्रत्ययान्त पदों को प्रयोग
(क) आचार्य: ………………… (वि + हस् + ल्यप्) अवदत्-राम! त्वं धन्योऽसि।
(ख) शङ्करः नदीं :……………. (प्र + विज्ञ + ल्यप्) मातरम् अवदत् – नक्र: मां न मुञ्चति, मातः! किं करवाणि।

6. तव्यत्-प्रत्ययान्त पदों का प्रयोग
(क) भारतीयैः वीराणां सम्मान …………… (कृ + तव्यत्)।
(ख) अस्माभिः एतानि पत्राणि ………….. (लिखु + तव्यत्)।

7. अनीयर् प्रत्ययान्त पदों का प्रयोग
(क) त्वया धर्म: ……………. (आ + चर् + अनीयर्)।
(ख) मया सत्कर्म …………… (कृ + अनीयर्)।

8. क्तिन् प्रत्ययान्त पदों का प्रयोग
(क) सः ………….. (मुच् + क्तिन्) वाञ्छति।
(ख) विषयेषु ……………. (रम् + क्तिन्) परित्यजत।

9. शतृ प्रत्ययान्त पदों का प्रयोग
(क) स…………… (नृत् + शतृ) मूर्च्छितोऽभवत्।
(ख) लेखका: ……………… (लिख् + शतृ) उत्साहं प्रादर्शयन्।

10. शानच् प्रत्ययान्त पदों का प्रयोग
(क) …………… (सेव् + शानच्) जनः स्वार्थं न गणयेत्।
(ख) ……………….. (वृध् + शानच्) वृक्षाः छायावन्तः सन्ति।

(ख) तद्धित प्रत्ययान्त पदों का प्रयोग

1. मतुप् प्रत्ययान्त पदों का प्रयोग
(क) ………… (बुद्धि + मतुप्) बालिका परीक्षायां सफला अभवत्।।
(ख) ……….. (विद्या + मतुप्) भव।

2. इन् प्रत्ययान्त पदों का प्रयोग
(क) ते ………….. (ज्ञान + इन्) सन्ति।
(ख) वयं ………… (सुख + इन्) स्याम।

3. ठक् प्रत्ययान्त पदों का प्रयोग
(क) श्रीपाद दामोदर सातवलेकरमहोदयः ” (वेद + ठक्) विद्वान् आसीत्।
(ख) किं त्वम् …………… (अस्ति + ठक्) ने असि?

4. ठञ् प्रत्ययान्त पदों का प्रयोग
(क) इदं वस्त्रं साप्ततिकम् (सप्तति + ठञ्) अस्ति।
(ख) इदम् अन्नं प्रास्थिकम् (प्रस्थ + ठञ्) अस्ति।
(ग) इदम् अध्यायः आह्निकः (अहन् + ठञ्) अस्ति।
(घ) स राजा श्वेतच्छत्रिकः (श्वेतच्छत्र + ठञ्) अस्ति।

5. त्व-प्रत्ययान्त पदों का प्रयोग
(क) सः गुणै: …………. (देव + त्व) प्राप्तवान्।
(ख) फलस्य ………… (मृदु + त्व)पश्यत।

6. तल् प्रत्ययान्त पदों का प्रयोग
(क) तस्य गायने ………….. (मधुर + तल्) न आसीत्।
(ख) तस्य प्रश्नपत्रे …………….. (सरल + तल्) न आसीत्।

अभ्यासार्थ :

निम्नलिखित वाकयेषु रेखाकितानां पदेषु प्रकृति-प्रत्ययानां संयोजनं विभाजनं वा कुरुत

1. …………….. च …………… वाकू। (चतुर्थी + ङीप्), (सुनृत + टाप्)
उत्तर:
चतुर्थी, सुनृता

2. सत्येन पन्थाः ………….. देवयानः। (वि + तन् + क्त)
उत्तर:
विततः

3. उत्तिष्ठत, जाग्रत, …………….. वरान् निबोधत। (प्र + आप् + ल्पय्)
उत्तर:
प्राप्य

4. क्षुरस्य धारा ……………. दुरत्यया। (निशित + टाप्)
उत्तर:
निशिता

5. सूर्यः एव ……………. आधारः। (प्र + कृ + क्तिन्)
उत्तर:
प्रकृतेः

6. अस्माकं ………….. आधारः कः? (सृष् + क्तिन्)
उत्तर:
सृष्टेः

7. सूर्योदयस्य ……………… अद्भूतं च शिवराजविजये वर्णनम् उपलभ्यते। (रम् + यत्)
उत्तर:
रम्यम्।

8. निजपर्णकुटीरात् …………… गुरुसेवन पटुः बटुः उदेष्यन्तं भास्वन्तं ……………… तस्य महिमानं वर्णयति। (निर + गम् + ल्यप्), (प्र + नम् + शतृ)
उत्तर:
निर्गत्य, प्रणमन्

9. एष पूर्वस्यां ……………….. अरुण प्रकाशः। (भग + मतुप्), (मरीचिमाल + इन्)
उत्तर:
भगवतः, मरीचिमालिनः

10. …………… खेचरचक्रस्य। (चक्रवर्ती + इन्)
उत्तर:
चक्रवर्ती

11. ……………. पुण्ड रीक पटलस्य। (प्रिय + ईयसुन्)
उत्तर:
प्रेयान्

12. अनेन एव युगभेदाः (सम् + पद् + णिच् + क्त)
उत्तर:
सम्पादिताः

13. अनेन एव …………. कल्पभेदाः। (कृ + क्त)
उत्तर:
कृताः

14. एनम् एव …………… ……………. परार्द्धसंख्या भवति। (आ + श्रि + ल्यप्), (परमेष्ठ + इन्)
उत्तर:
आश्रित्य, परमेष्ठिनः

15. वेदाः एतस्य एव ………… (वन्द + इन्)
उत्तर:
वन्दिनः

16. ………….. एषः विश्वेषाम्। (प्र + नम् + यत्)
उत्तर:
प्रणम्यः

17. राष्ट्रचिन्ता ……………..। (गुरु + ईयसुन् + ङीन्)
उत्तर:
गरीयसी

18. एषः अंशः ‘मुद्राराक्षस’ इति संस्कृतनाटकात् । (सम् + कल् + क्त)
उत्तर:
सङ्कलित

19. ……………, आकाशम् ………….. । (परि + क्रम् + ल्यप्), (उत् + वि + ईक्ष् + ल्यप्)
उत्तर:
परिक्रम्य, उद्वीक्ष्य

20. कौमुदीमहोत्सव-…………… अतिरमणीयं कुसुम पुरम् …………… इच्छामि। (कारण + तसिल्), (अवलोकयितुम्)
उत्तर:
कारणतः, अव + लोक् + तुमुन्

21. अतः सुगाङ्गप्रासादस्य उपरि …………… प्रदेशाः संस्क्रियन्ताम्। (स्था + क्त)
उत्तर:
स्थिताः

22. कौमुदीमहोत्सव ………….. ? (प्रति + सिध् + क्त)
उत्तर:
प्रतिषिद्धः

23. यथा ………….. , चन्दनवारिणा भूमिं शीघ्र सिञ्चन्तु। (आ + दिश् + क्त)
उत्तर:
आदिष्टः

24. इदम् अनुष्ठीयते देवस्य …………….. इति। (शास् + ल्युट्)
उत्तर:
शासनम्

25. अयम् ……………. एव देव चन्द्रगुप्तः। (आ + गम् + क्त)
उत्तर:
आगतः

26. राज्यं हि नाम धर्म …………….. परकस्य नृपस्य कृते महत् कष्टदायकम्। (वृत् + क्तिन्)
उत्तर:
वृत्ति

27. नाट्येन …………….। (आ + रुह् + ल्यप्)
उत्तर:
आरुह्य

28. तत्कथं कौमुदीमहोत्सवः न ……………….. ? (प्र + आ + रभ् + क्त)
उत्तर:
प्रारब्धः

29. देव! न अतः परं ……………… …………….। (वि + ज्ञाप् + णिच् + तुमुन्), (शक्य:)
उत्तर:
विज्ञापयितुम्, शक् + यत्

30. न खलु आर्यचाणक्येन …………. प्रेक्षकाणाम् अतिशय ………….. चक्षुषो विषय:?(अप + हृ + क्त), (रम् + अनीयर्)
उत्तर:
अपहृतः, रमणीयः

31. आर्य! आचार्यचाणक्यं …………. इच्छामि। (दृश् + तुमुन्)
उत्तर:
द्रष्टुम्

32. ततः स्वभवनगतः आसनस्थः चिन्तां …………….: चाणक्यः प्रविशति। (नट् + शतृ)
उत्तर:
नाट्यन्

33. आकाशे लक्ष्य ……………। (बध् + क्त्वा)
उत्तर:
बद्ध्वा

34. अहो राजाधिराजमन्त्रिणो …………… (वि + + + क्तिन्)
उत्तर:
विभूति:

35. इतः शिष्यैः: ……………. दर्भाणाम् स्तूपः। (आ + नी + क्त)
उत्तर:
आनीतः

36. अत्र ………….. समिभिः अतिनमितः छदिप्रान्तः। (शुष् + शानच्)
उत्तर:
शुष्यमाणैः

37. ………………. भित्तयः। (जीर् + क्त्)
उत्तर:
जीर्णाः

38. अतएव निस्पृह ………….. एतादृशैः जनैः राजा तृणवद् गण्यते। (त्याग + इन्)
उत्तर:
त्यागिभिः

39. भूमौ ……………. । (नि + पत् + ल्यप्)
उत्तर:
निपत्य

40. देव: आर्यचन्द्रगुप्तः आर्य शिरसा ……………… विज्ञापयति। (प्र + नम् + ल्यप्)
उत्तर:
प्रणम्य

41. तर्हि आर्य …………. इच्छामि। (दृश् + तुमुन्)
उत्तर:
द्रष्टुम्

42. किं ……………. कौमुदीमहोत्सव-प्रतिषेधः?। (ज्ञा + क्त)
उत्तर:
ज्ञातः

43. स्वयमेव देवेन ……………. । (अव + लोक् + क्त)
उत्तर:
अवलोकितम्

44. आः …………….. भवद्भिः एव …………….. बृषलः। (ज्ञा + क्त), (प्र + उत् + सह् + णिच् + ल्यप्) (कोपित:)
उत्तर:
ज्ञातम्, प्रोत्साह्य, कुप् +णिच् + क्त

45. भयं …………….। (नट् + शतृ)
उत्तर:
नटायन्

46. आर्य, देवेन एव अहम् आर्यस्य चरणयो ………………. । (प्र + इष् + क्त)
उत्तर:
प्रेषित:

47, नाट्येन …………………. अवलोक्य च। (आ + रुढ् + ल्यप्)
उत्तर:
आरुह्य

48. ……………….. विजयताम् वृषलः। (उप + सृ + ल्यप्)
उत्तर:
उपसृत्य

49. वृषल! किमर्थं वयं ……………. ? (आ + ह् + क्त)
उत्तर:
आहूताः

50. आर्यस्य दर्शनेन आत्मानम् ……………… । (अनु + ग्रह् + तुमुन्)
उत्तर:
अनुग्रहीतुम्

51. (स्मितं कृत्वा ) …………………. तर्हि वयम् आहूताः। (उप + आ + लभ् + तुमुन्)
उत्तर:
उपालब्धुम्

52. नहि, नहि, ……………….. (विज्ञापयितुम्)
उत्तर:
वि + ज्ञाप् + णिच् + तुमुन्

53. यदि एवं तर्हि शिष्येण गुरोः आज्ञा ………… (पाल् + अनीयर् + टाप्)
उत्तर:
पालनीया

54. किन्तु न कदाचित् आर्यस्य (निस् + प्रयोजन + टाप्), (प्रवृत्तिः)
उत्तर:
निष्प्रयोजना, प्र + वृत् + क्तिन्

55. अतएव ………………. इच्छामि। (श्रु + तुमुन्)
उत्तर:
श्रोतुम्

56. वैहीनरे! तिष्ठ तिष्ठ। न …………… (गम् + तव्यत्)
उत्तर:
गन्तव्यम्

57. आर्येण एव सर्वत्र निरुद्धचेष्टस्य में …………………. इव राज्यं, न राज्यम् इव। (बन्ध् + ल्युट्)
उत्तर:
बन्धनम्

58. पितृवधात् ……………….. राक्षसोपदेशप्रवण ………………. म्लेछबलेन ………………. (क्रुध् + क्त) (महत् + ईयसुन्), (परिवृत:)
उत्तर:
क्रूद्धः, महीयसा, परि+ वृत् + क्त

59. पर्वतक पुत्र: मलयकेतुः अस्मान् ……………. उद्यतः। (अभि + युज् + तुमुन्)
उत्तर:
अभियोक्तुम्

60. अतः इदानीं दुर्गसंस्कारः …………….. (प्र + आ + रभ् + तव्यत्)
उत्तर:
प्रारब्धव्यः

61. अस्मिन् समये किं कौमुदी-महोत्सवेन इति (प्रति + सिध् + क्त)
उत्तर:
प्रतिसिद्ध

62. राष्ट्रचिन्ता ननु ………………. (गुरु + ईयसुन् + ङीप्)
उत्तर:
गरीयसी

63. प्रथम राष्ट्र ……………….. ततः उत्सवाः इति। (सम् + रक्ष् + ल्युट्)
उत्तर:
संरक्षणम्

64. ……………… फलप्रदा भवति। (दूर + दृश् + क्तिन्)
उत्तर:
दूरदृष्टिः

65. जीवने दूरदृष्टिं विना ………………. न लभ्यते। (सफल + तल्)
उत्तर:
सफलता

66. प्रायः वयं ……………… सुखमेव पश्यामः। (तत्काल + ठक्)
उत्तर:
तात्कालिकम्

67. यः जन ……………….. भवति, सः सर्वान् पक्षान् ……………. निर्णयं करोति। (दूरदर्श + इन्), (गण् + क्त्वा)
उत्तर:
दूरदर्शी, गणयित्वा

68. पञ्चतन्त्रे विष्णुशर्मा राजपुत्रान् एतदेव ……………. प्रयत्नं करोति। (शिक्ष + तुमुन्)
उत्तर:
शिक्षयितुम्

69. परन्तु यद्भविष्य: भविष्यम् कथं विनश्यति। (आ + श्रि + ल्यप्)
उत्तर:
आ + श्रि + ल्यप्

70. कस्मिश्चित् जलाशये अनागत ………….. प्रत्युत्पन्न ……….. यद् भविष्यश्च त्र्योमत्स्याः वसन्ति स्म।। (वि + धा + तृच्), (मन् + क्तिन्)
उत्तर:
विधाता, मतिः

71. तं जलाशयं दृष्ट्वा …………… मत्स्य जीविभि. ………. (गम् + शतृ), ( वच् + क्त)
उत्तर:
गच्छभिः , उक्तम्

72. बहुमत्स्योऽयं हृदः कदापि न अस्माभिः (अनु + इष् + क्त)
उत्तर:
अन्विष्टः

73. अद्य …………………. (आहारवृत्ति + क्तिन्), (सम् + जन् + क्त + टाप्)
उत्तर:
आहारवृत् + क्तिन्, सञ्जाता

74. सन्धया-समयः अपि । (सम् + वृत् + क्त)
उत्तर:
संवृत्तः

75. ततः प्रभाते अत्र ………………… इति निश्चयः। (आ + गम् + तव्यत्)
उत्तर:
आगन्तव्यम्

76. तेषां तद् वचः ……………. अनागतविधाता सर्वान् : अवदत्। (सम् + आ + कर्ण + ल्यप्), (आ + वे + ल्यप्)
उत्तर:
समाकर्त्य, आहूय

77. अहो! ………….. भवद्भिः यत् मत्स्य जीविभिः ……………? (श्रुतम्), (अभि + धा + क्त्)
उत्तर:
श्रु + क्त्, अभिहितम्

78. ……………… च। (वच् + क्त)
उत्तर:
उक्तम्

79. …………….. ……………… शत्रोः कर्तव्यम् ………………. (अशक् + क्त), (बल + इन्), (प्रपलायनम्)
उत्तर:
अशक्तैः, बलिनः, प्र + पल् + णिच् + ल्युट्

80. ………………… अथवा दुर्ग: नान्या तेषां ……………. भवेत्। (आश्रयितव्यः), (गम् + क्तिन्)
उत्तर:
आ + श्रि + तव्यत्, गतिः

81. प्रभातसमये ………… अत्र ……….. मत्स्यसंक्षयं करिष्यन्ति। (मत्स्यजीविनः), (सम् + आ + गम् + ल्यप्)
उत्तर:
मत्स्यजीव् + इन्, समागम्य

82. तन्न ………………… क्षणम् अपि अत्र ……………………. (युज् + क्त), (अव+ स्था + तुमुन्)
उत्तर:
युक्तं, अवस्थातुम्

83. ………………… गतिः येषाम्। (विद्यमान + टाप्)
उत्तर:
विद्यमाना

84. अहो! सत्यम् ……………… भवता। (अभि + धी + क्त)
उत्तर:
अभिहितम्

85. उच्चैः …………… यद्भविष्यः उवाच। (वि + हस् + ल्यप्)
उत्तर:
विहस्य

86. भवद्भ्यां न सम्यक् …………… (मन्त्र् + क्त)
उत्तर:
मन्त्रितम्

87. किं वाङ्मात्रेण अपि एतत् सर: ……………. युज्यते? (त्यज् + तुमुन्)
उत्तर:
त्यक्तुम्

88. अन्यत्र …………… अपि मृत्युः भविष्यति एव। (गम् + क्त)/गम् + ल्युट्)
उत्तर:
गते/गमनात्

89. अरक्षितं तिष्ठति दैव ………………. (रक्ष् + क्त)
उत्तर:
रक्षितम्

90. सुरक्षितं दैव …………….. विनश्यति। (हन् + क्त)
उत्तर:
हतम्

91. जीवति अनाथोऽपि वने ……………..। (वि + सृज् + क्त)
उत्तर:
विसर्जितः

92. भवद्भ्यां यत् प्रतिभाति, तत् ……………… (कृ + तव्यत्)
उत्तर:
कर्तव्यम्

93. अनागतविधाता प्रत्युत्पन्न मतिश्च परिजनेन सह ……………… (निस् + क्रम् + क्त)
उत्तर:
निष्क्रान्तौ

94. तैः मत्स्य ………………………. जालैस्तं जलाशयम्। …………… यद्भविष्येण सह (जीव + इन्), (आलोड्य)
उत्तर:
जीविभिः, आ + लोड् + ल्यप्

95. तत् सरो …………….: नीतम्। (निर् + मत्स्य + तल्)
उत्तर:
निर्मत्स्यताम्

96. अहो राजते ……………. इयं …………… (कीदृश + ङीप्), (हिमान + ङीप्)
उत्तर:
कीदृशी, हिमानी

97. विद्यालयस्य वार्षिकपत्रिकायां पर्वत …………… ……………… (आ + रोह् + ल्युट्), (कृ + शतृ)
उत्तर:
आरोहणम्, कुर्वताम्

98. छात्राणां चित्राणि ……………… । (प्र + दृश् + णिच् + क्त)
उत्तर:
प्रदर्शितानि

99. सर्वे छात्राः शिक्षिकाम् ………….. तस्याः यात्रायाः ………….. वृत्तान्तं अनुरोधं कुर्वन्ति। (उप + इ + ल्यप्), (ज्ञातुम्)
उत्तर:
उपेत्य, ज्ञा + तुमुन्

100. शिक्षिका तेषाम् आग्रह यात्रायाः …………….. अनुभवं प्रसतौति। (मन् + क्त्वा), (रोमञ्चकारिणम्)
उत्तर:
मत्वा, रोमाञ्चकार + इन्।

101. ……………… किन्नु खलु असाध्यम्। (दृढ़संकल्प + मतुप्)
उत्तर:
दृढ़संकल्पवताम्

102. गतवर्षे द्वादशकक्षायाः छात्रा: मम ……………. लेह-लद्दाखनगर …………….. (सम् + रक्षक + त्व), (प्रयाता:)
उत्तर:
संरक्षकत्वे, प्र + या + क्त

103. चित्राणि ………………… प्रतिभाति यत् अभियानं अतीव ……………… आसीत्। (वि + ईक्ष् + ल्यप्), (साहस + ठक्)
उत्तर:
वीक्ष्य, साहसिकम्

104. तत् सर्वं यात्रावृत्तान्तं यूयं ………………. वाञ्छथ? (दृश् + तुमुन्)
उत्तर:
द्रष्टुम

105. ……………… उपविशत। (शुभ् + ल्युट्)
उत्तर:
शोभनम्

106. एते लद्दाख ………… गिरयः अतीव शोभन्ते। (प्रदेश + ईयसुन्)
उत्तर:
प्रदेशीयाः

107. सा उपत्यका ………….. सिन्धुनदी अस्ति। (वि + भज् + शतृ + ङीप्)
उत्तर:
विभजन्ती

108. बालुका …………. सर्वां भूमिम् आवृणोति। (उत् + डी + ल्यप्)
उत्तर:
उड्डीय

109. चित्रे …………….: कानि एतानि स्थलानि? (दृश् + शानच्)
उत्तर:
दृश्यमानानि

110. सिन्धूनद्या ……………… लेहनगरस्य दक्षिणपूर्वभागे एते …………….. बौद्धमठा: सन्ति। (पूर्व + तसिल्), (प्रख्याताः)
उत्तर:
पूर्वत:, प्र + ख्यै + क्त

111. किं-किम् ………………. ? ज्ञातुम् इच्छामः। (अव + लोक् + क्त), (भवत् + ङीप्)
उत्तर:
अवलोकितम्, भवत्या

112. मठाना ……………. भव्यता च प्रेक्षका आकर्षतः। (विशाल + तल्)
उत्तर:
विशालता

113. …………….. बुद्धस्य विशालकाया मूर्तिः पर्यटकानाम् आकर्षण केन्द्रम्। (भग + मतुप्)
उत्तर:
भगवतः

114. बौद्धानां …………….: जीवनं कीदृशम्? (समाज + ठक्)
उत्तर:
सामाजिक

115. मठेषु ……………. लेखाः तिब्बतशैल्या: परिचायकाः। (उत् + कीर् + क्त)
उत्तर:
उत्कीर्णाः

116. प्राकृतिकस्थलानां विषये किमपि ब्रवीतु ……………। (भवत् + ङीप्)
उत्तर:
भवती

117. भारतीयैः वीरैः यत् शौर्य ……………. यूयं जानीथ एव। (प्र + दृश् + णिच् + क्त)
उत्तर:
प्रदर्शितम्।

118. ग्रीष्मे ………….. स ………….. कृषकाणां भूमि-सेचने …………. उपकरोति। (सम् + आप् + क्त), (द्रवी + भू + ल्यप्), (भूयस् + इष्ठन्)
उत्तर:
समाप्ते, द्रवीभूय, भूयिष्ठम्

119. पर्वत …………….. ‘लिकिर’ ‘स्टाक’ नाम्नी स्थले ……………. स्तः। (आ + रोह + ल्युट्), (उप + युज् + क्त)
उत्तर:
आरोहणाय, उपयुक्ते

120. हिमं न सौभाग्य ……………. ……………..। (वि + लोप + इन्), (जन् + क्त)
उत्तर:
विलोपि, जातम्

121. विपदि …………… मानवाः सुभाषितैः आश्वासनं प्राप्नुवन्ति। (नि + पत् + क्त)
उत्तर:
निपतिताः

122. संस्कृत वाङ्मयं समधुरवचनैः सम्यग् ……………. वर्तते। (अलम् + कृ + क्त)
उत्तर:
अलङ् कृतं

123. केषाञ्चित् ” मधुरवचनानां सङ्कलनं अत्र प्रस्तूयते। (अमृतवर्षा + इन्)
उत्तर:
अमृतवर्षिणाम्।

124. एतादृशमेव सङ्क लनं यूयमपि ……………. शक्नुथ। (कृ + तुमुन्)
उत्तर:
कर्तुम्

125. …………. प्रसाद सदनम्। (वद् + ल्युट्)
उत्तर:
वदनम्।

126. ……………… परोपकरणं येषां केषां न ते ……………. (कृ + ल्युट्), (वन्द् + यत्)
उत्तर:
करणं, वन्द्याः

127. …………… च ……………. च सदैव तिष्ठति। (+ क्त), (दा + क्त)
उत्तर:
हुतं, दत्तं

128. ये मायाविषु ………….. न भवन्ति (माया + इन्)
उत्तर:
मायिनः

129. शठाः हि निशिताः इषवः इव तथाविधान् असंवृत्ताङ्गान् ………………….घ्नन्ति। (प्रविश्य)
उत्तर:
प्र + विश् + ल्यप्

130. यदि …………. अस्ति पातकैः किम्? (पिशुन + तल्)
उत्तर:
पिशुनता

131. यदि …………… गुणैः किम्? (सुजन + ण्यत्)
उत्तर:
सौजन्यम्

132. सः चारुदत्त: उज्जयिनी …………………….. सङ्गीतविद्यायाः च प्रेमी अस्ति। (वास + इन्), (गुण + मतुप्)
उत्तर:
वासी, गुणवान्।

133. किन्नु खलु संविधा ……………. न वेति। (वि + धा + क्त + टाप्)
उत्तर:
विहिता

134. यावत् …………… शब्दापयामि। (आर्य + टाप्)
उत्तर:
आर्याम्

135. एवं …………… भोजनानां ………………… भव। (शुभ + ल्युट्), (दा + तृच् + ङीप्)
उत्तर:
शोभनानां, दात्री

136. अहम् पर्वताद् दूरम्……………. अस्मि। (आ + रोप् + ल्यप्), (पत् + णिच् + क्त)
उत्तर:
आरोप्य, पातितः

137. कञ्चिद् जनं ………………. इच्छामि। (नि + मन्त्र् + तुमुन्)
उत्तर:
निमन्त्रयितुम्

138. आर्य! ……………. असि। (नि + मन्त्र + क्त)
उत्तर:
निमन्त्रितः

139. सम्पन्नम् ……………… भविष्यति इति। (अश् + ल्युट्)
उत्तर:
अशनम्।

140. एष तत्र भवान् आर्यः चारुदत्त: गृहदैवतानि ……….. इत एवागच्छति। (अर्च् + शतृ)
उत्तर:
अर्चयन्

141. चरिका ……………. चेटी प्रविशति। (हस्त + टाप्)
उत्तर:
हस्ता

142. भो …………… खलु …………. पुरुषस्य सोच्छ्वासं …………….. (दरिद्र + ण्यत्), (मनस् + इन्), (मृ + ल्युट्)
उत्तर:
दारिद्र्यम्, मनस्विनः, मरणम्

143. अलम् इदानीं भवान् अतिमात्रं …………….। (सम् + तप् + तुमुन्)
उत्तर:
सन्तप्तुम्

144. न खल्वहं …………… श्रियम् अनुशोचामि। (नश् + क्त + टा)
उत्तर:
नष्टाम्

145. सुखं हि दु:खानि …………… शोभते। (अनु + भू + ल्यप्)
उत्तर:
अनुभूय

146. यथा अन्धकारात् इव दीप …………….. । (दृश् + ल्युट्)
उत्तर:
दर्शनम्

147. सुखात् तु यो याति दशां …………..। (दरिद्र + तल्)
उत्तर:
दरिद्रताम्

148. ‘दानं श्रेयस्करम्’ इति प्रत्ययात् एव ममार्थाः क्षीणाः ………………… (जन् + क्त)
उत्तर:
जाताः

149. धनविनाशदु:खस्य पुनः पुनः …………….. चिन्ताकुराः प्रादुर्भवन्ति। (चिन्त् + शानच्)
उत्तर:
चिन्त्यमानस्य

150. ……………… विज्ञानजगत्। (आश्चर्य + मयट्)
उत्तर:
आश्चर्यमयं

151. विद्यालयस्य सूचनापट्टे (वि + ज्ञाप् + क्तिन्)
उत्तर:
विज्ञप्तिः

152. प्राचीनभारतीय …………… सङ्गोष्ठी भविष्यति। (वि + ज्ञा + ल्युट्), (अधि + कृ + ल्यप्), (एक + टाप्)
उत्तर:
विज्ञानम्, अधिकृत्य, एका

153. अस्माकं छात्रैः यद् विशिष्टाध्ययनं …………… (कृ + क्त)
उत्तर:
कृतम्

154. सभागारस्य …………… इदम् । (दृश् + यत्)
उत्तर:
दृश्यम्

155. अस्माकं मध्ये अद्य सर्वोत्तमान् अङ्ङ्कान्। (लभ् + क्तवतु), (सम् + उप + स्था + क्त)
उत्तर:
लब्धवन्तः, समुपस्थिताः

156. करतलध्वनिना एतेषां स्वागतम् ……………..: च कुर्वन्तु। (अभि + नन्द् + ल्युट्)
उत्तर:
अभिनन्दनम्

157. तृतीयाभागस्सञ्चारस्त्वन्तरिक्षे भवेत् “……………… (स्व + तसिल्)
उत्तर:
स्वतः

158. विमानशास्त्रे एतादृशः लेपः अपि वर्णित: यस्य लेपनेन विमानं …………….. गच्छति। (अदृश्य + तल्)
उत्तर:
अदृश्यताम्

159. करतलध्वनिना अभिनवस्य उत्साह – …………… कुर्वन्तु। (वृध् + ल्युट्)
उत्तर:
वर्धनम्

160. इदानीं …………… सुश्रुतस्य शल्यक्रियाम् …………. अस्माकं ज्ञानवृद्धिं करिष्यति। (शालिन् + ङीप्), (अधि + कृ + ल्यप्)
उत्तर:
शालिनी, अधिकृत्य

161. प्रिय सह ………….. ! यदि वयम् उत्तमचिकित्सका …………….. इच्छामः। (पाठ + इन्), (भू + तुमुन्)
उत्तर:
पठिनः, भवितुम्

162. तर्हि सुश्रुत …………….. सुश्रुतसंहिता अवश्यमेव ……………… (विरचित + टाप्), (पठ् + अनीयर् + टाप्)
उत्तर:
विरचिता, पठनीया

163. तत्र अष्टविधं शल्यकार्यं ……………….. यथा- ………….. ………………. आदयः। (वर्ण + क्त), (छिद् + यत्), (भिद् + यत्)
उत्तर:
वर्णितम्, छेद्यम्, भेद्यम्

164. ……………. खलु ………………. सुश्रुतस्य शल्यक्रिया। (बहुप्रसिद्ध + टाप्), (भग + मतुप्)
उत्तर:
बहुप्रसिद्धा, भगवतः

165. ……………. एव …………….. प्रस्तुतिः । (प्र + शंस् + अनीयर् + टाप्), (भवत् + ङीप्)
उत्तर:
प्रशंसनीया, भवत्याः

166. मिहिरः ………….. भूमेः अध्: कुत्र कुत्र जलं भवति इति सूचनां प्रदास्यति। (बृहत्संहिता + तसिल्)
उत्तर:
बृहत्संहितात:

167. मिहिरेण …………….. सामग्री ……………। (बहुमूल्य + टाप्), (सम् + कल् + क्त + टाप्)
उत्तर:
बहुमूल्या सङ्कलिता

168. भारती आर्यभटीयम् इति ग्रन्थस्य ……………….. वर्णयिष्यति। (विशिष्ट + ण्यत्)
उत्तर:
वैशिष्ट्यं

169. यथा मनुष्यः वृक्षादीन् पृष्ठः प्रति ……………… पश्यति। (गम् + शतृ)
उत्तर:
गच्छतः

170. तथा नक्षत्रादयः नरस्य कृते पश्चिमं प्रति ……………. प्रतीयन्ते। (धाव् + शतृ)
उत्तर:
धावन्तः

171. सः स्तम्भ: ……………. विना तथैव तिष्ठति। (वि + कृ + क्तिन्)
उत्तर:
विकृतिम्

172. अहमदनगरे ……………. स्तम्भा: भारतीयवैज्ञानिकानां गौरवगाथा वर्णयन्ति। (कम्प् + शानच्)
उत्तर:
कम्पमानाः

173. इदानीम् गोष्ठी ………………….. याति। (एतत् + टाप्), (सम् + आप् + क्तिन्)
उत्तर:
एषा, समाप्ति

NCERT Solutions for Class 12 Sanskrit

The post CBSE Class 12 Sanskrit व्याकरणम् प्रकृति-प्रत्यय-विभाग appeared first on Learn CBSE.

Notice Writing Class 12 Format, Examples, Topics, Exercises

Notice Writing Class 12 Format, Examples, Topics, Exercises

Notice is a written or printed information or news announcement. Notices are either displayed at prominent places or published in newspapers/ magazines. It is meant only for a select group. Since a notice contains a formal announcement or information, its tone and style is formal and factual. Its language should be simple and formal. Notice is always brief and to the point.

♦ Important Tips to be followed while writing a Notice

  • Adhere to the specified word limit of 50 words.
  • Write the word NOTICE at the top.
  • Name and place of the school, organisation or office issuing the notice should be mentioned.
  • Give an appropriate heading.
  • Write the date of issuing the notice.
  • Clearly mention the target group (for whom the notice is to be displayed).
  • Purpose of the notice.
  • Mention all the relevant details (date, venue, time).
  • Signature, name and designation of the person issuing the notice.
  • Put the notice in a box.

Format of a Notice
A NOTICE is always written in a box.

Notice:
Name of the Institution/Organization/Office, Place Suitable Heading

29 June 2012
Contents. The target group for whom the notice is. Date, time, venue and all-important details and any extra information that is needed (Body of the Notice, 50 words)

Signature
(NAME)
Designation

♦ Marking Scheme of Notice
format title (Notice/Heading/Place/Date/Signature with Name and Designation) 1 mark
Contents
(a) Where?
(b) When
(c) Target group
(d) Agenda
(All within 50 words) 2 marks
ExpressionCoherence, spellings and grammatical accuracy 2 marks

♦ Previous Years’ CBSE Examination Questions

♦ Short Answer Type Question s

Question 1.
As Principal of Sardar Patel Vidyalaya, Lucknow, draft notice in not more than 50 words informing students of the change in school timings with effect from the 1st of October. State valid reasons for the change. (Delhi 2009)
Answer:

Sardar Patel Vidyalaya, Lucknow
NOTICE

22nd September, 20xx

Change in School Timings
All students are hereby informed about a change in school timings from 1st October, 20xx. The school will now start at 9 a.m. & end at 3 p.m. In the past few years, it has been seen that winters are rather severe and it becomes quite difficult to start early due to the extreme cold weather and the dense fog. So these new timings will be followed till further notice.

Principal

Question 2.
You are Secretary of the History Club of Vidya Mandir School. Draft a notice in not more than 50 words informing students of a proposed visit to some important historical sites in your city. (Delhi 2009)
Answer:

Vidya Mandir School
History Club
Notice

22nd September, 20xx

Visit Historical Sites
Members of the club are hereby informed of a proposed educational visit to a few important historical sites in our city which is likely to be scheduled between the 28th and the 30th of Sep. Interested members are required to pay ? 550 (inclusive of transport and snacks) during the zero periods to the undersigned by the 25th of September.

Secretary
History Club

Question 3.
As Sports Secretary of G.D.G. Public School, Pune, draft notice in not more than 50 words for your school notice board informing the students about the sale of old sports goods of your school. You are Rohini/Rohit. (Delhi 2010)
Answer:

G.D.G. Public School
Pune
Notice

29th July, 20xx

Sale of Old Spoils Goods
Students are hereby informed that our school is organising a sale of its old sports goods like cricket bats, badminton & lawn teams rackets, footballs, cricket & football gear etc. in the P.E. Room on 2K1 August, 2Oxx. Those interested in purchasing these can visit the P.E. Room on the assigned date during their free periods or recess time.

Rohit
Sports Secretary.

Question 4.
You have found an expensive geometry box in the school playground. Draft a notice in not more than 50 words for the school notice board. You are Ra’Rani, Class XII, Angel School, Faridabad. (Delhi 2010)
Answer:

Angel School, Faridabad
Notice

31 October, 2Oxx

Found A Geometry Box
Have found a red-coloured ‘Faber-Castle’ Geometry Box in the school playground during the 8th period yesterday. Owner may please contact the undersigned in her class between the 6th and the 71 periods.

Rani
XII-C

Question 5.
You are SrinivafSrinidhi of D.P. Public School, Nagpur. As Student Editor of your school magazine, draft notice in not more than 50 words for your school notice board inviting article sketches from students of all classes. (Delhi 2011)
Answer:

D.P. Public School, Nagpur
Notice

Bring in your Articles and Sketches

29e” July, 2Oxx
Students of all classes are hereby invited to submit their articles and sketches for the school magazine to the undersigned Latest by the 6th of August in Room No. 102 during the V period. Please ensure that your articles are neat and legible and your sketches are clearly drawn.

Srinidhi
Editor
School Magazine

Question 6.
You are Secretary of Gymkhana Club, Madurai. Write a notice in not more than 50 words informing the members to attend an
an extraordinary meeting of the governing body. Include details like date, time, venue etc. Sign as PrabhufPratibha. (All India 2011)
Answer:

Gymkhana Club, Madurai
Notice

Extraordinary Meeting of Governing Body
22nd September, 20xx

All members of the governing body of the Club are hereby informed to attend an extraordinary meeting on 2S” September, 20xx at 4 p.m. in the Club Conference Room to discuss how to revise the Club’s policy for defaulter members for long outstanding dues.

Prabhu
Secretary

Question 7.
Due to a sudden landslide and inclement weather, St. Francis School, Vasco has to be closed for a week. As Principal of that school, draft notice in not more than 50 words to be displayed at the school main gate notice board. (All India 2011)
Answer:

St. Francis School, Vasco
Notice

School Closed for a Week

3rd January, 20xx
All are hereby informed that the school will remain closed for one week owing to the sudden landslide and inclement weather in the past few days. The school will re-open on the 10th of January, 20xx.

Principal

Question 8.
You are Secretary of J.P. Narain Housing Society, R.W.A., Meerut. Draft a notice in not more than 50 words stating that the second instalment of maintenance charges falls due on 31st March 2011, and requesting the members to pay before the due date. Sign as Anil/Anita. (All India 2011)
Answer:

J.P. Narain Housing Society
Meerut
Notice

All members of the Residents’ Welfare Association are hereby informed that the payment of the second instalment of maintenance charges is due on 31st March, 2011. Please pay before the due date to avoid late payment charges.

You are reminded that the amount of the same is? 550 only.

Anil
Secretary, R.W.A.

Question 9.
You are Scout Master/Guide Captain of K.R. Sagar Public School, Mysore. You have decided to send a troop of scouts and guides of your school to the jamboree to be held at Lucknow for a week. Draft a notice in not more than 50 words to be placed on the school notice board inviting the names of those scouts and guides who are interested to participate in the jamboree. Invent the necessary details. (All India 2011)
Answer:

K.R. Sagar Public School Mysore
Notice

Scouts & Guides Participation Needed
1st October, 20xx

Our school has decided to send a troop of scouts and guides to the jamboree to be held at Lucknow from the 20th to the 27th of October. Those scouts and guides interested to participate in the jamboree may give their names to the undersigned by the 7th of October.

Scout Master

Question 10.
You are Sports Secretary of Lalwani Public School, Udaipur. Draft a notice in not more than 50 words for your school notice board asking the students to give their names for participation in various events to be held on the Annual Sports Day of your school. Invent the details of the events. Sign as Lalit/Lalita. (All India 2011)
Answer:

Lalwani Public School
Udaipur
Notice

6th November, 20xx

Our school is celebrating its Annual Sports Day on 30th November, 20xx. Students are hereby informed to give their names for participation in various events to be held on the Sports Day.
The list of events is given below:

  • 100 metres race
  • high jump
  • 200 metres relay
  • short put
  • gymnastics
  • kho-kho

Please give your names to the undersigned latest by the 10th of November during the ‘0’ period in the Sports Room.

Lalit
Sports Secretary

Question 11.
Your school has planned an excursion to Lonavala near Mumbai during the autumn holidays. Write a notice in not more than 50 words for your school notice board, giving detailed information and inviting the names of those who are desirous to join. Sign as NaresfyNamita, Head Boy/Head Girl, D.V. English School, Thane, Mumbai. (Delhi 2012)
Answer:

D.V. English School
Thane, Mumbai
Notice

22nd September, 20xx

Excursion TO Lonavala Our school has planned a four-day excursion to Lonavala Hill Station during the autumn break from the 10th to 13th October. Activities such as trekking, mountain biking, nature walks are the highlights of the trip. The total cost of this trip is? 5,000. Students desirous to join this trip may give in their names along with the money and a consent letter from their parents to the undersigned latest by the 1st of October.

Namita
Head Girl
XII – A

Question 12.
You lost your Titan wrist-watch in your school. Draft a notice, in not more than 50 words, to be placed on your school notice board. You are a student of Class XII of Rani Ahalya Devi Senior Secondary School, Gwalior. Sign as Rani/Ram. (All India 2012)
Answer:

Rani Ahalya Devi Senior Secondary School Gwalior
Notice

Lost Wrist Watch
7th August, 20xx

Lost a black strap, white dial Titan wrist-watch in the school playground during the 7th period yesterday. Anyone who finds it please contact the undersigned in class XII-A. Finder will be rewarded with a treat in the canteen.

Ram
XII-A

Question 13.
As Student Editor, draft notice in not more than 50 words for your school notice board inviting articles from the students for your school magazine. You are RohaiVRupini of Vasant Vihar School, Pune. (All India 2012)
Answer:

Vasant Vihar School
Pune
Notice

Inviting Articles for School Magazine
5th July, 20xx

Students are hereby informed to submit their articles for our school magazine ‘Vasant Times’ latest by the 15th of July, 20xx. Articles can be neatly handwritten or typed on A4 size sheets with the name, class and section of the student-written under it. Please submit your articles to the under¬signed in Room no. 10 during the ‘0’ period.

Rupini
Student Editor

Question 14.
You are Smitha/Sunil, Secretary AVM Housing Society. You are going to organize a blood donation camp. Write a notice in not more than 50 words, urging the members of your society to come in large numbers for this noble cause. Invent all the necessary details. (Delhi 2013)
Answer:

AVM Housing Society
Notice

22nd September, 20xx

Blood Donation Camp Our society is organising a blood donation camp in the Community Centre on 1st October, 20xx, from 10 AM to 4 PM. All arrangements will be adequately made along with drinks and refreshments for all donors. The camp is being conducted under the expert guidance of the doctors and nurses of the Red Cross Society. Residents are hereby requested to ensure an overwhelming response to this camp by even persuading their friends and relatives to come forward for this noble task.

Smitha
Secretary

Question 15.
You are Vineeta/Vikram, School Pupil Leader of Rani Laxmi Bai Senior Secondary School, Gwalior. Draft a notice for your school notice board in not more than 50 words inviting the names of the students who want to participate in the cultural programme organised in aid of the victims of the recent Assam floods. (All India 2013)
Answer:

Rani Laxmi Bai Sr. Sec. School
Gwalior
Notice

12th September, 2013

Relief for Assam Flood Victims Our school is organizing a cultural programme in aid of victims of the recent Assam Floods on 30th September, 20xx. The different categories of the programme include singing, dance, skits and elocution. Students, who are interested to participate in this programme for a noble cause, may give their names to the undersigned latest by the 16th of September.

Vineeta Pupil Leader
2014

Question 16.
An interschool Kabaddi Competition is organized by your school. Write a notice, in . not more than 50 words, requesting the students to be present at the venue to encourage the players. Invent all the necessary details. You are Arjun, the sports captain of your school. (Delhi 2013)
Answer:

Abc Public School
Notice

22nd September, 20xx

Inter-School Kabaddi Competition
An interschool Kabaddi Competition is being organized by our school on 1st Oct., 20xx in the big field. 20 schools from different states will be participating in this mega event. The event will start at 8 am. All the students are hereby requested to be present at the venue to encourage the players.

Arjun
(Sports Captain)

Question 17.
You are Smrithi Saran of Victoria Public School, Hyderabad. Your school has organized a Science Exhibition in connection with the death anniversary of Ramanujam. Write a notice in not more than 50 words inviting students to participate in it. Provide all the necessary. (Delhi 2013)
Answer:

Victoria Public School
Hyderabad
Notice

25th February, 20xx

Science Exhibition

Our school is organizing a Science Exhibition to commemorate the death anniversary of the great mathematician Ramanujam on the 12th of March, 20xx in the school lawns. Students are hereby informed to participate in this exhibition and make it a success.

Smrithi Saran

Question 18.
You are Anoop/Arya, the cultural secretary of your school. As part of the national heritage programme, the school has decided to put up a show on ancient art forms. Write a notice to be put up on the school notice board inviting students to watch the show and encourage the artists. Write the notice in not more than 50 words. (Delhi 2013)
Answer:

Abc Public School
Notice

16th August, 20xx

Show on Ancient Art Forms As part of the national heritage programme, our school has decided to put up a show on ancient art forms on 24th August, 20xx in the school auditorium from 9:00 am onwards. Students are hereby invited to watch the show and encourage the artists.

Arya
Cultural Secretary

Question 19.
Every year in the central part of the city a flower show is held in the month of February. Your school has received a circular from the District Collector inviting your students to visit it. Write a notice in about 50 words informing the students about the show and advising them to go and enjoy it. You are Navtej/Navita, Head Boy/Head Girl, Sunrise Public School, Surat. (Delhi 2015)
Answer:

Sunrise Public School, Surat
Notice

1st February, 20xx

Flower Show Invitation:
Every year the central park of our city organises a flower show in the month of February. The District Collector has sent circular inviting students of our school to visit this show which will be held from the 6th to 25th February. Students are hereby advised to visit the central park and enjoy this show.

