CBSE Previous Year Question Papers Class 12 Physics 2019 Delhi
CBSE Previous Year Question Papers Class 12 physics 2019 Outside Delhi Set-I
Section-A
Question 1.
Draw the pattern of electric field lines, when a point charge – Q is kept near an uncharged conducting plate. [1]
Answer:
The positive charge will be induced on the uncharged conducting plate, kept near it. So, the lines of forces will start from metal plate and end -Q
Question 2.
How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant ?
Answer:
The mobility of electrons is given by the
As it’s independent of the applied potential difference, so it will not change if the applied potential difference will be doubled.
Question 3.
Define the term “threshold frequency” in the context of photoelectric emission.
OR
Define the term ‘Intensity’ in photon picture of electromagnetic radiation.
Answer:
The minimum frequency of the radiation incident on a metal surface below which there is no photoelectric emission is called threshold frequency. It is a characteristic property of a photosensitive material.
OR
The amount of light energy or photon energy incident per meter square per second is called intensity of radiation. SI unit of intensity of radiation is W/Sr
Question 4.
What is the speed of light in a denser medium of polarising angle 30°
Answer:
Question 5.
In sky wave mode of propagation, why is the frequency range of transmitting signals restricted to less than 30 MHz ?
OR
On what factors does the range of coverage in ground wave propagation depend ?
Section-B
Question 6.
Two bulbs are rated (P1, V) and (P2, V). If they are connected
(i) in series and
(ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P1and P2.
Answer:
(i) In series combination :
(ii) In parallel combination
Question 7.
Calculate the radius of curvature of an equiconcave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D ? [2]
OR
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index 4√2 / 5.
Answer:
In an equiconcave lens, radius of curvature of both surfaces are equal
Question 8.
An a-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic Field B, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. [2]
Answer:
Question 9.
State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. [2]
OR
Calculate the orbital period of the electron in the first excited state of hydrogen atom.
Answer :
According to Bohr’s quantization of angular momentum, the stationary orbits are those in which angular momentum of electron
Question 10.
Why a signal transmitted from a TV tower cannot be received beyond a certain distance? Write the expression for the optimum separation between the receiving and the transmitting antenna. [2]
Question 11.
Why is wave theory of electromagnetic radiation not able to explain photo electric effect ? How does photon picture resolve this problem ? [2]
Answer:
Wave theory cannot explain the following laws of photoelectric effect.
- The instantaneous emission