Navita (Flead Girl)

Question 20.
Sarvodaya Education Society, a charitable organisation is coming to your school to distribute books among needy students. As Head Boy/Head Girl, Sunrise Public School, Surat, write a notice in about 50 words asking such students to drop the lists of books they need in the box kept outside the Principal’s office. You are Navtej/Navita. (D) (Delhi 2015)
Answer:

Sunrise Public School, Surat
Notice

18th March, 20xx
Book Distribution

By Sarvodaya education society
Sarvodaya Education Society, a charitable organisation is coming to our school to distribute books among the needy students on 31st March. Interested students are hereby informed to drop the list of books they need in the box kept outside the principal’s office by the 24th of March.

Navtej (Head boy)

Question 21.
Your club is going to organise an interclass singing competition. Write a notice in about 50 words inviting names of the students who want to participate in it. Give all the necessary details. You are Navtej /Navita, Secretary, Music Club, Akash Public School, Agra. (All India 2015)
Answer:

Akash Public School, Agra
Notice

1st November, 20xx

Interclass Singing Competition
The Music Club of our school is organising an interclass singing competition for classes Ist to Xth on 19th November, 20xx. Those students who want to participate in the competition are hereby invited to give their names to the undersigned by 5th November in the Music Room.

Navita (Secretary, Music Club)

Question 22.
An inter-class drama competition is to be held in St. Stephens School, Visakhapatnam. As Akash, Head Boy of the school draft a notice to be put up on the notice board inviting entries. Provide all necessary information in about 50 words. (Comput. Delhi 2015)
Answer:

St. Stephens School,
Visakhapatnam
Notice

1st October, 20xx

Inter-Class Drama Competition:
An inter-class drama competition will be held in our school for Classes IXth to XIIth on 22nd Oct., 20xx. Students who are interested to participate may give in their names to the prefects of their respective classes latest by the 5th of October.

Akash (Head Boy)

Question 23.
You are Amar/Amrita, Secretary, Cultural Club, Aryamba Public School, Kochi. A charity show has been arranged in your school in aid of cancer patients. Write a notice to be displayed on the school notice board informing the students of the show and asking them to cooperate and make it a success. Draft the notice in about 50 words giving all necessary details. (Comput. Delhi 2015)
Answer:

Aryamba Public School, Kochi
Notice

22nd September, 20xx

Charity Show For Cancer Patients:
The Cultural Club of our school has arranged a charity show in aid of cancer patients on 7th October, 20xx at 9:00 a.m. in the school auditorium. The entry fee for the show is? 250 per person. Students are hereby requested to attend this show and thus cooperate in making this charitable and noble show a success.

Amar (Secretary, Cultural Club)

Question 24.
Water supply will be suspended for eight hours (10 am to 6 pm) on 6th of March for cleaning of the water tank. Write a notice in about 50 words advising the residents to store water for a day. You are Karan Kumar/Karuna Bajaj, Secretary, Janata Group Housing Society, Palam Vihar, Kurnool. (Delhi 2016)
Answer:

Janata Group Housing Society
Palam Vihar, Kurnool
Notice

4th March, 20xx

Suspension of Water Supply:
Residents are hereby informed that water supply will be suspended for eight hours (10 am to 6 pm) on 6th March for cleaning of the water tank. Residents are advised to make necessary arrangements to store water for a day. We regret the inconvenience caused.

Karan Kumar
Secretary

Question 25.
Yesterday, during lunch break you misplaced your notes on chemistry lectures. You want to get them back. Write a notice in about 50 words for the school notice board. You are Karuna/Karan, a student of class XII A. (All India 2016)
Answer:

Abc School
Notice

31st October, 20xx

Lost Chemistry Notes
Lost notes on Chemistry lectures of class XII during lunch break in the basketball ground on 31st October, 20xx. Notes are kept in a blue-coloured file. Anyone who finds it is requested to return it to the undersigned and get a treat in the canteen as a reward.

Karan
XII-A

Question 26.
You are Reshma/ Rajan Head Girl/Head Boy of Moonrise Public School, Chandigarh. A cooking competition is going to take place in your school. Write a notice to be displayed on the school notice board informing the students of the competition and inviting them to participate. Draft the notice in about 50 words giving all necessary details. (Comput. D 2016)
Answer:

Moonrise Public School
Chandigarh
Notice

22nd September, 20xx

Cooking Competition Students are hereby informed that a cooking competition will be taking place in our school Mini Auditorium on 1st October from 9 am to 11 am for students of classes Xth to XIIth. Those interested to participate, please give in your names to the undersigned by the 26th of September during the recess period in Room No. 101.

Reshma Head Girl

Question 27.
As a librarian, Moonlight Public School, Surat, write a notice in about 50 words informing the students of a book exhibition which is going to be organised in your school on Teacher’s Day. (Comput. All India 2016)
Answer:

Moonlight Public School, Surat
Notice

26th August, 20xx
Book Exhibition on Teachers’ Day

Students are hereby informed that our school is organising a Book Exhibition on 5th September on Teachers’ Day in the Football Ground from 9 am-7 pm. Re¬nowned educationist, Mrs. Rachna Kumar has kindly consented to be our chief guest. A special attraction of the book exhibition is a separate section of books in different languages and titles dedicated to teachers and educators.

Librarian

Question 28.
While walking in a park in your neighbourhood you found a small plastic bag containing some documents and some cash. Write a notice in about 50 words to be put on the park notice board asking the owner to identify and collect it from you. You are Amai/Amrita 9399123456. (Delhi 2017)
Answer:

Abc Park Notice Board
Notice

Lost And Found
17th January, 20xx

Found a small plastic bag in the park yesterday evening under the bench near the swings. It contains some documents, iden¬tity cards and H,500 cash. The owner is requested to identify and collect it from the undersigned between 12:00 pm to 5:00 pm.

Amrita
Contact: 9399123456

Question 29.
After the rains cases of dengue, Chikungunya etc., are on the rise in your city. As Principal, Sunshine Public School, Manu Vihar, you have decided to allow your students to wear full sleeve shirts and trousers in the school for a period of one month. Write the notice in about 50 words. (Delhi 2017)
Answer:

Sunshine Public School
Manu Vihar
Notice

12th September, 20xx

Wear Full Sleeve Shirts and Trousers After the rains, cases of Dengue, malaria, Chikungunya etc. have been on the rise in our city. So, the school keeping in mind the students’ health and safety has decided to allow the students to wear full sleeve shirts and trousers during school for a period of one month as a precaution against mosquito bites.

Principal

Question 30.
You are Health Secretary, Students Council, Citizens Public School, Ram Bagh, Varanasi. The Council had decided to start from the second of October week-long cleanliness drive around the school. Draft a notice in about 50 words asking the Class XI students to enrol for the drive. (Delhi 2017)
Answer:

Citizens Public School, Ram Bagh
Notice

1st October, 20xx

Cleanliness Drive:
The Students Council has decided to start week-long cleanliness drive around the school from the 7th to 14th October, 20xx. Students of Class XI are hereby requested to enrol for this drive. All interested volunteers are requested to give in your names to the undersigned latest by 4th October, 20xx.

Health Secretary:
Students Council

Question 31.
RJ Public School is located in a Central Government employees residential colony. Cultural Society of the school has decided to organise a fancy dress show on 25th of January in which each participant will wear the dress particular to higher region. The aim is to show the cultural diversity of India. As Secretary write a notice in about 50 words inviting the names of those who want to participate. (Delhi 2017)
Answer:

Rj Public School C.G. Employees
Residential Colony
Notice

5th January, 20xx

Fancy Dress Show
The Cultural Society of our school is organising a fancy dress show on 25th Jan., 20xx in which each participant has to wear the dress particular to his/her region. The aim of the show is to depict the cultural diversity of India. Those interested to participate are hereby informed to give in your names to the undersigned latest by 12th January, 20xx.

Secretary,
Cultural Society

Question 32.
An NGO has approached your school to offer book grants to needy students. As Head girl of Sunshine Public School, Aram Bagh, write a notice in about 50 words asking students who are in need to put their requests into the box kept outside the Principal’s office. (Delhi 2017)
Answer:

Sunshine Public School
Aram Bagh
Notice

7th March, 20xx

Book Grants for Needy Students
An NGO has approached our school to offer book grants to needy students. Students in need of this grant are hereby informed to put their requests into the box, with their name and class mentioned on it, outside the Principal’s office latest by 29th March, 20xx.

Head Girl

Question 33.
The Principal, Sunshine Public School, Dindigul has invited the Inspector of Police (Traffic) to deliver a lecture on ‘Road Safety’ in her school. Draft a notice in about 50 words informing the students to assemble in the school auditorium. (All India 2017)
Answer:

Sunshine Public School
Dindigul
Notice

29th July, 20xx

Lecture on Road Safety Our school has invited the Inspector of Police (Traffic) to deliver a lecture on ‘Road Safety’ in our school on 6th August, 20xx from 10 am to 11 am. Students are hereby informed to assemble in the school auditorium latest by 9:30 am with their respective class teachers.

Principal

RBSE Class 10 English Notes

The post Notice Writing Class 12 Format, Examples, Topics, Exercises appeared first on Learn CBSE.

Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Surface Areas and Volumes Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then calculate the rise of water level (in cm). (2011D)
Solution:
Volume of Cylinder = Volume of Sphere
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 1

Question 2.
Find the number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm. (2014D)
Solution:
Number of solid spheres
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 2

Question 3.
A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. Find the ratio of the volume of the smaller cone to the whole cone. (2012OD)
Solution:
Since the cone is cut from the middle,
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 3

Question 4.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? (2017D)
Solution:
Volume of hemisphere = Surface area of hemisphere …[Given
Image may be NSFW.
Clik here to view.
\frac{2}{3}
πr3 = 2πr2Image may be NSFW.
Clik here to view.
\frac{1}{3}
r = 1
r = 3
∴ Diameter of hemisphere = 27 = 2(3) = 6 cm

Surface Areas and Volumes Class 10 Important Questions Short Answer-I (2 Marks)

Question 5.
If the total surface area of a solid hemisphere is 462 cm2, find its volume. [Take π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2014OD)
Solution:
Total surface area of hemisphere = 462 cm2
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 4

Surface Areas and Volumes Class 10 Important Questions Short Answer-II (3 Marks)

Question 6.
Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid. (2011D)
Solution:
Length of resulting cuboid, I = 2(4) = 8 cm
Breadth of resulting cuboid, b = 4 cm
Height of resulting cuboid, h = 4 cm
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 5

Surface area of resulting cuboid
= 2(lb + bh +hl) = 2 [8(4) + 4(4) + 4(8)]
= 2 (32 + 16 + 32) = 2 (80) = 160 cm2

Question 7.
The radii of the circular ends of a bucket of height 15 cm are 14 cm and r сm (r < 14 cm). If the volume of bucket is 5390 cm3, then find the value of r. [Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2011D)
Solution:
Here h = 15 cm, R = 14 cm,'r' = r cm
Volume of bucket = 5390 cm3
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 6

⇒ r2 + 14r + 196 - 343 = 0
⇒ r2 + 14r - 147 = 0 = r2 + 21r - 77 - 147 = 0
⇒ r(r + 21) - 7 (r + 21) = 0
⇒ (r + 21) (r - 7) = 0
⇒ r + 21 = 0 or r - 7 = 0 =
⇒ r = -21 (rejected) or r = 7
..[∵ Radius cannot be negative
∴ Radius, r = 7 cm

Question 8.
Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid. (2011OD)
Solution:
Volume of a cube = 27 cm3
⇒ (Side)3 = (3)3 ∴ Side = 3 cm
Length of resulting cuboid, l = 2 × 3 = 6 cm
Breadth of resulting cuboid, b = 3 cm
Height of resulting cuboid, h = 3 cm
Surface area of resulting cuboid = 2(lb + bh + hl)
= 2(6 × 3 + 3 × 3 + 3 × 6)
= 2(18 + 9 + 18) = 2(45) = 90 cm2

Question 9.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder. (Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
) (2016D)
Solution:
Let the radius and height of cylinder be r and h respectively
r + h = 37 cm ...(i) [Given
Total surface area of cylinder = 1,628 cm2
2πr(r + h) = 1,628
⇒ 2πr(37) = 1,628
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Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 7

Question 10.
A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find the volume of the cone. (Use π = 3.14) (2016D)
Solution:
C.S. Area of cone = 47.1 cm2 ...[Given
...[Here, r = 3 cm
πrl = 47.1 = (3.14)(3)l = 47.1
Slant height, l =Image may be NSFW.
Clik here to view.
\frac{47.1}{3.14(3)}
= 5 cm
⇒ r2 + h2 = l2 ...[Pythagoras' theorem
⇒ (3)2 + h2 = (5)2 ⇒ h2 = 25 - 9 = 16
⇒ h = +4 cm
Volume of the cone = Image may be NSFW.
Clik here to view.
\frac{1}{3}
πr2h = Image may be NSFW.
Clik here to view.
\frac{1}{3}
(3.14)(3)2(4)
= 3.14(12) = 37.68 cm3

Question 11.
An icecream seller sells his icecreams in two ways: (2012OD)
(A) In a cone of r = 5 cm, h = 8 cm
(B) In a cup in shape of cylinder with r = 5 cm, h = 8 cm
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 8

He charges the same price for both but prefers to sell his icecream in a cone.
(a) Find the volume of the cone and the cup.
(b) Which out of the two has more capacity?
Solution:
Volume of type 'A'
Volume of cone + Volume of hemisphere
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 9

(a) ∴ Volume of a cone = 471.43 cm3
Volume of a cup = 628.57 cm3
(b) Cup has more capacity.

Question 12.
A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total (inner) suface area of the vessel. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2013D)
Solution:
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 10

r = Image may be NSFW.
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\frac{14}{2}
= 7 cm
Inner surface area of the vessel = C.S. area of Hemi-sphere + C.S. area of Cylinder
= 2πr2 + 2πrh = 2πr(r + h) ...C.S. area = curved surface area
= 2 × Image may be NSFW.
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\frac{22}{7}
× 77 + 6) = 44 × 13 = 572 cm2

Question 13.
A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 166Image may be NSFW.
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\frac{5}{6}
cm3. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹10 per cm2. [Use π = Image may be NSFW.
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\frac{22}{7}
] (2015D)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 11

Let the height of cone = h
Radius of cone = Radius of hemisphere = r = 3.5 cm
Volume of solid wooden toy = Volume of hemisphere + Volume of cone
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 12

Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 13

Question 14.
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs 100 per sq. m, find the amount, the associations will have to pay. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2015OD)
Solution:
Let l be the slant height of cone
h be the height of cone = 2.8 m
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 14

Question 15.
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ₹5 per 100 sq. cm. (Use π = 3.14) (2015OD)
Solution:
Let the side of cuboidal block (a) = 10 cm
Let the radius of hemisphere be r.
Side of cube = Diameter of hemisphere Largest possible diameter of hemisphere = 10 cm
∴ Radius, r = Image may be NSFW.
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\frac{10}{2}
= 5 cm
Total surface area = Total surface area of cube + Curved surface area of hemisphere- Area of base
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 15

Question 16.
In Figure, is a decorative block, made up of two solids-a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2016D)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 16

Solution:
Total surface area of the block
= Total surface area of cube + C.S. Area of hemisphere - Area of circle
= 6(side)2 + 2πr2 - πr2
= 6(side)2 + πr2
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 17

Question 17.
A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use π = Image may be NSFW.
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\frac{22}{7}
). (2016D)
Solution:
Let r and I be the radius and height of conical vessel i.e., 5 cm and 24 cm respectively.
Let H be the height of rise in water in cylindrical vessel and R be the radius, i.e., 10 cm.
Volume of water in cylindrical vessel = Volume of water in conical vessel
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 18

Question 18.
A milkman was serving his customers using two types of mugs A and B of inner diameter 5 cm to Mug‘A Mug‘B' serve the customers. The height of the mugs is 10 cm. (2012D)
He decided to serve the customers in 'B' type of mug.
(a) Find the volume of the mugs of both types.
(b) Which mathematical concept is used in the above problem?
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 19

Solution:
(a) Let the radius of cylinder, hemi-sphere and cone be r сm
Let the height of cylinder and cone h1 and h2 respectively.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 20

Question 19.
From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. [Use π = Image may be NSFW.
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\frac{22}{7}
] (2012D)
Solution:
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 21

Question 20.
A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy. [Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2013D)
Solution:
Here r = 3.5 cm = Image may be NSFW.
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\frac{7}{2}
cm, h = 10 cm
Volume of wood in the toy = Volume of cylinder - 2(Vol. of hemisphere)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 22

Question 21.
A solid cone of base radius 10 cm is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of the two parts of the cone. (2013OD)
Solution:
Let BC = r сm and
DE = R = 10 cm
B and C are the mid-points of AD and AE respectively. ...[Given
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 23

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 24

Question 22.
A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter Image may be NSFW.
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\frac{1}{12}
cm, find the length of the wire. (2014D)
Solution:
A solid cone AFE has been cut by BC || FE and
AD ⊥ FE.
∆ADF = ∆ADE (SSS)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 25

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 26

Question 23.
The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2014OD)
Solution:
Volume of the wood left = Volume of cube - Volume of sphere
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 27

Question 24.
In Figure, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use π = Image may be NSFW.
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\frac{22}{7}
and Image may be NSFW.
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 \sqrt{{3}}
= 2.236) (2015D)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 28

Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 29

Let AR = 4 cm,
RD = 12 - 4 = 8 cm
Height of frustum (RD),
h = 8 cm Lower radius of frustum (DC),
r1 = 6 cm
We know, ∆ARQ ≅ ∆ADC
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 30

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 31

Question 25.
In Figure, from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. (Use π =Image may be NSFW.
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\frac{22}{7}
) (2015D)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 32

Solution:
Let the length, breadth, height of cuboidal block be 15 cm, 10 cm and 5 cm respectively.
Total surface area of solid cuboidal block
= 2(lb + bh + lh)
= 2(15 × 10 + 10 × 5 + 15 × 5) cm2
= 2(150 + 50 + 75) = 2(275) = 550 cm2
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 33

Required area = (Area of cuboidal block – Area of two circular bases + Area of cylinder)
= (550 + 110 - 77) cm2 = 583 cm2

Question 26.
A toy is in the shape of a solid cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 21 cm and 40 cm respectively, and the height of cone is 15 cm, then find the total surface area of the toy. [π = 3.14, be taken] (2011D)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 34

Height of cylinder, H = 21 cm
Height of cone, h = 15 cm
Common radius,
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 35

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 36

Question 27.
A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere. (2011OD)
Solution:
Volume of the Sphere = Volume of Cone
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 37

∴ Diameter of the sphere = 2r1 = 2(5) = 10 cm

Question 28.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then find the radius and slant height of the heap. (2012D)
Solution:
Let x be radius of the conical heap.
Volume of conical heap = Volume of cyl. bucket
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 38

Question 29.
A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, find its width. (2013OD)
Solution:
Vol. of cylindrical wire = Volume of solid sphere
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 39

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 40

Question 30.
A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment. (2016D)
Solution:
Radius of well, r = Image may be NSFW.
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\frac{4}{2}
= 2 m
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 41

Radius of embankment, R = 2 + 3 = 5 m
Height of the well, h = 21 m
Required height raised
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 42

Question 31.
A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3Image may be NSFW.
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\frac{5}{9}
cm. Find the diameter of the cylindrical vessel. (2016D)
Solution:
Radius of sphere, r = Image may be NSFW.
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\frac{12}{2}
= 6 cm
Let h be the height of risen water level i.e., 3Image may be NSFW.
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\frac{5}{9}
cm or Image may be NSFW.
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\frac{32}{9}
cm and R be the radius of cylindrical vessel.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 43

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 44

Question 32.
A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2012 OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 45

Question 33.
The Image may be NSFW.
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\frac{3}{4}
th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel. (2017D)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 46

Question 34.
504 cones, each of diameter 3.5 cm and height 3cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2015OD)
Solution:
Let the radius of sphere (R) = ?
Let the radius of cone (r) = Image may be NSFW.
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\frac{3.5}{2}
= Image may be NSFW.
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\frac{35}{20}

Let the height of cone (h) = 3 cm
Volume of metal in 1 cone = Image may be NSFW.
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\frac{1}{3}
πr2h
Volume of metal in 504 cones
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 47

Question 35.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled completely? (2014D)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 48

Question 36.
Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/h. How much area will it irrigate in 10 minutes, if 8 cm of standing water is needed for irrigation? (2014OD)
Solution:
Speed of water = 4 km/hr
Canal, l = 4 × Image may be NSFW.
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\frac{10}{60}
= Image may be NSFW.
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\frac{2}{3}
km
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 49

Question 37.
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer. (2015OD)
Solution:
Let the radius of hemispherical bowl,
(R) = Image may be NSFW.
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\frac{36}{2}
= 18 cm
Let the radius of cylindrical bottle (r) = Image may be NSFW.
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\frac{6}{2}
= 3 cm
Let the height of cylindrical bottle be h = ?
Vol. of liquid in the hemispherical bowl = Image may be NSFW.
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\frac{2}{3}
πR3
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 50

Question 38.
An open metal bucket is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket at ₹30 per litre. [Use π = Image may be NSFW.
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\frac{22}{7}
] (2011OD)
Solution:
Here h = 21 cm, r = 10 cm, R = 20 cm
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 51

Question 39.
A toy is in the form of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2013OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 52

Radius = 7 cm
Vertical height of cone,
h = 31 - 7 = 24 cm
Slant height,
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 53

Question 40.
In Figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of ₹500/sq. metre. (Use π = Image may be NSFW.
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\frac{22}{7}
) (2016OD)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 54

Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 55

Surface Areas and Volumes Class 10 Important Questions Long Answer (4 Marks)

Question 41.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid. [Use π = Image may be NSFW.
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\frac{22}{7}
) (2012D)
Solution:
Radius, r = 7 cm
Height of cone, h = 2(7) = 14 cm
Volume of solid = Vol. of hemisphere + Volume of cone
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 56

Question 42.
A hemispherical tank, full of water, is emptied by a pipe at the rate of Image may be NSFW.
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\frac{25}{7}
litres per sec. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m? (2012OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 57

Question 43.
Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour. (2013D)
Solution:
Radius of tank, r1 = 40 cm
Internal radius of cylindrical pipe, r2 = Image may be NSFW.
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\frac{2}{2}
= 1 cm
Let the height of rises water, h1 = ?
Length of water flow in 1 second = 0.4 m
= Image may be NSFW.
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\frac{4}{10}
× 100 = 40 cm
∴ Length of water flow in 30 minutes, h2
= 40 × 60 × 30 = 72,000 cm
Volume of water in cylinder tank
= Volume of water flow from cylindrical pipe in half an hour
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 58

∴ Level of water in cylinder tank rises in half an hour, h1 = 45 cm

Question 44.
150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel. (2014OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 59

Question 45.
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Take π = Image may be NSFW.
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\frac{22}{7}
) (2014D)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 60

Question 46.
Water is flowing at the rate of 2.52 km/hr. through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe. (2015D)
Solution:
Let r be the internal radius of the pipe.
Radius of base of tank, R = 40 cm
= Image may be NSFW.
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\frac{40}{100}=\frac{2}{5}
m
Level of water raised in the tank (H) = 3.15 m
If the flow rate in an hour = 2.52 km = 2520 m
Then, the flow rate in half an hour
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 61

∴ Internal diameter of pipe = 2r = 2 × 2 = 4 cm
∴ Internal diameter of pipe = 4 cm

Question 47.
From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. [Use π = Image may be NSFW.
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\frac{22}{7}
] (2015OD)
Solution:
Let h be the height of cylinder = 10 cm
Let r be the radius of cylinder = 4.2 cm
Radius of cylinder
= Radius of hemispherical metal = r = 4.2 cm ..[Given
Total Volume of cylinder = πr2h
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 62

Question 48.
A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use π = 3.14) (2016D)
Solution:
Radius of the top of bucket, R = 20 cm
Radius of the bottom of bucket, r = 12 cm
Voulume of bucket = 12308.8 cm3
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 63

Question 49.
A container shaped like a right circular cylinder having base radius 6 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm and radius 3 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream. (2012D)
Solution:
Volume of cylinder = πr2h
Volume of ice-cream cone = 1/3 πr21h1
Volume of hemispherical top of ice-cream
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 64

Question 50.
A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas used in making the tent, if the breadth of the canvas is 1.5 m. (2012OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 65

Question 51.
A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment. (2015D)
Solution:
Let h be the height of well and r be the radius of well.
h = 14 m, r= Image may be NSFW.
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\frac{d}{2}=\frac{4}{2}
= 2 m
Volume of earth taken out after digging the well = πr2h
= Image may be NSFW.
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\left(\frac{22}{7} \times 2 \times 2 \times 14\right)
m3 = 176 m3 ...(i)
Let x be the width of embankment formed by using (i),
∴ Total width of well including embankment (R) = 2 + x
Height of embankment (H) = 40 cm = Image may be NSFW.
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\frac{40}{100}
m
Volume of well = Volume of embankment So, Volume of embankment = T(R2 - r2)H = 176 ...[From (i)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 66

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 67

Question 52.
A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal). (2014OD)
Solution:
Upper radius of frustum of a cone,
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 68

Question 53.
Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm? (2011OD)
Solution:
H = 15 km = 15000 m (1 km = 1000 m)
Radius of pipe (t) = Image may be NSFW.
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\frac{14}{2}
= 7 cm = Image may be NSFW.
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\frac{7}{100}
m
Volume of pipe (cylinder) = πr2H
∴ Volume of water flowing through the cylindrical pipe in an hour at the rate of 15 km/hr.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 69

Question 54.
Water is flowing at the rate of 10 km/hour through a pipe of diameter 16 cm into a cuboidal tank of dimensions 22 m × 20 m × 16 m. How long will it take to fill the empty tank? [Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2011OD)
Solution:
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 70

Question 55.
A drinking glass is in the shape of the frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. [Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2012OD)
Solution:
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 71

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 72

Question 56.
A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of 10 per 100 cm2. (Use π = 3.14) (2013D)
Solution:
Depth of bucket, h = 24 cm
Radius of top bucket,
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 73

Question 57.
The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. (2017OD)
Solution:
Let longer radius = R cm
Smaller radius = r cm
2πR = 18 and 2πr = 6
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 74

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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 75

Question 58.
Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which Image may be NSFW.
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\frac{2}{5}^{\text { th }}
of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. (2014D)
Solution:
Numbers of spherical balls
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 76

Question 59.
A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to form a platform 27 m × 11 m. Find the height of the platform. (Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
] (2015D)
Solution:
Let h be the height of well, h = 21 m
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 77

Question 60.
The dimensions of a solid iron cuboid are 4.4 m × 2.6 m × 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe. (2017OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 78

Inner radius, r = 30 cm = 0.30 m
Outer radius, R = (30 + 5) cm = 35 cm or 0.35 m
As per the question: (Volume of hollow Cylindrical pipe) = (Volume of solid iron cuboid)
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 79

Question 61.
In a rain-water harvesting system, the rain water from a roof of 22 m × 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfull in cm. (2017OD)
Solution:
Let the rainfall be x m.
Volume of water on the roof = Volume of cylindrical tank
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 80

Question 62.
A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket. (Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
) (2012D)
Solution:
Here R = 28 cm, r = 21 cm
Volume of Bucket = 28.49 litres
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 81

Question 63.
Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/h. (Use π = Image may be NSFW.
Clik here to view.
\frac{22}{7}
) (2013OD)
Solution:
Volume of water that flows for one hour
= (192.50 × 60) Its.
= 192.50 × 60 × 1000 cm3 ...[∵ 1 lt. 1000 cm3
Inner radius of the cylindrical pipe = Image may be NSFW.
Clik here to view.
\frac{7}{2}
cm
Let h cm be the length of the volume of water that flows in one hour.
Volume of water that flows in 1 hr. = πr2h
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 82

Hence the rate of flow of water in the pipe = 3 km/hr.

Question 64.
Due to heavy floods in a State, thousands were rendered homeless. 50 schools collectively offered to the State Government to provide place and the canvas for 1,500 tents to be fixed by the Government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents coşts ₹120 per sq. m, find the amount shared by 'each school to set up the tents. (2016OD)
Solution:
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 83

Let r and h be the radius and height of cylindrical part respectively and I be the slant height of conical part.
Slant height of conical part (l),
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 84

Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 85

Question 65.
A container, open at the top and made up of metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the cost of metal sheet used to make the container, if it costs 10 per 100 cm2. [Take a = 3.14] (2013OD)
Solution:
Height of frustum of a cone, h = 16 cm
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Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes 86

Important Questions for Class 10 Maths

The post Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes appeared first on Learn CBSE.

DOWNLOAD PUBG MOBILE ON PC WITH PUBG FREE DOWNLOAD

Free Download And Install PUBG on PC/Laptop [Windows 10, 8, 8.1, 7]

Well, if you want to get the official PUBG version for PC, then you will have to pay around $29.99 via Steam. But you are in luck today, with the help of this guide, you can get PUBG free for PC or PUBG Mobile PC. For sure, you are excited to player unknown battlegrounds free download for PC, but you should make it sure first that your PC or laptop is compatible to play PUBG on PC. So, let’s check out some PUBG PC requirements here:

PUBG PC Requirements

Minimum                              Recommended
Requires a 64-bit processor and operating systemRequires a 64-bit processor and operating system

 

OS: 64-bit Windows 7, Windows 8.1, Windows 10OS: 64-bit Windows 7, Windows 8.1, Windows 10
Processor: Intel Core i5-4430 / AMD FX-6300Processor: Intel Core i5-6600K / AMD Ryzen 5 1600
Memory: 8 GB RAMMemory: 16 GB RAM
Graphics: NVIDIA GeForce GTX 960 2GB / AMD Radeon R7 370 2GBGraphics: NVIDIA GeForce GTX 1060 3GB / AMD Radeon RX 580 4GB
DirectX: Version 11DirectX: Version 11
Storage: 30 GB available spaceStorage: 30 GB available space

Note: This process will allow you to download and play PUBG mobile on PC. If you don’t want to play PUBG mobile on your desktop, then you can go to the next process to get PUBG free for PC/laptop. Well, there is also PUBG Mobile 0.14.0 Beta update released with Helicopter, Companion, New Zombie Mode & More. It will come soon for Xbox One and PS4. Once after checking pubg pc requirements if you found that your PC is compatible, then you can go ahead.

How to Play PUBG Mobile On PC [Complete Free Guide]

Step #1: Download Bluestacks from their official site.

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How to Play PUBG Mobile On PC 2

Step #2: Run and Install it on your PC.

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PUBG Run and Install it on your PC 3

Step #3: Launch the Bluestacks app on your PC and set up your Google Play Store with your Gmail, sometimes it asks for phone verification.

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PUBG Google Play Store Download Set Up 4

Step #4: Once the Play Store established, search for “PUBG Mobile”

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Search for PUBG Mobile 5

Step #5: Click the Install button next to game icon.

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Install PUBG In Your Mobile 6

Step #6: Once it’s done, you can play PUBG Mobile in “All Apps” or in “My Applications” sections.

Image may be NSFW.
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Install PUBG in Mobile 7

Step #7: Once the game start, click on the mouse and keyboard icon at the bottom so that you can use it to play PUBG mobile on PC.

Image may be NSFW.
Clik here to view.
play PUBG mobile on PC 8

Step #8: After that, you can see what keys to control the game. You can customize it as well according to your preference.

This is how you can use Bluestacks to play PUBG Mobile on PC. In a case, if the Bluestacks doesn’t work well, then you can use the official PUBG mobile emulator which is known as the ‘Tencent Gaming Buddy’. The Tencent team has specially designed this emulator for PUBG user to allow them to play PUBG mobile on PC without using any unofficial tool.

How to Download PUBG Free For PC/Laptop

Don’t worry if you don’t want to spend your pennies on the official PUBG version. We will show you the step-by-step guide to get PUBG free on PC using the Playerunknown’s Battlegrounds free exe file that you will need to install. Here are some steps you should follow!

Step #1: Download the PUBG installer from the below link.

Step #2: Next, open the “exe” file and run to install PUBG on PC.

Step #3: While installing the files, you should follow some instruction that appears on the screen.

Step #4: Now the game starts download automatically on your PC.

Step #5: Once it’s done, you get pop up with download key and then activate the game.

Step #6: Once all done, enjoy the game!

That’s it!

This is how you can play PUBG on PC free. Once after start playing PUBG on PC, you can connect with your PUBG friends and play with them as squid. Make strategies, play in unity, battle hard and get the “Winner Winner Chicken Dinner” to fill up your hunger.

Best PUBG Settings For FPS and Visibility: PC

While playing PUBG on PC, the PUBG graphic settings matter a lot. It helps to improve the performance of the game and let the player perform every action smoothly. You may have seen much PUBG online streaming where they can efficiently do every task to get the chicken dinner. So, you might be wondering what PUBG settings pro players use. Let’s take a look at these best PUBG settings to improve FPS and visibility:

Option Recommended Setting
Window ModeFullscreen (Windowed)
Display resolution1920×1080
Camera FOV103
Brightness50-70
Overall QualityCustom
Screen Scale100
Anti-AliasingVery Low
Post-ProcessingVery Low / Low
ShadowsVery Low / Low
TexturesMedium – Ultra
Effects  VeryLow
FoliageVery Low
View DistanceMedium – Ultra
V-SyncOff
Motion BlurOff

Best PUBG Emulator To Play PUBG Mobile On PC For Free

Since the game has become very popular among people from all over the world, many player wish to play PUBG on PC free. But as we already said that PUBG PC free version is not officially available online, you will have follow some PUBG hack or PUBG trick to enjoy PUBG PC for free. Considering the need of people from all around the world, here we have collected world’s best emulator of PUBG PC to play the game for free without any lag and stoppage.

Before you head over to the list of best PUBG emulator, you should know about what is PUBG emulator? So, let me tell you that emulators specially designed to play some Android game on PC without any charge. In simple, if you want to play PUBG Mobile or Free Fire game on PC, then you can simply install the Android emulator also called (AVD) Android Virtual Device and enjoy all your favourite games on PC for free. Let’s check out which emulator is best for PC:

#1: Tencent Gaming Buddy (Official PUBG Emulator)

First we provide the most trustworthy and official PUBG emulator “Tencent Gaming Buddy” from the developer of the game. This emulator of PUBG is specially design to let PUBG geeks to play PUBG Mobile on PC for free. So, if you cannot afford the PUBG PC paid version of the game, then you can use this official PUBG PC emulator on your Windows PC to play PUBG Mobile on PC. Furthermore, you can also play other games as well through the Tencent Gaming Buddy PUBG emulator.

Tencent Gaming Buddy PC Requirements

  • OS: Windows 7, 8, 10(32bit & 64 bit)
  • RAM: 3 GB or Above
  • CPU: Dual-core from Intel or AMD at 1.8 GHz

#2: Bluestack

Before the development of Tencent Gaming Buddy, Bluestack was only the best emulator to play PUBG mobile on PC for free. Bluestack is not up to PUBG, you can enjoy every game or app from the Google Play Store. The great thing about the Bluestack is it can be installed on low end PC devices. This PUBG emulator also provides the great graphic experience and smooth control. So, whether you want play PUBG mobile on PC or any other favourite game, you can simply use Bluestack emulator to play every Android games on PC with better control.

Bluestack PC Requirements

  • OS: Windows 7, 8, 10(32bit & 64 bit)
  • RAM: 2 GB or Above
  • CPU: Intel or AMD Processor
  • Price: Free or Paid

#3: MemuPlay

Just like Bluestack, MemuPlay also every known for its fastest and free Android emulator service to play mobile games on PC with the great performance and amazing control system. Well, this PUBG emulator has some extra features compare to other Android emulator. As it is preloaded with Google Play Store, you can simply install any game or app on your PC and also open any apk file in the Memu. One greater thing is you can also run multiple games and apps simultaneously without any problem. There is also full screen and screen recording option available to help you do your task best.

Memu PC Requirements

  • OS: Windows 7, 8, 10(32bit & 64 bit)
  • RAM: 3 GB or Above
  • CPU: Intel or AMD CPU
  • Price: Free

#4: Nox Player

Let’s talk about one more popular PUBG emulator called Nox Player. With the help of this Android emulator, you can easily load any APK file on your PC without any restriction. It also offer screenshot and screen recording feature while playing the game. So, you specially moments will get captured or recorded. There is also a good option to customise the buttons of any game like PUBG to get better control. Multiple running of games and app is also a good function of this PUBG emulators.

Nox Player PC Requirements

  • OS: Windows 7, 8, 10(32bit & 64 bit)
  • RAM: 1.5 GB or Above
  • CPU: Intel or AMD Processor
  • Price: Free

#5: Remix OS

Remix OS is one of best PUBG emulator which allow you to play PUBG Mobile on PC with great experience as well as other games on your PC in full screen. As there is keyboard map button option available, you can easily customize and control buttons in a better manner. Well, you can play multiple games at the same time; it doesn’t support AMD chipset and also need Virtual Technology to activate BIOS of PC. It’s totally free and support on low end devices. So, it will be fun using this PUBG emulator.

Remix OS PC Requirements

  • OS: Windows XP, 7, 8, 10(32bit & 64 bit)
  • RAM: 2 GB or Above
  • CPU: Intel or AMD CPU
  • Price: Free

Well, there are more PUBG emulators available, but these were some of the best emulator for PUBG with great function. All these emulators are free and easy to use. Once after using any of these PUBG emulator, do not forget to share your opinion with us in comment. It will be pleasant.

The post DOWNLOAD PUBG MOBILE ON PC WITH PUBG FREE DOWNLOAD appeared first on Learn CBSE.


Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions

Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions

Surface Chemistry Class 12 Important Questions Very Short Answer Type

Question 1.
Define the term ‘Tyndall effect’. (Delhi 2009)
Answer:
Tyndall effect : When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.

Question 2.
What is the ‘coagulation’ process? (All India 2009)
Answer:
The process of settling of colloidal particles is called coagulation or precipitation of the solution.

Question 3.
What is an emulsion? (Delhi 2010)
Answer:
Emulsion is liquid-liquid colloidal system.

Question 4.
Give an example of ‘shape-selective catalyst’. (Delhi 2010)
Answer:
The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.

Question 5.
Define ‘electrophoresis’. (Delhi 2011)
Answer:
Electrophoresis : When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.

Question 6.
What is meant by ‘shape-selective catalysis’ of reactions? (All India 2011)
Answer:
The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.

Question 7.
What are lyophobic colloids? Give one example for them. (All India 2011)
Answer:
Lyophobic sols : Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can only be prepared by specific methods. They are not much hydrated and are irreversible in nature. They are also called extrinsic colloids.
Example : AS2S3 sol.

Question 8.
Define ‘peptization’. (All India 2012)
Answer:
Peptization may be defined as the process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of electrolyte.

Question 9.
What is meant by ‘shape selective catalysis’? (All India 2012)
Answer:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis.

Question 10.
Out of NH3 and CO2 which gas will be adsorbed more readily on the surface of activated charcoal and why? (Comptt. Delhi 2012)
Answer:
NH3 gas will be adsorbed more readily on activated charcoal. It has higher critical temperature than CO2 and is an easily liquifiable gas. Its van der Waals forces are stronger.

Question 11.
How can a colloidal solution and true solution of the same colour be distinguished from each other? (Comptt. Delhi 2012)
Answer:
A colloidal solution scatters a beam of light but a true solution does not.

Question 12.
How is a sol different from an emulsion ? (Comptt. All India 2012)
Answer:
A collidal sol contains solid as the dispersed phase and liquid as the dispersion medium e.g. paint, gold sol etc.
Emulsion : A colloidal dispersion in which the dispersed- phase and the dispersion medium are immiscible liquids, is known as emulsion e.g. milk, butter etc.

Question 13.
Write two applications of adsorption. (Comptt. All India 2012)
Answer:
Applications of adsorption :

  • In decolorisation of sugar.
  • In gas masks, charcoal is used which adsorbs poisonous gases in mines.

Question 14.
Why do true solutions not show Tyndall effect? (Comptt. All India 2012)
Answer: In true solution, the diameter of the dispersed particles is much smaller than the wavelength of the light used, hence there is no scattering of light.

Question 15.
Of physisorption or chemisorption, which has a higher enthalpy of adsorption? (All India 2013)
Answer:
Chemisorption has higher enthalpy of adsorption than physisorption due to chemical bond formation.

Question 16.
What is especially observed when a beam of light is passed through a colloidal solution? (All India 2013)
Answer:
Tyndall effect is observed when a beam of light is passed through a colloidal solution.

Question 17.
To which colloidal system does milk belong? (Comptt. All India 2013)
Answer:
Milk belongs to emulsion.

Question 18.
What is electrophoresis due to? (Comptt. All India 2013)
Answer:
Electrophoresis is due to electrical charge on the colloidal particles.

Question 19.
What is meant by the term peptization? (Comptt. All India 2013)
Answer:
Peptization may be defined as the process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of electrolyte.

Question 20.
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion. (Delhi 2014)
Answer:
Oil in water → Milk, vanishing cream
Water in oil → Butter, cold creams.

Question 21.
Give one example each of sol and gel. (Delhi 2014)
Answer:
Sol → Smoke, dust
Gel → Cheese

Question 22.
Give one example each of lyophobic sol and lyophilic sol. (Delhi 2014)
Answer:
Lyophobic sol — Metal sulphides
Lyophilic sol — Starch

Question 23.
What is the effect of temperature on chemisorption? (All India 2014)
Answer:
Chemisorption increases with increase of temperature.

Question 24.
Why is adsorption always exothermic? (All India 2014)
Answer:
Adsorption is accompanied by decrease of randomness. For the process to be spontaneous,
ΔG must be negative.
Hence, according to equation ΔG = ΔH – TΔS, ΔG can be -ve only if ΔH is negative.

Question 25.
What are the dispersed phase and dispersion medium in milk? (All India 2014)
Answer:
Milk : Dispersed phase → Fat (Liquid);
Dispersion medium → Liquid

Question 26.
What is the difference between lyophobic sol and lyophilic sol? (Comptt. Delhi 2014)
Answer:
Lyophobic sols: Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can only be prepared by specific methods. They are not much hydrated and are irreversible in nature. They are also called extrinsic colloids.
Example : AS2S3 sol.
Lyophilic sols: Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example : Starch sol, Gum sol, Gelatin sol

Question 27.
What is a ‘shape-selective catalyst’? (Comptt. Delhi 2014)
Answer:
The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.

Question 28.
What are emulsions? Name an emulsion in which water is a dispersed phase. (Comptt. All India 2014)
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids.
Water in oil → Butter, cold creams.

Question 29.
Define dialysis. (Comptt. All India 2014)
Answer:
Dialysis : The process of separating the particles of colloids from those of crystalloids by diffusion of the mixture through a parchment or an animal membrane is known as dialysis.

Question 30.
A delta is formed at the melting point of sea water and river water. Why? (All India 2015)
Answer:
Delta is formed at the meeting point of sea water and river water due to coagulation of colloidal clay particles.

Question 31.
In reference to surface chemistry, define dialysis. (Comptt. Delhi 2015)
Answer:
Dialysis : The process of removing the dissolved substances from a colloidal solution by means of diffusion through a suitable membrane is called dialysis.

Question 32.
What are emulsions? Give an example. (Comptt. All India 2015)
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids. For example, milk, cream.

Question 33.
Why is adsorption always exothermic? (Comptt. Delhi 2016)
Answer:
Due to the force of attraction/bond formation between adsorbate and adsorbent.

Question 34.
Why is Tyndall effect shown by colloidal solutions? (Comptt. All India 2016)
Answer:
It is so due to large size of colloidal particles. In colloidal solutions particle size of dispersed phase is comparable to the wavelength of light used.

Question 35.
What are associated colloids? Given an example. (Comptt. All India 2016)
Answer:
Colloids which act as electrolyte at low concentration and show colloidal behaviour at high concentration are called Associated colloids. Example : Soap solution, Detergents.

Question 36.
Write one similarity between physisorption and chemisorption. (Delhi 2017)
Answer:
Both physisorption and chemisorption increase with increase in pressure. Both increase with increase in surface area.

Question 37.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. (All India) 2017
Answer:
Gel is formed when a liquid is dispersed in a solid, e.g., Cheese, butter, etc.

Question 38.
What type of colloid is formed when a solid is dispersed in a liquid? Give an example. (All India 2017)
Answer:
Sol is formed when a solid is dispersed in a liquid, e.g., paints.

Question 39.
What type of colloid is formed when a gas is dispersed in a liquid? Give an example. (All India 2017)
Answer:
Foam is formed when a gas is dispersed in a liquid, e.g., Froth.

Question 40.
Which of the following is most effective in coagulating negatively charged hydrated ferric oxide sol? (Comptt. Delhi 2017)
(i) NaN03 (ii) MgSO4 (iii) AlCl3
Answer:
AlCl3 (Aluminium chloride) is most effective in coagulating negatively charged hydrated ferric oxide sol.

Question 41.
Which of the following is most effective in coagulating positively charged hydrated ferric oxide sol? (Comptt. Delhi) 2017
(i) NaNO3 (ii) Na2SO4 (Hi) (NH4)3PO4
Answer:
Ammonium phosphate (NH4)3PO4 is most effective in coagulating positively charged hydrated ferric oxide sol.

Question 42.
Which of the following is most effective in coagulating positively charged methylene blue sol? (Comptt. Delhi 2017)
(i) Na3PO4 (ii) K4[Fe(CN)6] (iii) Na2SO4
Answer:
Potassium ferrocyanide K4[Fe(CN)6]

Question 43.
What are emulsions? Give an example. (Comptt. All India 2017)
Answer:
An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids. For example, milk, cream.

Question 44.
Write the dispersion medium and dispersed phase in milk. (Comptt. All India 2017)
Answer:
Liquid fat is the dispersed phase and water is the dispersion medium.

Question 45.
Write the dispersed phase and dispersion medium in butter. (Comptt. All India 2017)
Answer:
Water is the dispersed phase and oil is the dispersion medium in butter.

Surface Chemistry Class 12 Important Questions Short Answer Type [SA – I]

Question 46.
Describe the following :
(i) Tyndall effect
(ii) Shape-selective catalysis (All India 2010)
Answer:
(i) Tyndall effect : When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.
(ii) Shape-selective catalysis : The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.

Question 47.
What is meant by coagulation of a colloidal solution? Name any method by which coagulation of lyophobic sols can be carried out. (All India 2010)
Answer:
Coagulation : The process of settling of colloidal particles is called coagulation or precipitation of the solution.
Name of method : The method which can be used for coagulation of lyophobic solutions is electrophoresis.

Question 48.
Define the following :
(i) Peptization (ii) Reversible sols (All India 2010)
Answer:
(i) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.
(ii) Reversible sols : The lyophilic solutions in which if the dispersed phase is separated from the dispersion medium, the sol can be made again by simply remixing with the dispersion medium. This shaking is called as Reversible sol.

Question 49.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved. (Delhi 2012)
Answer:
Homogeneous catalysis : When the reactants and the catalyst are in the same phase, the process is called homogeneous catalysis.
Example : 2SO2 (g) + O2 (g) Image may be NSFW.
Clik here to view.
\underrightarrow { NO\left( g \right) }
2SO3 (g)
Heterogeneous catalysis : When the reactants and the catalyst are in different phases, the process is called heterogeneous catalysis.
Example : 2SO2 (g) Image may be NSFW.
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\underrightarrow { Pt\left( s \right) }
2SO3 (g)

Question 50.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out. (Delhi 2012)
Answer:
The process of setting of colloidal particles is called coagulation or precipitation of the sol. Methods of coagulation :

  • By electrophoresis : The colloidal particles move towards oppositely charged electrodes.
  • By mixing two oppositely charged sols.
  • By addition of electrolytes : When excess of an electrolyte is added, the colloidal particles are precipitated. ‘

Question 51.
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol. of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol. (Delhi 2012)
Answer:
(i) Coagulation or precipitation of sol. takes place.
(ii) Scattering of light is observed when light is passed through a sol. whereas no scattering of light is observed in a solution of NaCl/ , Tyndall effect in the sol.

Question 52.
Explain the following terms giving one example for each :
(i) Miscelles (ii) Aerosol (Delhi 2012)
Answer:
(i) Miscelles : They are associated colloids showing colloidal behaviour at high concentration and strong electrolytes at low concentration.
Example : soap, detergent.
(ii) Aerosol : Aerosol is a colloidal solution of solid or liquid (dispersed phase) in gas (dispersion medium).
Example : smoke, dust, fog, mist, cloud.

Question 53.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk (Delhi 2013)
Answer:
(i) Smoke: Dispersed Phase → Solid;
Dispersed medium → Gas;
(ii) Milk: Dispersed Phase → Fat (Liquid);
Dispersed medium Liquid

Question 54.
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (Delhi 2013)
Answer:
Lyophobic sols : Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can only be prepared by specific methods. They are not much hydrated and are irreversible in nature. They are also called extrinsic colloids.
Example : AS2S3 sol.
Lyophilic sols : Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example : Starch sol, Gum sol, Gelatin sol

Question 55.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. (Delhi 2013)
Answer:
Difference between two types of emulsions are : Emulsions of oil in water in which oil is the dispersed phase and water is the dispersion medium.
Example : Milk is an emulsion of liquid fat dispersed in water.
Emulsions of water in oil in which water is the dispersed phase and oil is the dispersion medium. e.g. Cod liver oil is an emulsion of oil i.e. water is the dispersed phase and oil is the dispersion medium.
Two applications of emulsion are :

  • The digestion of fats in the intestines takes place by the process of emulsification.
  • Several oily drugs are prepared in the form of emulsion.

Question 56.
What is the difference between multi-molecular and macromolecular colloids? Give one example of each. (Delhi 2013)
Answer:
Multi-molecular colloid is aggregation of large number of atoms or smaller molecules of a substance having size in the colloidal range. Whereas macromolecular colloid is the solution containing macromolecules in the colloidal range.
Example: Multimolecular colloids: Gold sol,
Sulphur sol
Macromolecular colloids: Proteins,
Cellulose (any one)

Question 57.
Explain the following :
(a) Same substance can act both as colloids and crystalloids.
(b) Artificial rain is caused by spraying salt over clouds. (Comptt. Delhi 2013)
Answer:
(a) Sodium chloride behaves as a crystalloid when dissolved in water but behaves as a colloid when dissolved in benzene.
(b) Artificial rain is caused by spraying common salt over the clouds, as it is an electrolyte and brings about coagulation of water particles.

Question 58.
How are the following colloidal solutions prepared?
(a) Sulphur in water
(b) Gold in water (Comptt. Delhi 2013)
Answer:
(a) Sulphur in water : An alcoholic solution (true solution) of sulphur in excess of water is added, then colloidal solution is obtained. Or, Oxidation of H2S
HNO3 + H2 S → H2O + NO2 + S
(b) Gold in water : Gold solution can be prepared by reduction of AuCl3 solution with formaldehyde.
2AuCl3 + 3HCHO + 3H2O → 2Au Gold Sol. + 3HCOOH + 6HCl

Question 59.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.(Comptt. Delhi 2013)
Answer:
Adsorption isotherm : A graph between the amount of the gas adsorbed per gram of the adsorbent (x/m) and the equilibrium pressure of the adsorbate at constant temperature is called the adsorption isotherm.
The extent of adsorption is usually expressed as x/m,
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Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions 1

A relationship between the amount adsorbed (x/m) and the equilibrium pressure (P) can be obtained as follows :
At low values of P, the graph is nearly straight and sloping.
Image may be NSFW.
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\frac{x}{m}
∝ P or Image may be NSFW.
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\frac{x}{m}
= constant × P1
At high pressure, x/m becomes independent of the values of P. In this range of pressure,
Image may be NSFW.
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\frac{x}{m}
∝ P° or Image may be NSFW.
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\frac{x}{m}
= constant × P°
In the intermediate range of pressure, x/m will depend on P raised to powers between 1 and 0 i.e. fraction. For a small range of pressure values, we can write :
Image may be NSFW.
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\frac{x}{m}
∝ P1/n or Image may be NSFW.
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\frac{x}{m}
= KP1/n
Where n = Positive integer
K = constant depends upon the nature of adsorbate and adsorbent at a particular temperature.
The factor 1 / n has values between 0 and 1.
This relationship was given by Freundlich and is known as Freundlich adsorption isotherm.

Question 60.
What are emulsions? Describe different types of emulsions giving one example of each type. (Comptt. Delhi 2013)
Answer:
Emulsion : The colloidal solution in which both dispersed phase and medium are in liquid state is called emulsion. This is liquid-liquid colloidal system.
Different types of emulsions are :
Emulsions of oil in water in which oil is the dispersed phase and water is the dispersion medium.
Example : Milk is an emulsion of liquid fat dispersed in water.
Emulsions of water in oil in which water is the dispersed phase and oil is the dispersion medium. e.g. Cod liver oil is an emulsion of oil i.e. water is the dispersed phase and oil is the dispersion medium.
Two applications of emulsion are :

  • The digestion of fats in the intestines takes place by the process of emulsification.
  • Several oily drugs are prepared in the form of emulsion.

Surface Chemistry Class 12 Important Questions Short Answer Type – II [SA – II]

Question 61.
How are the following colloids different from each other in respect of dispersion medium and dispersed phase? Give one example of each type.
(i) An aerosol
(ii) A hydrosol
(iii) An emulsion (Delhi 2009)
Answer:
(i) An aerosol: It is a colloidal solution in which dispersed phase is a solid and dispersion medium is a gas.
Example : Smoke, haze.
(ii) A hydrosol : The colloidal solution where water is used as the dispersion medium is called hydrosol or aquasol.
Example : Starch sol.
(iii) An emulsion : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example : Milk, cream.

Question 62.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids? (Delhi 2009)
Answer:

Multimolecular colloidsMacromolecular colloidsAssociated colloids
1. They are formed by the aggregation of a large number of atoms or molecules which generally have diameter less than 1 mm, e.g. sols of gold, sulphur etc.They are mol­ecules of large size e.g. polymers like rubber, ny­lon, starch, pro­teins etc.They are formed by aggregation of a large num­ber of ions in concentrated so­lution e.g. soap sol.
2. Their molecular masses are not very high.They have high molecular masses.Their molecular masses are gen­erally high.
3. Their atoms or molecules are held together by weak Van der Waals forces.Due to long chain, the Van der Waals forces holding them are comparatively stronger.Higher is the concentration, greater are the Van der Waals forces.

Question 63.
What happens in the following activities and why?
(i) An electrolyte is added to a hydrated ferric oxide sol in water.
(ii) A beam of light is passed through a colloidal solution.
(iii) An electric current is passed through a colloidal solution. (All India 2009)
Answer:
(i) When NaCl is added to hydrated ferric oxide solution, then coagulation will take place. A negatively charged solution is obtained with absorption of OH ion.

(ii) When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.

(iii) When a beam of strong light is passed through a colloidal solution, scattering of light occurs by colloidal particles and the path of light becomes visible and the phenomenon is known as Tyndall effect.

Question 64.
What is the difference between multimolecular and macromolecular colloids? Give one example of »gach type. How are associated colloids different from these two types of colloids? (Delhi 2010)
Answer:

Multimolecular colloidsMacromolecular colloidsAssociated colloids
1. They are formed by the aggregation of a large number of atoms or molecules which generally have diameter less than 1 mm, e.g. sols of gold, sulphur etc.They are mol­ecules of large size e.g. polymers like rubber, ny­lon, starch, pro­teins etc.They are formed by aggregation of a large num­ber of ions in concentrated so­lution e.g. soap sol.
2. Their molecular masses are not very high.They have high molecular masses.Their molecular masses are gen­erally high.
3. Their atoms or molecules are held together by weak Van der Waals forces.Due to long chain, the Van der Waals forces holding them are comparatively stronger.Higher is the concentration, greater are the Van der Waals forces.

Question 65.
How are the following colloids different from each other in respect of their dispersion medium and dispersed phase? Give one example of each.
(i) Aerosol
(ii) Emulsion
(iii) Hydrosol (Delhi 2010)
Answer:
(i) An aerosol: It is a colloidal solution in which dispersed phase is a solid and dispersion medium is a gas.
Example : Smoke, haze.
(ii) A hydrosol : The colloidal solution where water is used as the dispersion medium is called hydrosol or aquasol.
Example : Starch sol.
(iii) An emulsion : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example : Milk, cream.

Question 66.
Explain how the phenomenon of adsorption finds application in each of the following processes :
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth Floatation process (Delhi 2011)
Answer:
(i) Production of vacuum : In Dewar flasks, activated charcoal is placed between the walls of the flask so that any gas which enters into annular space either due to glass imperfection or diffusion through glass is adsorbed and create a vacuum.

(ii) Heterogeneous catalysis : If the catalyst is present in a different phase than that of the reactants, it is called a heterogeneous catalyst and this type of catalysis is called Heterogeneous catalysis.
Example : Manufacture of NH3 from N2 and H2 by Haber’s process using iron as catalyst
N2 (g) + 3H2 (g) Image may be NSFW.
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\underrightarrow { Fe\left( s \right) }
2NH3 (g)
Reactants are gaseous whereas catalyst is solid.

(iii) Froth Floatation process : When KCl is added to hydrated ferric oxide sol, then a negatively charged sol is obtained with absorption of OH ion.

Question 67.
Define each of the following terms :
(i) Micelles
(ii) Peptization
(iii) Desorption (Delhi 2011)
Answer:
(i) Micelles : The substances, which when dissolved in a medium at low concentrations behave as normal, strong electrolytes but at higher concentration exhibit colloidal state properties due to the formation of aggregated particles and such aggregated particles thus formed are called micelles.

(ii) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.

(iii) Desorption : The removal of adsorbed substance from the surface of a solid or liquid by heating or by reducing pressure is called desorption.

Question 68.
Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes. (All India 2011)
Answer:
Colloids are classified into two types in the cases where dispersion medium is water. They are:
(i) Hydrophilic or Lyophilic colloids : Those substances, which when mixed with the dispersion medium, form directly the colloidal solution and are termed as hydrophilic colloids. They are reversible solutions, quite stable and cannot be easily precipitated.
Example : gum, gelatine, starch, rubber etc.

(ii) Hydrophobic or Lyophobic colloids : Those substances which do not form colloidal solution when simply mixed with the dispersion medium, are called hydrophobic colloids. They are irreversible solutions, unstable and can be easily precipitated. Example : Metals and their sulphides

Question 69.
Explain what is observed when
(i) an electric current is passed through a sol
(ii) a beam of light is passed through a sol
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol (All India 2011)
Answer:
(i) When NaCl is added to hydrated ferric oxide solution, then coagulation will take place. A negatively charged solution is obtained with absorption of OH ion.

(ii) When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.

(iii) When a beam of strong light is passed through a colloidal solution, scattering of light occurs by colloidal particles and the path of light becomes visible and the phenomenon is known as Tyndall effect.

Question 70.
Explain the following terms giving a suitable example for each :
(i) Aerosol
(ii) Emulsion
(iii) Micelle (All India 2012)
Answer:
(i) Aerosol : An aerosol is a colloid in which dispersed phase is a solid and dispersion medium is a gas.
Example : Smoke, dust
(ii) Emulsion : It is a colloidal solution of two immiscible liquids of which one is the dispersion medium while the other is the dispersed phase (liquid-liquid).
Example : Milk, cream
(iii) Micelles: They are associated colloids showing colloidal behaviour at high concentration and strong electrolytes at low concentration. Example : Soap, detergent.

Question 71.
Write three distinct features of chemisorptions which are not found in physisorptions. (All India 2012)
Answer:
Three distinct features of chemisorptions :

  • It is caused by chemical bond formation.
  • It is highly specific in nature.
  • It is irreversible.

Question 72.
(a) Adsorption of a gas on surface of solid is generally accompanied by a decrease in entropy, still it is a spontaneous process. Explain.
(b) Some substances can act both as colloids ’ and crystalloids. Explain.
(c) What will be the charge on Agl colloidal particles when it is prepared by adding small amount of AgNO3 solution to KI solution in water ? What is responsible for the development of this charge ? (Comptt. Delhi 2012)
Answer:
(a) We know ΔG = ΔH – TΔS for adsorption. ΔH and ΔS are negative and ΔH > TΔS. Thus from this equation ΔG is negative. Therefore, for adsorption ΔH, ΔG and ΔS all are negative.
(b) A crystalloid can be found to behave as a colloid under a different set of conditions and vice-versa.
Example ; NaCl behaves as a crystalloid when dissolved in water but behaves as a colloid when dissolved in benzene.
(c) When AgNO3 solution is added to Kl solution, the precipitated Agl adsorbs I ions from the dispersion medium and negatively charged colloidal solution results.

Question 73.
Explain the following:
(i) Deltas are formed when river and sea water meet.
(ii) Artificial rain is caused by spraying salt over clouds.
(iii) Physisorption is multi-layered, while chemisorption is mono-layered. (Comptt. Delhi 2012)
Answer:
(i) River water is a colloidal solution of day. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition with the formation of deltas.

(ii) Clouds are colloidal dispersion of water particles in air carrying some charge over them.
It is possible to cause artificial rain by throwing electrified sand or spraying a sol carrying charge opposite to the one on clouds from an aeroplane. The colloidal water particles present in the clouds will get neutralized and as result they will come closer and grow in size to form bigger water drops and ultimately cause artificial rain.

(iii) In physical adsorption, layers of the gas can be adsorbed one over the other by van der Waals forces. Multi-molecular layers are formed under high pressure. In chemical adsorption, chemical bond can be formed only with the layer of molecules coming in direct contact with the surface of the adsorbent, hence this type of adsorption is mono-layered.

Question 74.
How are the two types of emulsions different from one another? Give suitable examples to justify the difference. State two applications of emulsions. (Comptt. All India 2012)
Answer:
Difference between two types of emulsions are : Emulsions of oil in water in which oil is the dispersed phase and water is the dispersion medium.
Example : Milk is an emulsion of liquid fat dispersed in water.
Emulsions of water in oil in which water is the dispersed phase and oil is the dispersion medium. e.g. Cod liver oil is an emulsion of oil i.e. water is the dispersed phase and oil is the dispersion medium.
Two applications of emulsion are :

  • The digestion of fats in the intestines takes place by the process of emulsification.
  • Several oily drugs are prepared in the form of emulsion.

Question 75.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm. (Comptt. All India 2012)
Answer:
Adsorption isotherm : The variation of the amount of the gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve. This curve is termed as adsorption isotherm at a particular temperature.
Freundlich adsorption isotherm : A mathematical equation which describes the relationship between pressure (p) of the gaseous adsorbate and the extent of adsorption at any fixed temperature is called Freundlich adsorption isotherm.
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Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions 2

Question 76.
What are the characteristics of the following j colloids? Give one example of each.
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions (All India 2013)
Answer:
(i) Multimolecular colloids :

  • They are formed by the aggregation of a large number of atoms or molecules which generally have diameter less than 1 nm, Example : sols of gold, sulphur etc.
  • Their molecular masses are not very high.
  • Their atoms or molecules are held together by weak Van der Waals’ forces.

(ii) Lyophobic sols : Liquid hating colloids in which there is no affinity between disperse phase and dispersion medium.
Example: AS2S3sol. Fe(OH)3 sol.

(iii) Emulsions : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example : Milk, cream.

Question 77.
Define the following terms giving an example of each :
(i) Associated colloids
(ii) Lyophilic sol
(iii) Adsorption (All India 2013)
Answer:
(i) Associated colloids :

  • They are formed by aggregation of a large number of ions in concentrated solution Example : soap sol
  • Their molecular masses are generally high.
  • Fligher is the concentration, greater are the Vander Waals’ forces.

(ii) Lyophilic sol : Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example: Starch sol, Gum sol, Gelatin sol

(iii) Adsorption : The process of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into higher concentration of the molecules on the surface is called adsorption.

Question 78.
Define the following terms with an example in each case :
(i) Macromolecular sol
(ii) Peptization
(iii) Emulsion (All India 2013)
Answer:
(i) Macromolecular sol :

  • They are molecules of large size
    Example : polymers like rubber, nylon, starch proteins etc.
  • They have high molecular masses.
  • Due to long chain, the van der Waals’ forces holding them are comparatively stronger.

(ii) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.

(iii) Emulsions : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example : Milk, cream.

Question 79.
Explain what is observed when :
(i) A beam of light is passed through a colloidal solution.
(ii) NaCl solution is added to hydrated ferric oxide sol.
(iii) Electric current is passed through a colloidal solution. (Comptt. All India 2013)
Answer:
(i) When NaCl is added to hydrated ferric oxide solution, then coagulation will take place. A negatively charged solution is obtained with absorption of OH ion.

(ii) When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.

(iii) When a beam of strong light is passed through a colloidal solution, scattering of light occurs by colloidal particles and the path of light becomes visible and the phenomenon is known as Tyndall effect.

Question 80.
(a) How can we get the following colloidal solutions :
(i) Silver in water
(ii) Fe(OH)3 in water
(b) List two applications of adsorption. (Comptt. All India 2013)
Answer:
(a) (i) Silver in water : Colloidal sol of Ag in water (Reduction with dil Sn(Cl2)
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Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions 3

(b) Applications of adsorption :

  • All gas devices contain suitable adsorbent so that poisonous gases present in the atmosphere are adsorbed and the air for breathing is purified.
  • Sugar is decolourized by treating sugar solution with charcoal powder which adsorbs the undesirable colours present.

Question 81.
(a) In reference to Freundlich adsorption isotherm write the expression for adsorption of gases on solids in the form of an equation.
(b) Write an important characteristic of lyophilic sols.
(c) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid. (Delhi 2014)
Answer:
(a) The equation representing adsorption of gases on solids is
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\frac{x}{m}
= KP1/n
where
x / m = amount of gas adsorbed per unit mass of adsorbent
K = Constant
n = Positive integer
(b) Lyophilic sols are reversible and self stable.
(c) Associated colloid : Soap sol
Multimolecular colloid : Gold sol

Question 82.
Define the following terms:
(i) Adsorption
(ii) Peptization
(iii) Sol (Comptt. Delhi 2014)
Answer:
(i) Adsorption : It is a surface phenomenon which occurs only at the surface of the adsorbent.
(ii) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.
(iii) Sol : The colloids in which a solid is dispersed in the liquid, or, In a colloidal sol, the dispersed phase is a solid and the dispersion medium is a liquid.

Question 83.
Define the following terms :
(i) Sorption
(ii) Tyndall effect
(iii) Electrophoresis (Comptt. Delhi 2014)
Answer:
(i) Sorption : It is a physical and chemical process by which one substance becomes attached to another.
or, In order to distinguish between adsorption and absorption a common term sorption is used.
(ii) Tyndall effect: When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.
(iii) Electrophoresis : When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.

Question 84.
Giving appropriate examples, explain how the two types of processes of adsorption (physisorption and chemisorption) are influenced by the prevailing temperature, the surface area of adsorbent and the activation energy of the process? (Comptt. All India 2014)
Answer:

PhysisorptionChemisorption
1. Temperature: It decreases with increase in temperature.1. It increases with increase in temperature initially after adsorption it is decreasing
2. Surface area of adsorbent. The adsorption area ofincreases with increase in surface area of adsorbent2. It also increases with inadsorbent.
3. Activation energy: It requires very low or no activation energy.3. It requires high activation energy.

Question 85.
Explain clearly how the phenomenon of adsorption finds application in
(i) production of vacuum in a vessel
(ii) heterogeneous catalysis
(iii) froth floatation process in metallurgy. (Comptt. All India 2014)
Answer:
(i) Production of vacuum : In Dewar flasks, activated charcoal is placed between the walls of the flask so that any gas which enters into annular space either due to glass imperfection or diffusion through glass is adsorbed and create a vacuum.

(ii) Heterogeneous catalysis : If the catalyst is present in a different phase than that of the reactants, it is called a heterogeneous catalyst and this type of catalysis is called Heterogeneous catalysis.
Example : Manufacture of NH3 from N2 and H2 by Haber’s process using iron as catalyst
N2 (g) + 3H2 (g) Image may be NSFW.
Clik here to view.
\underrightarrow { Fe\left( s \right) }
2NH3 (g)
Reactants are gaseous whereas catalyst is solid.

(iii) Froth Floatation process : When KCl is added to hydrated ferric oxide sol, then a negatively charged sol is obtained with absorption of OH ion.

Question 86.
Give reasons for the following observations :
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. (Delhi 2015)
Answer:
(i) Due to mutual coagulation of leather by tanning. [Positively charged animal hyde (leather) with negatively charged colloidal particles of tannin].
(ii) Lyophilic sols are more stable because there is strong interaction between dispersed phase and dispersion medium.
(iii) Because CO acts as a poison for catalyst.

Question 87.
Write any three differences between Physisorption and Chemisorption. (All India 2015)
Answer:

BasisPhysisorplionChemisorption
(i) SpecificityIt is not specific in nature i.e. all gases are adsorbed on all solids to some extent.It is highly specific in nature and occurs only when there is some possibility of compound formation between the gas being adsorbed and the solid being adsorbent.
(ii) Temperature dependenceIt is independent of temperature as it occurs at low temperature and deceases with increase in temperature.It is temperature dependent and increase with increase in temperature.
(iii) ReversibilityIt is reversible i.e. desorption of gas takes place by increasing the temperature or decreasing the pressure.It is irreversible in nature as it involves formation of compound instead of release of gas.
(iv) Enthalpy changeIt has low enthalpy of adsorption i.e., 20-40 kj mol-1.It has high enthalpy of adsorption i.e., 40-240 kj mol-1.

Question 88.
Define the following terms :
(i) Electrophoresis
(ii) Adsorption
(iii) Shape selective catalysis (Comptt. Delhi 2015)
Answer:
(i) Electrophoresis : When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. This phenomenon is known as electrophoresis.
(ii) Adsorption : The phenomenon of accumulation or higher concentration of molecular species (gases or , liquids) at the surface rather than in the bulk of a solid -or liquid is called adsorption.
(iii) Shape selective catalysis : The catylytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. .

Question 89.
Describecribe the following processes:
(i) Dialysis
(ii) Tyndall effect (Comptt. All India 2015)
Answer:
(i) Dialysis : The process of removing the dissolved substance from a colloidal solution by diffusion of the mixture through a semi-permeable membrane is known as dialysis.
(ii) Tyndall effect: Colloidal particles scatter light in all directions in space. When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.

Question 90.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl3 which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? (Delhi 2016)
Answer:

Adsorption

Absorption

(a) It is the phenomenon by which one substance gets concentrated mainly on the surface of the other substance rather than in the bulk of a solid or liquid.

(b) It is a surface phenomenon.

(c) Its concentration is different at surface from the bulk.

(a)  It is the phenomenon by which one substance gets uniformly distributed thoughout the body of the other substance.

(b)  It is a bulk phenomenon.

(c)  Its concentration is same throughout the bulk.

(ii) AlCl3 is more effective than MgCl2 in causing coagulation of negatively charged sol because coagulating power of an electrolyte is directly proportional to the valency of the active ion i.e., Al3+ > Mg2+.
(iii) Sulphur sol forms multimolecular colloids.

Question 91.
Define the following terms:
(i) Lyophilic colloid
(ii) Zeta potential
(iii) Associated colloids (All India 2016)
Answer:
(i) Lyophilic sol : Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example: Starch sol, Gum sol, Gelatin sol
(ii) Zeta potential. When one type of the ions of the electrolyte are adsorbed on the surface of colloidal particles it forms a fixed layer which attracts another layer of opposite ions thus forming a Helmholtz electrical double layer whose potential difference between the two layers is termed as zeta potential.
(iii)  Associated colloids :

  • They are formed by aggregation of a large number of ions in concentrated solution Example : soap sol
  • Their molecular masses are generally high.
  • Fligher is the concentration, greater are the Vander Waals’ forces.

Question 92.
What are emulsions? What are their different types? Give an example of each type. (Comptt. Delhi 2016)
Answer:
A liquid-liquid colloidal systems is called emulsions. They are of two types :
(i) Oil in water
(ii) Water in oil.
Examples : o/w—Vanishing cream, milk
w/ o—Butter, cold cream, cod-liver oil

Question 93.
Explain the following terms :
(i) Peptization
(ii) Lyophobic colloids
(iii) Dialysis (Comptt. All India 2016)
Answer:
(i) Peptization : The conversion of freshly prepared ppt into colloidal solution by shaking with disperson medium containing small amount of electrolyte.
(ii) Lyophobic colloids : Solvent hating colloids * are called lyophobic colloids. For example, Gold sol.
(iii) Dialysis : It is the process of removing a dissolved substance from a colloidal solution by means of diffusion through a membrane.

Question 94.
Write one difference in each of the following:
(i) Lyophobic sol and Lyophilic sol
(ii) Solution and Colloid
(iii) Homogeneous catalysis and Heterogeneous catalysis (Delhi 2017)
Answer:
(i) Lyophobic sol and Lyophilic sol. Lyophobic solutions are liquid (dispersion medium)— hating and lyophilic solutions are liquid (dispersion medium)—loving colloids.

(ii) Solution and Colloid. Solution is a homogenous solution whose particle size is less than 10-9 m while colloid is a heterogenous solution whose particle size is in between 10-9 to 10-6 m.

(iii) Homogeneous catalysis and Heterogeneous catalysis. Homogeneous catalysis is the phenomenon of changing the rate of reaction when catalyst has same phase as the reactants while in heterogeneous catalysis, the catalyst has different phase than that of the reactants.

Question 95.
Write one difference between each of the following:
(i) Multimolecular colloid and Macromolecu- lar colloid
(ii) Sol and Gel
(iii) O/W emulsion and W/O emulsion (Delhi 2017)
Answer:
(i) Multimolecular colloid. A large number of atoms or smaller molecules of a substance aggregate together to form species having size in the colloidal range.
Macromolecular colloid. Large sized molecules whose particle size lies in the colloidal range.
(ii) Sol is a colloidal system in which dispersed phase is a solid and dispersion medium is a liquid. Gel is a colloidal system in which dispersed phase is a liquid and dispersion medium is a solid.
(iii) In O/W emulsion, water acts as dispersion medium while in W/O emulsion oil acts as dispersion medium.

Question 96.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Homogeneous catalysis and Hetero¬geneous catalysis. (All India 2017)
Answer:
(a) Multimolecular colloid and Associated colloid. Multimolecular colloids are formed by the aggregation of a large number of atoms or molecules which generally have diameters less than 1 nm, eg., sols of gold, etc. while Associated colloids are formed by the aggregation of a large number of ions in concentrated solutions, e.g., micelles in soap.

(b) Coagulation and Peptization. Coagulation is a process of aggregating together the colloidal particles into large sized particles to form their precipitate while peptization is a process of converting fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.

(c) Homogeneous catalysis and Heterogeneous catalysis. Homogeneous catalysis is the ‘ phenomenon of changing the rate of reaction when catalyst has same phase as
the reactants while in heterogeneous catalysis, the catalyst has different phase than that of the reactants.

Question 97.
(a) Write the dispersed phase and dispersion medium of milk.
(b) Write one similarity between physisorption and chemisorption.
(c) Write the chemical method by which Fe(OH)3 sol is prepared from FeCl3. (All India 2017)
Answer:
(a) In milk dispersed phase is liquid fat and dispersion medium is water.
(b) Both physisorption and chemisorption increase with increase in pressure.
(c) Fe(OH)3 is prepared from FeCl by hydrolysis.
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Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions 4

Question 98.
Define the following terms :
(i) Desorption
(ii) Critical micelle concentration
(iii) Shape selective catalysis (Comptt. Delhi 2017)
Answer:
(i) Desorption is a phenomenon whereby an adsorbed substance is removed from or through a surface.
(ii) Critical micelle concentration: The formation of micelles takes place only above a particular temperature i.e., Kraft temperature (Tk) and above a particular concentration.
(iii) Shape-selective catalysis : The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is know n as shape-selective catalysis.

Question 99.
Define the following terms :
(i) Kraft temperature
(ii) Peptization
(iii) Electrokinetic potential (Comptt. Delhi 2017)
Answer:
(i) Kraft temperature : The formation of micelles takes place only above a particular temperature known as Kraft temperature (Tk).
(ii) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable, electrolyte.
(iii) The potential difference between the fixed layer and the diffused layer is known as electrokinetic potential.

Question 100.
Explain the following phenomenon giving reasons :
(i) Tyndall effect
(ii) Brownian movement
(iii) Physical adsorption decreases with increase in temperature. (Comptt. All India 2017)
Answer:
(i) Tyndall effect: When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called^ Tyndall cone and the phenomenon is called Tyndall effect. The phenomenon is due to scattering of light by the colloidal particles. .

(ii) Brownian movement is the continuous and random zig-zag motion of the colloidal particles. This motion is due to kinetic motion of striking colloidal particles which produces a resultant force to cause motion.

(iii) Physical adsorption decreases with increase in temperature due to Le- Chatelier’s principle which shifts the equilibrium in forward direction to complete the reaction. As the adsorption is an exothermic process, it decreases with increase in temperature.

Important Questions for Class 12 Chemistry

The post Important Questions for Class 12 Chemistry Chapter 5 Surface Chemistry Class 12 Important Questions appeared first on Learn CBSE.

NTSE Andaman and Nicobar 2019-20 for Class X | Admit Card, Dates and Application Process

NTSE Andaman and Nicobar 2019-20: The Directorate of Educational Research and Training (DERT) will conduct NTSE Andaman and Nicobar Stage 1 on 2nd November. The application form will be released in July 2019. Students studying in class X can apply for NTSE Andaman and Nicobar till September 2019. Candidates can download the NTSE Andaman and Nicobar Stage 1 exam admit card on October 2019. Students who obtained qualified marks in the NTSE Andaman and Nicobar stage1 will be invited to appear for the stage 2 exam to be held on June 2020. The NTSE final merit list will be prepared and published on the official website of NCERT.

NTSE is conducted by NCERT to award scholarships to deserving candidates till the doctorate level of their studies. Candidates should read this article till the end to get detailed information about NTSE Andaman and Nicobar 2019-20. Information such as exam dates, the application process, admit card, eligibility, and syllabus, etc.

NTSE Andaman and Nicobar Exam Dates

EventsDates
Starting Date of ApplicationJuly 2019
Closing of ApplicationSeptember 2019
Admit Card for Stage 1October 2019
NTSE Andaman and Nicobar Stage 1 ExamNovember 2, 2019
NTSE Andaman and Nicobar Answer KeyNovember 2019
NTSE Andaman and Nicobar Stage 1 Result and Cut Off ScoresApril 2020
NTSE Admit Card for Stage 2May 2020
NTSE Stage 2June 2020
NTSE Stage 2 ResultSeptember 2020

NTSE Andaman and Nicobar Eligibility Criteria

The following are the eligibility criteria to appear for NTSE Andaman and Nicobar 2019-20.

  • Candidate belongs to the Reserved category should obtain at least 55% marks in class IX.
  • Candidate belongs to the general category should obtain at least 65% marks in class IX.
  • Candidates studying in class X in the academic year of 2019-20 are eligible for appearing in the exam.
  • Age limit for NTSE Andaman and Nicobar Stage 1 should not be above 18 years as on 1st July 2019.
  • Candidates from any recognized or top schools of Andaman and Nicobar are eligible.
  • Open schooling or distance learning candidates within the age limits are also eligible.
  • NMMS Scholarship holders could also be eligible for the exam.

NTSE Andaman and Nicobar Application Form

Students can collect an application form from the school authorities or the state liaison officer in July 2019. The last date for applying to NTSE Andaman and Nicobar Stage 1 exam is in September 2019. The following are the important points to apply for the NTSE Andaman and Nicobar exam.

  • All the required details should be filled correctly to avoid rejection of the form.
  • Application form should be rechecked thoroughly to avoid the chances of rejection.
  • Make sure to keep ready required documents including two passport size photos and copies of class IX mark sheets.
  • Submit the filled application form along with the required documents at the Principal’s office of your respective school.
  • From your school, the application will be forwarded to the state liaison’s officer for NTSE Andaman and Nicobar Stage 1 exam.
  • Reserved candidates should submit a copy of their Caste Certificates if required.
  • Physically Challenged candidates should submit the Disability Certificates if required.
  • Open schooling and distance learning students could submit it directly to the state liaison officer.

NTSE Andaman and Nicobar Application Fee

No application fee will be charged from the applicants to appear for the NTSE Andaman and Nicobar Stage 1 exam.

NTSE Andaman and Nicobar Admit Card

Andaman and Nicobar students can collect their admit card from the school authorities in October 2019. Admit card contains the name of the candidate, roll number, date and exam center details. Admit card is the most important document for students who are going to attend the exam. Without showing the admit card students will not be allowed to enter into the examination hall.

NTSE Andaman and Nicobar Question Pattern

Candidates have to attempt two types of question papers in the NTSE Andaman and Nicobar Stage 1 exam. These are the Mental Ability Test (MAT) and Scholastic Ability Test (SAT). MAT and SAT contains a total of 200 questions. Refer to the below-given table to know the details of the exam pattern for NTSE Andaman and Nicobar.

PapersNumber of questionsDuration
Mental Ability Test (MAT)100120 minutes
Scholastic Ability Test (SAT)100120 minutes
  • Each right answer carries 1 mark and no negative marking for wrong attempts.
  • Candidates can prepare from the syllabus of class IX and X with the aim of qualifying in the exam.
  • MAT evaluates the candidates reasoning, series, analogies, hidden figures, and problem-solving skills, etc.
  • MAT contains 100 questions with a duration of 120 minutes.
  • SAT evaluates the candidate’s subject-wise knowledge and skills.
  • SAT contains 100 questions with a duration of 120 minutes.

NTSE Andaman and Nicobar Answer Key

The official answer keys will not be released by DERT, Andaman and Nicobar. Some coaching institutions will release the answer keys after some days of the NTSE Andaman and Nicobar Stage 1 exam. That’s why candidates can check their answer keys from the websites of coaching institutions. The following points are given to know the detailed information about  the NTSE Andaman and Nicobar answer key:

  1. The answer key consists of all the answers to all the questions asked in each paper of the exam.
  2. The answer keys help the candidates to determine their performance in Stage 1 exam based on the expected scores.
  3. The correct marking approach should be followed.
  4. Candidates from Andaman and Nicobar can check the answer keys to calculate their marks and get the idea of qualifying for NTSE Stage 2.

NTSE Andaman and Nicobar Exam Result

NTSE Andaman and Nicobar exam result will be declared in offline mode by DERT in April 2020. The final merit list and cut off will be declared along with the result. DERT will declare the merit list of shortlisted candidates for the NTSE stage 2 to be held on June 2020. The following points contain some information about the NTSE Andaman and Nicobar exam results:

  • Students can visit their respective schools to know their exam result.
  • In school, they have to contact the concerned person to get the result.
  • In result, students can check their name, roll number and marks obtained in MAT and SAT papers.
  • Students can get a printed copy of the result sheet from the school and keep it safely for future reference.

Details Mentioned on NTSE Andaman and Nicobar Exam Result

Students can check the NTSE Andaman and Nicobar Stage 1 result in the respective schools. The NTSE Stage1 result sheet contains the following information of the selected candidates.

  • Name of the student and Roll number
  • Students School Address, Gender, and Date of Birth
  • Rank and Total marks of the students
  • Students scores in MAT and SAT

NTSE Andaman and Nicobar Stage 1 Qualifying Scores

Qualifying scores means the minimum scores obtained by a student to get selected for the NTSE stage 2 exam. NTSE Andaman and Nicobar qualifying scores will be announced along with the result by DERT on April 2020. The below mentioned are the various factors of qualifying scores :

  • Highest marks obtained by students
  • Lowest marks obtained by students
  • Total number of students belongs to each category
  • Previous year’s cut-off marks
  • The difficulty level of the exam pattern

The following table shows the cut-off marks for NTSE Andaman and Nicobar Stage 1 category candidates

CategoryCut-Off Marks
SC122
ST113
General150

The following table shows the qualifying percentage for NTSE Andaman and Nicobar category candidates

PapersGeneral CategoryReserved Categories (ST, SC, and PH)
MAT40%32%
SAT40%32%

NTSE Andaman and Nicobar Stage 1 Reservation Criteria

The Andaman and Nicobar students will be awarded scholarships as mentioned below:

CategoryReservation Percentages
SC15%
ST7.5%
PH4%
OBC27%

NTSE Andaman and Nicobar Stage 1 Syllabus

The NTSE Andaman and Nicobar Stage 1 has no specific syllabus. So, students can prepare from class IX and X syllabus which covers various topics on the subjects which will be asked in this exam. The below-mentioned table shows some important topics for NTSE Andaman and Nicobar Stage 1 2019-20.

PapersSubjectsSyllabus
SATMathQuadratic Equation, Surface Area & Volume, Algebraic Expression, Arithmetic Progression, Square Roots & Cube Roots, Percentage, Simple and Compound Interest, Exponent, Linear Equation, Number System, Probability, Triangles, Trigonometry, and Statistics.
ScienceMetals and non-metals, Acid, Bases and Salt, Human Body, Food Production & Management, Motion & Force, Source of Energy, Fibers & Plastics, Structure of an Atom and Measurements, Air, Carbon & its Components, Magnetism and Electricity, Cellular Level of Organisms, Environment, Light, Evolution, Living Organisms, Water, Physical and Chemical changes in matter.
Social SciencesIndustrial Revolution, Early Medieval Period, Resources & Development, New Empires and Kingdoms, The Mughal Empire, Delhi Sultanate, British Raj, Indian Constitution, Democracy & Elections, Atmosphere, The Judiciary, Map & Globe, Natural Vegetation and Solar System, Agriculture, Map and Globes, Motion of the Earth, Democracy, Diversity & Livelihood, State Government, Union Government, UN & International Agencies.
MATReasoningHypothetical Situations, Word Sequencing, Coding-Decoding, Distance & Direction, Word Problems, Calendar, Time & Clock, Mirror and Water Images, Blood Relations, Ranking and Arrangements, Alphabetic test, Venn Diagrams, Questions on Directions, Truth Verification, Cubes and Dice Problems, Classification, Embedded Figures, Analogy, Folding/Cutting Paper, Analytic Reasoning and Dot Fixing.

NTSE 2019-20 Scholarship Amounts

Students have to qualify in both the stages of NTSE with minimum cut-off marks to benefited with the amount of the scholarships every month.

Education LevelScholarship Amount
Class XI to XIIRs. 1250/-
UndergraduateRs. 2000/-
PostgraduateRs. 2000/-
PhDAccording to the UGC norms

For further information, leave your queries in the comment section to get in touch with you. Hope this NTSE Andaman and Nicobar 2019-20 article will be helpful to you.

The post NTSE Andaman and Nicobar 2019-20 for Class X | Admit Card, Dates and Application Process appeared first on Learn CBSE.

HP ePASS Scholarship | List of Scholarships, Application Procedure, Eligibility

HP ePASS Scholarship: HP ePASS scholarship is an online portal dedicated for students of Himachal Pradesh invented by the Department of Higher Education(DHE), Government of Himachal Pradesh. The state has been actively allotting scholarships to learners who are economically and financially weak and in need of higher education. The start of HP ePASS (HP Electronic Payment and Application System of Scholarships) gateway is a gradual step in the appropriate allocation of scholarships to scholars.

Every year, the state expends scholarships to up to 1.40 Lakh bonafide Himachal Pradesh students for persevering education in government and private institutions through HP ePASS scholarship portal. It has been meticulously designed to assure the scholarship prize money gets transferred to the actual beneficiary in a time-constrained manner. This article defines each and every important detail on HP ePASS scholarship, including the list of scholarships, application procedure, eligibility criteria and rewards given to each scholar.

List of HP ePASS Scholarships

The HP ePASS scholarships are broadly categorized into three different divisions named as, Central Sponsored Scheme, State-Sponsored Scheme, and Technical Education Schemes. These accomplishments not only allow the students to seek an education in different institutes of Himachal Pradesh but also across the country. Below are the highlights of the complete list of scholarships open on HP ePASS portal, along with their essential information like category and tentative application period.

Central Sponsored Scheme-14 Scholarships

Scholarship NameApplication Duration
National Scheme of Incentive to Girls for Secondary Education (NSIGSE) (CSS)October to December
Pre-Matric Scholarship Scheme for SC/ST Students (9thand10th) CSSJuly to October
Pre-Matric Scholarship for OBC Students (1st to 10th) CSSJuly to October
Post-Matric Scholarship to SC/ST Students (CSS)July to October
Post-Matric Scholarship to OBC Students (CSS)July to October
Dr. Ambedkar Post-Matric Scholarship for Economically Backward Classes (EBC)October to December
Upgradation of Merit of SC/ST Students (CSS) (Only for GSSS Bharmour, Chamba)October to December
Merit Cum Means Scholarship Scheme for Students belonging to Minority Community (CSS)October to December
Post-Matric Scholarship Scheme to Students belonging to Minority CommunityJuly to October
Maulana Azad National Scholarship Scheme for Meritorious girl students belonging to MinoritiesJuly to October
National Means Cum Merit ScholarshipsOctober to December
Scholarship to Disabled StudentsJuly to October
Dr. Ambedkar Pre-Matric Scholarship for DNTs (CSS)July to October
Dr. Ambedkar Post-Matric Scholarship for DNTs (CSS)July to October

State-Sponsored Scheme-12 Scholarships

Scholarship NameApplication Duration
Dr. Ambedkar Medhavi Chattervriti Yojana for SC StudentsJuly – September
Dr. Ambedkar Medhavi Chattervriti Yojana for OBC StudentsJuly – September
Swami Vivekanand Utkrisht YojanaJuly – September
Thakur Sen Negi Utkrisht Chattervriti YojanaJuly – September
Kalpana Chawala Chattervriti YojanaJuly – September
Mukhiya Mantri Protsahan Yojana (One-time Incentive)July – September
Maharshi Balmiki Chattervriti YojanaJuly – September
IRDP Scholarship SchemeJuly – September
Sainik School Sujanpur Tihara ScholarshipJuly – September
Rashtriya Indian Military College Scholarship (Only for RIMC Dehradun)July – September
NDA Scholarship Scheme (Only for NDA Khadakwasla)July – September
Financial Assistance to the Children of the Armed Forces personnel killed/ disabled during the different war/ operationsJuly – September
Mukhya Mantri Gyandeep YojanaJuly – September
Indira Gandhi Utkrisht Chattervriti Yojana for Post plus two StudentsJuly – September
Lahaul Spiti PatternJuly – September
Girls Attendance ScholarshipJuly – September

Technical Education Scheme-3 Scholarships

Scholarships for PG/Degree and Engineering Level CoursesNA
Scholarships for Diploma CoursesNA
Scholarships for ITI StudentsNA

Eligibility Criteria And Rewards for HP ePASS Scholarships

Let us discuss here the eligibility criterion required for all the schemes of HP ePASS scholarships one by one.

National Scheme of Incentive to Girls for Secondary Education (NSIGSE) (CSS)

  • Only for girl candidate.
  • Student who want to take admission in 9th standard
  • Student should not have completed 16 years of age, as on 31st March of the year.
  • All girls who have passed Middle Standard Examination from Kasturba Gandhi Balika Vidyalaya irrespective of caste/religion are eligible for this scholarship.
  • The amount of incentive under this scheme is Rs 3,000/- and will be given in the shape of a Time Deposit.

Pre-Matric Scholarship Scheme for SC/ST Students (9th and 10th) CSS

  • Students of class 9th and 10th are eligible.
  • Only for SC/ST candidates
  • Parents or guardians annual income should not exceed 2.5 lacs.
  • the scholarship will be awarded for 10 months in an academic year

Pre-Matric Scholarship for OBC Students (1st to 10th) CSS

  • This scholarship is available for students from 1st to 10th classes.
  • Only for OBC students.
  • Candidates whose parents Annual Income does not exceed 2.5 lacs.
  •  the scholarship will be awarded for 10 months in an academic year.

Post Matric Scholarship for SC/ST Students (CSS)

  • This scholarship is available for students who want to pursue higher studies after their matriculation.
  • Only for SC/ST candidates
  • Family annual income should not be more than 2.5 lacs.

Post Matric Scholarship for OBC Students (CSS)

  • This scholarship is available for students who want to pursue higher studies after their matriculation.
  • Only for OBC candidates
  • Family annual income should not be more than 1.5 lacs.

Dr. Ambedkar Post-Matric Scholarship for Economically Backward Classes (CSS)

  • Available for General Candidates
  • Family income should not be more than 1 lac.

Upgradation of Merit of SC/ST Students (CSS) (Only for GSSS Bharmour, Chamba)

  • Under this Scheme, Scholarship is awarded to Seven students (SC-6 and ST-1) of the State, admitted in Class 9th at the GSSS Bharmour, Chamba, on the basis of merit list of middle standard exams. The Scheme is tenable from Class 9th to 10+2.

Merit Cum Means Scholarship Scheme for Students belonging to Minority Community (CSS)

  • This Scholarship is given to pursue Degree / Postgraduate level Technical Professional Courses from Recognized Institutions
  • Students belonging to the Minority Groups(Muslim – 37, Sikh – 22, Christian – 2, Buddhist – 24, Total = 85 Students)
  • Students should have not less than 50% marks
  • The annual income of the parent or guardian of the beneficiary should not exceed Rs 2.50 Lacs.

Post-Matric Scholarship Scheme to Students belonging to Minority Community

  • Students belonging to the Minority Groups (Muslim – 309, Sikh – 187, Christian – 20, Buddhist – 196,Jain – 4,Parsi – 2, Total = 718 Students).
  • The annual income of the parent or guardian of the beneficiary should not exceed Rs 2.00 Lacs.
  • Student should not have less than 50% marks in the previous final examination.

Maulana Azad National Scholarship Scheme for Meritorious girl students belonging to Minorities

  • Girl students belonging to the Minority Groups (Muslim – 309, Sikh – 187, Christian – 20, Buddhist – 196,Jain – 4,Parsi – 2, Total = 718 Students).
  • The annual income of the parent or guardian of the beneficiary should not exceed Rs 2.00 Lacs.
  • Students should not have less than 50% marks in the previous final examination.

Scholarship to Disabled students

  • Trust Fund
  • National Fund

National Means Cum Merit Scholarships

  • Students who want to pursue further studies class 9th to class 10th.
  • Students whose parental income from all sources is not more than Rs 150000/- are eligible.
  • Scholarship of Rs 6,000/- p.a. ( Rs 500/- p.m. ) per student is awarded to the selected students every year, for study in classes from IX to XII in Government, Government – Aided and local body schools. The Scholarship will be renewed for class X – XII after the student passes the previous class, subject to good conduct and regularity in attendance.

Dr. Ambedkar Pre-Matric Scholarship for DNTs (CSS)

  • Applicable for those Denotified, Nomadic and Semi-Nomadic Tribes, (DNTs) whose family income is less than Rs. 2.0 lacs per annum will be eligible.
  • For DNTs who are not covered under the SC/ST/OBC category.

Dr. Ambedkar Post-Matric Scholarship for DNTs (CSS)

  • Applicable for those Denotified, Nomadic and Semi-Nomadic Tribes, (DNTs) whose family income is less than Rs. 2.0 lacs per annum will be eligible.
  • For DNTs who are not covered under the SC/ST/OBC category.

Dr. Ambedkar Medhavi Chattervriti Yojana for SC Students

  • The Scholarship is given to top 1250 meritorious students of Scheduled Caste category.
  • The scholarship is given Rs 12,000/- p.a.

Dr. Ambedkar Medhavi Chattervriti Yojana for OBC Students

  • The Scholarship is given to top 1000 meritorious students of OBC.
  • The scholarship is given Rs 10,000/- p.a.

Swami Vivekanand Utkrisht Yojana

  • The Scholarship is given to top 2000 meritorious students of General category.
  • The scholarship amount given is Rs 10,000/- p.a.

Thakur Sen Negi Utkrisht Chattervriti Yojana

  • The Scholarship is given to top 100 Girls and top 100 Boys students belonging to the Tribal Community of HP.
  • The scholarship is given Rs 11,000/- p.a.

Indira Gandhi Utkrisht Chattervriti Yojana for Post plus two Students

  • Awarded to the Ten Male Toppers each from the merit list of 10+2 Arts, Science and Commerce, supplied by the H.P. Board of School Education Dharamshala, and to the Ten Toppers (both Male and Female) from the merit lists of B.A. / B.Sc. / B.Com., supplied by the HPU, Shimla, provided they join any academic/professional stream.
  • The rate of the scholarship would be Rs 10,000/- p.a. per student.

Kalpana Chawala Chattervriti Yojana

  • Top 2000 meritorious girl students of 10+2 of all study groups i.e. Science, Arts and Commerce streams.
  • Scholarship of Rs 15000/- per student p.a.

Mukhiya Mantri Protsahan Yojana (One-time Incentive)

  • All Himachali students, who are selected and take admission for a degree course in any Indian Institute of Technology (IITs) or All India Institute of Medical Sciences (AIIMSs) or for a Post Graduate Diploma Course in any Indian Institute of Management (IIMs). Indian School of Mines (ISM) Dhanbad at Jharkhand and Indian Institute of Science (IISc) at Bangalore.
  • a one-time incentive/award of Rs 75000/- will be given.

Maharshi Balmiki Chattervriti Yojana

  • BonafideHimachali Girl Students belonging to Balmiki Families, engaged in unclean occupation, beyond Matric to College Level, for studies from any Govt. or Private Recognizedschool / college situated within Himachal Pradesh.
  • Scholarship of Rs 9,000/- p.a. is given.

IRDP Scholarship Scheme

Students belonging to IRDP families and studying in 9th to University level are eligible for this scholarship provided they are pursuing their study in Govt. / Govt. Aided Institutions in H.P.

Sainik School Sujanpur Tihara Scholarship

The students who are studying in Sainik School SujanpurTihra and are Bonafide residents of Himachal Pradesh from class VI to XII.

Rashtriya Indian Military College Scholarship (Only for RIMC Dehradun)

  • This award is made to the Ten Students who are Bonafide residents of H.P. and are studying from VIII to XII in Rashtriya Indian Military College, Dehradun. Two students from each class are eligible for the scholarship.
  • The amount of scholarship is Rs 20,000/- p.a.

NDA Scholarship Scheme (Only for NDA Khadakwasla)

Cadets of Himachal Pradesh, who are getting training at National Defence Academy (NDA), Khadakwasla, Pune.

Financial Assistance to the Children of the Armed Forces personnel killed/ disabled during the different war/ operations

  • Children of Armed Forces Personnel killed / disabled in the different wars /operations are eligible
  • In case of disability is below 50%, the children will get half scholarship.

Mukhya Mantri Gyandeep Yojana

  • All Those Students who Bonafide of the HP State pursuing Professional/Technical course and Higher Education Course from recognized Institutions in India.
  • Under this Scheme, the Interest subsidy is admissible from any Bank on Education loan availed up to the maximum of Rs. 10 (Ten lakh only) Interest subsidy to the extent of 4% p.a. on Education loan will be allowed.

Scholarships for PG/Degree and Engg Level and Diploma Courses

  • The Scholarship shall be awarded strictly according to merit to be determined with reference to (i) Income of parents or guardian and (ii) Marks obtained in the Examination / Class by virtue of which a candidate is eligible for admission provided the students have secured at least 60% of marks in the said exam. This condition will however not be applicable to the SC and ST Students.
  • The Scholarship shall be awarded only if the total income of the sources of parents or guardian is not more than Rs 36000/- p.a. For PG/Degree and Engg Level and Diploma Courses.
  • Scholarships @ Rs.250/- p.m. shall be awarded to the ITI students belonging to the SC, ST and OBC Categories and For General Category the value of Scholarship is @ Rs.100/- p.m.

HP ePASS Application Procedure

To apply for all the HP ePASS scholarships, students have to visit the official website and follow the registration procedure. A candidate can apply for up to 13 scholarships at a time. To know the complete details of the application process, follow the steps given below.

  • Log on to the ePASS website at http://hpepass.cgg.gov.in .
  • Click on the “Apply Online” button below Official login.
  • Read the Instructions carefully and Mark the checkbox at the end of the page certifying “I have read the Instructions. Proceed to Registration / Login Page.”
  • First-time users click on “Register”, which would take you to the “Student Registration Form”.
  • Enter your basic details (like AADHAARDetails (UID / EID), Name, Date of Birth, Father and       Mother Name, Category, Mobile No. and email) on the form, match the Security Captcha, and click on “Register”. (Please note that either Mobile No. or email address would be mandatory, as the computer-generated Password would be sent through SMS / e-Mail on Registration).
  • Once registered, then enter your login details “Login Id/ AADHAAR and Password”, and click on “Login”.
  • After Login, the student needs to fill the various details in the following order:
    • Primary Details
    • Candidate Details
    • Present Institution Details
    • Contact Details
    • Bank Details
  • Certificates Details
    • The Student can edit / update or fill his form in parts, then take a printout of Online Application Form, and submit the same along with the Scheme Specific Document(s) to the respective Institution, within the stipulated time as mentioned in the Schedule above.
    • After successful submission of his Online Application Form, it would be automatically forwarded to his Institute for verification.
    • After that, the Student can check the status of his online application, any time, by logging in at the ePass Website, and keep a track of all progress.

Note: As soon as the students submit the application online, the details will be automatically sent to his/her institute for verification. The students can review and check the status of their application by logging in to the HP ePASS Portal.

Documents Required To Apply for HP ePASS Scholarship

The documents that need to be associated while filling the application form at the HP ePASS website, differ according to the scheme chosen by the scholar. A general list of documents that a candidate needs to present in support of his/her application has been mentioned below.

  • A passport size photo
  • Birth certificate
  • Aadhar card number
  • Himachali Bonafide Certificate
  • Income certificate issued by the authority, not below the tehsildar
  • Receipts and fee structure of the university (if applicable)
  • Marks card of the last qualifying examinations
  • Bank statement of the student’s bank account
  • Caste certificate, tribal certificate (if applicable)

FAQ’s

Question 1.
Who are eligible to apply for HP ePASS Central and State-Sponsored Scholarships?

Answer:
H.P. bonafide students, studying in Class IX onwards, within the state or outside the state, within India, can apply online under these schemes.

Question 2.
Can a Student apply as a fresh candidate, if he/she is a renewal candidate?

Answer:
No, a student cannot apply as a fresh if he/she is a Renewal candidate. Their application will be rejected in that case.

Question 3.
Which fields in the application form, students can edit?

Answer:
Student can edit all fields except a few like parental income, mobile no. email Id etc. It may be noted that once you click on “Finalize and submit” button your application will be forwarded to the next level and then you cannot edit further.

Question 4.
What to do, when a student detect mistakes after forwarding the application form to the next level?

Answer:
The student should separately inform the mistakes detected by them to the Institute/District/Region/State. The software provides the facility at the level of the Institute and State to edit and correct limited information.

The post HP ePASS Scholarship | List of Scholarships, Application Procedure, Eligibility appeared first on Learn CBSE.

PUBG Mobile APK Download

PUBG MOBILE APK +OBB DATA: FREE DOWNLOAD, INSTALL, PLAY, TIPS & TRICKS, SETTINGS, AND MORE

Nowadays, everyone is talking about the free-to-play battle royal game PlayerUnknown’s Battleground aka PUBG.  The Bluehole Studio has done a great job by converting the PC experience of the game into mobile. They joined hands with Tencent to release PUBG mobile version on Google Play Store and Apple App Store for free. So, you can download PUBG mobile free from your respected stores. So, are you excited to play PUBG mobile on your Android or iOS device? Check out the following procedure to download latest apk of PUBG Mobile.

Before heading to the process to install PUBG on your mobile, you should make it sure that your Android and or iOS have the capacity tosupport PUBG Mobile Apk. If you don’t have any idea about the PUBG Mobile requirements, then here’s the information.

PUBG Mobile Requirements

Whether you want to play PUBG mobile on Android or iPhone, here are some necessities of PUBG mobile.

PUBG Mobile Android System Requirements 

The PUBG mobile is compatible with more than 500 Android devices. Make sure that your Android device has 2GB of RAM and running 5.1.1 lollipop or later version.

PUBG Mobile iOS System Requirements 

If you have iPhone and want to play PUBG on your iOS device, then you need iPhone 5S and above version running iOS 9.0,10, 11,12, 13 or later. Well, you may get the lag issue on your iPhone 5S. But, it will run smoothly on 5S following devices.

PUBG Mobile Maps

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PUBG Mobile Maps 2

If you are new on this battle royal games then you should also know about four PUBG mobile maps names so that you will be ready to play with your friends on your favourite map. There is a total of four maps available in PUBG Mobile-ErangelMiramarSanhok, and newly added Vikendi snow map.

How to Download PUBG Mobile APK Android or iPhone

You can free download PUBG mobile APK from Google Play Store for Android and App Store for iOS devices. There are three different login options available while playing PUBG mobile- As Guest, use Facebook, or Twitter. It will help you to keep your PUBG mobile record sync whenever you play PUBG APK form different smartphone. So, here can find every Apk of PUBG Mobile inclusing latest betas.

PUBG Mobile 0.14.0 Beta

PUBG Mobile 0.13.5 APK

PUBG Mobile 0.13.0 APK

PUBG Mobile 0.12.5 APk

PUBG Mobile 0.12.0 APK+OBB

PUBG Mobile 0.14.5 APK (Chinese Lightspeed)

PUBG Mobile 0.11.5 APK

PUBG Mobile 0.13.5 Chinese New Year APK

PUBG Mobile 0.11 APK (Zombie Mode)

PUBG Mobile 0.10.5 APK

PUBG MOBILE 0.10.0 APK

PUBG MOBILE 0.9.0 APK

PUBG MOBILE 0.8.0 APK

PUBG MOBILE 0.7.0 APK

How To Install PUBG Mobile APK on Android and iPhone

Step #1: Once you have downloaded the latest PUBG Mobile APK file from the above link, you need to go to the settings, and then go to Security.
Step #2: Once you tap on security, scroll down you will find unknown sources just tick mark that option.
Step #3: Now go to Downloads and tap on PUBG Mobile APK version and click on install and then follow the instructions as asked to enjoy the game on your smartphone.

How to Play PUBG Mobile Cross-player Multiplayer

If you want to know you can play PUBG mobile cross multiplayer or not after installing PUBG APK, then let me tell you that you can play PUBG mobile in cross-player multiplayer on your iOS and Android device. For instance, If you have logged PUBG on your iOS device with Facebook, then you see all Facebook friend playing PUBG on Android and iPhone. So, whether you are playing on an iOS device or Android, you can play PUBG mobile in cross-platform.

How to Play PUBG Mobile with Friends (Add Friends in PUBG Mobile)

Once after successfully installing PUBG mobile on your phone, you can add your Friends to play in the squad. Well, you can play with one friend or with three friends from PUBG mobile. Here’s how to add friends in PUBG mobile on Android and iPhone.

Step #1: Launch the PUBG mobile on your smartphone.

Step #2: Next, tap on Friend icon at the left side of the screen.

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Step #3:  Now tap on “Add Friend” on the right side of the screen and search by name in AdHow to Play PUBG Mobile with Friends – Enter the Name 5vance search.

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Step #4: Enter the name and hit the yellow Search button.

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Clik here to view.

Step #5: Once you get the friend suggestions, tap on Add button next to the profile to a sent request.

Step #6: Type a message and hit the Send button to send the request.

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How to Play PUBG Mobile with Friends - Type a Message 6

Step #7: Once they accept the request, you can add them from the Invite list once they come online and play with them.

How to Accept Friend Request in PUBG mobile?

If you get a request from your friends, then you will need to accept the request to play with them. You will see the friend request in the friend icon. Here’s the process to accept the friend request in PUBG mobile.

Step #1: Open the PUBG mobile on your device.

Step #2: Tap on friend icon on the left side of the screen with a friend request.

Step #3: Now tap on Game Friend and then Request List.

Step #4: Here you need to tap Accept button next to your friend’s profile to accept PUBG friend request.

How to Turn on Voice Chat on PUBG Mobile

To keep connected with your friends, you can use the PUBG mobile voice chat to keep on a conversation with your squad. With the PUBG mobile 0.7.0 updates, the voice chat feature has been clear and noise-free. To enable voice chat on PUBG mobile, you will need to tap on speaker and mic icon at the button on the home page and right top side when the game start.

How to Send Message to Your PUBG Mobile Friend

In the PUBG mobile, there is also an option to message your friend. So, if you want to message your friend to invite play PUBG with you, then here’s how to message your friends in PUBG mobile.

Step #1: Open the PUBG mobile main menu.

Step #2: Next, tap on the friend list at the left side of the screen and select friend you want to message.

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How to Send Message to Your PUBG Mobile Friend 7

Step #3: Now tap on “Start a Chat”, Enter the message and hit the Send button.

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How to Send Message to Your PUBG Mobile Friend - Start Chat 8

How to Select Solo, Due, or Squad play in PUBG Mobile

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How to Select Solo, Due, or Squad play in PUBG Mobile 9

The PUBG Mobile allows up to four friend play together. You can also play Duo if you want to play with only one friend. Well, in the updated version of PUBG mobile, you can select Solo, Due and Squad mode by tapping on the Mode and select single, due, or squad person icon under the Team section on the pop-up screen as per your requirement.

How to Select TPP (third-person view) or FPP (first person view) Mode in PUBG Mobile?

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How to Select TPP third-person view or FPP first person view Mode in PUBG Mobile 10

The PUBG Mobile June update included first-person view mode in the game. So, you can play PUBG mobile in FPP mode on your Android device. To choose TPP or FPP mode, you will need to tap on the mode on the main menu and select Third-Person Perspective (TPP) or First- Person Perspective (FPP) at the top of the pop-up screen. Here you can also select Classic and Arcade mode to play as per your choice.

Top 10 Best PUBG Mobile Tips & Trick You Should Know To Be The Last Man Standing

PUBG mobile is all among surviving in the long island by killing everyone else to get chicken dinner. But, there are some tactics and PUBG mobile tips you can follow to get Chicken Dinner.  Here are those:

 #1: Choose Mode Friendly Clothes

PUBG mobile is the game of survival where you have a battle against other players. So, when it comes to wearing cloth, you should keep in mind that it isn’t a fashion show or party. You can choose the right clothes as per the mode you are playing. You can wear a sand colour t-shirt if you are going to play Miramar. Likewise, you can choose Erangle map friendly wild t-shirt so that your enemies can’t track you easily.

#2: Choose Lower Competitive and More Loot Area

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PUBG Mobile Choose Lower Competitive and More Loot Area 11

The first thing is to try to land in the area where fewer people jump and more weapons, consumables, and other necessary stuff available. It will help you to get every needed item quickly without being shot by other players. Personally, I prefer to land on Rozhok, Mylta Power, School in Erangle map.

#3: Follow Your Map Wisely

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PUBG Mobile Follow Your Map Wisely 12

Once you start playing PUBG mobile on your smartphone, the left top corner side of the screen you will always see live map. This map helps you to track your enemies by gunshot and footsteps. And also, show you the blue and safe zone. So, also keep your eyes on the map.

#4: Enable Peek & Fire

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PUBG Enable Peek and Fire 13

The Peek & Fire feature is not enabled by default in the game. You have to turn it in the Basic setting to play by just revealing your head to shot by hiding your full body. Once you activate Peek & Fire, you will see two upper body icons at the left side of the screen.

#5: Use Headphones and Connected With Your Squad

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PUBG Use Headphones and Connected With Your Squad 14

To get a better sound of gunshot and enemy footstep, we suggest you always use headphones while playing the game with or without a squad. It will give you the bright idea about from where the gun triggered to track the enemy. Also, you can use headphone to keep communicating with your squad to make strategies and plans to play like a team to get chicken dinner.

#6: Keep Moving

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Keep Moving PUBG Lite 15

Do not stick to one point; it will give other players a perfect headshot of you. So, try to keep moving to avoid getting killed. There are a lot of snipers roaming around you, so keep moving by taking good covers. It will help you to survive for long.

#7: Avoid Gun Fights

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Avoid Gun Fights PUBG 166

Sometimes, involving in other fights won’t the right choice. You may get killed in the battle. So, you need to take your action correctly. If you hear two groups or persons are fighting, then hiding is an excellent idea. After that, you can kill the reaming one. Agree?

#8: Silencers Are Good Friends

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PUBG Silencers Are Good Friends 17

In the PUBG mobile game, there are differently typed of guns available-AR, Sniper, SMG, etc. When pressing the trigger to shoot someone, it makes noise, and you will get easily tracked. Therefore, there are silencers are available for different type of weapons.  So, once you found a suitable silencer for your gun, just attacked it on your weapon to kill rivals silently.

#9: Lie Down While Looting Enemy Crates

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PUBG Lie Down While Looting Enemy Crates 18

Once you kill someone, for you would want to raid the crate, right? So, our advice is always raid crates while being lying down so that your enemy can locate you even if they see the green light popping on the person you have just killed.

#10: Prefer To Play On The Edge

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If there are only two or three circles left, then try to play on the edge. It will give you the unobstructed view of the whole ring especially if you have 4x or 8x scope. It also minimizes the chances of getting killed from behind.  Therefore, always prefer to play on the edge of the play zone where our blue line is near to the white.

Finishing

This was all about PUBG mobile on your Android and iPhone. Hope here you have got everything about the PUBG mobile apk. The lot more are coming in upcoming PUBG mobile updates. Till then, keep playing PUBG mobile.

Happy Playing!

Download PUBG Mobile On PC

 

 

 

 

 

 

 

 

 

 

 

The post PUBG Mobile APK Download appeared first on Learn CBSE.

Important Questions for Class 10 Maths Chapter 15 Probability

Important Questions for Class 10 Maths Chapter 15 Probability

Probability Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
In a family of 3 children calculate the probability of having at least one boy. (2014OD)
Solution:
S = {bbb, bbg, ggb, ggg}
Atleast 1 boy = {bbb, bbg, ggb}
∴ P(atleast 1 boy) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 2.
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. (2015D)
Solution:
Total English alphabets = 26
Number of consonants = 21
∴ P (letter is a consonant) = Image may be NSFW.
Clik here to view.
\frac{21}{26}

Question 3.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. (2016OD)
Solution:
S = 52
P(neither a red card nor a queen)
= 1 – P(red card or a queen)
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 1

Question 4.
A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square. (2013OD)
Solution:
Total number of cards = 50 – 6 + 1 = 45
Perfect square numbers are 9, 16, 25, 36, 49, i.e.,
5 numbers
∴ P(a prefect square) = Image may be NSFW.
Clik here to view.
\frac{5}{45}=\frac{1}{9}

Question 5.
Cards marked with number 3, 4, 5, …, 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. (2016D)
Solution:
Total no. of cards = 50 – 3 + 1 = 48
Perfect square number cards are 4, 9, 16, 25, 36, 49 i.e., 6 cards
∴ P(perfect square number) = Image may be NSFW.
Clik here to view.
\frac{6}{48}=\frac{1}{8}

Question 6.
A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1? (2017D)
Solution:
(-3)2 = 9; (-2)2 = 4; (-1)2 = 1; (0)2 = 0
(1)2 = 1; (2)2 = 4; (3)22 = 9
∴ P(Sq. of nos. of ≤ 1) = Image may be NSFW.
Clik here to view.
\frac{3}{7}

Question 7.
If two different dice are rolled together, calculate the probability of getting an even number on both dice. (2014D)
Solution:
Two dice can be thrown as 6 × 6 = 36 ways
Even numbers on both the dice can be obtained as
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Important Questions for Class 10 Maths Chapter 15 Probability 2

∴ P(even numbers) = Image may be NSFW.
Clik here to view.
\frac{9}{36}=\frac{1}{4}

Question 8.
Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6. (2015OD)
Solution:
Total outcomes = 6= 62 = 36
Possible outcomes having the product of the two numbers on the top of the dice as 6 are
(3 × 2, 2 × 3, 6 × 1, 1 × 6), i.e, 4
P(Product of two nos. is 6) = Image may be NSFW.
Clik here to view.
\frac{4}{36}=\frac{1}{9}

Question 9.
The probability of selecting a rotten apple ran domly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap? (2017OD)
Solution:
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Important Questions for Class 10 Maths Chapter 15 Probability 3

∴ No. of rotten apples = 900 × 0.18 = 162

Probability Class 10 Important Questions Short Answer-I (2 Marks)

Question 10.
A coin is tossed two times. Find the probability of getting both heads or both tails. (2011D)
Solution:
S = {HH, HT, TH, TT) = 4
P (both heads or both tails)
= P (both heads) + P (both tails)
= Image may be NSFW.
Clik here to view.
\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}

Question 11.
In a simultaneous toss of two coins, find the probability of getting:
(i) exactly one head,
(ii) atmost one head. (2012D)
Solution:
The sample space is given by
S = {HH, HT, TH, TT}
Total events n(S) = 4
(i) exactly one head = {HT, TH} = 2
P(exactly one head) = Image may be NSFW.
Clik here to view.
\frac{2}{4}=\frac{1}{2}

(ii) atmost one head = {HT, TH, TT} = 3
P(atmost one head) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 12.
Two coins are tossed simultaneously. Find the probability of getting at least one head. (2013OD)
Solution:
S = {HH, HT, TH, TT}, i.e., 4
∴ P (atleast one head) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 13.
Three coins are tossed simultaneously. Find the probability of getting exactly two heads. (2013OD)
Solution:
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8
P(exactly two heads) = Image may be NSFW.
Clik here to view.
\frac{3}{8}

Question 14.
Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail. (2014D)
Solution:
S = {HH, HT, TH, TT}, i.e., 3
P(at least one tail) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 15.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king. (2011OD)
Solution:
P (neither an ace nor a king)
= 1 – P (either an ace or a king)
= 1 – [P (an ace) + P (a king)]
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Important Questions for Class 10 Maths Chapter 15 Probability 4

Question 16.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen. (2013D)
Solution:
P (neither a king nor a queen)
= 1 – P (king or queen)
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 5

Question 17.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a jack nor an ace. (2013D)
Solution:
Total number of cards = 52
Numbers of jacks = 4
Numbers of aces = 4
Card is neither a jack nor an ace
= 52 – 4 – 4 = 44
∴ Required probability = Image may be NSFW.
Clik here to view.
\frac{44}{52}=\frac{11}{13}

Question 18.
Two dice are thrown simultaneously. Find the probability of getting a doublet. (2013OD)
Solution:
Two dice can be thrown as 6 x 6 = 36 ways
“a doublet” can be obtained by (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways
P(a doublet) = Image may be NSFW.
Clik here to view.
\frac{6}{36}=\frac{1}{6}

Question 19.
A dice is tossed once. Find the probability of getting an even number or a multiple of 3. (2013D)
Solution:
S = {1, 2, 3, 4, 5, 6} = 6
‘an even number or multiple of 3’ are 2, 3, 4, 6, i.e., 4 numbers
∴ Required probability = Image may be NSFW.
Clik here to view.
\frac{4}{6}=\frac{2}{3}

Question 20.
Two different dice are tossed together. Find the probability. (2014OD, 2015D)
(i) that the number on each die is even.
(ii) that the sum of numbers appearing on the two dice is 5.
Solution:
Two dice can be thrown as 6 × 6 = 36 ways
(i) The probability of number on each die is even are
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 6

Probability Class 10 Important Questions Short Answer-II (3 Marks)

Question 21.
A coin is tossed two times. Find the probability of getting at least one head. (2011D)
Solution:
S = {HH, HT, TH, TT}
Total number of ways = 4
Atleast one head = {HH, HT, TH}, i.e., 3 ways
∴ P (atleast one head) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 22.
A coin is tossed two times. Find the probability of getting not more than one head. (2011D)
Solution:
S = {HH, HT, TH, TT} = 4
Favourable cases are HT, TH, TT i.e., 3 cases
∴ P (not more than 1 head) = Image may be NSFW.
Clik here to view.
\frac{3}{4}

Question 23.
Three distinct coins are tossed together. Find the probability of getting (2015 D)
(i) at least 2 heads
(ii) at most 2 heads.
Solution:
Total number of possible outcomes = 21 = 23 = 8
(HHH, TIT, HHT, THH, THT, HTH, TTH, HTT)
(i) Possible outcomes of at least two heads = 4
(HHT, THH, HHH, HTH)
∴ P(at least two heads) = Image may be NSFW.
Clik here to view.
\frac{4}{8}=\frac{1}{2}

(ii) Possible outcomes of at most two heads = 7
(HHT, TTT, THH, THT, HTH, TTH, HTT)
∴ P(at most two heads) = Image may be NSFW.
Clik here to view.
\frac{7}{8}

Question 24.
Three different coins are tossed together. Find the probability of getting
(i) exactly two heads
(ii) at least two heads
(iii) at least two tails. (2016OD)
Solution:
S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 8
(i) Exactly two heads
= HHT, HTH, THH, i.e., 3 ways
∴ P (exactly two heads) = Image may be NSFW.
Clik here to view.
\frac{3}{8}

(ii) Atleast two heads
= HHT, HTH, THH, HHH i.e., 4 ways
∴ P (atleast two heads) = Image may be NSFW.
Clik here to view.
\frac{4}{8}=\frac{1}{2}

(iii) Atleast two tails
= TTH, THT, HTT,TTT i.e., 4 ways
∴ P (atleast two tails) = Image may be NSFW.
Clik here to view.
\frac{4}{8}=\frac{1}{2}

Question 25.
A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5. (2011OD)
Solution:
Total number of tickets = 40
‘A multiple of 5’ are
5, 10, 15, 20, … 40, i.e., 8 tickets
∴ P (A multiple of 5) = Image may be NSFW.
Clik here to view.
\frac{8}{10}=\frac{1}{5}

Question 26.
A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. (2012 D)
Solution:
Multiples of 3 & 4 are 12, 24, 36, 48, i.e., 4 nos.
P(a multiple of 3 and 4) = Image may be NSFW.
Clik here to view.
\frac{4}{50}=\frac{2}{5}

Question 27.
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is Image may be NSFW.
Clik here to view.
\frac{1}{4}
. The probability of selecting a blue ball at random from the same jar is Image may be NSFW.
Clik here to view.
\frac{1}{3}
. If the jar contains 10 orange balls, find the total number of balls in the jar. (2015OD)
Solution:
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 7

Question 28.
A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is Image may be NSFW.
Clik here to view.
\frac{3}{10}
and that of a black ball is Image may be NSFW.
Clik here to view.
\frac{2}{5}
, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag. (2015OD)
Solution:
Let W, B and R denote the White, Black and Red balls respectively,
We know, Total probability = 1
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 8

Question 29.
A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. (2017OD)
Solution:
Let the number of black balls = x
the number of white balls = 15
∴ Total number of balls = x + 15
∴ P(black ball) = 3P(White balls)
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 9

Question 30.
In a single throw of a pair of different dice, what is the probability of getting
(i) a prime number on each dice?
(ii) a total of 9 or 11? (2016D)
Solution:
Two dice can be thrown in 6 × 6 = 36 ways
(i) “a prime number on each dice” can be obtained as (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5), i.e., 9 ways.
∴ P(a prime no. on each dice) = Image may be NSFW.
Clik here to view.
\frac{9}{36}=\frac{1}{4}

(ii) “a total of 9 or 11” can be obtained as (3, 6),(6, 3),(4, 5),(5, 4) (5, 6),(6, 5).
Total ‘9’ Total ’11’ i.e., 6 ways
∴ P(a total of 9 or 11) = Image may be NSFW.
Clik here to view.
\frac{6}{36}=\frac{1}{6}

Question 31.
A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that:
(i) Ramesh will buy the selected shirt?
(ii) ‘Kewal will buy the selected shirt? (2016D)
Solution:
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 10

No. of good shirts = 88
(i) P(Ramesh buys a shirt)
= P(good shirts) = Image may be NSFW.
Clik here to view.
\frac{88}{100}=\frac{22}{25}

(ii) No. of shirts without major defect = 96
P(Kewal buys a shirt)
= P(shirts without major defect)
= Image may be NSFW.
Clik here to view.
\frac{88+8}{100}=\frac{96}{100}=\frac{24}{25}

Question 32.
A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of getting: (2012 OD)
(i) a red king.
(ii) a queen or a jack.
Solution:
Cards in a pack = 52
Number of kings = 4
Number of red kings = 2
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 11

Question 33.
All red face cards are removed from a pack of playing cards. The remaining cards were well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (2015D)
(i) a red card
(ii) a face card
(iii) a card of clubs.
Solution:
Number of red face cards removed = 6
∴ Remaining cards = 52 – 6 = 46
Hence, Total no. of outcomes = 46
(i) Possible outcomes of red cards = 26 – 6 = 20
∴ P(a red card) = Image may be NSFW.
Clik here to view.
\frac{20}{46}=\frac{10}{23}

(ii) Possible outcomes of face cards = 6
∴ P(a face card) = Image may be NSFW.
Clik here to view.
\frac{6}{46}=\frac{3}{23}

(iii) Possible outcomes of card of clubs = 13
∴ P(a card of clubs) = Image may be NSFW.
Clik here to view.
\frac{13}{46}

Question 34.
From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:
(i) a black King
(ii) a card of red colour
(iii) a card of black colour (2016OD)
Solution:
Total cards in the pack = 52
Cards removed
= 2(Jacks) + 2(Queens) + 2(Kings) = 6
∴ Remaining cards = 52 – 6 = 46
(i) Number of black Kings = 2
∴ P(a black King) = Image may be NSFW.
Clik here to view.
\frac{2}{46}=\frac{1}{23}

(ii) Total red cards in the pack = 26
Red cards removed = 6
Remaining red cards = 26 – 6 = 20
∴ P(a card of red colour) = Image may be NSFW.
Clik here to view.
\frac{20}{46}=\frac{10}{23}

(iii) Total black cards in the pack = 26
∴ P(a card of black colour) = Image may be NSFW.
Clik here to view.
\frac{26}{46}=\frac{13}{23}

Question 35.
There are 100 cards in a bag on which num bers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card
(i) is divisible by 9 and is a perfect square
(ii) is a prime number greater than 80. (2016OD)
Solution:
Total cards = 100
(i) Numbers which are “Divisible by 9 and perfect squares” are 9, 36, 81, i.e., 3 .
∴ P(Divisible by 9 & perfect square) = Image may be NSFW.
Clik here to view.
\frac{3}{100}

(ii) Numbers which are “prime numbers greater than 80” are 83, 89, 97, i.e., 3
∴ P(Prime nos. > 80) = Image may be NSFW.
Clik here to view.
\frac{3}{100}

Question 36.
Two different dice are rolled together. Find the probability of getting: (2015 D)
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice.
Solution:
Total possible outcomes = 6n = 62 = 36
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) when the sum of numbers on two dice is 5, i.e., 4
∴ Required Probability, P(E) = Image may be NSFW.
Clik here to view.
\frac{4}{36}=\frac{1}{9}

(ii) The possible outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) for even numbers on both dice; 9
∴ Required Probability, P(E) = Image may be NSFW.
Clik here to view.
\frac{9}{36}=\frac{1}{4}

Question 37.
Two different dice are thrown together. Find the probability of:
(i) getting a number greater than 3 on each die
(ii) getting a total of 6 or 7 of the numbers on two dice (2016D)
Solution:
Two dice can be thrown in 6 x 6 = 36 ways
(i) “getting a number > 3 on each die” can be obtained as (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), i.e., 9 ways.
P(number > 3 on each die) = Image may be NSFW.
Clik here to view.
\frac{9}{36}=\frac{1}{4}

(ii) “a total of 6 or 7 can be obtained as (2, 4),(4, 2),(3, 3), (1, 6),(6, 1),(2, 5) (1, 5),(5, 1) (5, 2),(3, 4),(4, 3), i.e., 11 ways
Total 6 Total “7′
P(a total of 6 or 7) = Image may be NSFW.
Clik here to view.
\frac{11}{36}

Question 38.
Two different dice are thrown together. Find the probability that the numbers obtained.
(i) have a sum less than 6
(ii) have a product less than 16
(iii) is a doublet of odd numbers. (2017D)
Solution:
Two dice can be thrown in (6 × 6) = 36 ways
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 12

(i) Sum less than ‘6’ are (1, 1) (1, 2) (1, 3) (1, 4), (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) i.e., 10 ways
∴ P(sum < 6) = Image may be NSFW.
Clik here to view.
\frac{10}{36}=\frac{5}{18}

(ii) Product less than ’16’ are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2,5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3,5)(4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5,3) (6, 1) (6, 2) i.e., 25 ways
∴ P(product less than 16) = Image may be NSFW.
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\frac{25}{36}

(iii) (1, 1) (3, 3) (5, 5) i.e., 3 ways
∴ P(doublet of odd nos.) = Image may be NSFW.
Clik here to view.
\frac{3}{36}=\frac{1}{12}

Probability Class 10 Important Questions Long Answer (4 Marks)

Question 39.
A survey has been done on 100 people out of which 20 use bicycles, 50 use motorbikes and 30 use cars to travel from one place to another. Find the probability of persons who use bicycles, motorbikes and cars respectively? (2011D)
Solution:
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 13

Question 40.
A box contains 35 blue, 25 white and 40 red ‘marbles. If a marble is drawn at random from the box, find the probability that the drawn marble is (i) white (ii) not blue (iii) neither white nor blue. (2012D)
Solution:
No. of blue marbles = 35
No. of white marbles = 25
No. of red marbles = 40
Total no. of marbles = 35 + 25 + 40 = 100
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 14

Question 41.
A box contains 70 cards numbered from 1 to 70. If one card is drawn at random from the box, find the probability that it bears
(i) a perfect square number.
(ii) a number divisible by 2 and 3. (2012OD)
Solution:
(i) Perfect squares upto 70 are
12, 22, …, 82 = 8
∴ P(a perfect square) = Image may be NSFW.
Clik here to view.
\frac{8}{70}=\frac{4}{35}

(ii) Numbers divisible by 2 and 3 are:
6, 12, 18, …, 66, i.e., 11 nos.
∴ P(a no. divisible by 2 and 3) = Image may be NSFW.
Clik here to view.
\frac{11}{70}

Question 42.
A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (2013D)
(i) extremely patient
(ii) extremely kind or honest.
Solution:
Extremely patient = 3
Extremely honest = 6
Extremely kind = 12 – 3 – 6 = 3
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 15

Question 43.
A box contains cards numbered 3, 5, 7, 9, …, 35, 37. A card is drawn at random from the box. Find the probability that the number on the drawn card is a prime number. (2013OD)
Solution:
Total number of cards = 18
Prime numbers are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, i.e., 11
∴ P(Prime number) = Image may be NSFW.
Clik here to view.
\frac{11}{180}

Question 44.
Find the probability that a leap year selected at random, will contain 53 Mondays. (2013OD)
Solution:
In a leap year, total number of days = 366
∴ 366 days
= 52 complete weeks + 2 extra days
Thus, a leap year always has 52 Mondays and extra 2 days.
Extra 2 days can be,
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Let E be the event that a leap year has 53 Mondays.
∴ E = {Sun and Mon, Mon and Tues}
∴ P(E) = Image may be NSFW.
Clik here to view.
\frac{2}{7}

Question 45.
A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number (2014D)
Solution:
Total number of cards = 49
(i) Odd numbers are 1, 3, 5, …., 49, i.e., 25
∴ P(an odd number) = Image may be NSFW.
Clik here to view.
\frac{25}{49}

(ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9
∴ P(a multiple of 5) = Image may be NSFW.
Clik here to view.
\frac{9}{49}

(iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7
∴ P(a perfect square number) = Image may be NSFW.
Clik here to view.
\frac{7}{49}=\frac{1}{7}

(iv) “An even prime number” is 2, i.e., only one number
∴ P(an even prime number) = Image may be NSFW.
Clik here to view.
\frac{1}{49}

Question 46.
A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (2015D)
(i) divisible by 2 or 3,
(ii) a prime number.
Solution:
(i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13
Total outcomes = 20
Possible outcomes = 13
∴ P(divisible by 2 or 3) = Image may be NSFW.
Clik here to view.
\frac{13}{20}

(ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8
Total Outcomes = 20
Possible outcomes = 8
∴ P(a prime number) = Image may be NSFW.
Clik here to view.
\frac{8}{20}=\frac{2}{5}

Question 47.
A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is: (2015D)
(i) divisible by 3 or 5
(ii) a perfect square number.
Solution:
Total number of outcomes = 25
(i) Possible outcomes of numbers divisible by 3 or 5 in numbers 1 to 25 are (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) = 12
∴P(No. divisible by 3 or 5) = Image may be NSFW.
Clik here to view.
\frac{12}{25}

(ii) Possible outcomes of numbers which are a perfect square = 5, i.e., (1, 4, 9, 16, 25)
∴ P(a perfect square no.) = Image may be NSFW.
Clik here to view.
\frac{5}{25}=\frac{1}{5}

Question 48.
A game of chance consists of spinning an arrow on a 3 circular board, divided into / 4 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, …, 8 (Figure), which are equally likely outcomes. What is the probability that the arrow will point at
(i) an odd number
(ii) a number greater than 3
(iii) a number less than 9. (2016 D)
Image may be NSFW.
Clik here to view.
Important Questions for Class 10 Maths Chapter 15 Probability 16

Solution:
Total numbers = 8
(i) “Odd numbers” are 1, 3, 5, 7, i.e., 4
∴ P(an odd number) = Image may be NSFW.
Clik here to view.
\frac{4}{8}=\frac{1}{2}

(ii) “nos. greater than 3” are 4, 5, 6, 7, 8, i.e., 5
∴ P(a number > 3) = Image may be NSFW.
Clik here to view.
\frac{5}{8}

(iii) “numbers less than 9” are 1, 2, 3, …8 i.e., 8
∴ P(a number < 9) = Image may be NSFW.
Clik here to view.
\frac{8}{8}
= 1

Question 49.
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. (2016OD)
Solution:
X can be any one of 1, 2, 3, and 4 i.e., 4 ways
Y can be any one of 1, 4, 9, and 16 i.e., 4 ways
Total no. of cases of XY = 4 × 4 = 16 ways
No. of cases, where product is less than 16 (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4,1) i.e., 8 ways
∴ P (product x & y less then 16) = Image may be NSFW.
Clik here to view.
\frac{8}{16}=\frac{1}{2}

Question 50.
The probability of guessing the correct answer to a certain question is Image may be NSFW.
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\frac{x}{12}
. If the probability of guessing the wrong answer is Image may be NSFW.
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\frac{3}{4}
, find x. If a student copies the answer, then its probability is Image may be NSFW.
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\frac{2}{6}
. If he doesn’t copy the answer, then the probability is Image may be NSFW.
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\frac{2y}{3}
. Find the value of y. (2011OD)
Solution:
P (guessing) + P (guessing wrong) = 1
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 17

Question 51.
A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16. (2016OD)
Solution:
x can be any one of 1, 4, 9, or 16, i.e., 4 ways
y can be any one of 1, 2, 3, or 4, i.e., 4 ways
Total number of cases of xy = 4 × 4 = 16 ways
Number of cases, where product is more than 16
(9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4), i.e., 6 ways
∴ Required probability = Image may be NSFW.
Clik here to view.
\frac{6}{16}=\frac{3}{8}

Question 52.
A game consists of tossing a coin 3 times and noting its outcome each time. Hanif wins if he gets three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. (2011D)
Solution:
The possible outcomes on tossing a coin 3 times are,
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT) = 8
Outcomes when Hanif wins = {HHH, TTT} = 2
∴ P (Hanif wins) = Image may be NSFW.
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\frac{2}{8}=\frac{1}{4}

∴ P (Hanif will lose) = 1 – Image may be NSFW.
Clik here to view.
\frac{1}{4}=\frac{3}{4}

Question 53.
Two dice are rolled once. Find the probability of getting such numbers on the two dice, whose product is 12. (2011OD)
Solution:
Two dice can be thrown in 6 × 6 = 36 ways
‘Product is 12′ can be obtained as (2, 6), (6, 2), (3, 4), (4, 3), i.e., in 4 ways
∴ P(Product is 12) = Image may be NSFW.
Clik here to view.
\frac{4}{36}=\frac{1}{9}

Question 54.
A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card. (2012D)
Solution:
No. of red cards = 100
No. of yellow cards = 200
No. of blue cards = 50
Total no. of cards = 100 + 200 + 50 = 350
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 18

Question 55.
Two different dice are thrown together. Find the probability that the numbers obtained have (2017OD)
(i) even sum, and
(ii) even product.
Solution:
Two dice can be thrown as 6 × 6 = 36 ways.
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 19

Question 56.
A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) the queen of diamonds. (2012 D)
Solution:
(i) Total cards in a deck = 52
Total no. of kings = 4
Total no. of red kings = 2
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Important Questions for Class 10 Maths Chapter 15 Probability 20

Question 57.
All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card. (2012OD)
Solution:
Total no. of cards = 52
No. of cards removed = (4 + 4 + 4) = 12
Remaining cards = 40
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Important Questions for Class 10 Maths Chapter 15 Probability 21

Question 58.
Red kings and black aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card. (2012OD)
Solution:
Total nos. of cards = 52
Cards removed = 2 + 2 = 4
Remaining cards = 52 – 4 = 48
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Important Questions for Class 10 Maths Chapter 15 Probability 22

Question 59.
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a:
(i) face card
(ii) red card
(iii) black card
(iv) king (2014D)
Solution:
Total number of cards = 52
Black face cards = 6
Remaining cards = 52 – 6 = 46
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 23

Question 60.
Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is: (2014D)
(i) an odd number
(ii) a perfect square number
(iii) divisible by 5
(iv) a prime number less than 20
Solution:
Total number of cards = 60 – 11 + 1 = 50
(i) Odd nos, are 11, 13, 15, 17, …. 59 = 25 no.
∴ P(an odd number) = Image may be NSFW.
Clik here to view.
\frac{25}{50}=\frac{1}{2}

(ii) Perfect square numbers are 16, 25, 36, 49 = 4 numbers
∴ P(a perfect square no.) = Image may be NSFW.
Clik here to view.
\frac{4}{50}=\frac{2}{25}

(iii) “Divisible by 5” numbers are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, = 10 numbers
∴ P(divisible by 5) = Image may be NSFW.
Clik here to view.
\frac{10}{50}=\frac{2}{25}

(iv) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8 numbers
∴ P(a prime no. less than 20) = Image may be NSFW.
Clik here to view.
\frac{8}{50}=\frac{4}{25}

Question 61.
Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is (2014OD)
(i) a king
(ii) of red colour
(iii) a face card
(iv) a queen
Solution:
Number of red queens = 2
Number of black jacks = 2
Remaining cards = 52 – 2 – 2 = 48
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Important Questions for Class 10 Maths Chapter 15 Probability 24

Question 62.
All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is (2014OD)
(i) of red colour
(ii) a queen
(iii) an ace
(iv) a face card
Solution:
Total number of cards = 52
Red face cards = 6
Remaining cards = 52 – 6 = 46
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 25

Question 63.
Five cards, the ten, jack, queen, king and ace of diamonds, are well shuffled with their faces downwards. One card is then picked up at random. (2014OD)
(a) What is the probability that the drawn card is the queen?
(b) If the queen is drawn and put aside, and a second card is drawn, find the probability that the second card is (i) an ace (ii) a queen.
Solution:
(a) Total events = 5; P(queen) = Image may be NSFW.
Clik here to view.
\frac{1y}{5}

(b) Now total events = 4
(i) P (an ace) = Image may be NSFW.
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\frac{1}{4}

(ii) P (a queen) = Image may be NSFW.
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\frac{0}{4}
= 0 …[As there is no queen left

Question 64.
A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is (2015OD)
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king.
(iv) either a king or a queen.
Solution:
Total no. of outcomes = 52
Image may be NSFW.
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Important Questions for Class 10 Maths Chapter 15 Probability 26

Important Questions for Class 10 Maths

The post Important Questions for Class 10 Maths Chapter 15 Probability appeared first on Learn CBSE.

NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत

NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत (उठो, जागो)

पाठपरिचयः सारांशः च :
प्रथमः पाठः ‘उत्तिष्ठत जाग्रत’ (उठो, जागो) के शीर्षक में उपनिषदों के निम्न मन्त्र के पहले दो शब्द हैं

उत्तिष्ठत, जाग्रत, प्राप्य वरान् निबोधत।
क्षुरस्य धारा निशिता दुरत्यया, दुर्गं पथस्तत् कवयो वदन्ति॥

इस पाठ के सभी पद्य मनुस्मृति, धर्मसूत्र तथा उपनिषदों से लिए गए हैं।
प्रस्तुत पाठ में उपर्युक्त मन्त्र सहित पाँच पद्य. हैं। प्रथम पद्य (मनुस्मृति) में सज्जनों के घर की विशेषता बताई गई है। सज्जनों के घर में ये चार चीजें सदा होती हैं-(1) अतिथि को दिया जानेवाला तिनकों से बना आसन (चटाई आदि), (2) अतिथि के बैठने तथा रहने के लिए आश्रयस्थान, (3) पेय जल, एवं (4) सत्य तथा मधुर वाणी। कभी भी इनमें से किसी एक का भी अभाव नहीं होता। उनके घर जो भी अतिथि आता है उसे ये चार चीजें अवश्य दी जाती हैं।

दूसरे मन्त्र (बौधायन धर्मसूत्र) में बताया गया है कि शरीर, बुद्धि, आत्मा और मन की शुद्धि किस प्रकार होती है। शरीर की शुद्धि जल से, बुद्धि की शुद्धि ज्ञान से, आत्मा की शुद्धि अहिंसा से तथा मन की शुद्धि सत्य से होती है। अतः बाह्य तथा आन्तरिक शुद्धि के लिए ज्ञान, अहिंसा और सत्य की आवश्यकता होती है। इन तत्त्वों के धारण करने से मनुष्य अन्तर्बाह्य सब प्रकार से पवित्र हो जाता है।
तीसरे पद्य (मुण्डकोपनिषद्) में सत्य की महिमा बताई गई है। अन्त में सत्य की ही विजय होती है। महापुरुषों का मार्ग सत्य से भरा होता है। सब कामनाएँ पूर्ण होने के पश्चात् ऋषि जिस धाम में जाते हैं वह सत्य का परम निधान चौथे पद्य (ईशोपनिषद्) में बताया गया है कि सब प्राणियों को अपने जैसा समझकर व्यवहार करना चाहिए। जो सब प्राणियों को अपने में तथा सब प्राणियों में अपने-आप को देखता है, वह किसी से कभी कोई घृणा नहीं करता, सबकी रक्षा में तथा सबके प्रति परोपकार की भावना से भरा रहता है।

अन्तिम पद्य (कठोपनिषद्) में कहा गया है कि सदा ज्ञान के लिए प्रयत्नशील रहो। अज्ञान की निद्रा का परित्याग करो। महापुरुषों के पास जाकर ज्ञान प्राप्त करने का प्रयत्न करो। जिस प्रकार छुरे की धार तीखी होती है, उस पर नंगे पैरों से चलना कठिन होता है, ज्ञान का मार्ग भी वैसा ही कठिन होता है। कवि तथा ऋषि ऐसा वर्णन करते हैं। अतः हमें एकाग्र मन से, ध्यान से सन्मार्ग का आश्रय लेना चाहिए। ‘उत्तिष्ठत जाग्रत’ स्वामी विवेकानन्द का राष्ट्रीय वाक्य है।

मूलपाठः, अन्वयः, शब्दार्थः, भावार्थः, सरलार्थश्च :

1. तृणानि भूमिरुदकं वाक्चतुर्थी च सूनृता।
एतान्यपि सतां गेहे नोच्छिद्यन्ते कदाचन॥ (मनुस्मृतिः)

पदच्छेद – तृणानि भूमिः उदकम् वाक् चतुर्थी च सूनृता। एतानि अपि सताम् गेहे न उच्छिद्यन्ते कदाचन। अन्वयः- तृणानि, भूमिः, उदकम्, चतुर्थी च सूनृता वाक्। सताम् गेहे एतानि कदाचन अपि न उच्छिद्यन्ते।
शब्दार्थः, पर्यायवाचिशब्दाः टिप्पण्यश्च – तृणानि (नुपं०) – आसनम् (नपुं०) (तृण, प्रथमा, बहुवचनम्), तिनके, तिनकों से बना बैठने का आसन। भूमिः (स्त्री०)- आश्रयः (भूमि, प्र०, ए०व०), रहने के लिए स्थान, निवास स्थान।

उदकम् (नपुं०) – जलम्, पीने के लिए पानी। चतुर्थी च – और चौथी। सूनृता सत्या मधुरा च- सत्येन माधुर्येण युक्ता, सच तथा मिठास से भरी हुई। वाक्- वाणी, बोली, बोलचाल। सताम् – सज्जनानाम् (सत्, षष्ठी, बहुवचनम्), अच्छे लोगों के। गेहे (नपु०) – गृहे, सप्तमी, घर में। एतानि- इमानि एकवचन, ये चारों चीजें। कदाचन – कदापि, कभी अपि – अपि, भी। न – नहि, नहीं। उच्छिद्यन्ते – विनाश्यन्ते, उत्क्षिप्यन्ते, नष्ट होती, (उत् + छिद्, लट्, प्र०पु०, बहुवचनम्)।
भावार्थ – सज्जनानां गृहेषु अतिथीनां कृते आसनं, वासाय स्थानम्, जलम्, मधुरवाण्या सत्कारः इति भावानाम् कदापि अभार : न भवति। ते सर्वदा अतिथिसत्काराय उद्यताः भवन्ति।
सरलार्थ – सज्जनों के घरों में अतिथियों के लिए आसन, रहने के लिए स्थान, (पीने के लिए) पानी, मधुरवाणी के द्वारा सत्कार (सम्मान)-इन पदार्थों का कभी अभाव (कमी) नहीं होता। वे (सज्जन) सदा अतिथि – सत्कार के लिए तैयार (तत्पर) रहते हैं।

2. अद्भिः शुध्यन्ति गात्राणि, बुद्धिः ज्ञानेन शुध्यति।
अहिंसया च भूतात्मा मनः सत्येन शुध्यति॥ (बौधायन धर्मसूत्र)

पदच्छेद – अद्भिः शुध्यन्ति गात्राणि, बुद्धिः ज्ञानेन शुध्यति। अहिंसया च भूतात्मा, मनः सत्येन शुध्यति।।
अन्वय – गात्राणि अद्भिः शुध्यन्ति, बुद्धिः ज्ञानेन शुध्यति, भूतात्मा च अहिंसया (शुध्यति), मनः सत्येन शुध्यति।
शब्दार्थः, पर्यायवाचिशब्दाः टिप्पण्यश्चः- ‘गात्राणि- शरीराणि (गात्र, प्रथमा, बहुवचनम्), शरीर, अंग। अद्भिःजलैः (अप्, तृतीया, बहुवचनम्), पानी से। शुध्यन्ति- शुद्धानि भवन्ति (शुद्ध, लट्, प्र०पु०, बहुवचनम्), शुद्ध होते हैं बुद्धिः- विचारः, बुद्धिः, विचार, बुद्धि। ज्ञानेन- ज्ञान द्वारा, ज्ञान के द्वारा। शुध्यति- पवित्रा भवति (शुध्, लट् लकार, प्र०पु०, एकवचनम्), पवित्र होती है। भूतात्मा च- जीवात्मा (भूतानाम् आत्मा च षष्ठी बहुवचनम्), जीवात्मा, आत्मा। अहिंसया- अहिंसा-माध्यमेन (अहिंसा, तृतीया, एकवचनम्), अहिंसा के द्वारा। मनः- मनः, संकल्प-आश्रयः, संकल्पों का आश्रय, मन। सत्येन- सत्यवचनेन, सत्यव्यवहारेण (सत्य, तृतीया, एकवचनम्), सच बोलने से, सही व्यवहार से।
भावार्थ – अस्माकम् शरीरम् जलेन स्वच्छं/पवित्रं भवति, अस्माकं बुद्धिः ज्ञानेन शुद्धा भवति, यथा यथा च वयं ज्ञानं प्राप्नुमः, अस्माकं बुद्धिः पवित्रा भवति, मनुष्यस्य आत्मा अहिंसया शुद्धः भवति, सत्यस्य आचरणेन मनः पवित्रम् भवति।
सरलार्थ- हमारा शरीर जल से स्वच्छ/पवित्र होता है, हमारी बुद्धि ज्ञान से शुद्ध होती है, और जैसे-जैसे हम ज्ञान प्राप्त करते हैं, हमारी बुद्धि पवित्र होती जाती है। मनुष्य की आत्मा अहिंसा से पवित्र (शुद्ध) होती है, सत्य के आचरण से मन पवित्र होता है।

3. सत्यमेव जयति नानृतम्,
सत्येन पन्था विततो देवयानः।
येनाक्रमन्त्षयो ह्याप्तकामाः,
यत्र तत् सत्यस्य परमं निधानम्॥ (मुण्डकोपनिषद्)

पदच्छेद – सत्यम् एव जयति न अनृतम्, सत्येन पन्थाः विततः देवयानः। येन आक्रमन्ति ऋषयः हि आप्तकामाः, यत्र तत् सत्यस्य परमम् निधानम्।।
अन्वय – सत्यम् एव जयति अनृतम् न। देवयानः पन्थाः सत्येन विततः। आप्तकामाः ऋषयः येन यत्र आक्रमन्ति तत् हि सत्यस्य परमम् निधानम्।।शब्दार्थः, पर्यायवाचिशब्दाः टिप्पण्यश्च – सत्यम्- सत्यम्, सच। एव- एव, ही। जयति- जयते, विजयं प्राप्नोति, जय को प्राप्त करता है। अनृतम्- असत्यम् (न ऋतम्), झूठ। न- नहि, नहीं। देवयानः- देवानां यानः (महापुरुषाणां मार्गः), देवताओं, महापुरुषों का मार्ग। पन्थाः – मार्गः (पथिन्, पु०, प्र०, ए०व०), मार्ग। सत्येन- सत्य द्वारा, सत्य के द्वारा। वितत – विस्तृतः, परिपूर्ण (वि + तन् + क्त), फैला हुआ, भरा हुआ। आप्तकामाः (आप्ता: कामाः ते यैः)-कृतकृत्याः, सफलमनोरथयुक्ताः, जिनके मनोरथ पूरे हो गए हैं वे। ऋषयः- मन्त्रद्रष्टारः, ऋषि लोग, मन्त्रद्रष्टा। येन- येन मार्गेण, जिस

रास्ते से। अत्र- अस्मिन् स्थाने, इस स्थान पर। आक्रमन्ति- गच्छन्ति (आ + क्रम्, लट्, प्र०पु०, बहुवचनम्), जाते हैं। तत्- (सः मार्गः), असौ एव, तदेव, वह ही, वही। हि- एव, निश्चयेन, ही, निश्चय से। सत्यस्य- सत्यस्य, तत्त्वस्य, सत्य का। परमम्- महत्, महत्तमम्, बड़ा, सबसे बड़ा। निधानम्- धाम (नि + धा + ल्युट्), स्थान।
भावार्थ – सत्यस्य एव सर्वदा जयः भवति, अनृतम् असत्यं कदापि अन्ते जयं न आप्नोति। महापुरुषाणां मार्गः सत्येन एव परिपूर्णः। सफलमनोरथाः ऋषयः येन मार्गेण गच्छन्ति यत्र च प्राप्नुवन्ति सः मार्गः सत्यस्य एव मार्गः, सत्यस्य परमम् धाम तदेव वर्तते। सरलार्थ – सत्य की ही सदा जीत होती है। झूठ (जो सच या ऋत-शाश्वत् नहीं है) अन्त में विजय कदापि प्राप्त नहीं करता। जिनके मनोरथ सफल (पूर्ण) हो गए हैं, ऐसे ऋषि जिस मार्ग से जाते हैं और जहाँ पर पहुँचते हैं, वह मार्ग सत्य का ही मार्ग है, सत्य का सबसे बड़ा स्थान (आश्रय) वही है।

4. यस्तु सर्वाणि भूतान्यात्मन्येवानुपश्यति,
सर्वभूतेषु चात्मानम् ततो न विजुगुप्सते॥ (ईशोपनिषद्)

पदच्छेदः- यः तु सर्वाणि भूतानि आत्मनि एव अनुपश्यति, सर्वभूतेषु च आत्मानम् ततः न विजुगुप्सते।।
अन्वयः- यः तु सर्वाणि भूतानि आत्मनि एव अनुपश्यति, सर्वभूतेषु च आत्मानम् (अनुपश्यति) ततः न विजुगुप्सते।
शब्दार्थ:- पर्यायवाचिशब्दाः टिप्पण्यश्चः- यः- यः मनुष्यः, प्राणी, जो मनुष्य/प्राणी। तु- निश्चयेन, निश्चय से।

सर्वाणि – सम्पूर्णानि, अखिलानि (सर्व, द्वितीया, बहुवचनम्), समस्त। भूतानि- मनुष्यान् प्राणिनः (भूत, द्वितीया, बहुवचनम्), मनुष्यों/प्राणियों को। आत्मनि- आत्म-मध्ये, निजे अन्त:करणे, स्वकीये, हृदये (आत्मन्, सप्तमी, एकवचनम्), आत्मा में, अपने हृदय में। एव- एव, ही। अनुपश्यति- अन्वीक्षणं करोति, पश्यति, (अनु + दृश्, लट्, प्र०पु०, एकवचनम्), देखता है। सर्वभूतेषु- सर्वेषु प्राणिषु, सर्वेषु मनुष्येषु (सर्वभूत, सप्तमी, बहुवचनम्), सब प्राणियों में। च- तथा, और। आत्मानम्- (अनुपश्यति) निजम् (आत्मन्, द्वितीया, एकवचनम्, स्वरूपम्), अपने-आप को देखता है। तत – तदनन्तर, तस्मात् परं, उसके बाद (वह)। न- नहि, नहीं। विजुगुप्सते- घृणां करोति (वि. गुप् + सन्, लट्, एकवचनम्), घृणा करता है।

भावार्थ – यः मनुष्यः सर्वान् प्राणवतः जनान् स्वकीये हृदये पश्यति, सर्वान् आत्मवत् पश्यति, सर्वे जनाः मत्सदृशाः एव, सर्वेषु स एव आत्मा विराजते यः मयि अस्ति इति यदा ज्ञानं भवति. तत्पश्चात् सः केनापि सह घृणां कर्तुं न शक्नोति, सर्वेषां रक्षणे परोपकरणे च उद्यतः भवति।

सरलार्थ – जो व्यक्ति सब प्राणधारी जीवों को अपने हृदय में देखता है, अर्थात् सबको अपने समान देखता है, सब लोग मेरे जैसे ही हैं, सबमें वह ही आत्मा विराजमान है जो मुझमें है इस प्रकार का ज्ञान रखता है, इस प्रकार की अनुभूति करने लगता है, उसके बाद वह मनुष्य किसी के साथ घृणा नहीं कर सकता। वह सबकी रक्षा में तथा परोपकार में सदा उद्यत रहता है।

5. उत्तिष्ठत, जाग्रत, प्राप्य वरान् निबोधत।
क्षुरस्य धारा निशिता दुरत्यया,
दुर्ग पथस्तत् कवयो वदन्ति॥ (कठोपनिषद्)

पदच्छेद – उत्तिष्ठत, जाग्रत, प्राप्य वरान् निबोधत। क्षुरस्य धारा निशिता दुरत्यया, दुर्गम् पथः तत् कवयः वदन्ति।
अन्वय – उत्तिष्ठत, जाग्रत, वरान् प्राप्य निबोधत। क्षुरस्य धारा निशिता दुरत्यया। कवयः तत् पथः दुर्गम् वदन्ति।
शब्दार्थ – पर्यायवाचिशब्दाः टिप्पण्यश्च – उत्तिष्ठत – ज्ञानाय प्रयत्नं कुरुत (उत् + स्था, लोट, म०पु०, बहुवचनम्), ज्ञानार्थ प्रयत्न करो। जाग्रत – जागृत, अज्ञानस्य निद्रां त्यजत (वैदिक प्रयोग) (जागृ, लोट, म०पु०, बहुवचनम्), अज्ञान की नींद को छोड़ो। वरान् – श्रेष्ठान् जनान् (वर, द्वितीया, बहुवचनम्), श्रेष्ठ पुरुषों के। प्राप्य – गृहीत्वा, प्र + आप् + ल्यप्, समीप पाकर। निबोधत – जानीत, (नि, बुध्, लोट, म०पु०, बहुवचनम्), जानो। क्षुरस्य – छुरिकायाः (क्षुर, पु०, षष्ठी, एकवचनम्), छुरिका की। धारा – धारा, धार। निशिता – तीक्ष्णा, तीखी। दुरत्यया – दु:खे गन्तुम् शक्या (दु: + अति + अया), कष्टपूर्वक जाने योग्य। कवयः- विद्वांसः, (कवि, प्रथमा, बहुवचनम्), विद्वान् लोग। तत् (नपुं०) – अदस्, उपर्युक्तम्,

उस (उपर्युक्त)। पथः (नपुं०) – मार्गम्, (पथः वैदिकप्रयोगः), मार्ग को। दुर्गम् (नपुं०) – कठिनम्, कठिन। वदन्तिकथयन्ति, वर्णयन्ति (वद्, लट्, प्र०पु०, बहुवचनम्), वर्णन करते हैं।

भावार्थ – हे जनाः! यूयं ज्ञानं प्राप्तुं तत्पराः उद्यताः भवत। अज्ञाननिद्रां त्यक्त्वा उत्तिष्ठत। महापुरुषाणां समीपे गत्वा ज्ञानं प्राप्तुं प्रयत्नं कुरुत। ज्ञानमार्गः सरल: नास्ति। छुरिकायाः धारा यथा तीक्ष्णा, पद्भ्याम् गन्तुम् अशक्या भवति तथैव महापुरुषाः ज्ञानस्य मार्गम् अतीव कठिनम् इति वर्णयन्ति। तर्हि एकाग्रमनसा ध्यानेन सन्मार्गम् आश्रित्य चलत।

सरलार्थ – हे लोगो! तुम सब ज्ञान को पाने के लिए तैयार हो जाओ। अज्ञान की नींद को छोड़कर उठो। महापुरुषों के पास जाकर ज्ञान प्राप्त करने का प्रयत्न करो। ज्ञानमार्ग सरल नहीं है। जैसे छुरिका (छुरी) की धार तीक्ष्ण होती है, उस पर पैदल नहीं चला जा सकता, वैसे ही महापुरुष- ‘ज्ञान का मार्ग बहुत कठिन है’- ऐसा वर्णन करते हैं तो एकाग्र मन से ध्यानपूर्वक सन्मार्ग का आश्रय (सहारा) लेकर चलो।

अनुप्रयोगस्य प्रश्नोत्तराणि 

प्रश्न 1.
अधोलिखिताः पंक्तीः मेलयत
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 1

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 2

प्रश्न 2.
विलोमपदानि मेलयत
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 3

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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 4

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 5

प्रश्न 3.
कर्तृपदैः सह क्रियापदानि योजयत –
(i) गात्राणि अद्भिः ……………
(ii) यूयम् सर्वे वरान प्राप्य ………………..
(iii) सत्यम् एव ………………
(iv) कवयः तत् मार्ग दुर्गं ……………
(v) एतानि सज्जनानां गृहेषु कदापि न …………
(vi) बुद्धिः ज्ञानेन ……………
(vii) आत्मवत् सर्वान् दृष्ट्वा नरः न …………….
उत्तरम्:
(i) गात्राणि अद्भिः शुध्यन्ति।
(ii) यूयम् सर्वे वरान् प्राप्य निबोधत।
(iii) सत्यम् एव जयति।
(iv) कवयः तत् मार्गं दुर्गं वदन्ति।
(v) एतानि सज्जनानां गृहेषु कदापि न उच्छिद्यन्ते।
(vi) बुद्धिः ज्ञानेन शुध्यति।
(vii) आत्मवत् सर्वान् दृष्ट्वा नरः न विजुगुप्सति।

प्रश्न 4.
सन्धिच्छेदमधिकृत्य अधोलिखितां तालिकां पूरयत –
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 6

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 7

प्रश्न 5.
(i) अत्र कुत्र तृतीया विभक्तिः न अस्ति ?
अद्भिः, सत्येन, अहिंसया, निशिता
(ii) अत्र कुत्र षष्ठीबहुवचनम् अस्ति?
क्षुरस्य, सत्यस्य, सताम्, कवयः
(iii) एतेषु किम् अव्ययपदम् न अस्ति?
ततः, कदाचन, वाक्, एव
(iv) एतेषु किं क्रियापदम् न अस्ति?
विततः, उच्छिद्यन्ते, उत्तिष्ठत, विजुगुप्सते
(v) एतेषु किं क्रियापदं बहुवचने न अस्ति?
आक्रमन्ति, वदन्ति, निबोधत, अनुपश्यति
उत्तरम:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 8

प्रश्न 6.
अधोलिखितसूक्तिभिः सह सम्बद्धां पंक्ति मेलयत
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 9

उत्तरम:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 10

प्रश्न 7.
पाठं पठित्वा अशुद्धं तथ्यं चिनुत
(i) अतिथये आसनम्, जलम् च दातव्यम्।
(ii) सत्यस्य एव अन्ते जयः भवति।
(iii) आत्मा हिंसया दूष्यते।
(iv) वसुधैव कुटुम्बकम् इति ज्ञात्वा नरः घृणां करोति।
(v) ऋषयः सत्यस्य मार्गे चलन्ति।
(vi) ज्ञानमार्गः अतीव सरलः।
(vii) वयम् अज्ञाननिद्रां त्यक्त्वा उत्तिष्ठाम।
(viii) ज्ञानिनां समीपे गत्वा ज्ञान प्राप्तव्यम्।
(ix) सज्जनानां गृहे मधुरवचनानाम् अभावः भवति।
(x) यदा वयं सत्यं वदामः तदा अस्माकं मनः पवित्रम् भवति।
उत्तरम्:
(iv) वसुधैवकुटुम्बकम् इति ज्ञात्वा नरः घृणां करोति।
(vi) ज्ञानमार्गः अतीव सरलः।
(ix) सज्जनानां गृहे मधुरवचनानाम् अभावः भवति।

प्रश्न 8.
अधोलिखितानां पदानां स्थाने पाठे किं पदं प्रयुक्तम्
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 11

उत्तरम:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 12

प्रश्न 9.
रिक्तस्थानानि पूरयत
वाक्यानि
(i) बुद्धिः ……………… शुध्यति।
(ii) अद्भिः …………….. शुध्यन्ति।
(iii) …………. आत्मानं दृष्ट्वा घृणां न करोति।
(iv) ……………… गेहे अतिथिभ्यः सर्वदा आसनं जलं, स्थानं मधुरवचनानि च विराजन्ते।
(v) …………….. सत्यमार्गम् आश्रयन्ति।
उत्तरम्:
(i) बुद्धिः ज्ञानेन शुध्यति।
(ii) अद्भिः गात्राणि शुध्यन्ति।
(iii) सर्वभूतेषु आत्मानं दृष्ट्वा घृणां न करोति।
(iv) सताम् गेहे अतिथिभ्यः सर्वदा आसनं जलं, स्थानं मधुरवचनानि च विराजन्ते।
(v) आप्तकामाः ऋषयः सत्यमार्गम् आश्रयन्ति।

प्रश्न 10.
अधोलिखितविशेष्यैः सह विशेषणानि विशेष्याणि वा योजयत
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 13

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 14

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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 15

पाठ-विकासः

(क) संदर्भ-ग्रन्थ परिचय
हमारा वैदिक साहित्य बहुत विस्तृत है। वेद, उपनिषद्, ब्राह्मण ग्रन्थ, आरण्यक ग्रन्थ, स्मृतियाँ, वेदाङ्ग धर्मसूत्र इत्यादि।
इस पाठ में उपनिषदों से तथा धर्मसूत्र व मनुस्मृति से पद्य संकलित किए गए हैं
उपनिषद् – प्रमुख उपनिषद् 11 हैं – (1) ईश (2) केन (3) कठ (4) प्रश्न (5) मुण्डक (6) माण्डूक्य (7) छान्दोग्य (8) ऐतरेय (9) तैत्तिरीय (10) बृहदारण्यक (11) श्वेताश्वेतर।

संदर्भ – (1) तृणानि भूमिः ……………. (मनुस्मृतिः)
मनुस्मृति – यह हमारा प्राचीनतम संविधान (Constitution) है। इसमें 12 अध्याय हैं।
अद्भिः गात्राणि (बौधयन-धर्मसूत्र) कल्पसूत्र, श्रौतसूत्र, गृह्यसूत्र, शुल्वसूत्र तथा धर्मसूत्र ये प्रसिद्ध सूत्रग्रन्थ हैं। सूत्रग्रन्थों में नीति, सदाचार, शासनव्यवस्था, प्रजा के अधिकार, कर्तव्य, सामाजिक आचार-विचार, सामाजिक व्यवस्था इत्यादि विषय हैं। स्वयं भूखे रहकर भी सेवक को भोजन देना चाहिए। इस प्रकार आदेश आपस्तम्बीय धर्मसूत्र (259) का है। शुल्वसूत्रों में रेखागणित (ज्यॉमेट्री) के सिद्धान्तों का वर्णन है।
(2) यस्तु सर्वाणि ……………. (ईशोपनिषद्)
ईशोपनिषद यजुर्वेद का चालीसवाँ अध्याय है।
(3) सत्यमेव जयति ………… (मुण्डकोपनिषद्)
यह हमारा राष्ट्रीय लक्ष्य भी है (सत्यमेव जयते)
(4) उत्तिष्ठत जाग्रत …………….. (कठोपनिषद्)
यह मन्त्र सब नागरिकों के प्रति आह्वान है। स्वामी विवेकानन्द के द्वारा इसे राष्ट्रीय वाक्य के रूप में स्वीकार किया गया है।

(ख) भाव-विस्तार
समानान्तर सूक्तियाँ
मूल पाठ्यपुस्तक से देखें।

(ग) भाषा-विस्तार
(1) ‘आत्मा’ शब्द पुंल्लिङ्ग में प्रयुक्त होता है; जैसे – यह आत्मा बलवान् होनी चाहिए। अयम् आत्मा बलवान् भवेत्। यहाँ अयम् तथा बलवान् विशेषण पद भी पुंल्लिङ्ग में है।

(2) पथिन् शब्द भी पुंल्लिङ्ग में है। इसके रूप प्रथमा विभक्ति में इस प्रकार चलते हैं
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 16

(3) अप् शब्द स्त्रीलिङ्ग बहुवचन में ही प्रयुक्त होता है। रूप इस प्रकार हैं-आपः (प्रथमा), अपः (द्वितीया), अद्भिः (तृतीया), अद्भ्यः (चतुर्थी-पञ्चमी), अपाम् (षष्ठी), अप्सु (सप्तमी)।
प्रयोग – इस जल में = एतासु अप्सु।

सन्धि परिचय :

(1) दीर्घ सन्धिः – दीर्घ से दीर्घ स्वर आ, ई, ऊ, ऋ को लिया जाता है। नियम है-अक् (अ, आ, इ, ई उ, ऊ, ऋ ऋ, ल) से परे सवर्ण (समान वर्ण अर्थात् अ आ/ इ ई/ उ ऊ /ऋ ऋ) हों तो दीर्घ (आ, ई, ऊ, ऋ) हो जाता है।

(2) गुण सन्धिः – ए, ओ तथा अर् को गुण कहा जाता है। अ, आ इन दो वर्णो से परे इनसे भिन्न अर्थात् इ-ई/उ-ऊ/ऋ-ऋ के आ जाने पर मेल के द्वारा क्रमशः ए/ओ/अर् हो जाते हैं; जैसे-न + उच्छिद्यन्ते = नोच्छिद्यन्ते।

(3) यण् सन्धिः – य, व, र, ल् को यण् कहते हैं। इ, ई/उ, ऊ/ऋऋ के परे असमान स्वर आ जाने पर क्रमशः – य /व् /र् हो जाते हैं। उदाहरण –
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 17

अतिरिक्त-अभ्यासः

प्रश्न: 1.
अधोलिखितम् पद्यम् पठित्वा तदाधारितानाम् प्रश्नानाम् उत्तराणि लिखत।
(क) उत्तिष्ठत, जाग्रत, प्राप्य वरान्निबोधत।
क्षुरस्य धारा निशिता दुरत्यया। दुर्गं पथस्तत् कवयो वदन्ति॥

I. एकपदेन उत्तरत –
(i) कस्य धारा निशिता?
(ii) वरान् प्राप्य किं कुरुत?
(iii) पथः कीदृशम् अस्ति?
(iv) कान् प्राप्य मार्ग निबोधत?
उत्तर:
(i) क्षुरस्य
(ii) निबोधत
(iii) दुर्गम्
(iv) वरान्

II. पूर्णवाक्येन उत्तरत –
कवयः पन्थानं कथं वदन्ति?
उत्तर:
कवयः पन्थानं दुर्गं वदान्ति।

III निर्देशानुसारम् उत्तरत
(i) ‘निशिता’ कस्याः विशेषणम् आस्ति? –
(ii) ‘वदन्ति’ पदस्य कर्तृपदं लिखत।
उत्तर:
(i) ‘धारा’ पदस्य
(ii) कवयः

(ख) यस्तु सर्वाणि भूतान्यात्मन्येवानुपश्यति।
सर्वभूतेषु चात्मानम् ततो न विजुगुप्सते॥
एकपदेन उत्तरत –
(i) भूतानि कस्मिन् अनुपश्यति?
(ii) आत्मनि कानि अनुपश्यति?
(iii) आत्मानम् कुत्र अनुपश्यति?
(iv) ततः स किं न करोति?
उत्तर:
(i) आत्मनि
(ii) भूतानि
(iii) सर्वभूतेषु
(iv) विजुगुप्सते

(ग) उत्तिष्ठत, जाग्रत, प्राप्य वरान् निबोधत।
क्षुरस्य धारा निशिता दुरत्यया, दुर्गं पथस्तत् कवयो वदन्ति॥

I. एकपदेन उत्तरत –
(i) कस्य धारा निशिता वर्तते?
(ii) किं कृत्वा वरान् निबोधत?
उत्तर:
(i) क्षुरस्य
(ii) उत्थाय

II. पूर्व वाक्येन उत्तरत कवयः तत् पथः कीदृशं वदन्ति?
उत्तर:
कवयः तत् पथः दुर्गं वदन्ति।

III. निर्देशानुसारेण उत्तरत
(i) श्लोके ‘धारा’ पदस्य किं विशेषणं वर्तते?
(ii) जानीत’ इति पदस्य अर्थे श्लोके किं पदं प्रयुक्तम्?
उत्तर:
(i) निशिता
(ii) निबोधत

IV. श्लोके ‘वदन्ति’ इत्यस्य क्रियापदस्य कर्तृपदं किम्?
(घ) सत्यमेव जयति नानृतम्,
सत्येन पन्था विततो देवयानः।
येनाक्रमन्त्य॒षयो ह्याप्तकामाः,
यत्र तत् सत्यस्य परमं निधानम्॥
उत्तर:
कवयः

I. एकपदेन उत्तरत
(i) ऋषीणाम् मार्गः कस्य परमं निधानं भवति?
(ii) कस्य जयः न भवति?
उत्तर:
(i) सत्यस्य
(ii) अनृतस्य

II. पूर्ण वाक्येन उत्तरत
देवयानः पन्था कीदृशो भवति?
उत्तर:
देवयानः पन्थाः सत्येन विततों भवति।

III. निर्देशानुसारेण उत्तरत –
(i) ‘असत्यम्’ अस्य पदस्य अर्थे श्लोके किं पदं प्रयुक्तम्?
(ii) श्लोके ‘आक्रमन्ति’ अस्याः क्रियायाः कर्तृपदं किम्?
उत्तर:
(i) अनृतम्
(ii) ऋषयः

IV. ‘परमं निधानम्’ अत्र विशेष्यपदं किम्?
(ङ) तृणानि भूमिरुदकं वाक्चतुर्थी च सूनृता।
एतान्यपि सतां गेहे नोच्छिद्यन्ते कदाचन॥
उत्तर:
विधानम्

I. एकपदेन उत्तरत (1/2 x 2 = 1)
(i) अस्मिन् श्लोके तृणानि पदं कस्य कृते आगतम्?
(ii) अस्माकं वाक् कीदृशी भवेत्?
उत्तर:
(i) आसनस्य
(ii) सूनृता

II. पूर्ण वाक्येन उत्तरत (2 x 1 = 2) सतां गृहेषु के गुणाः कदापि न उच्छिद्यन्ते?
उत्तर:
सतां गृहेषु तृणानि भूमि उदकं सूनृता च वाक् एते गुणाः कदापि उच्छिद्यन्ते।

III. निर्देशानुसारेण उत्तरत (1/2 x 2 = 1)
(i) ‘जलम्’ इति पदस्य अर्थे श्लोके किं पदं प्रयुक्तम्?
(ii) ‘वाक्चतुर्थी’ इति पदयोः विशेषण पदं किम्?
उत्तर:
(i) उदकम्
(ii) चतुर्थी

IV. श्लोकात् एकम् अव्ययपदं चित्त्वा लिखत।
उत्तर:
अपि

प्रश्न: 2.
ग्रन्थस्य लेखकस्य च नामनी लिखत –
(i) तृणानि भूमिरुदकं वाक्चतुर्थी च सूनृता।
(ii) सत्यमेव जयति नानृतम्।
(iii) उत्तिष्ठत, जाग्रत, प्राप्य वरान् निबोधत।
उत्तर:
(i) ग्रन्थः- मनुस्मृतिः लेखक:- महर्षिः मनु
(ii) ग्रन्थः- मुण्डकोपनिषदः लेखक:- अज्ञात ऋषि
(iii) ग्रन्थ:- कठोपनिषद लेखकः- कठ ऋषि

प्रश्न: 3.
भावार्थ लेखनम् भावचयनम् वा (1/2 x 4 = 4)
(I) तृणानि भूमिरुदकं वाक् चतुर्थी …………… कदाचन’। अस्य श्लोकस्य भावोऽस्ति यत्-ये जनाः ……..(1)…… भवन्ति तेषां गृहे अतिथि-सत्काराय आसनम्, निवासाय च ……(2)…….. पिपासा-शान्त्यै जलम् मधुरा च ………(3)……. कदापि ……….(4)…….. न भवन्ति अर्थात् एतानि वस्तूनि सदैव सज्जनानां गृहेषु अतिथि-सत्कारार्थ सज्जितानि भवन्ति।
उत्तर:
(1) सज्जनाः
(2) भवनम्
(3) वाणी
(4) समाप्ताः

(II) ‘सत्यमेव जयति नानृतम् ………….. निधानम्’। अर्थात्- संसारे सदैव ……..(1)……… एव जयते कदापि असत्यस्य जयः न भवति। महापुरुषाणां पन्थानः सदैव ………(2)……… एव आच्छादिताः भवन्ति। अतः श्रेष्ठमनोरथयुक्ताः ………(3)……. येन मार्गेण गच्छन्ति सः मार्गः एव …….(4)……… मार्गः भवति तेन एव सदैव गन्तव्यम् अन्येन मार्गेण न।
उत्तर:
(1) सत्यम्
(2) सत्येन
(3) जनाः
(4) सत्यस्य

(III) ‘उत्तिष्ठत जाग्रत, प्राप्य ……………. वदन्ति’। अस्य श्लोकस्य भावोऽस्ति यत् यमाचार्य: नचिकेताय कथयति यत् वत्स! यूयं ज्ञानस्य प्राप्यै तत्पराः भवत। अज्ञानस्य निद्रा ………..(1)….. ज्ञान-प्राप्तुं महापुरुषाणा ……..(2)…… कुरु। यतः ज्ञान मार्गः …….(3)…….. नास्ति। अयं तु छुरिकायाः धारा इव कठिनः वर्तते अतः ……..(4)…….. सद्ज्ञान प्राप्तयर्थ प्रत्यनं कुरुत।
उत्तर:
(1) परित्यज्य
(2) अनुसरणं
(3) सरलः
(4) यूयम्

(IV) ‘सर्वभूतेषु चात्मानम् ततो न विजुगुप्सते’ अर्थात्
(i) जनः सर्वजनान् स्व-आत्मनि पश्यति।
(ii) जनः सर्वप्राणिषु स्वात्मानम् दृष्ट्वा केनापि सह घृणां न करोति।
(iii) जनः स्वात्मनि सर्वप्राणिनं न पश्यन्नपि केनापि सह घृणां न करोति।
उत्तर:
(ii) जनः सर्वप्राणिषु स्वात्मानम् दृष्ट्वा केनापि सह घृणां न करोति।

प्रश्न: 4.
निम्नलिखितानां श्लोकानाम् रिक्तस्थानपूरयन् अन्वयं लिखतु (1 x 4 = 4)

(I) यस्तु सर्वाणि भूतान्यात्मन्येवानुपश्यति।।
सर्व भूतेषु चात्मानम् ततो न विजुगुप्सते॥
अन्वयः
यः तु …….(1)…….. भूतानि आत्मनि एव …….(2)……… सर्वभूतेषु च …….(3)………. (पश्यति) ………(4)….. न विजुगुप्सते।
उत्तर:
(1) सर्वाणि
(2) अनुपश्यति
(3) आत्मानम्
(4) ततः

(II) सत्यमेव जयति नानृतम्,
सत्येन पन्था विततो देवयानः।
येनाक्रमन्त्यूषयो ह्याप्तकामाः,
यत्र तत् सत्यस्य परमं निधानम्॥
अन्वयः
सत्यम् एव जयति ……..(1)………. न। ………(2)………. पन्था सत्येन विततः। आप्तकामाः ………..(3)………. येन यत्र आक्रमन्ति तत् हि ……..(4)……. परमम् निधानम् (भवति)।
उत्तर:
(1) अनृतम्
(2) देवयानः
(3) ऋषयः
(4) सत्यस्य

(II) तृणानि भूमिरुदकं वाक्चतुर्थी च सूनृता।
एतान्यपि सतां गेहे नोच्छिद्यन्ते कदाचन।
अन्वय –
तृणानि भूमिः ……..(1)………. चतुर्थी च ……….(2)……… वाक्। सताम् गेहे …………(3)……. कदाचन अपि ……(4)……. न उच्छिद्यन्ते।
उत्तर:
(1) उदकम्
(2) सूनृता
(3) एतानि
(4) कदाचन

प्रश्नः 5.
निम्नलिखितानां पङ्क्तिनां सार्थकं संयोजनं क्रियताम्
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NCERT Solutions for Class 12 Sanskrit Chapter 1 उत्तिष्ठत जाग्रत 18

उत्तर:
(i) (ङ)
(ii) (ग)
(iii) (ज)
(iv) (च)
(v) (क)
(vi) (छ)
(vii) (ख)
(vi) (घ)

प्रश्नः 6.
रेखाकितानां श्लिष्टपदानाम् उचितम् अर्थमेलनम् कुरुत
I. दुर्ग पथस्तत् कवयो वदन्ति। .
(i) कठिनम्
(ii) सेनायाः निवासम्
(iii) विशिष्टं स्थानम्।
उत्तर:
(i) कठिनम्

II. यत्र तत् सत्यस्य परमं निधानम्।
(i) धनम्
(ii) अन्नम्
(iii) स्थानम्।
उत्तर:
(iii) स्थानम्

II. अद्भिः गात्राणि शुध्यन्ति।
(i) गायत्र्यादि छन्दांसि
(ii) अंगानि
(iii) पात्राणि।
उत्तर:
(ii) अंगानि

IV. एतान्यपि सतां गेहे नोच्छिद्यन्ते कदाचन।
(i) सज्जनानाम्
(ii) सत्यानाम्
(iii) साधूनाम्।
उत्तर:
(i) सज्जनानाम्

v. सर्वभूतेषु चात्मानम् ततो न विजुगुप्सते।
(i) अन्तर्हितो भवति
(ii) लीनो भवति
(iii) घृणां करोति।
उत्तर:
(iii) घृणां करोति

NCERT Solutions for Class 12 Sanskrit

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Advertisement Writing Class 12 Format, Examples

Advertisement Writing Class 12 Format, Examples

An Advertisement is a kind of public notice asking for or offering services or buying and selling property, goods etc., or providing information about missing persons, pets, etc.

♦ There are two kinds of advertisements:

  1. Classified;
  2. Commercial

♦ Classified Advertisements:
You will come across classified advertisements in the columns of newspapers and magazines. Important features of a Classified Advertisement are:

  • No blocks, no designs and the language to be factual.
  • Simple, formal and to the point.
  • Comprehensive, yet must not leave any important matter.
  • Never be too lengthy (confine to the word limit).

♦ Important Tips to be followed
♦ Classified advertisements

  1. Clearly state the category at the top, e.g., Tor Sale’, ‘To Let’, etc.
  2. Give all necessary details in points using commAnswer:
  3. Give contact address, name and telephone number.
  4. Put the matter in a box.

♦ Kinds of Classified Advertisements

  1. Situation Vacant/Wanted
  2. Lost and found
  3. Sale and purchase
  4. Accommodation wanted
  5. Educational
  6. Placement services
  7. Matrimonial
  8. To Let
  9. Tuitions
  10. Packers and Movers
  11. Kennel
  12. Travels and Tours

♦ Necessary Information
♦ Situation Vacant

  • Specify the post and number of vacancies
  • Qualities of the person required
  • Name of the Company (optional)
  • Age and sex of the candidate
  • Payscale and perks
  • Mode of applying
  • Contact address and phone number

♦ To-Let

  • Type of accommodation, Number of rooms/ floor
  • Whether it is independent or an apartment.
  • Rent expected
  • Type of tenant required: Bank employee / small family.
  • Whom and when to contact.
  • Contact address/phone number

♦ Kennel;

  • Breed of dog
  • Age
  • Colour
  • Training
  • Price expected
  • Contact address and phone number

♦ Packers And Movers

  • Specify the services
  • Give reason why you should be given a chance
  • Area of work
  • Whom to contact
  • Contact address and phone number

♦ Vehicles For Sale

  • Make Maruti/Hyundai/Honda, etc.
  • Model/colour/accessories/year of manufacture/mileage
  • Condition
  • Ownership details
  • Price expected
  • Contact address/phone number

♦ Property For Sale
♦ Plot for sale:

  • Location: where it is
  • Area: In sQuestion metres/yards
  • Name of the development authority
  • Price expected
  • Contact address/ phone number

♦ Lost And Found

  • Begin with LOST/FOUND
  • Description of the article
  • When & where the article was lost/found
  • Reward for finding it
  • Contact address and phone number

♦ Travels And Tours

  • Name of the agency
  • Destinations and durations
  • Details of the package
  • Discounts, if any
  • Contact address/phone number

♦ Tuitions

  • Classes and subjects
  • Special qualities of the tutor
  • Qualifications and experience of the tutor
  • Previous results of his/her students
  • Contact address/ phone number

♦ Housi/Flat For Sale

  • Type of accommodation. No. of rooms/floor
  • Whether it is independent or an apartment
  • Price expected/negotiable
  • Location—where it is
  • Area: In sq. metres/ yards
  • Name of the development authority
  • Whom and when to contact
  • Contact address/phone number

♦ Commercial Advertisements:
♦ Commercial or Display Advertisements

  • These are designed for commercial purposes.
  • Require more space, hence are costly. Must be attractive with visuals, catchy phrases and slogans

♦ Main Features:

  • Must be attractive with a catchy caption, heading or sub-headings.
  • Figurative language (alliteration and metaphors especially).
  • The proportionate spacing of fonts with different sizes.
  • Usually attractive with catchy slogans, punch lines, witty expressions, pictures or sketches.
  • Special offers or discount, if any.
  • Details of the product or event given in a clear, precise way.
  • Give name, contact number and address of the advertiser.
  • Present the matter in a box.

♦ Previous Years’ CBSE Examination Questions

♦ Short Answer Tvpe Questions

Question 1.
You are Manisha of 10, Rajaji Nagar, Bangalore. You want a Maths teacher for your son who is a class 10 student. Draft a suitable advertisement in not more than 50 words stating your requirements. (Delhi 2009)
Answer:

Wanted

An experienced male Maths teacher as a private tutor for an Xth standard boy. Candidate should have at least 10 years experience of teaching Maths in a public school. He should be a strict disciplinarian to be able to deal with the student on a stern note. Remuneration no constraint for the right candidate. Apply with complete details to Manisha, 10, Rajaji Nagar, Bengaluru.

Question 2.
You want to sell your newly built flat. Draft a suitable advertisement in not more than 50 words to be inserted in the classified columns of ‘The Hindu’ giving all necessary details. You are Niranjan, 247, J.P. Nagar, Bengaluru. (All India 2009)
Answer:

New flat for sale

Newly built, ideally located flat for sale on M.G. Road. Two bedrooms with attached toilets, huge drawing-cum-dining and servant room, spacious and well-lit. Barely 10 minutes drive from the airport. In close vicinity of reputed schools and multi-speciality hospital. Expected price: ?25 lacs. Contact Niranjan, 247, J.P. Nagar, Bengaluru.

Question 3.
You are General Manager of Ivy Software Solutions, Agra Cantt, Agra. You need a software engineer for your organisation. Draft an advertisement in not more than 50 words to be published in ‘The Times of India’ under the classified columns. (Delhi 2010)
Answer:

Situation Vacant

Wanted software engineer for Ivy Software Solutions, a leading name in computers. Candidate should possess a Master’s Degree in Computer Software Programming and at least 3 years experience with a known computer concern. Remuneration no constraint for suitable candidate. Apply with detailed resume within ten days to General Manager, Ivy Software Solutions, Agra Cantt, Agra.

Question 4.
You are the General Manager of E.V.L. Company which requires posh bungalows in company lease, as guest houses. Draft an advertisement in not more than 50 words under classified columns to be published in ‘The New Indian Express’.
Answer:

Accommodation Required

EVL Company requires posh bungalows to be used as guest houses by their executives and guests. Accommodation in a centrally located, posh area, with power back-up facility and uninterrupted water supply is a must. Close proximity to the market will be preferred. Owners please contact: General Manager, EVL Company, Delhi.

Question 5.
Your younger brother aged 5 has been missing for the last three days. Draft an advertisement in not more than 50 words for the Missing Persons column of a local newspaper. You are Ram/Rama. Contact number 931070000. (Comptt. Delhi)
Answer:

Missing Person

General Public is hereby informed about the missing of a 5-year-old boy from Central Park in Connaught Place three days ago. The child responds to the name ‘Sonu’, is fair complexioned and was wearing a red shirt and denim shorts. Anyone knowing anything about his whereabouts please contact: Ram, contact no. 931070000. A cash prize of? 50,000 for the finder.

Question 6.
You are Ramanuj am/Revathi, a student of Class XII, St. Xavier’s School, Jhansi. You are interested in giving tuition in Maths to one or two students of class VIII. Draft an advertisement in not more than 50 words for a local newspaper. (Comptt. All India)
Answer:

Maths Tuition Available

Tuition of Maths is available for students of class VIII. Not more than two students will be taught five days a week (Mon. to Fri.) between 4.30 pm to 6 pm. Assurance of good result. Those interested may please contact: Ramanujam, student of Class XII at St. Xavier’s School, Jhansi

Question 7.
You are Uttarq/Umesh, a visually challenged person, running a telephone booth in the Central Market, Delhi. Give a suitable advertisement in not more than 50 words for a telephone operator in ‘Situation Vacant’ column of Delhi Times, offering attractive remuneration. (Comptt. All India 2011)
Answer:

Situation Vacant Telephone

Operator required for a telephone booth, run by a visually challenged man, in the Central Market, Delhi. The person must be fluent in Hindi and English. Work timings: 12 p.m. to 9 p.m. (Monday to Saturday). Remuneration no constraint for suitable candidate. Contact: Umesh, 9818182838 (between 3-5 p.m.)

Question 8.
You want to rent out your newly constructed flat in the heart of the city. Draft an advertisement in not more than 50 words to be published in ‘The Deccan Herald’, Bangalore under classified columns. Give all the necessary details. You are Mohan/Mahima of Jayanagar, Bengaluru. (Delhi 2011)
Answer:

Available For Rent

Newly constructed flat on M.G. Road with 24 hrs. water and electricity back-up facility. Has two bedrooms with attached bathrooms, one huge drawing-cum-dining. Expected rent? 24,000 p.m. Company lease only. Contact: Mohan, Jayanagar, Bengaluru.

Question 9.
You are Mohan/Mohini, General Manager of PJC. Industries, Hyderabad. You need an accountant for your company. Draft, in not more than 50 words, an advertisement to be published in ‘The Hindu’ in classified columns. (All India 2012)
Answer:

Situation Vacant

Wanted Accountant with a Master’s Degree in Commerce and minimum of three years experience in a reputed firm. Candidate should not be more than 30 years of age. Remuneration no bar for a suitable candidate. Apply with detailed resume within seven days to Mr. Mohan, General Manager, P.K. Industries, Hyderabad.

Question 10.
You are Ratan/Rani, General Manager of Hotel Green Park, Lucknow. You need a receptionist for your hotel. Draft an advertisement in not more than 50 words to be published in ‘Hindustan Times’, Lucknow, calling for applications. (All India 2012)
Answer:

Situation Vacant

Wanted male/ female receptionist in the age group of 22-28 years for a reputed four-star hotel. Candidate should be a graduate having fluency in speaking English and should be computer literate. Apply with detailed resume within seven days to General Manager, Hotel Green Park, Lucknow.

Question 11.
You are Keshav/Karuna, interested in purchasing a house in Bengaluru. Draft in not more than 50 words an advertisement to be published in the classified column of a local daily giving the details of your requirement. (Comptt. Delhi 2012)
Answer:

Accommodation Wanted

Need a 300 square yards newly built bungalow in a prime location in Bengaluru for immediate purchase. Three bedrooms with attached bathrooms, drawing room, dining room preferably in close vicinity to a market-place. Contact: Mr. Keshav; Mobile: 9812345678

Question 12.
You are Jay/Jaya, interested in purchasing a second-hand flat. Draft an advertisement in not more than 50 words to be published in the classified column of a local daily giving the details of your requirement. Your contact number is 9012341234. (Comptt. All India 2012)
Answer:

Accommodation Wanted

Wanted for immediate purchase a second-hand flat having two bedrooms, one with attached bathroom in close vicinity to a market place and bus-stop. Price range between 15-20 lakhs. Owners, please contact Jay mobile no. 9012341234 between 3 p.m. to 8 p.m.

Question 13.
You have lost your leather wallet containing your Examination Entry Ticket for Class XII while travelling by bus from Banashankari to M.G.Road in Bangalore. Write a notice in not more than 50 words, to be published in ‘Deccan Herald’. You are Pranav/Parveen, 12 Gandhi Road, Bengaluru. (All India 2012)
Answer:

Lost And Found

Lost a black leather ladies wallet containing my Examination Entry Ticket for class XII travelling on Bus Route no. 123 from Banashankari to M.G.Road on 15th Feb., 20xx. Anyone who finds it please contact urgently Parveen, 12, Gandhi Road, Bengaluru. Suitable reward for finder.

Question 14.
You are General Manager, Hotel Dosa, Gur- gaon. You need a lady Front Office Assistant with sound knowledge of computers. She must be a graduate and good in communication skills with pleasing manners. Draft an advertisement in not more than 50 words to be published in Gurgaon Times. (Delhi 2013)
Answer:

Situation Vacant

Wanted a Lady Front Office Assistant for a reputed hotel in Gurgaon. Candidates should be between 25-35 years of age, graduate, good in communication skills with pleasing manners and also have sound knowledge of computers. Interested candidates can send their resume to the General Manager, Hotel Dosa, Gurgaon within ten days.

Question 15.
You have a three-bedroom flat in Dwarka, which you want to let out on rent. Draft an advertisement in not more than 50 words to be published in The Times of India’ under classified columns. Contact 2758902. (All India 2012)
Answer:

Available For Rent

Newly constructed flat on M.G. Road with 24 hrs. water and electricity back-up facility. Has two bedrooms with attached bathrooms, one huge drawing-cum-dining. Expected rent? 24,000 p.m. Company lease only. Contact: Mohan, Jayanagar, Bengaluru.

Question 16.
You possess an acre of land in the heart of the city. You want to dispose of this property since you have decided to buy a flat. Write an advertisement to be published in a national daily, giving all the necessary details. You are Krishan of Moti Nagar, Delhi. (Delhi 2014)
Answer:

Land For Sale

Available for sale one acre of land in the heart of Delhi. Owner is interested in selling land as early as possible. Interested parties can contact Krishan, Moti Nagar, Delhi, Mobile No. 1234 0.

Question 17.
You would like to let out a portion of your independent house for office use. Write an advertisement for the classified columns of the local newspaper giving all the necessary details. Write the advertisement in not more than 50 words. (Delhi 2014)
Answer:

Office Space On Rent

a portion of an independent house is available for rent for office use. Located in Central Delhi, 1km from Metro station, 2000 sq.m area. Parking facility for over twenty cars. Expected rent? 35,000 pm. Interested parties can contact: ABC, contact no. 12340.

Question 18.
You want to sell off your motorbike which you have been using for five years since you have decided to buy a car. Write an advertisement, in not more than 50 words, to be published under the classified columns of a national daily. Furnish all the necessary details. (Delhi 2014)
Answer:

Motor Bike For Sale

Yamaha 100 cc motorbike fully loaded with accessories for sale. Done 8,000 km, excellent condition, 2013 Model, Price expected ?28,000 (non-negotiable). Genuine buyers, contact Raj Kumar, Mobile No. 1234 0.

Question 19.
Your school, Akash Public school, Agra needs a canteen manager. On behalf of the Principal, write an advertisement in about 50 words to be published in the classified columns of a local daily. Mention the educational and professional qualifications, other qualities required in the manager, who to apply to and the last date for the receipt of applications. (All India 2015)
Answer:

Situation Vacant

Wanted a pleasant and friendly canteen manager for a reputed public school in Agra. Candidate should be a graduate with a Diploma in Food and Beverage and at least 5 years experience of working in a canteen. Interested candidates, please contact. Principal, Akash Public School, Agra with complete bio-data within 10 days of publishing of this advertisement. PH: 98100xxxxx.

Question 20. You require a teacher to teach maths and science to your son at home who is in class 10. Draft an advertisement in not more than 50 words giving all your requirements. You are Arun/Aruna. Contact No. 93xxxxxxxx.
(Comptt. All India 2014)
Answer:

Wanted

An experienced male Maths teacher as a private tutor for a Xth standard boy. Candidate should have at least 10 years experience of teaching Maths in a public school. He should be a strict disciplinarian to be able to deal with the student on a stem note. Remuneration no constraint for the right candidate. Apply with complete details to Manisha, 10, Rajaji Nagar, Bangalore.

Question 21.
You have retired from a bank after 30 years of service and are looking for a job. Draft an advertisement in about 50 words for the situation wanted column of a local daily giving your qualifications, experience and the kind of job expected. You are Sunil/Sita, 4, Bank Enclave, (Delhi 2014)
Answer:

Situation Wanted

Retired bank official seeks a job as Senior Accountant in a firm. Have a master’s degree in Commerce and 30 years experience of working in a nationalised bank. Job profile can include managing and maintaining logbooks and account balance. Contact: Sunil 4, Bank Enclave, Delhi.

Question 22.
You have lost an expensive watch probably in the market. Write an advertisement for the ‘Lost and Found’ column of a local newspaper giving all the relevant details. Offer a reward also. Write the advertisement in about 50 words. You are Gopal/Gopa, 4 Manav Road, Kanpur. (Comptt. Delhi 2014)
Answer:

Lost And Found

Lost a Casio watch, black dial and black leather strap, on Bus Route No. 205 on 1st February, 20xx at 10:00 a.m. Anyone who finds the watch please contact Gopal, 4 Manav Road, Kanpur. Finder will be rewarded with? 2,500. Gopal

Question 23.
You have cleared your IIT Entrance Exam and so want to sell off the reading material you have with you. Write an advertisement to be placed in the ‘For Sale’ columns of a local daily giving all details of the material you have with you in not more than 50 words. You are Mohan/Mohini. Contact No. 9811111111. (Comptt. Delhi 2014)
Answer:

For Sale

IIT Entrance Exam books of FIITJEE and Brilliant worth? 6,000 for sale. All books are in mint condition. A rare chance to prepare for IIT Entrance Exam with all the relevant study material at half the price. Those interested please contact Mohini, Contact no. 98111 11111.

2016
Question 24.
Principal, Sunrise Global School, Agra requires a receptionist for her school. Draft a suitable advertisement in about 50 words to be published in the classified columns of a national newspaper giving all the necessary details of qualifications and experience required in the receptionist. (Delhi 2014)
Answer:

Situation Vacation

Required a female receptionist for a reputed public school in Agra. Candidate should be a graduate with at least three years experience in a similar position. Knowledge of computers, fluency in English and good communication skills are essential requirements. Salary negotiable. Interested candidates may apply with complete biodata within seven days to Principal, Sunrise Global School, Agra.

Question 25.
You need to buy a flat. Draft a suitable advertisement in about 50 words to be published in the classified columns of a local newspaper giving all the necessary details of your requirement. You are Karan/Karuna, M 114, Mall Road, Delhi. (Delhi 2014)
Answer:

Flat Wanted

Wanted 1300 sq. ft., three-bedroom flat on the ground/first floor with separate servant quarters in a ready-to-move-in society in the NCR region. Rent negotiable. East facing, twenty-four-hour water supply, power backup, close proximity to metro station preferred. Contact: Karan, M 114, Mall Road, Delhi.

Question 26.
You are Karan Kumar/Karuna Bajaj, a leading lawyer practising in Surat. You want to buy an independent house at City Light Road to be used as office-cum-residence. Draft an advertisement in about 50 words for the classified columns of a local newspaper. You can be contacted at 45645678. (All India 2014)
Answer:

Accommodation Wanted

A leading lawyer practising in Surat wants to buy an independent house to be used as office-cum-residence at City Light Road. House should have sufficient parking space and should be located in close proximity to the main road. Contact: Karan Kumar on 45645678 between 10 am to 2 pm.

Question 27.
You are Karan/Karuna of M 114, Mall Road, Delhi. You are a civil engineer and have recently returned from UAE. You are looking for a suitable job in India. Draft an advertisement for the same in about 50 words. Give details of your qualifications, experience, nature of job and expected remuneration. (All India 2014)
Answer:

Situation Wanted

A qualified civil engineer recently returned from UAE seeks a suitable job in India. Done Civil Engineering from Jamia Milia University. Fifteen years experience of working in the construction field (commercial) in reputed firms in India and UAE. Expected remuneration—negotiable. Contact: Karan, M 114, Mall Road, Delhi.

Question 28.
Situation Wanted Qualified civil engineer recently returned from UAE seeks suitable job in India. Done Civil Engineering from Jamia Milia University. Fifteen years experience of working in construction field (commercial) in reputed firms in India and UAE. Expected remuneration—negotiable. Contact: Karan, M 114, Mall Road, (Delhi 2014)
Answer:

House For Sale

3 BHK with attached baths, modular kitchen, 2,000 square feet house built three years ago for immediate sale in Greenwood Society, Sector-52, Gurgaon. Owner going abroad. Price expected is as per the ongoing market rate. Must sell soon. Contact no.: XXXXXX

Question 29.
Write an advertisement, as Sanjay Gupta, offering Do not exceed 50 words. Contact: xxxxxx (Comptt. All India 2014)
Answer:

To-Let

3 BHK Flat available on rent in Dwarka, Delhi. In close proximity to Metro Station, Sector 10, Dwarka. Society flat, with complete security and ample parking space. 12-hour electricity backup available. Expected rent—?25,000 p.m. Contact Sanjay Gupta, Mobile no.: 9891 xxxxxx.

Question 30.
You want to sell your car. Write an advertisement for the ‘sale and purchase ‘ column of a local newspaper giving all relevant details. Write the advertisement in about 50 words. Contact no. xxxxxx. (Comptt. All India 2014)
Answer:

Hyundai Santro For Sale

2016 May Hyundai Santro for sale, white colour, done 20,000 kms, single-owner driver, Top-end model, Excellent condition, owner going abroad, 3.00 lacs non- negotiable price. Contact Sushil, 21, Ram Nagar, Delhi between 3 p.m. to 5 p.m.

Question 31.
You VikranySonia, an Hon’s graduate in history with specialization in Medieval India. You are well acquainted with places of historical interest in Delhi, Agra and Jaipur. You are looking for the job of tourist guide. Write an advertisement in about 50 words for the situations wanted column of a local newspaper. Your contact no. 999751234. (Delhi 2014)
Answer:

Situation Wanted

A Hon’s graduate in History with specialization in Medieval India and well acquainted with places of historical interest in Delhi, Agra and Jaipur seeks the job of a tourist guide. Can speak fluent English and Hindi. Expected salary? 15,000 for a personal interaction please contact Vikram at 999751234.

Question 32.
You are Vikrain/Sonia, an electronics engineer who has recently returned from the U.S. and looking for a suitable job in the IT industry. Draft an advertisement in about 50 words for
Answer:

Situations Wanted

An electronic engineer with a first-class degree from reputed US college seeks a suitable job in IT industry. Has a master’s degree in Electrical Systems Engineering and 7 years of work experience with a reputed IT firm in New Jersey. Please contact: Vikram L Mobile no.: 91930 10203

Question 33.
You are Ran/RaJani. Draft a classified advertisement, in not more than 50 words, for the purchase of a house,, giving all necessary details of your requirement. You can be contacted at 45678900. (Comptt. All India 2014)
Answer:

Purchase Of House

Advertiser requires an independent newly-built house, with latest amenities in or around the NCR in Gurugram. Close proximity to the metro station is a must. The house should be in an open area with power back-up and uninterrupted water supply. Owners, please contact: Ram/Rajani at 45678900

Question 34.
You are Ran/Rajani. Draft a classified advertisement, in not more than 50 words, to be published in India Times for the sale of a used motor car giving all the necessary details. You can be contacted at 12345679. (Comptt. Delhi 2014)
Answer:

Honda City For Sale

2015 August, Honda City for Sale, Silver colour, done 24,000 km. single-owner drove, top-end model. Excellent condition. Price expected 4.5 lacs, Non-negotiable. For any other queries contact: Ram, Mobile no: 123456789 between 9 am to 5 pm.

RBSE Class 10 English Notes

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Poster Writing Class 12 Format, Examples, Samples, Topics

Poster Writing Class 12 Format, Examples, Samples, Topics

A Poster is a large notice announcing or advertising something. It generally creates social awareness about any problem or needs. It also conveys a social message in an eye-catching way.

♦ Main Features of a Poster

  • Layout: A creative title in the form of a slogan or rhyming words.
    For example, Speed thrills but kills; Pollution needs a solution; Green is clean.
  • You can use Match Stick figures for illustration.
  • Proper spacing and proportionate headings and illustrations.
  • Date, time and venue must be mentioned in case of an event.
  • A poster displays a message or an appeal to create awareness.
  • The content should be organised in an appropriate style.
  • Use appropriate expressions and correct language.
  • The theme should be clear.
  • The poster should be creative and related to the content and theme should not be vague.
  • Do not spend much time on illustration. Written language and expressions will only help you score better.
  • Name of the issuing authority or the organizers is a must.

♦ Previous Years’ CBSE Examination Questions

♦ Short Answer Type Questions

Question 1.
Your school, Kendriya Vidyalaya, Burdwan is going to organise a Diwali Mela. Design a poster to inform the students about various activities connected with it. Ask them to participate in the Mela. You are Divya/Dewan. (Comptt. Delhi 2010)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 1

Question 2.
As President of Lion’s Club of Vasundhra City, design a poster in not more than 50 words for . promoting good health through ‘Health Mela’ to be held at Central Community Hall of the city. Mention some of its attractions. (Comptt. Delhi 2011)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 2

Question 3.
As President of the Residents Welfare Association of Mayur Colony, Delhi, design a poster in not more than 50 words for promoting cleanliness in the surroundings of your colony. (Comptt. Delhi)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 3

Question 4.
Prepare a poster in not more than 50 words on kindness to animals to be displayed in the city at public places appealing to people to show kindness to animals. You are Secretary of the Society for Prevention of Cruelty to Animals, Delhi. (Comptt. All India 2011)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 4

Question 5.
You were very upset about the reports on communal riots in various parts of the country. As a concerned social worker, design a poster in not more than 50 words, highlighting the importance of communal harmony. You are Vinay/Vineeta. (Comptt. All India 2014)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 5

Question 6.
You are a member of the S.P.C.A. Draft a poster in not more than 50 words, to create awareness on the need to prevent cruelty to animals. You are Suhas/Suhasini. (Comptt. All India 2014)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 6

Question 7.
You are a fitness trainer in a health club. Design a poster in not more than 50 words, to emphasize the importance of exercise in maintaining mental and physical fitness. You are Prem/Priya. (All India 2014)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 7

Question 8.
Fireworks and crackers are known to create pollution during festivals. As an environ-mentalist design a poster in about 50 words to create awareness of their ill effects. (Comptt. All India 2015)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 8

Question 9.
Design a poster in about 50 words to create awareness among the people of your city on the importance of following traffic rules. (Comptt. All India 2015)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 9

Question 10.
Your school is going to organise a Diwali Mela. Design a poster to inform the students about it (50 words). (Comptt. Delhi 2015)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 10

Question 11.
Draw a poster on the importance of cleanliness in our life. (Comptt. Delhi 2015)
Answer:
Image may be NSFW.
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Poster Writing Class 12 Format, Examples, Samples, Topics 11

Question 12.
Open drains are death traps, risky for old persons and children. They are also breeding grounds for rats, cockroaches etc. Design a poster highlighting the danger of open drains. (Comptt. All India 2015)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 12

Question 13.
Publicly we proclaim that dowry is an evil. Privately we want out sons to fetch good dowries. Right from our school days we should be taught that demanding and even giving dowry is not only illegal but immoral too. Draw a poster in about 50 words highlighting dowry as a curse. You are Vikram/ Sonia. (All India 2017)
Answer:
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Poster Writing Class 12 Format, Examples, Samples, Topics 13

RBSE Class 10 English Notes

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State-level National Talent Search Exam 2019-20 | Eligibility, How to Apply

State-level National Talent Search Exam 2019-20: State-level National Talent Search Exam is conducted every tear only for class 10 students. Under the procedure, the selection of candidates is done in the the the two-tier process. In the first tier, the NTSE exam is conducted by states as well as union territories for screening. This exam is known as a State-level National Talent Search Exam. Each union territory, as well as state, has to select and recommend a stipulated number of students for the national level exam which is conducted for 3000 students by NCERT.

NTSE exam stage 1 follows the national pattern that includes SAT and MAT. The MAT exam which is also known as a mental ability test consists of 100 multiple-choice questions. Scholastic aptitude test or SAT also consists of 100 questions that contain multiple-choice questions from physics, biology, chemistry, history, maths, political science, geography, and economics.

State Level National Talent Search Exam (NTSE) 2019 – 2020

State-level National Talent Search Exam Eligibility Criteria

All students that are studying in class 10 in recognized schools including Navodaya Vidyalaya, Kendriya Vidyalaya, and Sainik school, etc are eligible to apply for this exam. The states may impose their respective eligibility conditions for appearing in the screening exam like certain qualifying percentage marks in the last annual exam or something. Students that have registered under open distance learning platform can also appear for this exam. For these students, the maximum age limit is 18 years. Also, the students that are appearing for the exam should be appearing in class 10 for the first time.

Age Criteria: Students should be below the age of 18 years (as on 1st July)

Attempts: Students should be appearing in class X for the first time.

Exam Preparation: NTSE Preparation

How To Apply for This Exam?

Students studying in class 10 in India should look for any sort of advertisement in the newspapers or circulars by the respective government of the state. The filled-in application by students should be submitted to the respective state liaison officer and duly signed by the principal of the school before the final date.

State-level National Exam Pattern

The state-level exam will be divided into 2 sections which are mental ability test and scholastic aptitude test. Below is the exam pattern for the state level exam

SectionExam Pattern and TimeQualifying scores
Mental Ability Test (MAT)100 questions & 120 MintsGeneral category: 40% of the maximum marks.
SC, ST, and PH: 32% of the maximum marks.
Scholastic Aptitude Test (SAT)100 questions & 120 MinsGeneral category: 40% of the maximum marks.
SC, ST, and PH: 32% of the maximum marks.

State-level National Talent Search Exam Results

The state or union territory will prepare a merit list of students that ensures that minimum qualifying marks are scored in both the sections. For general and OBC category students, these marks are 40%. While for SC, ST, and PH students, these marks are 32%. The results of the State-level National Talent Search Exams are usually declared in January or February months by union territories and States themselves. This exam is taken to find the eligible candidates for the stage 2 exam conducted by the NCERT.

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions

Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions

Ray Optics and Optical Instruments Class 12 Important Questions Very Short Answer Type

Question 1.
Aglasslensof refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the liquid in order to mark the lens disappear? (Delhi 2008)
Answer:
In order to make the lens disappear the refractive index of liquid must be equal to 1.5 i.e. equal to that of glass lens.

Question 2.
A converging lens of refractive index 1.5 is kept in a liquid medium having same refractive index. What would be the focal length of the lens in this medium? (Delhi 2008)
Answer:
The lens in the liquid will act like a plane sheet of glass
∴ Its focal length will be infinite (∞)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 1

Question 3.
How does the power of a convex lens vary, if the incident red light is replaced by violet light? (Delhi 2008)
Answer:
According to Lens Maker’s formula
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 2

∴ power of the lens will be increased.

Question 4.
How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with red light? (All India 2008)
Answer:
We know that λ red > λ violet, therefore µ red < µ violet and hence δ red < δ violet.
When incident violet light is replaced with red light, the angle of minimum deviation of a glass decreases.

Question 5.
Why does the bluish colour predominate in a clear sky? (All India 2008)
Answer:
Blue colour of the sky : The scattering of light by the atmosphere is a colour dependent. According to Rayleigh’s law, the intensity of scattered light Image may be NSFW.
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\mathrm{I} \propto \frac{1}{\lambda^{4}}
, blue light is scattered much more strongly than red light. Therefore, the colour of sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of the blue sky.

Question 6.
How does the angle of minimum deviation of a glass prism of refractive index 1.5 change, if it is immersed in a liquid of refractive index 1.3? (All India 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 3

Hence angle of deviation is decreased.

Question 7.
You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? (Delhi 2009)

LensesPower (P)Aperture
L13D8 cm
L26D1 cm
L310D1 cm

Answer:
Objective – Less power and more aperture. So L1
Eyepiece – More power and less aperture. So L3.

Question 8.
Two thin lenses of power + 4D and – 2D are in contact. What is the focal length of the combination? (All India 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 4

Question 9.
Two thin lenses of power + 6D and – 2D are in contact. What is the focal length of the combination? (All India 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 5

Question 10.
A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of refractive index of the liquid? (Delhi 2010)
Answer:
The value of refractive index of the liquid should be 1.45 so that the glass lens of refractive index 1.45 disappears when immersed in a liquid.

Question 11.
State the conditions for the phenomenon of total internal reflection to occur. (Delhi 2010)
Answer:
Two essential conditions for total internal reflection are :

  1. Light should travel from an optically denser medium to an optically rarer medium.
  2. The angle of incidence in the denser medium must be greater than the critical angle for the two media.

Question 12.
Calculate the speed of light in a medium whose critical angle is 30°. (Delhi 2010)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 6

∴ Speed of light, v = 1.5 × 108 ms-1

Question 13.
A converging lens is kept coaxially in contact with a diverging lens — both the lenses being of equal focal lengths. What is the focal length of the combination? (Delhi 2010)
Answer:
Focal length of the combination is Infinity.

Question 14.
When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. (All India 2010)
Answer:
No, the energy carried by the lightwave remains the same.
Reason : As energy E = hv
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 7

Here frequency remains same.

Question 15.
When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same. Explain. (Delhi 2011)
Answer:
If v1 and v2 denote the velocity of light in medium 1 and medium 2 respectively and λ1 and λ2 denote the wavelength of light in medium 1 and medium 2. Thus
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 8

The above equation implies that when a wave gets refracted into denser medium (v1 > v2) the wavelength and the speed of propagation decreases but the frequency v (Image may be NSFW.
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=v / \lambda
) remains the same.

Question 16.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? (Delhi 2012)
Answer:
When the refractive index of glass of biconvex lens is equal to the refractive index of the liquid in which lens is immersed
or µ1 = µg

Question 17.
For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? (All India 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 9

∴ Velocity of light is minimum in medium A.

Question 18.
How would a biconvex lens appear when placed in a trough of liquid having the same refractive index as that of the lens? (Comptt. Delhi 2011)
Answer:
A biconvex lens appears plane glass when placed in a trough of liquid having the same refractive index as that of the lens.

Question 19.
Two thin lenses of power -4D and 2D are placed in contact coaxially. Find the focal length of the combination. (Comptt. All India 2011)
Answer:
Power of combination = – 4D + 2D = – 2D
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 10

∴ Focal length, f = – 50 cm

Question 20.
Two thin lenses of power -2D and 2D are placed in contact coaxially. What is the focal length of the combination? (Comptt. All India 2011)
Answer:
Power of combination = -2D + 2D = 0
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 11

Question 21.
Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviation for a triangular prism. (Delhi 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 12

where [δm is angle of minimum deviation]

Question 22.
When red light passing through a convex lens is replaced by light of blue colour, how will the focal length of the lens change? (Comptt. All India 2013)
Answer:
Focal length of lens will decrease Image may be NSFW.
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\mu_{v}>\mu_{r}

Question 23.
If the wavelength of light incident on a convex lens is increased, how will its focal length change? (Comptt. All India 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 13

Question 24.
A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? ‘ (Delhi 2014)
Answer:
Focal length of lens = 20 cm
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 14

(Hint: Rays coming out of lens are incident normally on plain mirror and hence reflected rays will trace the path of incident ray, hence forming image on the object itself, thus object and image overlapping each other at F of convex lens.)

Question 25.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (All India 2014)
Answer:
The lens will behave as a diverging lens, because -1)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 15

The value of (µ – 1) is negative and ‘f’ will be negative.

Question 26.
A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (All India 2014)
Answer:
The lens will behave as a converging lens because
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 16

Hence value of ‘f’ will be positive.

Question 27.
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens? (Delhi 2015)
Answer:
Converging.

Question 28.
Why does bluish colour predominate in a clear sky? (All India 2015)
Answer:
Blue colour of the sky : The scattering of light by the atmosphere is a colour dependent. According to Rayleigh’s law, the intensity of scattered light Image may be NSFW.
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\mathrm{I} \propto \frac{1}{\lambda^{4}}
, blue light is scattered much more strongly than red light. Therefore, the colour of sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of the blue sky.

Question 29.
When an object is placed between f and 2f of a concave mirror, would the image formed be
(i) real or virtual and
(ii) diminished or magnified? (Comptt. Delhi 2015)
Answer:
(i) Real
(ii) magnified

Question 30.
How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason. (Delhi 2017)
Answer:
The angle of minimum deviation decreases, when violet light is replaced by red light because refractive index for violet light is more than that for red light.

Question 31.
An object is kept in front of a concave lens. What is the nature of the image formed? (Comptt. Delhi 2017)
Answer:
When an object is kept in front of a concave lens, the nature of image formed is virtual, erect and diminished.

Question 32.
When light travels from a rarer medium to denser medium, the speed of light decreases. Does the reduction in speed imply a reduction in the energy? (Comptt. Delhi 2017)
Answer:
The reduction in speed, due to light travelling from a rarer to denser medium does not imply reduction in the energy.

Question 33.
The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1: 2. Compare the resolving powers of the two telescopes. (Comptt. All India 2017)
Answer:
Ratio of resolving power = 1 : 1
Resolving power is same because it does not depend on focal length of the objective.

Question 34.
Why must both the objective and the eye piece of a compound microscope have short focal lengths? (Comptt. All India 2017)
Answer:
For getting higher magnification in compound microscope, both objective and eyepiece must have short focal length, because
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 17

Ray Optics and Optical Instruments Class 12 Important Questions Short Answer Type SA-I

Question 35.
Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting telescope. (Delhi 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 18

(ii) Advantages of reflecting telescope over a refracting telescope:

  1. Due to large aperture of the mirror used, the reflecting telescopes have high resolving power.
  2. This type of telescope is free from chromatic aberration (formation of coloured image of a white object).
  3. The use of paraboloidal mirror reduces the spherical aberration (formation of non-point, blurred image of a point object).
  4. Image formed by reflecting telescope is brighter than refracting telescope.
  5. A lens of large aperture tends to be very heavy and therefore difficult to make and support by its edges. On the other hand, a mirror of equivalent optical quality weights less and can be supported over its entire back surface.

Question 36.
Draw a ray diagram of an astronomical telescope in the normal adjustment position. State two drawbacks of this type of telescope. (Delhi 2008)
Answer:
(i) Magnifying power m = Image may be NSFW.
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-\frac{f_{0}}{f_{e}}
. It does not change with increase of aperature of objective lens, because focal length of a lens has no concern with the aperature of lens.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 19

(ii) Drawbacks :

  • Images formed by these telescopes have chromatic aberrations.
  • Lesser resolving power.
  • The image formed is inverted and faintes.

Question 37.
Draw a ray diagram of a compound microscope. Write the expression for its magnifying power.
(Delhi 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 20

When the final image is formed at the least distance of distinct vision
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 21

Question 38.
Draw a labelled ray diagram of an astronomical telescope in the near point position. Write the expression for its magnifying power. (All India 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 22

Question 39.
Draw a labelled ray diagram, showing the image formation of an astronomical telescope in the normal adjustment position. Write the expression for its magnifying power.(All India 2008)
Answer:
(i) Magnifying power m = Image may be NSFW.
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-\frac{f_{0}}{f_{e}}
. It does not change with increase of aperature of objective lens, because focal length of a lens has no concern with the aperature of lens.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 19

(ii) Drawbacks :

  • Images formed by these telescopes have chromatic aberrations.
  • Lesser resolving power.
  • The image formed is inverted and faintes.

Question 40.
Draw a ray diagram for the formation of image in a compound microscope. Write the expression for its magnifying power. (All India 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 20

When the final image is formed at the least distance of distinct vision

Question 41.
A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism. (Delhi 2008)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 23

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 24

Question 42.
Calculate the distance of an object of height h from a concave mirror of focal length 10 cm, so as to obtain a real image of magnification 2. (Delhi 2008)
Answer:
Given : f = -10 cm; Magnification, m = 2
To calculate : u = ?
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 25

Question 43.
Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence. (All India 2009)
Answer:
Refractive index of a transparent medium is the ratio of the speed of light in free space to the speed in the given medium.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 26

Question 44.
(i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain. (All India 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 27

Question 45.
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens. (Delhi 2010)
Answer:
Given : R1 = 10 cm,
R2 = -15 cm,
f = 12 cm
Using lens maker’s formula, we have
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 28

Refractive index of the material of the lens :

Question 46.
(a) The bluish colour predominates in clear sky.
(b) Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism.
State reasons to explain these observations. (Delhi 2010)
Answer:
(a) The scattering of light by the atmosphere is colour dependent. According to Rayleigh’s law, the intensity of scattered light, Image may be NSFW.
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I \propto \frac{1}{\lambda_{4}}

Blue light is scattered much more strongly than red light. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of the blue sky.

(b) As refractive index of prism is different for different colours, therefore, different colours deviate through different angles on passing through the prism. As λviolet < λred therefore µviolet > µred. Hence δviolet > δred maximum deviation is of violet colour. That is why violet colour, is seen at the bottom of the spectrum when white light is dispersed by a prism.

Question 47.
A biconvex lens has a focal length 2/3 times the radius of curvature of either surface. Calculate the refractive index of lens material.
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 29

Question 48.
(i) Why does the Sun appear reddish at sun-set or sun-rise?
(ii) For which colour the refractive index of prism material is maximum and minimum? (Delhi 2009).
Answer:
(i) During Sunrise or sunset, the Sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the Sun appears reddish.
(ii) Refractive index of prism material is maximum for violet colour and refractive index of prism material is minimum for red colour.

Question 49.
Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5. (Delhi 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 30

∴ Radius of curvature = -15 cm.

Question 50.
(i) Out of blue and red light which is deviated more by a prism? Give reason.
(ii) Give the formula that can be used to determine refractive index of material of a prism in minimum deviation condition. (Delhi 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 31

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 32

Question 51.
Two convex lenses of same focal length but of aperture Ar and A2 (A2 < A-,), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. (Delhi 2009)
Answer:
Resolving power of a telescope is given by R.P.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 33

From the given condition, the ratio of resolving power of two astronomical telescopes will be R.P, A,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 34

Telescope with large aperture (A1) should be preferred as it increases the resolution by collecting more light.

Question 52.
A ray of light, incident on an equilateral glass prism (µg = Image may be NSFW.
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\sqrt{3}
) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 35

∴ Angle of incidence, i = 60°

Question 53.
An object AB is kept in front of a concave mirror as shown in the figure.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 36

(i) Complete the ray diagram showing the image formation of the object.
(ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black? (All India 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 37

(ii) When the lower half of the mirror is painted black, the image formed is still of the same size as that with unpainted mirror but the intensity of the image has now reduced.

Question 54.
Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope. (All India 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 38

Two advantages over the refracting telescope :

  1. There is no chromatic aberration as the objective is a mirror.
  2. Spherical aberration is reduced using mirror objective in the form of a paraboloid.
  3. Image is brighter compared to that in a refracting type telescope.
  4. Higher resolving power. (any two)

Question 55.
(a) Plane and convex mirrors are known to produce virtual images of the objects. Draw a ray diagram to show how, in the case of convex mirrors, virtual objects can produce real images.
(b) Why are convex mirrors used as side view mirrors in vehicles? (Comptt. Delhi 2012)
Answer:
(a) When rays incident on a plane. mirror or convex mirror are tendmg to converge to a point behind the mirror, they are reflected ‘ to a point on a screen in front of the mirror. Hence, a real image is formed (when the object is virtual).
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 216

(b) A convex mirror is used as side view mirror in vehicles because it has larger field of view as compared to other mirror. The image formed is small and erect.

Question 56.
(a) Draw a ray diagram for a convex minor showing the image formation of an object placed anywhere in front of the minor.
(b) Use this ray diagram to obtain the expression for its linear magnification. (Comptt. All India 2012)
Answer:
(a) Ray diagram
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 39

Question 57.
(a) Draw a ray diagram for a concave minor showing the image formation of an object placed anywhere in front of a minor.
(b) Using the ray diagram, obtain the expression for its linear magnification. (Comptt. All India 2012)
Answer:
(a) Ray diagram :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 40

Question 58.
Deduce, with the help of ray diagram, the expression for the mirror equation in the case of convex minor. ‘ (Comptt All India 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 41

Question 59.
A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature? (Delhi 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 42

The system will be a diverging lens as it has negative power.

Question 60.
Draw a ray diagram showing the image formation by a compound microscope. Hence obtain the expression for total magnification when the image is formed at infinity. (Delhi 2013)
Answer:
Ray Diagram:
(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

(i) The focal length of the objective lens has no effect on the resolving power of microscope.
(ii) When the wavelength of light is increased, the resolving power of a microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

Expression for total magnification when image is formed at infinity:
Magnification of object,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 43

Angular magnification due to eye piece,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 44

Total magnification when image is formed at infinity
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 45

Question 61.
A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focal length 40 cm. Determine the power of the combination. Will the system be converging or diverging in nature? (Delhi 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 46

The system is converging in nature.

Question 62.
A convex lens of focal length f1 is kept in contact with a concave lens of focal length fr Find the focal length of the combination. (All India 2013)
Answer:
f1 ➝ is focal length of convex lens
f1 ➝ is focal length of concave lens.
fnet ➝ be the focal length of the combination. The power of the combination is the sum of the two powers.
So, net focal length of combined lens is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 47

Question 63.
Draw a schematic arrangement of a reflecting telescope (Cassegrain) showing how rays coming from a distant object are received at the eye-piece. Write its two important advantages over a refracting telescope. (Comptt. Delhi 2013)
Answer:
Reflecting telescope. Telescope with mirror objectives is called reflecting telescope. This is also known as Cassegrain telescope / Newtonian telescope. The ray diagram of reflecting type telescope is shown in figure.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 48.

Advantage over refracting telescope :

  1. Since reflecting telescope has mirror objective, so the image formed is free from chromatic aberration.
  2. Since the spherical mirrors are parabolic mirrors, free from spherical aberration, they produce a very sharp and distinct image.

Question 64.
Draw a labelled ray diagram of refracting type telescope in normal adjustment. Write two main considerations required of an astronomical telescope. (Comptt. Delhi 2013)
Answer:
Refracting telescope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 125

Magnifying power. It is defined as the ratio of angle (β) subtended by the final image on the eye to the angle (α) subtended by object on eye.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 126

Limitations of refracting telescope over a a reflecting type telescope :

  1. It suffers from chromatic aberration due to refraction and hence the image obtained is multicoloured and blurred.
  2. As a lens of large apparatus can’t be manufactured easily, its light gathering power is low and hence can’t be used to see faint stars.

Question 65.
Draw a labelled ray diagram of a compound microscope. Why are the objective and the eye-piece chosen to have small focal length? (Comptt. Delhi 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 49

  1. If the focal lengths are less, then their magnifying power will be more.
  2. To avoid any aberration in refraction due to larger bend on passing through the eyepiece.

Question 66.
A ray of light passes through an equilateral prism in such a way that the angle of incidence is equal to the angle of emergence and each of these angles is 3/4 times the angle of the prism. Determine
(i) the angle of deviation and
(ii) the refractive index of the prism. (Comptt. All India 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 50

We know, that δ + A = i + e
=> δ = z + e – A
∴ 8 = 45° + 45° – 60° = 30°
(i) angle of deviation = 30°
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 51

Question 67.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays entering through the prism. (All India 2014)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 52

Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 53

Since the angle of incidence (45°) is less than the critical angle (48°), the ray will be refracted.
(ii) For the ray 2,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 54

Since the angle of incidence (45°) is more than the critical angle (43°), the ray will be total internally reflected.

Question 68.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘T and ‘2’ are respectively 1.3 and 1.5. Trace the path of these rays after entering through the prism. (All India 2014)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 55

Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 56

Since the angle of incidence (45°) is more than the critical angle (42°), the ray will be total internally reflected.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 57

(ii) For the ray 1,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 58

Since the angle of incidence (45°) is less than the critical angle (50°) the ray will be total refracted.

Question 69.
Draw a schematic diagram of a reflecting telescope (Cassegrain). Write its two advantages over a refracting telescope. (Comptt. Delhi 2014)
Answer:
Reflecting telescope. Telescope with mirror objectives is called reflecting telescope. This is also known as Cassegrain telescope / Newtonian telescope. The ray diagram of reflecting type telescope is shown in figure.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 48.

Advantage over refracting telescope :

  1. Since reflecting telescope has mirror objective, so the image formed is free from chromatic aberration.
  2. Since the spherical mirrors are parabolic mirrors, free from spherical aberration, they produce a very sharp and distinct image.

Question 70.
Draw a ray diagram for the formation of image by a compound microscope. Write the expression for total magnification when the image is formed at infinity. (Comptt. Delhi 2014)
Answer:
Compound Microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 169

Magnifying power : The magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final virtual image to the angle subtended at the eye by the object, when both are at the least distance of distinct vision from the eye.

Numerical:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 170

The -ve sign shows that the final image is an inverted image.

The expression magnifying power of compound microscope when the final image is formed at infinity is :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 59

Question 71.
Write the conditions for observing a rainbow. Show, by drawing suitable diagrams, how one understands the formation of a rainbow. (Comptt. All India 2014)
Answer:
The conditions for observing a rainbow are:

  1. The Sun comes out after a rainfall.
  2. The observer stands with the Sun towards his/her back.

Formation of a rainbow
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 60

  • The rays of light reach the observer through a refraction, followed by a reflection followed by a refraction.
  • Figure shows red light from drop 1 and violet light from drop 2, reaching the observer’s eye.

Question 72.
Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (Delhi 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 61

Hence, image distance v ≥ -2f
Since, v is negative therefore the image is real.

Question 73.
You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece. (All India 2014)
Answer:
(a) When the image lies at infinity
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 62

(b) When the image is formed at the near point
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 63

30 × 1.25 = L × 6
L = 5 × 1.25 = 6.25 cm

Question 74.
A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (All India 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 64

Question 75.
Why does white light disperse when passed through a glass prism?
Using lens maker’s formula, show how the focal length of a given lens depends upon the colour of light incident on it. (Comptt. Delhi 2014)
Answer:
(i) The white light disperses when passed through a prism, because the refractive index of the glass of the prism is different for different wavelengths (colours). Hence, different colours get bent along different directions.

(ii) Using lens maker’s formula,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 65

As the refractive index of the medium (n2) (glass) with respect to air (n1) depends on the wavelength or colour of light, therefore focal length of the lens would change with colour.

Question 76.
A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 66

Answer:
(i) The ray will emerge from the face AC as shown. A
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 67

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 68

The angle of incidence (i) on the face AC is 30°, which is < ic, hence the ray will emerge as shown in the diagram, and will NOT be reflected back.

Question 77.
Draw a ray diagram to show how a right angled isosceles prism may be used to “bend the path of light rays by 90°”.
Write the necessary condition in terms of the refractive index of the material of this prism for the ray to bend to 90°. (Comptt. Delhi 2014)
Answer:
(a) Ray diagram
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 69

Question 78.
The image of an object, formed by a combination of a convex lens (of focal length f) and a convex mirror (of radius of curvature R), set up, as shown is observed to coincide with the object.
Redraw this diagram to mark on it the position of the centre of curvature of the mirror.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 70

Obtain the expression for R in terms of the distances, marked as a and d, and the focal length, l, of the convex lens. (Comptt. Delhi 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 218

Question 79.
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain. (Delhi 2016)
Answer:
Definition : Magnifying power of a compound microscope is defined as “the angle subtended at the eye by the image to the angle subtended (at the unaided eye) by the object”.
Reason : To increase the magnifying power, both the objective and the eyepiece must have short
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 71

Question 80.
Why should the objective of a telescope have large focal length and large aperture? Justify your answer.
Answer:
Large focal length : to increasing magnifying power.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 72

Question 81.
A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index upto a height H. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut-off the light from the bulb. (Comptt. Delhi 2016)
Answer:
It is only the light coming out from a cone of
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 73

Question 82.
A ray of light is incident on a glass prism of refractive index and refractive angle A. If it just suffers total internal reflection at the other face, obtain an expression relating the angle of incidence, angle of prism and critical angle. (Comptt. Delhi 2017)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 74

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 75

Question 83.
(i) Define refractive index of a medium.
(ii) In the following ray diagram, calculate the speed of light in the liquid of unknown refractive index. (Comptt. All India 2017)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 76

Answer:
(i) The refractive index of a medium is defined as the ratio of speed of light in vacuum (c) to the speed of light in the medium (v)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 77

(ii) Given : Width of liquid surface (w) = 30 cm
Depth of liquid (d) = 40 cm
Since it is a critical case of total internal reflection, when refracted ray gazes along the liquid surface.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 78

From the given ray diagram, the length of ray inside liquid is 50 cm
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 79

Ray Optics and Optical Instruments Class 12 Important Questions Short Answer Type SA-II

Question 84.
With the help of a suitable ray diagram, derive the mirror formula for a concave mirror. (All India 2009)
Answer:
Consider a concave mirror of focal length/, radius of curvature R receiving light from an object AB placed between F and C as shown in the figure. The image will be formed as shown in the ray diagram.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 80

Using Cartesian sign convention, we find
Object distance, BP = – u
Image distance B’P = – v
Focal length, FP = – l
Radius of curvature, CP = -R = -2f
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 81

This proves the mirror formula for a concave mirror.

Question 85.
Three light rays red (R), green (G) and blue (B) are incident on a right B angled prism ‘abc’ Q at face ‘ab’. The R refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Out of the three which colour ray will emerge out of face ‘ac’? Justify your answer. Trace the path of these rays after passing through face ‘ab’. (Delhi 2009)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 214

Answer:
Critical angle ic for total internal reflection is related to refractive index µ as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 82

Critical angle for :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 83

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 84

Incident angle in the surface ac is 45° for all the three colours. So red colour will undergo refraction while the other two colours will undergo total internal reflection in a.c. It is indicated in the figure. All the three colours will undergo total internal reflection if they are incident normally on one of the faces of an equilateral prism as shown in Figure 3. This is due to the reason that the incident angle on the second surface will be greater than critical angle for all the colours.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 85

Question 86.
(i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working.
(ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment? (All India 2009)
Answer:
(i) Magnifying power m = Image may be NSFW.
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-\frac{f_{0}}{f_{e}}
. It does not change with increase of aperature of objective lens, because focal length of a lens has no concern with the aperature of lens.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 19

(ii) Drawbacks :

  • Images formed by these telescopes have chromatic aberrations.
  • Lesser resolving power.
  • The image formed is inverted and faintes.

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 86

Question 87.
(i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working.
(ii) Why must both the objective and the eye-piece of a compound microscope have short focal lengths? (All India 2010)
Answer:
(i)
(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

(i) The focal length of the objective lens has no effect on the resolving power of microscope.
(ii) When the wavelength of light is increased, the resolving power of a microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

(ii) The magnifying power of a compound microscope is given by,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 87

Angular magnification (m0) of objective will be large when u0 is slightly greater than f0. Since microscope is used for viewing very close objects, so u0 is small. Consequently f0 has to be small.
Moreover, the angular magnification (me) of the eyepiece will be large if f0 is small.

Question 88.
An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object. (All India 2010)
Answer:
According to the question,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 88

Nature of lens : Convex lens of focal length 20 cm is required.

Question 89.
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances. (All India 2010)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 89

Putting the value of u in eqn (i), we get v = – 60 cm
∴ Object distance, u = 15 cm and
Image distance, v = 60 cm.

Question 90.
A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. If the magnification is 19, find the focal length of the lens. (All India 2010)
Answer:
Given : u = -10 m, m = 19
For real image m = -19
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 90

Question 91.
Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.
Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation. (Delhi 2011)
Answer:
Ray diagram : The minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base, we have
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 91

Question 92.
Use the mirror equation to show that
(a) an object placed between / and 2/ of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. (All India 2011)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 92

This implies that v < 0, formed on left. Also the above inequality implies 2f > v
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 93

i.e., the real image is formed beyond 2f.
(b) For a convex mirror, f > 0 and for an object on left, u < 0. From the mirror formula, Image may be NSFW.
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\frac{1}{v}=\frac{1}{f}-\frac{1}{u}
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\frac{1}{v}
> 0 or v > 0 v
This shows that whatever be the value of u, a convex mirror forms a virtual image on the right.
(c) From mirror formula : Image may be NSFW.
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\frac{1}{v}=\frac{1}{f}-\frac{1}{u}

For a concave mirror, f < 0 and for an object located between the pole and focus of a concave mirror, f < u < 0
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 94

i.e., image is enlarged

Question 93.
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope. (All India 2011)
Answer:
Given f0 = 4 cm, fe = 10 cm, u0 = -6 cm
Magnifying power of microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 95

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 96

Question 94.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 106 m. (All India 2011)
Answer:
Given : f0 = 15 m, fe = 1.0 cm = 0.01 m
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 97

(ii) Let d be the diameter of the image in metres Then angle subtended by the moon will be
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 98

Angle subtended by the image formed by the objective will also be equal to a and is given by
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 99

Question 95.
A convex lens made up of glass of refractive index 1.5 is dipped, in turn,
(i) a medium of refractive index 1.6,
(ii) a medium of refractive index 1.3.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media? (All India 2011)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 100

Let fair be the focal length of the lens in air
According to lens maker formula :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 101

As the sign of fB is opposite to that of fair, the lens will behave as a diverging lens.
(b) When lens is dipped in medium B :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 102

As the sign of fB is same as that of /air, the lens will behave as a converging lens.

Question 96.
A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length. (All India 2011)
Answer:
According to lens maker formula :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 103

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 104

New focal length, fw = 52 cm

Question 97.
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
(i) a medium of refractive index 1.65,
(ii) a medium of refractive index 1.33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media? (All India 2011)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 105

Hence the lens will behave as a diverging lens.
(ii) When lens is dipped in medium B : Image may be NSFW.
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a_{\mu_{\mathrm{B}}}=1.33

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 106

Hence the lens will behave as a converging lens.

Question 98.
You are given three lenses L1 L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus T of L3. Find the separations between L1, L2 and L3.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 107

Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 108

This shows that for lens L2 the object should be at focus of that lens.
Hence the distance between L1 and L2 = v1 + f2 = 40 + 20 = 60 cm
It clearly indicates that the distance between L2 and L3 can have any value.

Question 99.
You are given three lenses L1 L2 and L3 each of focal length 15 cm. An object is kept at 20 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 109

Answer:
Let f1 f2 and f3 be the focal length of three lenses.
For lens L1 : u = 20 cm
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 110

It shows that lens infinite.
Hence for lens L1, image is formed at a distance of 15 cm from L2
∴ the focus of L2 i.e. u2 = 15 cm
Now, to calculate the distance between L1 and L2,
u1 + H2 = 60 + 15 = 75 cm
Distance between L2 and L3 = v2 + v3 = ∞ or can be any value.

Question 100.
A fish in a water tank sees the outside world as if it (the fish) is at the vertex of a cone such that the circular base of the cone coincides with the surface of water. Given the depth of water, where fish is located, being ‘h’ and the critical angle for water-air interface being ‘ic‘, find out by drawing a suitable ray diagram the relationship between the radius of the cone and the height ‘h’. (Comptt. Delhi 2012)
Answer:
Let the fish beat point B, and OA is the base of the water
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 111

Question 101.
Draw a ray diagram to show the formation of the image of an object placed on the axis of a convex refracting surface, of radius of curvature ‘R’, separating the two media of refractive indices “n1 and ‘n2‘ (n2 > n1). Use this diagram to
deduce the relation Image may be NSFW.
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\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{\mathbf{R}}
, where u and v represent respectively the distance of the object and the image formed. (Comptt. Delhi 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 112

Suppose all the rays are paraxial
Then the angles i, r, a, P and y will be small
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 113

Using new Cartesian sign convention, we find
Object distance, OP = – u,
Image distance, PI = tv
Radius of curvature, PC = + R
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 114

Question 102.
Answer the following:
(i) Do the frequency and wavelength change when light passes from a rarer to a denser medium?
(ii) Why is the value of the angle of deviation for a ray of light undergoing refraction through a glass prism different for different colours of light? (Comptt. Delhi 2012)
Answer:
(i) Frequency remains same.
While wavelength changes by λ/m
(ii) Deviation produced by small angles prism,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 115

Question 103.
Define power of a lens. Write its S.L units. Two thin convex lenses of focal lengths f1 and f2 are placed in contact coaxially. Derive the expression for the effective focal length of the combination. (Comptt. Delhi 2012)
Answer:
Power of a lens is defined as the ability to converge a beam of light facing on the lens
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 116

Its S.I. unit is dioptre (D)
Let C1, C2 be the optical centres of two thin convex lenses L1 and L2 held co-axially in contact with each other in air.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 117

Suppose f1 and f2 are their respective focal lengths Let a point object O be placed on the common principal axis at a distance OC1 = u
The lens L1 alone would form its image at I’ where C1I’ = v’
From the lens formula for L1,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 118

I’ would serve as a virtual object for lens L2, which forms a final image I at a distance of
C1I = v
As the lenses are thin, therefore, for the lens L2,
u = i2I’ = C1I’ = v’
From the lens formula for L2,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 119

Adding equations (i) and {ii), we get
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 120

Let the two lenses be replaced by a single lens of focal length f which forms image I at distance v, of an object at distance u from the lens
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 121

Question 104.
Draw a ray diagram showing the path of a ray of light entering through a triangular glass prism. Deduce the expression for the refractive index of glass prism in terms of the angle of minimum deviation and angle of the prism. (Comptt. All India 2012)
Answer:
A ray PQ is incident on the face AB of prism at ∠i and effected along QR at ∠r. The angle of incidence (from glass to air) to the second face is Y and the angle of refraction or emergence is i’.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 122

The angle between the emergent ray RS and incident ray in the direction PQ is called the angle of deviation δ.
In the quadrilateral AQNR,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 123

Question 105.
Draw a ray diagram showing the image formation by a compound microscope when the final image is formed at the near point.
Define the resolving power of a microscope. Write two factors by which resolving power can be increased. (Comptt. All India 2012)
Answer:
(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

(i) The focal length of the objective lens has no effect on the resolving power of microscope.
(ii) When the wavelength of light is increased, the resolving power of a microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

Definition of resolving power: Resolving power of compound microscope is defined as reciprocal of the smallest distance between two point objects at which they can be just resolved when seen through microscope.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 124

It can be increased by

  • increasing the diameter of objective lens upto a certain limit; “
  • using light of smaller wavelength.
  • having a medium of higher refractive index.

Question 106.
Draw a ray diagram to show the formation of image by an astronomical telescope when the final image is formed at the near point. Answer the following, giving reasons:
(i) Why the objective has a larger focal length and a larger aperture than the eyepiece?
(ii) What would be the effect on the resolving power of the telescope if its objective lens is immersed in a transparent medium of higher refractive index?(Comptt. All India 2012)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 22

  1. Objective of larger focal length and large aperture gives higher magnification.
  2. Resolving power will increase.

Question 107.
Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it.
Write two important limitations of a refracting telescope over a reflecting type telescope. (All India 2013)
Answer:
Refracting telescope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 125

Magnifying power. It is defined as the ratio of angle (β) subtended by the final image on the eye to the angle (α) subtended by object on eye.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 126

Limitations of refracting telescope over a a reflecting type telescope :

  1. It suffers from chromatic aberration due to refraction and hence the image obtained is multicoloured and blurred.
  2. As a lens of large apparatus can’t be manufactured easily, its light gathering power is low and hence can’t be used to see faint stars.

Question 108.
A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of 80 cm. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of the refractive index of water to be 4/3. (Comptt. Delhi 2013)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 127

Question 109.
(a) A small telescope has an objective lens of focal length 140 cm and an eye-piece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when
(i) the telescope is in normal adjustment,
(ii) the final image is formed at the least distance of distinct vision.
(b) Also find the separation between the objective lens and the eye-piece. (Comptt. All India 2013) Answer:
(a) (i) For normal adjustment :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 128

(ii) When the final image is formed at the least distance of distinct vision,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 129

(b)
(i) For normal adjustment separation = (f0 + fe)
(ii) For least distance of distinct vision

Question 110.
An equiconvex lens of refractive index µ1, focal length ‘f’ and radius of curvature ‘R’ is immersed in a liquid of refractive index µ2. For
(i) µ2 µ1, and
(ii) µ2 < µ1, draw the ray diagrams in the two cases when a beam of light coming parallel to the principal axis is incident on the lens. Also find the focal length of the lens in terms of the original focal length and the refractive index of the glass of the lens and that of the medium.
(Comptt. All India 2013)
Answer:
(i) The ray diagrams are as shown below:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 130

Question 111.
(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and the eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece. (Delhi 2014)
Answer:
(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

(i) The focal length of the objective lens has no effect on the resolving power of microscope.
(ii) When the wavelength of light is increased, the resolving power of a microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 131

Hence Focal length of objective = 3.5 cm and Focal length of eyepiece = 5 cm

Question 112.
(a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is not uniform.
(b) Suppose the lower half of the concave mirror’s reflecting surface is covered with an opaque material. What effect will this have on the image of the object? Explain (Delhi 2014)
Answer:
(a) The formation of image is shown in the
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 132

The magnification is not uniform and image is distorted because of the reason that the parts of mobile are situated at different distances from the mirror.
(b) When lower half of the concave mirror’s reflecting surface is covered with an opaque material, the image will be of the whole object, i.e. mobile. However, as the area of the reflecting surface has been reduced, the intensity of image (brightness) will be reduced to half.

Question 113.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 133

The image I1 is formed behind the mirror and acts as a virtual object for the convex mirror; and finally image I2 is formed, which is virtual between focus and pole of mirror.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 134

Question 114.
A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image. (All India 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 135

Final image is real and lies between F and C of concave mirror.

Question 115.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm apart. A point object is placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw the ray diagram showing the image formation. (All India 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 136

The final image is virtual and lies between pole and focus of convex mirror.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 137

Question 116.
A ray PQ is incident normally on the face AB of a triangular prism of refracting angle of 60°, made of a transparent material of refractive index Image may be NSFW.
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2 / \sqrt{3}
, as shown in the figure. Trace the path of the ray as it passes through the prism. Also calculate the angle of emergence and angle of deviation. (Comptt. Delhi 2014)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 138

Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 139

Angle of incidence at face AC of the prism = 60°
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 140

Hence, refracted ray grazes the surface AC as ic = i
∴ Angle of emergence = 90°
and Angle of deviation = 30°

Question 117.
(i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 m. (Delhi 2015)
Answer:
(i) Let : f0 = focal length of objective lens = 15 cm
fe = focal length of eye lens = 1.0 cm
Angular magnification (m)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 141

Question 118.
Define the term ‘critical angle’ for a pair of media.
A point source of monochromatic light ‘S’ is kept at the centre of the bottom of a cylinder of radius 15.0 cm. The cylinder contains water (refractive index 4/3) to a height of 7.0 cm. Draw the ray diagram and calculate the area of water surface through which the light emerges in air.
Answer:
(a) Critical Angle : For an incident ray, travelling from an optically denser medium to optically rarer medium, the angle of incidence, for which angle of refraction is Image may be NSFW.
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90^{\circ}
, is called the critical Angle.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 142

∴ The Area of water surface through which the light emerges in air is 63π m2

Question 119.
Which two of the following L1, L2 and L3 will you select as objective and eyepiece for constructing best possible
(i) telescope
(ii) microscope? Give reason to support your answer. (Comptt. Delhi 2015)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 143

Answer:
(i) Telescope : L2 : objective, L3 = eyepiece
Reason : Light gathering power magnifying power will be larger.

(ii) Microscope : L3 : objective, L1 = eyepiece
Reason : Angular magnification is more for short focal length of objective and eyepiece.

Question 120.
(a) Write the factors by which the resolving power of a telescope can be increased.
(b) Estimate the angular separation between first order maximum and third order minimum of the diffraction pattern due to a single slit of width 1 mm, when light of wavelength 600 nm is incident normal on it. (Comptt. All India 2015)
Answer:
(a) Factors for increasing the resolving power of telescope : The resolving power of a telescope is given by,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 144

(iii) Resolving power is independent of focal length of objective lens.

(b) Given : a = 1 mm = 1 × 10-3m,
λ = 600 nm = 600 × 10-9m
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 145

Putting the value of λ and a, we have
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 146

Question 121.
(a) Calculate the distance of an object of height ‘h’ from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also.
(b) Using mirror formula, explain why does a convex mirror always produce a virtual image. (Delhi 2016)
Answer:
(a) Given : R = -20 cm and magnification
m = -2
Focal length of the mirror f = Image may be NSFW.
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\frac{\mathrm{R}}{2}
= -10 cm
Magnification (m) = Image may be NSFW.
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-\frac{v}{u}

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 147

∴ f is positive and u is negative,
⇒ v is always positive, hence image is always virtual.

Question 122.
Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eye-piece. Write its two important advantages over a refracting telescope. (Delhi 2016)
Answer:
Reflecting telescope. Telescope with mirror objectives is called reflecting telescope. This is also known as Cassegrain telescope / Newtonian telescope. The ray diagram of reflecting type telescope is shown in figure.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 48.

Advantage over refracting telescope :

  1. Since reflecting telescope has mirror objective, so the image formed is free from chromatic aberration.
  2. Since the spherical mirrors are parabolic mirrors, free from spherical aberration, they produce a very sharp and distinct image.

Question 123.
Explain the following, giving reasons :
(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.
(ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?
(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light? (All India 2016)
Answer:
(i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency of reflected and refracted light both have the same frequency as the incident frequency.

(ii) No, energy carried by a wave depends on the amplitude of the wave and not on the speed of wave propagation.

(iii) For a given frequency, intensity of light in the photon picture is determined by the number of photons incident normally on crossing a unit area per unit time.

Question 124.
A convex lens, of focal length 25 cm, and a convex mirror, of radius of curvature 20 cm, are placed co-axially 40 cm apart from each other. An incident beam, parallel to the principal axis, is incident on the convex lens. Find the position and nature of the image formed by this combination. (Comptt. All India 2016)
Answer:
The given ‘setup’ is as shown
The object, being at infinity, the image formed by the convex lens, is at its focus, i.e. 25 cm from the lens. This image becomes the ‘object’ for convex mirror.
Now, for the convex mirror
Object distance = (40 – 25) cm = 15 cm
Radius of curvature, R = 20 cm
u = -15 cm, R = +20 cm
Using the mirror formula :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 148

The final image is, therefore, a vitrual image that appears to be formed (behind the convex mirror) at a distance of 6 cm from it.

Question 125.
(i) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 149

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC. (Delhi 2016)
Answer:
(i) Given : For equilateral glass prism ∠A = 60°, ∠δm = 30°
We know
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 150

At face AC, let the angle of incidence be r2 For grazing ray e = 90°
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 151

Question 126.
(a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. If p for water is 1.33, find the wavelength, frequency and speed of the refracted light.
(b) A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm. (All India 2017)
Answer:
(a) Given : Wavelength of monochromatic
light, λ = 589 nm = 589 × 10-9m,
Refractive index of water, µ = 1.33,
Speed of light in air, c = 3 × 108 ms-1
Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the emergent ray in water will be equal to the frequency of the incident or emergent light in air.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 152

Speed of light in water is related to the refractive index of water, as :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 153

Wavelength of light in water is given by the relation,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 154

Question 127.
(a) Draw a ray diagram depecting the formation of the image by an astronomical telescope in normal adjustment.
(b) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? Give reason. (All India 2017)

LensesPower (D)Aperture (cm)
L138
L261
L3101

Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 155

(b) Objective Lens : Lens L1
Eyepiece Lens : Lens L2
Reason : The objective should have large aperture and large focal length; while the eyepiece should have small aperture and small focal length.

Question 128.
(a) Draw a ray diagram showing the formation of image by a reflecting telescope.
(b) Write two advantages of a reflecting telescope over a refracting telescope. (All India 2017)
Answer:
Reflecting telescope. Telescope with mirror objectives is called reflecting telescope. This is also known as Cassegrain telescope / Newtonian telescope. The ray diagram of reflecting type telescope is shown in figure.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 48.

Advantage over refracting telescope :

  1. Since reflecting telescope has mirror objective, so the image formed is free from chromatic aberration.
  2. Since the spherical mirrors are parabolic mirrors, free from spherical aberration, they produce a very sharp and distinct image.

Question 129.
(a) Draw a ray diagram for the formation of image by a compound microscope.

LensesPower (D)Aperture (cm)
L138
L261
L3101

(b) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?
(c) Define resolving power of a microscope and write one factor on which it depends.
(All India)
Answer:
(a) Compound microscope :

(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Objective : Lens L3 Eyepiece : Lens L2

(c)
Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

  1. The focal length of the objective lens has no effect on the resolving power of microscope.
  2. When the wavelength of light is increased, the resolving power of a microscope
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    Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

Question 130.
An optical instrument uses eye-lens of power 16 D and objective lens of power 50 D and has a tube length of 16.25 cm. Name the optical instrument and calculate its magnifying power if it forms the final image at infinity. (Comptt. Delhi 2017)
Answer:
The optical instrument is Compound Microscope.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 156

Question 131.
Answer the following questions:
(a) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a vitrual image, we are obviously bringing it on to the ‘screen’ (i.e. the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparatus depth of a tank of water change if viewed obliquely? If so, does the apparatus depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter? (Comptt. Delhi 2017)
Answer:
(a) Rays converging to a point ‘behind’ a plane or convex mirror are reflected to a point in front of the mirror on a screen. In other words, a plane or convex mirror can produce a real image if the object is virtual. Convince yourself by drawing an appropriate ray diagram.

(b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged on to a screen by means of an appropriate converging lens. The convex lens of the eye does just that. The virtual image here serves as an object for the lens to produce a real image. Note, the screen here is not located at the position of the virtual image. There is no contradiction.

(c) Taller

(d) The apparent depth for oblique viewing decreases from its value for near-normal viewing. Convince yourself of this fact by drawing ray diagrams for different positions of the observer.

(e) Refractive index of a diamond is about 2.42, much larger than that of ordinary glass (about 1.5). The critical angle of diamond is about 24°, much less than that of glass. A skilled diamond-cutter exploits the larger range of angles of incidence (in the diamond), 24° to 90°, to ensure that light entering the diamond is totally reflected from many faces before getting out, thus producing a sparkling effect.

Question 132.
An optical instrument uses eye-lens of power 20 D and the objective lens of power 50 D. It has a tube length of 15 cm. Name the optical instrument and calculate its magnifying power if it forms the final image at infinity. (Comptt. Delhi 2017)
Answer:
The optical instrument is Compound Microscope.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 156

Hint  : Compound microscope, m = 37.5

Question 133.
An optical instrument uses eye-lens of power 12.5 D and object lens of power 50 D and has a tube length of 20 cm. Name the optical instrument and calculate its magnifying power, if it forms the final image at infinity. (Comptt. Delhi 2017)
Answer:
The optical instrument is Compound Microscope.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 156

Hint : Compound microscope, m = 31.25

Ray Optics and Optical Instruments Class 12 Important Questions Long Short Answer Type

Question 134.
(a) For a ray of light travelling from a denser medium of refractive index n1 to a rarer medium of refractive index n2, prove that Image may be NSFW.
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\frac{n_{2}}{n_{1}}
, where ic is the critical angle of incidence for the media.
(b) Explain with the help of a diagram, how the above principle is used for transmission of video signals using optical fibres. (Delhi 2008)
Answer:
(a) When i = ic, r = 90°
Using Snell’s law of refraction : nx sin ic = n2 sin 90°
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 157

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 158

(b) When a video signal is directed into an optical fibre at a suitable angle, it undergoes internal reflections repeatedly along the length of the optical fibre and comes out of it with almost neglible loss of intensity.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 159

Question 135.
Derive the lens formula, Image may be NSFW.
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\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
for a concave lens, using the necessary ray diagram. Two lenses of powers 10 D and – 5 D are placed in contact.
(a) Calculate the power of the new lens.
(b) Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 160

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 161

Question 136.
Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n1 and n2. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface.
Hence derive the expression of the lens maker’s formula. (Delhi 2009)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 162

By applying Cartesian sign convention,
OM = -M, MI = -v, MC = +R
Substituting these values in (iii), we get
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 163

This equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of the curvature of the curved spherical surface. It holds for any curved spherical surface.
Lens maker’s formula:

(a) Lens maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1. Here, n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature and R1 and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively. For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 172

For refraction at surface ADC, we can write the relation between the object distance v1, image distance v and radius of curvature R2, as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 173

Question 137.
Draw the labelled ray diagram for the formation of image by a compound microscope.
Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.(Delhi 2009) Answer:
Compound Microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 164

Magnifying power : The magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final virtual image to the angle subtended at the eye by the object, when both are at the least distance of distinct vision from the eye.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 165

As the object AB is placed close to the focus f0 of the objective
∴ u0 = -fo
Also image A’B’ is formed close to the eyelens whose focal length is short, therefore v0 = L = the length of the microscope tube or the distance between the two lenses
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 166

(i) If the focal lengths are less, then their magnifying power will be more.
(ii) To avoid any aberrations in refraction due to larger bend on passing through the eye-piece.

Question 138.
(i) Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object.
(ii) Write three distinct advantages of a reflecting type telescope over a refracting type telescope.
(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system. (All India 2009)
Answer:
(a)
(i) Magnifying power m = Image may be NSFW.
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-\frac{f_{0}}{f_{e}}
. It does not change with increase of aperature of objective lens, because focal length of a lens has no concern with the aperature of lens.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 19

(ii) Drawbacks :

  • Images formed by these telescopes have chromatic aberrations.
  • Lesser resolving power.
  • The image formed is inverted and faintes.

(ii)

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 18

(ii) Advantages of reflecting telescope over a refracting telescope:

  1. Due to large aperture of the mirror used, the reflecting telescopes have high resolving power.
  2. This type of telescope is free from chromatic aberration (formation of coloured image of a white object).
  3. The use of paraboloidal mirror reduces the spherical aberration (formation of non-point, blurred image of a point object).
  4. Image formed by reflecting telescope is brighter than refracting telescope.
  5. A lens of large aperture tends to be very heavy and therefore difficult to make and support by its edges. On the other hand, a mirror of equivalent optical quality weights less and can be supported over its entire back surface.

(b) For convex lens :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 167

Since concave lens is at 5 cm distance, a virtual object for concave lens is said to be at a distance of 10 cm.
For concave lens :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 168

∴ Image will be at infinite distance or the ray will emerge parallel to the axis.

Question 139.
Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope. (Delhi 2010)
Answer:
Compound Microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 169

Magnifying power : The magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final virtual image to the angle subtended at the eye by the object, when both are at the least distance of distinct vision from the eye.

Numerical:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 170

The -ve sign shows that the final image is an inverted image.

Question 140.
(a) Obtain lens makers formula using the expression
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 171

Here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2) is incident on the convex side of spherical refracting surface of radius of curvature R.
(b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed. (Delhi 2011)
Answer:
(a) Lens maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1. Here, n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature and R1 and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively. For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 172

For refraction at surface ADC, we can write the relation between the object distance v1, image distance v and radius of curvature R2, as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 173

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 174

Question 141.
Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eyepiece. (Delhi 2012)
Answer:
The magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the final image formed at the least distance of distance vision to the angle subtended at the eye by the object at infinity, when seen directly
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 175

Angle substended by the 100 m tall tower at 3 km aways is,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 176

Let h be the height of the image of tower formed by the objective.
Then angle subtended by the image produced by the objective will also be equal to h and is given by ‘
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 217

Magnification produced by the eyepiece
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 178

Question 142.
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment. (Delhi 2012)
Answer:

TelescopeMicroscope
1.Resolving power should be higher for certain magni­fication.Resolving power is not so large but the magnification should be higher.
2.Focal length of objective should be kept larger while eyepiece focal length should be small for better magni­fication.Both objective and eyepiece should have less focal length for better magnification.
3.Objective should be of large aperture.Eyepiece should be of large aperture.
4.Distance between objective and eyepiece is adjusted to focus the object at infinity.Distance between objective and eyepiece is fixed, for focusing an object the distance of the objective is changed.

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 179

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 180

Question 143.
Draw a ray diagram showing the formation of the image by a point object on the principal axis of a spherical convex surface separating two media of refractive indices n1 and n2, when a point source is kept in rarer medium of refractive index nv Derive the relation between object and image distance in terms of refractive index of the medium and radius of curvature of the surface.
Hence obtain the expression for Lens-maker’s formula in the case of thin convex lens. (Comptt. Delhi 2014)
Answer:
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 162

By applying Cartesian sign convention,
OM = -M, MI = -v, MC = +R
Substituting these values in (iii), we get
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 163

This equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of the curvature of the curved spherical surface. It holds for any curved spherical surface.
Lens maker’s formula:

(a) Lens maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1. Here, n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature and R1 and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively. For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 172

For refraction at surface ADC, we can write the relation between the object distance v1, image distance v and radius of curvature R2, as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 173

Question 144.
(a) A point object is placed in front of a double convex lens (of refractive index n = n2/n1 with respect to air) with its spherical faces of radii of curvature R1 and R2. Show the path of rays due to refraction at first and subsequently at the second surface to obtain the formation of the real image of the object.
Hence obtain the Lens-maker’s formula for a thin lens.
(b) A double convex lens having both faces of the same radius of curvature has refractive index 1-55. Find out the radius of curvature of the lens required to get the focal length of 20 cm. (Comptt. All India)
Answer:
(a) Lens-maker’s formula. The image of point object O is formed in two steps. The first reflecting surface ‘c1 forms the image I1 of the object O. The image I1 acts as a virtual object for formation of image I by the second surface ‘c2‘.
For the first surface ABC ‘c1‘,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 181

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 182

Question 145.
(a) Draw a labelled ray diagram showing the image formation of a distant object by a refracting telescope.
Deduce the expression for its magnifying power when the final image is formed at infinity.
(b) The sum of focal lengths of the two lenses of a refracting telescope is 105 cm. The focal length of one lens is 20 times that of the other. Determine the total magnification of the telescope when the final image is formed at infinity. (Comptt. All India 2015)
Answer:
(a) Refracting telescope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 125

Magnifying power. It is defined as the ratio of angle (β) subtended by the final image on the eye to the angle (α) subtended by object on eye.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 126

Limitations of refracting telescope over a a reflecting type telescope :

  1. It suffers from chromatic aberration due to refraction and hence the image obtained is multicoloured and blurred.
  2. As a lens of large apparatus can’t be manufactured easily, its light gathering power is low and hence can’t be used to see faint stars.

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 183

Question 146.
(a) A ray ‘PQ’ of light is incident on the face AB of a glass prism ABC (as shown in the figure) and emerges out of the face AC. Trace the path of the ray. Show that
∠i + ∠e = ∠A + ∠δ
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 184

where δ and e denote the angle of deviation and angle of emergence respectively.
Plot a graph showing the variation of the angle of deviation as a function of angle of incidence. State the condition under which ∠δ is minimum.
(b) Find out the relation between the refractive index (µ) of the glass prism and ∠A for the case when the angle of prism (A) is equal to the angle of minimum deviation (δm). Hence obtain the value of the refractive index for angle of prism A = 60°. (All India 2015)
Answer:
(a) The figure shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and rv while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation δ.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 185

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 186

From the triangle
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 187

Comparing these two equations, we get
r1 + r2 = A
The total deviation 8 is the sum of deviations at the two faces,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 188

A plot between the angle of deviation and angle of incidence is shown in the figure. In general, any given value of δ, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in equation (i) above, i.e., δ remains the same if i and e are. interchanged. Physically, this is related to the fact that the path of ray in the diagram above can be traced back, resulting in the same angle of deviation. At the minimum deviation dm, the refracted ray inside the prism becomes parallel to its base. We have,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 189

When angle of incidence (i) and angle of emergence (e) are equal, ie,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 190

Question 147.
(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain expression for total magnification when the images is formed at infinity.
(b) How does the resolving power of a compound microscope get affected, when
(i) focal length of the objective is decreased.
(ii) the wavelength of light is increased? Give reasons to justify your answer.
Answer:
(a) Ray diagram of a compound microscope : A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 191

Magnification due to a compound microscope.
The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 192

Here h’ is the size of the first image, the object size being h and f0 being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for the simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 193

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 215

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 194

(b) Resolving power of a microscope :
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 195

(i) The focal length of the objective lens has no effect on the resolving power of microscope.
(ii) When the wavelength of light is increased, the resolving power of a microscope
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 196

Question 148.
(a) A point object ‘O’ is kept in a medium of refractive index n1 in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index n2 from the first one, as shown in the figure.
Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n1 n2 and R.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 197

(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n2 from n2 (n2 > n1), draw this ray diagram and write the similar [similar to (a)] relation. Hence obtain the expression for the lens maker’s formula. (Delhi 2015)
Answer:
(a) For small angles
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 198

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 199

This equation gives us a relation between object and image distance interms of refractive index of the medium and the radius of the curvature of the curved Spherical surface. It holds for any curved Spherical surface.

(b) Len’s maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1.

Here n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature R1and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively.

For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 200

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 201

Question 149.
(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light? What is its cause?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in figure. What must be the minimum value of refractive index of glass? calculations.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 202

Answer:
(i)
(a) The figure shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and rv while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation δ.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 185

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 186

From the triangle
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 187

Comparing these two equations, we get
r1 + r2 = A
The total deviation 8 is the sum of deviations at the two faces,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 188

A plot between the angle of deviation and angle of incidence is shown in the figure. In general, any given value of δ, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in equation (i) above, i.e., δ remains the same if i and e are. interchanged. Physically, this is related to the fact that the path of ray in the diagram above can be traced back, resulting in the same angle of deviation. At the minimum deviation dm, the refracted ray inside the prism becomes parallel to its base. We have,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 189

When angle of incidence (i) and angle of emergence (e) are equal, ie,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 190

(ii) The white light disperses when passed through a prism, because the refractive index of the glass of the prism is different for different wavelength (colours). Hence, different colours get bent along different directions.

(iii) For total internal reflection, Zi > Zi (critical angle)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 203

Hence, the minimum value of refractive index must be Image may be NSFW.
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\sqrt{2}
= 1.41

Question 150.
(i) Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2.
Hence, derive lens maker’s formula.
(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed?
Answer:
(a) For small angles
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 198

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 199

This equation gives us a relation between object and image distance interms of refractive index of the medium and the radius of the curvature of the curved Spherical surface. It holds for any curved Spherical surface.

(b) Len’s maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1.

Here n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature R1and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively.

For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 200

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 201

⇒ v = 100 cm a real image on the other side, 100 cm away from the surface.

Question 151.
(a) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
(b) You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.
(i) Which lenses should be used as objective and eyepiece? Justify your answer.
(ii) Why is the aperture of the objective preferred to be large? (All India 2016)
Answer:
(a)
(i) Magnifying power m = Image may be NSFW.
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-\frac{f_{0}}{f_{e}}
. It does not change with increase of aperature of objective lens, because focal length of a lens has no concern with the aperature of lens.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 19

(ii) Drawbacks :

  • Images formed by these telescopes have chromatic aberrations.
  • Lesser resolving power.
  • The image formed is inverted and faintes.

(b) Definition : It is the ratio of the angle subtended at the eye, by the final image, to the angle which the object subtends at the lens, or the eye.
(i) Objective = 0.5 D and eye lens = 10 D
∵ This choice would give higher
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 204

(ii) High resolving power/Brighter image/ lower limit of resolution.

Question 152.
(a) A point object, O is on the principal axis of a spherical surface having a radius of curvature, R. Draw a diagram to obtain the relation between the object and image distances, the refractive indices of the media and the radius of curvature of the spherical surface.
(b) Write the Lens Maker’s formula and use it to obtain the range of values of µ (the refractive index of the material of the lens) for which the focal length of an equiconvex lens, kept in air, would have a greater magnitude than that of the radius of curvature of its two surfaces. (Comptt. Delhi 2016)
Answer:
(a) For small angles
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 198

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 199

This equation gives us a relation between object and image distance interms of refractive index of the medium and the radius of the curvature of the curved Spherical surface. It holds for any curved Spherical surface.

(b) Len’s maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n1.

Here n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature R1and R2 be the radii of curvature of the two lens surfaces ABC and ADC, respectively.

For refraction at surface ABC, we can write the relation between the object distance u, image distance v1 and radius of curvature R1 as
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 200

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 201

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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 205

Hence required range is 1.0 < µ < 1.5

Question 153.
The relation, between the angle of incidence
(i) and the corresponding, angle of deviation (δ), for a certain optical device, is represented by the graph shown in the figure. Identify this device. Draw a ray diagram for this device and use it for obtaining an expression for the refractive index of the material of this device in terms of an angle characteristic of the device and the angle, marked an 8m, in the graph.(Comptt. All India 2016)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 206

Answer:

  1. The device identified is a ‘PRISM’.
  2. For ray diagram and derivation :

(a) The figure shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and rv while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation δ.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 185

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 186

From the triangle
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 187

Comparing these two equations, we get
r1 + r2 = A
The total deviation 8 is the sum of deviations at the two faces,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 188

A plot between the angle of deviation and angle of incidence is shown in the figure. In general, any given value of δ, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in equation (i) above, i.e., δ remains the same if i and e are. interchanged. Physically, this is related to the fact that the path of ray in the diagram above can be traced back, resulting in the same angle of deviation. At the minimum deviation dm, the refracted ray inside the prism becomes parallel to its base. We have,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 189

When angle of incidence (i) and angle of emergence (e) are equal, ie,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 190

Question 154.
(a) Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of focal lengths of the lenses.
(b) A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is Image may be NSFW.
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\frac{3}{4}
th of the angle prism. Calculate the speed of light in the prism. (All India 2017)
Answer:
(a) Two thin lenses in contact: Two thin lenses, of focal length f1 and f2 are kept in contact. Let O be the position of object and let u be the object distance. The distance of the image (which is at I1), for the first lens is v1.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 207

This image serves as object for the second lens.
Let the final image be at I, We then have
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 208

Question 155.
(a) Explain with reason, how the power of a diverging lens changes when
(i) it is kept in a medium of refractive index greater than that of the lens,
(ii) incident red light is replaced by violet light.
(b) Three lenses L1 L2, L3 each of focal length 30 cm are placed co-axially as shown in the figure. An object is held at 60 cm from the optic centre of Lens L1. The final real image is formed at the focus of L3. Calculate the separation between
(i) (L3 and L2) and
(ii) (L2 and L3). (Comptt. All India 2017)
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 209

Answer:
(a) The power of a lens is the reciprocal of its focal length.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 210

which means that the power of lens increases on changing to violet light from red light.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 211

Since final image (I3) is formed at the focus of L3, the rays emerging from L2 and incident on L3 have to be parallel to principal axis.
Since the object is placed at a distance of 60 cm from L1 i.e., at 2F; the image will be formed at 2F on the other side of L1 (60 cm).
This image I1 will be at the focus of L2, because rays emerging out from L2 are parallel to principal axis.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 212

(ii) L2L3 can be any distance.

Question 156.
(a) Deduce the expression, by drawing a suitable ray diagram, for the refractive index of triangular glass prism in terms of the angle of minimum deviation (D) and the angle of prism (A).
Draw a plot showing the variation of the angle of deviation with the angle of incidence.
(b) Calculate the value of the angle of incidence when a ray of light incident on one face of an equilateral glass prism produces the emergent ray, which just grazes along the adjacent face. Refractive index of the prism is Image may be NSFW.
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\sqrt{2}
. (Comptt. All India 2017)
Answer:
(a) The figure shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and rv while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation δ.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 185

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 186

From the triangle
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 187

Comparing these two equations, we get
r1 + r2 = A
The total deviation 8 is the sum of deviations at the two faces,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 188

A plot between the angle of deviation and angle of incidence is shown in the figure. In general, any given value of δ, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in equation (i) above, i.e., δ remains the same if i and e are. interchanged. Physically, this is related to the fact that the path of ray in the diagram above can be traced back, resulting in the same angle of deviation. At the minimum deviation dm, the refracted ray inside the prism becomes parallel to its base. We have,
Image may be NSFW.
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 189

When angle of incidence (i) and angle of emergence (e) are equal, ie,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 190

(b) Given µ = Image may be NSFW.
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\sqrt{2}
, ∠i = ?
Since the emergent ray just grazes along the adjacent face of an equilateral glass prism,
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Important Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Class 12 Important Questions 213

Important Questions for Class 12 Physics

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Invitation and Replies Class 12 Format, Examples

Invitation and Replies Class 12 Format, Examples

♦ Invitation
To invite someone for an occasion, we use the written form Invitation.
Invitations are generally printed cards through which we invite our guests on some auspicious occasions like wedding, birthday, wedding anniversary, house warming, the inauguration of a shop/factory, etc.

♦ Invitations are of two types:
(a) Formal and
(b) Informal.

These can be printed on cards or can be drafted in the form of letters.

♦ Main Characteristics:
An invitation is a complete information. It answers the questions: who, whom, when, where, what time and for what. The important components of an invitation, therefore, are:

  • Occasion
  • Name(s) of the invitee(s)
  • Name(s) of the host(s)
  • Date, time and venue.
  • Name(s) of the chief guest or special invitees, in case of an official invitation.

♦ Format of Formal Invitations:
In case of formal invitations, each of the following is written in a separate line with fonts of varying sizes.

  • Names of the hosts
  • Name of the invitee (in case of a formal letter of invitation)
  • Formal phrase of invitation, for example:
    Request the pleasure your benign presence/company Seek your auspicious presence
    Solicit your gracious presence on the auspicious occasion
  • Date, time and venue of the event
  • Occasion/ reason of the invitation.

♦ Characteristics of Formal Invitations:
1. Meant for a lot of invitees:

  • These are written in the third person
  • In case a VIP is invited as the chief guest, the name of the VIP must appear prominently.
  • Name of the invitee is not to be included. The addressee’s address is to be written only on the envelope.
  • Simple present tense is to be used.
  • The date of writing is not to be given.
  • There is no signature of the host.
  • The abbreviation RSVP (French: repondez silvers plait) i.e. ‘Please reply’ is written below on the left side with name(s), address and phone number of the host(s).
  • Put the invitation into a box.
  • Do not exceed 50 words.

2. Meant for an individual (a formal letter of invitation)

  • Include the name of the invitee.
  • These are to be written on run-on lines. The sentence is not broken into different words/phrases.
  • Other details are similar to the mass-scale invitations.

♦ Writing Informal Invitations:
Informal

  • Written in a letter form, in an informal format. Such letters are very persuasive in nature.
  • Written in the first person.
  • Salutation is ‘dear + name’.
  • Complimentary close ‘Yours sincerely’.
  • Date of writing the invitation is given.
  • Sender’s address appears on the left-hand side.
  • Various tenses used to suit the sense.

♦ Replies

Replies Accepting or Declining
Formal Follow a set formula:

  • formal words: ‘kind invitation’, ‘great pleasure’, ‘regret’, etc.
  • Use third person (‘they’) instead of first person (T, ‘We’)
  • Address of the writer and the date to be written.

♦ Informal – Accepting or Declining

  • Like an ordinary letter
  • Do not use any formal expressions, but use informal words and expressions
  • Use first person (‘I’, ‘We’).

♦ Previous Years’ CBSE Examination Questions

♦ Short Answer Type Questions

Question 1.
As Secretary of the Literary Club of St. Anne’s School, Ahmedabad, draft a formal invitation in not more than 50 words for the inauguration of the club in your school. (Delhi 2009)
Answer:

St. Anne’s School
Ahmedabad

We cordially invite all staff, students and parents for the inauguration of the Literary Club of the School on 29th July between 8.30 a.m. to 11.30 a.m. within the school premises. Well, known novelist, Mr Sandeep Kumar will be our Chief Guest. You all are also requested to join us in the auditorium for tea and snacks after the inauguration ceremony.
Secretary

Question 2.
The literary club of your school is putting up the play ‘Waiting for Godot’. As secretary of the club, draft an invitation inviting the famous writer Sudeesh Gupta to be the guest of honour at the function. Write the invitation in not more than 50 words. You are Govind/Gauri. (All India 2014)
Answer:

ABC School
Shastri Nagar
Mumbai
February 20, 20xx
Mr. Sudesh Gupta
XYZ Lane
PQR Colony
Mumbai
Dear SirSubject: Invitation as Chief Guest
The Literary club of our school is putting up the play ‘Waiting for Godot’ on 17th January, 20xx in our school auditorium from 9-10 am. On behalf of our club, I would like to extend a cordial invitation to you to be the guest of honour at the function. It would be our privilege if you consent to grace this occasion with your esteemed presence.Gauri
Secretary (Literary Club)

Question 3.
As the principal of a reputed college, you have been invited to inaugurate a Book Exhibition in your neighbourhood. Draft a reply to the invitation in not more than 50 words, expressing your inability to attend the function. You are Tarun/Tanvi. (All India 2014)
Answer:

Reply To Invitation:

The Principal
ABC College
March 25th, 20xx

Subject: Inability to accept the Invitation Sir
I would like to express my gratitude to the Civil Lines Book Club for inviting me to inaugurate the first edition of the Civil Club Book Exhibition. But I regret my inability to attend the function due to a prior commitment. I have to attend a meeting with the governing body of our college on the same day for which the inauguration is scheduled.

Yours faithfully
Tanvi
Principal

Question 4.
You have received an invitation to be the judge for a literary competition in St. Ann’s School. Send a reply in not more than 50 words, confirming your acceptance. You are Mohan/Mohini. (All India)
Answer:

Confirming Acceptance:

16, XYZ Lane
ABC Nagar,
Chennai
St. Ann’s School
TVS Nagar
Chennai

Subject: Acceptance of invitation for judging the literary competition.

Sir,
I would like to express my gratitude to St. Ann’s School for inviting me to be the judge for a literary competition in your school. I confirm my acceptance for the same and will make sure that I am present there at the scheduled date and time.

Yours sincerely
Mohini

Question 5.
Sunrise Global School, Agra is going to organize a one-act play competition in the school auditorium. You have decided to invite noted stage artiste, Nalini to grace the occasion. Draft a formal invitation for her in about 50 words. You are Karuna/Karan, Cultural Secretary. (Delhi 2016)
Answer:

The Principal, Staff And Teachers
Of
Sunrise Public School
take pleasure in inviting
noted stage artist
MS. NALINI
to grace the one-act play competition
in the School Auditorium
on Tuesday, 1st April, 20xx
From 8:30 am to 11:00 am RSVP
Awaiting a favourable response from your end.

RSVP
Karuna, Cultural Secretary
987xxxxx00

Question 6.
On 30th November your school is going to hold its annual sports day. You want Mr. Dhanraj Pillai, a noted hockey player to give away the prizes to the budding sportspersons of the school. Write a formal invitation in about 50 words requesting him to grace the occasion. You are Karun^/Karan, Sports Secretary, Sunrise Global School, Agra. (All India)
Answer:

Sunrise Global School
takes pleasure in inviting
Noted Hockey Player
MR, DHANRAJ PILLAI
to grace the occasion of its
Annual Sports Day
and to give away prizes to our budding sportspersons
on 30th November, 20xx
From 8:00 am to 12:00 pm
Awaiting a favourable response from your end.

Koruna
Sports Secretary

Question 7.
Your friend, P.V. Sathish, has invited you to attend the wedding of his sister, Jaya. You find that you have an important paper of pre-board examination on the day of the wedding. Thus you cannot attend the event. Write in about 50 words a formal reply to the invitation expressing your regret. You are PuneeV Puneeta Vij, M 114, Fort Road, Chennai. (All India 2017)
Answer:

15th January, 20xx
Thank you P.V. Sathish for your kind invitation to the wedding of your sister, Jaya on 23rd January, 20xx. I regret my inability to attend as I have an important pre-board examination on the day of the wedding. Wishing the newly wedded couple a very happy married life. Best wishes Puneet
M-114, Fort Road, Chennai

RBSE Class 10 English Notes

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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः

NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः (सूर्य ही पृथ्वी का आधार है)

पाठपरिचयः सारांशः च

प्रस्तावना :

सूर्य ही प्रकृति का आधार है। ऋग्वेद, उपनिषद् तथा कालिदास व भास आदि महाकवि एवं पण्डित अम्बिकादत्त व्यास सरीखे आधुनिक कवियों ने सूर्य की आध्यात्मिक, लौकिक तथा वैज्ञानिक दृष्टि से महत्ता बताई है। प्रस्तुत पाठ का उद्देश्य भी यह बताने के लिए है कि सूर्य किस प्रकार प्रकृति का आधार है।

पाठ-सन्दर्भ :

प्रस्तुत पाठ पण्डित अम्बिकादत्त व्यास के गद्य उपन्यास ‘शिवराजविजय’ के प्रथम अध्याय से लिया गया है। शिवाजी महाराज के गुरु सेवा में तत्पर एक शिष्य अपनी पर्णकुटी से बाहर निकलकर उदित होते हुए तथा चमकते हुए सूर्य को प्रणाम करता हुआ उसकी महिमा का वर्णन करता है। यह पाठ निश्चित रूप से न केवल ज्ञानवर्द्धक ही है अपितु भाषा के सौन्दर्य का भी उत्कृष्ट उदाहरण है।

पाठ-सार :

क्या आप जानते हैं कि हमारी सृष्टि का आधार क्या है? किसके चारों ओर यह पृथ्वी नित्य भ्रमण कर रही है? निश्चय ही यह सूर्य ही है। आधार के बिना पृथ्वी आकाश में कैसे ठहरती है? कैसे ऋतुओं का तथा दिन-रात का परिवर्तन होता है। वनस्पतियों का तथा औषधियों का भी सूर्य के बिना अस्तित्व समाप्त हो जाता है। वेदों तथा उपनिषदों में सर्वत्र ही सूर्य की महिमा का वर्णन है। शिवराजविजय के प्रारम्भ में भी रम्य, अद्भुत, विज्ञानमय वर्णन उपलब्ध होता है जो यहाँ इस पाठ में दिया गया है।

मूलपाठः, शब्दार्थः, भावार्थः, सरलार्थश्च

1. अरुण एष प्रकाशः पूर्वस्यां भगवतो मरीचिमालिनः। एष भगवान् मणिराकाश-मण्डलस्य, चक्रवर्ती खेचरचक्रस्य, कुण्डलम् आखण्डलदिशः, दीपकः ब्रह्माण्डभाण्डस्य, प्रेयान् पुण्डरीकपटलस्य, शोकविमोकः कोकलोकस्य, अवलम्बो रोलम्बकदम्बस्य, सूत्रधारः सर्वव्यवहारस्य, इनश्च दिनस्य।

शब्दार्थः, पर्यायवाचिशब्दाः टिप्पण्यश्चः- अरुणः- रक्तवर्णः, लाल, अरुण (पुं०, प्रथमा विभक्तिः, एकवचनम् विशेषणपदम्। एष- एषः, यह, एतत् (पुं०), प्रथमा विभक्तिः , एकवचनम्। प्रकाश:- ज्योतिः, प्रकाश, प्र० वि०, ए०व०। पूर्वस्याम्-पूर्वदिशायाम्, प्राच्याम् दिशि, पूर्व दिशा में, पूर्व (स्त्री०), सप्तमी विभक्ति, एकवचनम्। भगवतः- ऐश्वर्य सम्पन्नस्य, भगवत्, षष्ठी विभक्तिः , एकवचनम्, विशेषणम्, भगवान् (ऐश्वर्य सम्पन्न) का। मरीचिमालिनः- मरीचिमालिन्, षष्ठी विभक्तिः, एकवचनम्, सूर्यस्य, सूर्य का, मरीचिनां किरणानां माला यस्य सः, मरीचिमालिन्, मरीचिमाला + इन्। मणिः (संज्ञा), मणिः, पुं०, प्रथमा विभक्तिः , एकवचनम्, द्युतिमत् रत्नम्, चमकता हुआ रत्न। आकाशमण्डलस्य- आकाशवृत्तस्य, आकाशस्य, आकाश का, आकाश प्रदेश का। चक्रवर्ती- चक्रवर्तिन्, प्र०वि०, ए०व०, सम्राट, राजचक्रं राजसमूहः यस्य सः। खेचर-चक्रस्य-खे चरति इति खेचरम् नक्षत्रम् उपपद तत्पुरुषः, खेचराणाम् चक्रम् समूहः, षष्ठी तत्पुरुषः, नक्षत्रमण्डलस्य, नक्षत्रमण्डल का।

कुण्डलम् (संज्ञा) – कुण्डल, नपुं०, प्रथमा विभक्तिः, एकवचनम्, कर्णाभूषणम्। आखण्डल-दिशः (संज्ञा)-आखण्डनस्य इन्द्रस्य दिशः, पूर्वदिशः, इन्द्र की दिशा का, पूर्ण दिशा का। दीपकः-दीपः, दीपयति (जगत्) इति, दीपक। ब्रह्माण्ड-भाण्डस्य-ब्रह्माण्डम् एव भाण्डम् तस्य, विश्व-सदनस्य, विश्वरूपी सदन (घर) का। प्रेयान्- प्रियतरः, प्रियः। पुण्डरीक-पटलस्य- पुण्डरीकाणां पटलम्, षष्ठी तत्पुरुषः, तस्य, कमलसमूहस्य, कमलों के समूह (कुल) का। शोकविमोकः-शोकस्य विमोकः, षष्ठी तत्पुरुषः, चिन्ताहरः, शोक-विनाशक, शोक (चिन्ता) को हरनेवाला। कोकलोकस्यकोकानाम् चक्रवाकानाम् लोकस्य समूहस्य, चकवों के समूह का। अम्बलम्बः- आश्रयः, सहारा। रोलम्ब-कदम्बस्य- रोलम्बानां भ्रमराणां कदम्बस्य समूहस्य, भौंरों के समूह का। सूत्रधारः- सूत्रम् धारयति इति, सूत्र को धारण करनेवाला। व्यवस्थापकः-प्रबन्धक। सर्व-व्यवहारस्य-सर्वस्य व्यवहारस्य कार्यजातस्य, सब कार्यों का। इनश्च-इनः + च, इनः (संज्ञा)- इन, पुं०, प्रथमा विभक्तिः, एकवचनम् स्वामी। दिनस्य- दिन का।

भावार्थ:- भगवान् सूर्य आकाश, नक्षत्र, पूर्वदिशा, ब्रह्माण्ड, चक्रवाक-भ्रमर-समूह, व्यवहार तथा दिन के प्राण हैं, ज्योति हैं, जीवन हैं, आनन्द हैं तथा व्यवस्थापक एवं स्वामी हैं।
सरलार्थ- पूर्व दिशा में भगवान् सूर्य का यह लाल प्रकाश है। ये भगवान् (सूर्य) आकाशमण्डल के रत्न हैं, नक्षत्र समूह के सम्राट हैं, इन्द्र की पूर्व दिशा के कर्णाभूषण (कुण्डल) हैं, ब्रह्माण्ड रूपी घर के दीपक हैं, चकवों के समूह के प्रिय हैं, भौंरों के समूह के आश्रय हैं। सब व्यवहारों के व्यवस्थापक हैं तथा दिन के स्वामी हैं।

2. अयमेव अहोरात्रं जनयति। अयम् एव वत्सरं द्वादशसु भागेषु विभनक्ति। अयम् एव कारणं षण्णाम् ऋतूणाम्। एष एव अङ्गीकरोति उत्तरं दक्षिणं चायनम्। अनेन एव सम्पादिताः युगभेदाः। अनेन एव कृताः कल्पभेदाः। एनम् एव आश्रित्य भवति परमेष्ठिनः परार्द्धसङ्ख्या। वेदा एतस्य एव वन्दिनः। गायत्री अमुम् एव गायति। धन्य एष कुलमूलं श्रीरामचन्द्रस्य। प्रणम्यः एषः विश्वेषाम्।

शब्दार्थः, पर्यायवाचिशब्दाः टिप्पण्यश्चः- अहोरात्रम्- अहः च रात्रिः स; समाहार द्वन्द्व समासः दिनं च निशा च तम् तयोः समाहार: दिन-रात को। जनयति – जन्, णिच्, लट्, प्र०पु०, ए०व०, उत्पादयति, उत्पन्न करता है। वत्सरम्-वर्षम्, वर्ष को। द्वादशसु- द्वादश-भागेषु, बारह (भागों) में। बारह महीनों में, द्वादश मासेषु। विभनक्ति – वि, भज्, लट्, प्र० पु०, ए०व०, विभाजयति, विभक्त करता है, विभजते। कारणम्- हेतुः, निमित्तम्, कारण। षण्णाम्- षट् सङ्ख्यात्मकानां छह संख्या वाली ऋतुओं) का। ऋतूणां- ऋतु, षष्ठी, बहुवचनम्-वसन्त, ग्रीष्म, वर्षा, शरद्, हेमन्त, शिशिर, प्रत्येक ऋतु दो-दो मास तक रहती है।

पृथ्वी की सूर्य के चारों ओर गति से इनका निर्माण होता है। पृथ्वी का अपने गिर्द घूमने से दिन-रात का निर्माण होता है। कोणार्क के सूर्य मन्दिर में एक विशाल सूर्य के रथ के माध्यम से इसको स्पष्ट किया गया है। अङ्गीकरोति – अङ्ग च्चि कृ, लट् लकार, प्रथम पुरुषः, एकवचनम्, स्वीकरोति, स्वीकार करता है। उत्तरम् अयनम्उत्तरायणम्, मकर संक्रान्ति से कर्क संक्रान्ति तक छह मासों तक का काल उत्तरायण कहलाता है। इसमें दिन अपेक्षाकृत बड़े होते हैं तथा रातें छोटी होती हैं, दक्षिणम् अयनम्- दक्षिणायनम्, कर्क संक्रान्ति से मकर संक्रान्ति का छह मास का काल दक्षिणायन कहलाता है।

इसमें दिन छोटे होते हैं तथा रात्रि बड़ी होती है। इन दोनों में उत्तरायण के काल को श्रेष्ठ माना जाता है। हर एक अयन में तीन-तीन ऋतुएँ होती हैं। अनेन-अनेन सूर्येण, इस सूर्य में द्वारा। युगभेदाः- सत्युग, द्वापर, त्रेता तथा कलियुग-ये चार युग भी सूर्य की अन्य ग्रहों के गिर्द गति के परिणाम से बनते हैं। सम्पादिताः- सम् + पद् + णिच्, क्त, पुं०, प्रथमा वि०, बहुवचनम् रचिताः, बनाए गए हैं। कृताः-कृ + क्त। कल्पभेदाः- कल्पों के भेद अनादि सृष्टि में, सर्ग, प्रतिसर्ग की परम्परा से विविध कल्पों का निर्माण भी सूर्य की कृपा से होता है। परमेष्ठिन:-परमेष्ठिन, षष्ठी विभक्तिः, एकवचनम्।

ब्राह्मण, विधातुः, ब्रह्मा को विधाता की। परार्द्ध-सङ्ख्या- परार्द्धा सङ्ख्या, कर्मधारय समास, अन्तिमा परार्द्धनामा सङ्ख्या, गिनती में सबसे अन्तिम सङ्ख्या। एक के साथ सत्रह बिन्दु लगाकर परार्द्ध सङ्ख्या बनती है। सङ्कल्प पाठ का आरम्भ ब्रह्मणः परार्द्ध (ब्रह्मा की परार्द्ध सङ्ख्या ) से होता है। वेदाः- चतुर्वेदाः, चारों वेद, ऋग्वेद, यजुर्वेद, अथर्ववेद तथा सामवेद। अस्य-सूर्यस्य, इस सूर्य की। वन्दिन:- स्तोतः, स्तुति करनेवाले। गायत्री- गायत्री मन्त्रः, गायत्री मन्त्र, यह सवितृदेव की स्तुति में गाया जाता है। सवितृदेव भी सूर्य ही है। अमुम्- एनम्, इसको। गायति- गै,

लट् लकार, प्र०पु, ए०व०। धन्यः- स्तुत्यः, स्तुति के योग्य। कुलमूलम्- कुलस्य आधारः अधिष्ठातृदेवः, भगवान् राम सूर्यवंशी थे। सूर्य से ही उनकी वंश परम्परा आरम्भ होती है। प्रणम्यः- प्रणाम-योग्यः, प्रणमनीयः, प्रणाम के योग्य। विश्वेषाम्सर्वेषाम् सत्त्व के द्वारा। सबको भगवान् सूर्य को अवश्व प्रणाम करना चाहिए।

सरलार्थ – यह (सूर्य) ही दिन-रात को पैदा करता है। यह ही वर्ष को बारह भागों में बाँटता है। यह ही छः ऋतुओं का कारण है। यह ही दक्षिण और उत्तर इन दानो अयनों को स्वीकार करता है। इसके द्वारा ही युगों के भेदों का सृजन किया गया है। इससे ही कल्पभेदों की रचना हुई है। इसी पर आश्रित होकर ब्रह्मा जी की परार्द्ध संख्या होती है। वेद इसकी वन्दना करनेवाले हैं। गायत्री इसी का गान करती है। धन्य है यह श्रीरामचन्द्र जी के कुल का आधार। यह सबके द्वारा प्रणाम योग्य है।

अनुप्रयोगः

प्रश्न: 1.
अधोलिखितशब्दानां स्थाने पाठे ये शब्दाः प्रयुक्ताः तेषां मेलनं कुरुत’
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 1

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 1

प्रश्न: 2.
कोष्ठकात् शुद्धम् उत्तरं चित्वा लिखत
(i) भगवतः मरीचिमालिनः उदयः कस्यां दिशायां भवति? ……………………….. (पूर्वस्याम्/उत्तरस्याम्/पश्चिमदिशायाम्)
(ii) सूर्यः वर्ष कतिभागेषु विभजते?………………….. (चतुर्भागेषु द्वयोः भागयो: द्वादशभागेषु)
(iii) अहोरात्रं कः जनयति?…………………. (ब्रह्मा/सूर्य:/चन्द्रमा)
(iv) ‘आदित्यो ह वै प्राणः’ इति कस्मिन् उपनिषदि वर्णितम्?………………. (तैत्तिरीये/छान्दोग्ये प्रश्ने)
(v) सूर्यः केषाम् आत्मा?………….. (जड़वस्तूनाम् चेतनानाम्/जड़चेतनानाम्)
उत्तरम्:
(i) पूर्वस्याम्
(ii) द्वादशभागेषु
(iii) सूर्यः
(iv) तैत्तिरीये
(v) जड़चेतनानाम्

प्रश्न: 3.
अधोलिखितानि पदानि कस्य पदस्य पर्यायवाचिनः इति पाठात् चित्वा कोष्ठके लिखतयथा-
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 3

उत्तरम्:
(ii) रात्रेः
(iii) परमेष्ठिनः
(iv) सूत्रधारस्य
(v) मरीचेः
(vi) भ्रमरस्य
(vii) प्रशंसायोग्यस्य
(viii) आधारस्य

प्रश्न: 4.
विशेषणानि विशेष्यैः सह मेलयतविशेषणानि
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 4

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 5

प्रश्नः 5.
समस्तपदानि रचयत
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 6

उत्तरम्:
(i) मरीचिमालिनः
(ii) खेचरचक्रस्य
(iii) ब्रह्माण्डमाण्डस्य
(iv) पुण्डरीकपटलस्य
(v) रोलम्बकदम्बस्य
(vi) अहोरात्रम्
(vii) कल्पभेदाः
(viii) पर्णकुटीरात्

प्रश्न: 6.
समानध्वन्यात्मकशन्दान् मेलयत
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 7

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 8

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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 9

प्रश्नः 7.
अधोलिखितवाक्येषु क्रियापदानि योजयत –
(i) अयमेव अहोरात्रम् …………………।
(ii) एष एव उत्तरं दक्षिणं च अयनम् ………………।
(iii) परमेष्ठिनः परार्द्धसंख्या एनम् एव आश्रित्य ……………….।
(iv) प्रजानां प्राणः एषः सूर्यः ………………….।
(v) सूर्यः वत्सरं द्वादशसु भागेषु ……………………..।
उत्तरम्:
(i) अयमेव अहोरात्रम् जनयति।
(ii) एष एव उत्तरं दक्षिणं च अयनम् अङ्गीकरोति।
(iii) परमेष्ठिनः परार्द्धसंख्या एनम् एव आश्रित्य भवति।
(iv) प्रजानां प्राणः एषः सूर्यः प्रकाशते।
(v) सूर्यः वत्सरं द्वादशसु भागेषु विभनक्ति।

प्रश्नः 8.
शब्दानां मूलशब्दं विभक्तिं च प्रदर्श्य तालिकां पूरयत
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 10

उत्तरम्:
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 11

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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 12

प्रश्नः 9.
सन्धि कृत्वा लिखत
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 15

उत्तरम्:
(i) भगवतो मरीचिमालिनः
(ii) मणिराकाशमण्डलस्य
(iii) अवलम्बो रोलम्बकदम्बस्य
(iv) इनश्च
(v) चायनम्
(vi) परमेष्ठिन्

प्रश्न: 10.
अधोलिखितवाक्येषु अव्ययपदानि योजयत –
(i) सूर्यः ……………… आत्मा जगतः तस्थुषः च।
(ii) इनः …………………………. दिनस्य।
(iii) सूर्यस्य वर्णनं ……………….. केवलं ज्ञानवर्धकम् ……………… भाषायाः सौन्दयस्य उत्कृष्टम् उदाहरणम्।
(iv) अयम् ………………. कारणं षण्णाम् ऋतूनाम्।
(v) निजपर्णकुटीरात् …………………. बटुः सूर्यस्य महिमानं वर्णयति।
उत्तरम्:
(i) सूर्यः एव आत्मा जगतः तस्थुषः च।
(ii) इनः च दिनस्य।
(iii) सूर्यस्य वर्णनम् न केवलं ज्ञानवर्धकम् अपितु भाषायाः सौन्दर्यस्य उत्कृष्टम् उदाहरणम्।
(iv) अयम् एव कारणं षण्णाम् ऋतूनाम्।
(v) निजपर्णकुटीरात् निर्गत्य बटुः सूर्यस्य महिमानं वर्णयति।

पाठ विकासः

(क) लेखक व ग्रंथ का परिचय
यह पाठ विविध उपनिषदों से तथा शिवराजविजय नामक गद्यकाव्य के प्रथम अध्याय से संकलित व संपादित किया गया है। शिवराजविजय के रचयिता श्री अंबिकादत्त व्यास हैं। श्री अंबिकादत्त व्यास आधुनिक युग के ऐतिहासिक उपन्यास लेखन की विधा के प्रवर्तक हैं।

(ख) भावविकास
सूर्य ही गतिशील जंगम व स्थायी (स्थावर) की आत्मा है। ऐसा ऋग्वेद में वर्णन है। छान्दोग्य उपनिषद् में कहा गया है
‘आदित्यो ह वै प्राणः’ (सूर्य ही प्राण है)। तैत्तिरीय उपनिषद् की सूचना है –
‘आदित्येन वाव सर्वे लोकाः महीयन्ते’ (सूर्य से ही सब लोक महत्त्व को प्राप्त करते हैं) प्रश्नोपनिषद् में भी लिखा गया है कि जो हज़ार किरणोंवाला सौ प्रकार से प्रजाओं में विद्यमानप्राण का उदय होता है, यह सूर्य ही है

  1. नवो नवो भवसि जायमानः। (उत्पन्न होते हुए तुम बार-बार नए होते हो) – अथर्व० 7/81/2
  2. कः शक्तः सूर्यं हस्तेनाच्छादयितुम्? (सूर्य को कौन हाथ से ढक सकता है?) भासकृत-अविमारक।
  3. आनन्दमयो ज्ञानमयो विज्ञानमय आदित्यः। (सूर्य आनंदमय, ज्ञानमय के साथ विज्ञानमय भी है।) – सूर्योपनिषद्
  4. सहस्रगुणमुस्रष्टुमादत्ते हि रसं रविः।। (सूर्य हज़ार गुणा करके छोड़ने के लिए जल (रस) ग्रहण करता है।) – अभिज्ञानशाकुन्तलम्
  5. एकः श्लाघ्यः विवस्वान् परहितकरणायैव यस्य प्रयासः। (एक प्रशंसायोग्य सूर्य है जिसका प्रयत्न सदा दूसरों का हित करना है।) हर्ष-नागानन्द, 3/18

सूर्याष्टक :

आदिदेव नमस्तुभ्यं प्रसीद मम भास्कर
दिवाकर नमस्तुभ्यं प्रभाकर नमोऽस्तु ते॥1॥
सप्ताश्वरथमारूढं प्रचण्ड कश्यपात्मजम्
श्वेतपद्मधरं देवं तं सूर्य प्रणमाम्यहम्॥2॥
बृहितं तेजः पुजं च वायुमाकाशमेव च।
प्रभु च सर्वलोकानां तं सूर्य प्रणमाम्यहम्॥5॥
बन्धुकपुष्पसङ्काशं हारकुण्डलभूषितम्
एकचक्रधरं देवं तं सूर्य प्रणमाम्यहम्॥6॥

अर्थ – हे आदिदेव सूर्य (भास्कर) तुम्हें मेरा प्रणाम। आप कृपा करें। हे दिन को बनानेवाले तुम्हें प्रणाम, हे प्रकाश देनेवाले, तुम्हें प्रणाम। सात घोड़े के रथ पर चढ़े हुए, तीक्ष्ण, कश्यप के पुत्र, हे सफेद काम को धारण करनेवाले सूर्यदेव, मैं आपको प्रणाम करता हूँ। महत् तेज:पुञ्ज, वायु तथा आकाश रूप, तथा सब लोकों के स्वामी, आपको प्रणाम। गुडहल के फूल के समान, हार तथा कुंडल से सुभूषित, एकचक्र को धारण करनेवाले देव सूर्य को मैं प्रणाम करता हूँ।
भाव – अवबोधनम् –
जगतः तस्थुषश्च – जो चेतन जीव हैं तथा जो अचेतन (जड़) हैं, सूर्य उन सबकी आत्मा है।
चक्रवर्ती खेचरचक्रस्य – सब नक्षत्र समूहों का सम्राट भी सूर्य ही है क्योंकि सूर्य से ही सब नक्षत्र प्रकाशित होते हैं। सूर्य के चारों ओर ही सब नक्षत्र घूमते हैं।
कुण्डलम् आखण्डलदिश – इंद्र से संबंधित पूर्व दिशा नायिका है उसके कर्णाभूषण कुंडल के समान सूर्य शोभा देता है।
शोकविमोकः कोकलोकस्य – चकवा पक्षी रात को अपनी पत्नी से अलग हो जाता है। प्रातः उन दोनों का मेल होता है। प्रात:काल का जनक सूर्य उन दोनों के शोक को दूर करता है।
अवलम्बः रोलम्बकदम्बस्य – भौंरे रात में कमल के फूलों में बंद हो जाते हैं। सवेरे जब कमल खिलते हैं, सूर्य कमलों से उन्हें मुक्त कर पुनः जीवनदान देता है।
कुलमूलं श्रीरामचन्द्रस्य – सूर्य ही भगवान् श्रीराम के कुल का जनक है। इसी कारण श्रीराम सूर्यवंशी कहलाते हैं।
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 13

सी.बी.एस.ई. पाठ्यपुस्तक ऋतिक-2 पृष्ठ संख्या 22 पर प्रहेलिका देखें।

प्रहेलिकायाः प्रश्नाः उत्तराणि च॥
उपरिष्टात् अधः (ऊपर से नीचे) (वर्णसङ्ख्या )

प्रश्न: 1.
हेमन्तात् पश्चात् अयं ऋतुः आयाति।
(हेमंत के बाद यह ऋतु आती है।)
उत्तरम्:
वसन्तः।

प्रश्न: 2.
अस्मिन् युगे कृष्णस्य जन्म अभवत्।
(इस युग में कृष्ण का जन्म हुआ।)
उत्तरम्:
द्वापरे।

प्रश्न: 3.
अस्मिन् मासे होलिकोत्सवः भवति।
(इस महीने में होली का उत्सव होता है।)
उत्तरम्:
फाल्गुने।

प्रश्न: 4.
विक्रमाब्दस्य प्रारम्भः अस्मात् मासात् भवति।
(विक्रमसंवत् का आरंभ इस मास से होता है।)
उत्तरम्:
चैत्रात्।

प्रश्न: 5.
अस्मिन् ऋतौ .मेघान् दृष्ट्वा मयूराः नृत्यन्ति, नद्यः वेगेन वहन्ति।
उत्तरम्:
वर्षायाम्।

प्रश्न: 6.
अस्मिन् मासे रावणं हत्वा रामः विजयी अभवत्।
(इस महीने में रावण को मारकर राम विजयी हुए।)
उत्तरम्:
आश्विने।
वामतः दक्षिणम् (बायें से दाहिने)

प्रश्न: 7.
शप्तः यक्षः अस्मिन् मासे मेघम् अपश्यत्।
(शापग्रस्त यक्ष ने इस महीने में बादल को देखा) उत्तरम्- भाद्रपदः।

प्रश्नः 8.
वेदाध्ययनम् अस्मिन् मासे स्थगितं भवति।
(वेदों का अध्ययन इस मास में रोक दिया जाता है।)
उत्तरम्:
श्रावणे।

प्रश्न: 9.
मान्धाता महीपतिः आस्मिन युगे अभवत्।
(मान्धाता राजा इस युग में हुए।)
उत्तरम्:
कृतयुगे
उपरिष्टात् अधः (ऊपर से नीचे)

प्रश्नः 10.
वैशाखादनन्तरम् एष मासः भवति।
(वैसाख के बाद यह मास होता है।)
उत्तरम्:
ज्येष्ठः।

प्रश्न: 11.
अस्य मासस्य अमावस्यायां दीपावलिः मान्यते। .
(इस महीने की अमावस्या में दीपावली मनाई जाती है।)
उत्तरम्:
कार्तिकस्य।
वामतः दक्षिणम् (बायें से दाहिने)

प्रश्न: 12.
अस्य ऋतोः आदौ ‘हे’ भवति।
(इस ऋतु का आदि अक्षर ‘हे’ होता है।)
उत्तरम्:
हेमन्तः।

प्रश्न: 13.
अस्य मासस्य पूर्णिमायां चन्द्रमसः करेभ्यः अमृतवर्षणं भवति।
(इस महीने की पूर्णिमा में चन्द्रमा की किरणों से अमृत की वर्षा होती है।)
उत्तरम्:
आश्विने।

प्रश्न: 14.
अस्मिन् ऋतौ तापतप्ताः जनाः पर्वतीयस्थलेषु गच्छन्ति।
(गर्मी से बेहाल लोग इस ऋतु में पर्वतों पर जाते हैं।)
उत्तरम्:
ग्रीष्मे।

प्रश्न: 15.
कृष्णे दिवं प्रयाते एषः युगः प्रारब्धः।
(कृष्ण के स्वर्ग को प्रयाण करने पर यह युग प्रारंभ हुआ।)
उत्तरम्:
कलिः ।

प्रश्न: 16.
विशाखानक्षत्रम् अस्मिन् मासे भवति।
(विशाखा नक्षत्र इस महीने में होता है।)
उत्तरम्:
वैशाखे

प्रश्न: 17.
रामः अस्मिन् युगे जातः।
(राम इस युग में पैदा हुए।)
उत्तरम्:
त्रेता।
उपरिष्टात्अधः (ऊपर से नीचे)

प्रश्न: 18.
उत्तरायणादनन्तरम् एतत् अयनम् आगच्छति।
(उत्तरायण के बाद यह अयन आता है।)
उत्तरम्:
दक्षिणायनम्।
वामतः दक्षिणम् (बायें से दाहिने)

प्रश्न: 19.
माघमासात् पूर्वं भवति अयं मासः।
(माघ महीने से पहले यह मास होता है।)
उत्तरम्:
पौषः।

प्रश्न: 20.
(क) शिशुपालवधस्य लेखकस्य नाम अपि अस्य मासस्य नाम्नः सदृशः।
(शिशुपालवध के लेखक का नाम भी इस मास के नाम के सदृश है।)
उत्तरम्:
माघस्य।
उपरिष्टात्अधः (ऊपर से नीचे)
(ख) मृगशिरा नक्षत्रात् सम्बद्धः अयं मासः।
(मृगशिरा नक्षत्र से यह मास संबद्ध है।
उत्तरम्:
मार्गशीर्षः।
वामतः दक्षिणम् (बायें से दाहिने)

प्रश्न: 21.
भीष्मः अस्मिन् अयने प्राणान् अत्यजत्।
(भीष्म ने इस अयन में प्राण छोड़े।)
उत्तरम्:
उत्तरायण।

प्रश्न: 22.
दिनञ्च रात्रिश्च।
(दिन भी और रात भी)
उत्तरम्:
अहोरात्रम्।

अतिरिक्त-अभ्यासः

प्रश्न: 1.
अधोलिखितम् गद्यांशं पठित्वा तदाधारितानां प्रश्नानाम् उत्तराणि लिखत् –
(क) अरुण एष प्रकाशः पूर्वस्यां भगवतो मरीचिमालिनः। एष भगवान् मणिराकाश-मण्डलस्य, चक्रवर्ती खेचरचक्रस्य, कुण्डलम् आखण्डलदिशः, दीपकः ब्रह्माण्डभाण्डस्य, प्रेयान् पुण्डरीकपटलस्य, शोकविमोकः कोकलोकस्य, अवलम्बो रोलम्बकदम्बस्य, सूत्रधारः सर्वव्यवहारस्य, इनश्च दिनस्य।

I. एकपदेन उत्तरत (1/2 x 4 = 2)
1. मरीचिमालिनः प्रकाशः कीदृशः?
2. एषः भगवान् सूर्यः कस्य दीपकः?
3. सर्वव्यवहारस्य सूत्रधारः कः?
4. सूर्यः कस्याः नायिकायाः कुण्डलम्?
उत्तरम्:
1. अरुणः
2. बह्माण्डभाडस्य
3. सूर्यः
4. आखण्डलदिशः .

II. पूर्णवाक्येन उत्तरत (1x 1 = 1)
भगवान् मरीचिमाली कस्य चक्रवर्ती अस्ति?
उत्तरम्:
भगवान् मरीचिमाली खेचर चक्रस्य चक्रवर्ती अस्ति।

III. निर्देशानुसारेण उत्तरत (1 x 2 = 2)
1. तत् शब्दयुग्मं लिखतं यत्र ‘म्ब’ अक्षरस्य पुनरावृत्तिः अस्ति।
2. ‘पूर्वस्यां’ इति पदे का विभक्तिः ?
उत्तरम्:
1. अवलम्बों रोलम्बकदम्बस्य
2. सप्तमी विभक्तिः

(ख) अयमेव अहोरात्रं जनयति। अयम् एव वत्सरं द्वादशसु भागेषु विभनक्ति। अयम् एव कारणं षण्णाम्
ऋतूणाम्। एष एव अङ्गीकरोति उत्तरं दक्षिणं चायनम्। अनेन एव सम्पादिताः युगभेदाः। अनेन एव कृताः कल्पभेदाः। एनम् एव आश्रित्य भवति परमेष्ठिनः परार्द्धसङ्ख्या । वेदा एतस्य एव वन्दिनः, गायत्री अमुम् एव गायति। धन्य एष कुलमूलं श्रीरामचन्द्रस्य। प्रणम्यः एषः विश्वेषाम्।
I. एकपदेन उत्तरत (1/2 x 4 = 2)
1. सूर्येण के सम्पादिताः?
2. सूर्येण केषां भेदाः कृता?
3. का सूर्यं गायति?
4. अयं सूर्यः कस्य कुलमूलम्?
उत्तरम्:
1. युगभेदाः
2. कल्पानाम्
3. गायत्री
4. श्रीरामचन्द्रस्य

II. पूर्णवाक्येन उत्तरत (2 x 1 = 2)
सूर्यः वत्सरं केषु विभनक्ति?
उत्तरम्:
सूर्यः वत्सरं द्वादश भागेषु विभनक्ति।

III. निर्देशानुसारम् उत्तरत (1/2 x 2 = 1)
1. ‘परमेष्ठिनः’ पदे विभक्तिं लिखत।
2. ‘परार्द्धसङ्ख्या ‘ इत्यस्य विग्रहं कुरुत।
उत्तरम्:
1. षष्ठी विभक्तिः
2. परार्द्धा सङ्ख्या (कर्मधारयः)

(ग) एष भगवान् मणिराकाश-मण्डलस्य, चक्रवर्ती खेचर चक्रस्य, कुण्डलम् आखण्डलदिशः, दीपकः ब्रह्माण्ड भाण्डस्य, प्रेयान्, पुण्डरीकपटलस्य, शोकविमोकः कोकलोकस्य, अवलम्बो रोलम्बकदम्बस्य, सूत्रधारः सर्वव्यवहारस्य।

I. एकपदेन उत्तरत (1 x 2 = 2)
(i) सूर्यः कस्य शोकं विमोचयति?
(ii) भगवान् सूर्यः सर्वव्यवहारस्य कोऽस्ति?
उत्तरम्:
(i) कोकलोकस्य
(ii) सूत्रधारः

II. पूर्णवाक्येन उत्तरत (1 x 1 = 1)
एषः भगवान् सूर्यः कस्य मणिः वर्तते?
उत्तरम्:
एषः भगवान् सूर्यः आकाशमण्डलस्य मणिः वर्तते।

III. निर्देशानुसारेण उत्तरत (1/2 x 4 = 2)
(i) ‘एष भगवान्’ इति पदद्वयम् कस्य विशेषणं वर्तते?
(ii) ‘नक्षत्र समूहस्य’ इति पदस्य अर्थे किं पदं प्रयुक्तम्?
(iii) ‘प्रजा’ इति पदस्य किं विपर्यय पदम् अत्र अनुच्छेदे प्रयुक्तम्?
(iv) ‘प्रियः’ पदस्य कः पर्यायः अत्र प्रयुक्तः?
उत्तरम्:
(i) सूर्यस्य
(ii) खेचर चक्रस्य
(iii) चक्रवर्ती
(iv) प्रेयान्

(घ) अयम् एव कारणं षण्णाम् ऋतूणाम्। एष एव अङ्गीकरोति उत्तरं दक्षिणं चायनम्। अनेन एव सम्पादिताः युगभेदाः। अनेन एव कृताः कल्पभेदाः। एनम् एव आश्रित्य भवति परमेष्ठिनः परार्द्धसङ्ख्या। वेदाः एतस्य वन्दिनः, गायत्री अमुम् एव गायति। धन्यः एषः कुलमूलं श्रीरामचन्द्रस्य।
I. एकपदेन उत्तरत (1/2 x 2 = 1)
(i) केन कल्पानां भेदाः कृताः?
(ii) उत्तरं दक्षिणञ्च किं भवति?
उत्तरम्:
(i) सूर्येण
(ii) अयनम्

II. पूर्ववाक्येन उत्तरत (2 x 1 = 2)
(i) सूर्यम् आश्रित्य किं भवति?
उत्तरम्:
सूर्यम् आश्रित्य परमेष्ठिनः परार्द्ध सङ्ख्या भवति।

III. निर्देशानुसारेण उत्तरत (1/2 x 4 = 2)
(i) अनुच्छेदे ‘अमुम्’ पदस्य कः पर्यायः आगतः?
(ii) ‘षण्णाम् ऋतूणाम्’ अत्र विशेषण पदं किम?
(iii) ‘कृताः’ इति क्रिया पदस्य अनुच्छेदे कर्तृपदं किम्?
(iv) अनुच्छेदे ‘निराश्रित्य’ पदस्य कः विपर्ययः प्रयुक्तो वर्तते?
उत्तरम्:
(i) एनम्
(ii) षण्णाम्
(iii) अनेन
(iv) आश्रित्य

प्रश्न: 2.
ग्रन्थस्य लेखकस्य च नामनी लिखत (1 + 1 = 2)
(i) ‘अरुण एष प्रकाशः पूर्वस्यां भगवतो मरीचिमालिनः।”
(ii) ‘अयम् एव वत्सरं द्वादशसु भागेषु विभनक्ति।”
(iii) ‘धन्य एष कुलमूलं श्रीरामचन्द्रस्य। प्रणम्यः एषः विश्वेषाम्।
उत्तरम्:
(i) ग्रन्थः- शिवराजविजयः लेखकः- श्री पं० अम्बिकादत्तः व्यासः
(ii) ग्रन्थः- शिवराजविजयः लेखक:- श्री पं० अम्बिकादत्तः व्यासः
(iii) ग्रन्थः- शिवराजविजयः लेखक:- श्री पं० अम्बिकादत्तः व्यासः

प्रश्न: 3.
(क) अधोलिखितानां पङ्क्तिनां दत्तेषु भावार्थेषु शुद्ध भावं चित्त्वा लिखत (1 x 2 = 2)
I. दीपकः ब्रह्माण्डभाण्डस्य।
अर्थात्
(i) सूर्यः ब्रहमाण्ड पात्रं प्रकाशयति।
(ii) सूर्येण विश्व गृहं प्रकाशयति।
(iii) सूर्यः ब्रहमाण्ड पात्रस्य दीपकोऽस्ति!
उत्तरम्:
(ii) सूर्येण विश्व गृहं प्रकाशयति।

II. अयमेव अहोरात्रं जनयति।
अर्थात्
(i) सूर्येण एव दिवारात्रम् भवतः।
(ii) सूर्यः दिवारात्रम् प्रकाशयति।
(iii) सूर्यः एव दिवारात्र्योः कारणं वर्तते।
उत्तरम्:
(iii) सूर्य एव दिवारात्र्यो कारणं वर्तते।

(ख) दत्तासु पंक्तिषु रिक्तस्थानपूर्ति माध्यमेन उचितं भावं लिखत (½ x 4 = 2)
I. “वेदाः एतस्य एव वन्दिनः।” अस्य भावोऽस्ति यत् ऋग्वेदः, यजुर्वेदः सामवेदः अथर्ववेदश्च …..(i)…. वेदाः सदैव अस्य ………..(ii)……….. एव गुणगानं कुर्वन्ति। यतः अयमेव सम्पूर्ण संसारस्य एक मात्रमेव ………..(iii)………. वर्तते। एनं विना संसारस्य ………….(iv)………… अपि कर्तुं न शक्यते।
उत्तरम्:
(i) चत्वारः
(ii) सूर्यस्य
(iii) कारणम्
(iv) व्यवहारम्

II. “अयम् एव कारणं षण्णाम् ऋतूणाम्।” अर्थात्अयं ……….(i)………. एव वसन्त-ग्रीष्म-वर्षा-शरद्-हेमन्त-शिशिराणाञ्च ……..(ii)….. मूल कारणं वर्तते। तेन विना ऋतूणाम् …….(iii)……. एव न परिवर्तते। एतेषाम् ऋतूणाम् एव आधारेण ………(iv)……. अस्तित्त्वम् अस्ति।
उत्तरम्:
(i) सूर्यः
(ii) ऋतूणाम्
(iii) चक्रम्
(iv) जीवानाम्

प्रश्न: 4.
निम्न वाक्यांशानां सार्थक संयोजनं कृत्वा लिखत (1/2 x 8 = 4)
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NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः 14

उत्तरम्:
(i) (3)
(ii) (5)
(iii) (7)
(iv) (1)
(v) (8)
(vi) (4)
(vii) (6)
(viii) (2)

प्रश्नः 5.
प्रदत्त वाक्येषु रेखाङ्कितानां पदानां प्रसंगानुसारेण उचितार्थानां चयनं कुरुत (1 x 4 = 4)
1. चक्रवर्ती खेचर चक्रस्य
(i) पक्षिणां समूहस्य
(ii) नक्षत्राणां समूहस्य
(iii) कीटानां समूहस्य।
उत्तरम्:
(ii) नक्षत्राणां समूहस्य

2. इनश्च दिनस्य।
(i) स्वामी
(ii) गणक:
(iii) विघ्नः।
उत्तरम्:
(i) स्वामी

3. वेदा एतस्य एव वन्दिनः
(i) बन्दीजनाः
(ii) प्रशंसकाः
(iii) स्तोतारः।
उत्तरम्:
(iii) स्तोतारः

4. एष एव अङ्गीकरोति उत्तरं दक्षिणं चायनम्।
(i) स्वीकरोति
(ii) सम्मानयति
(iii) उत्पादयति।
उत्तरम्:
(i) स्वीकरोति

NCERT Solutions for Class 12 Sanskrit

The post NCERT Solutions for Class 12 Sanskrit Chapter 2 सूर्यः एव प्रकृतेः आधारः appeared first on Learn CBSE.

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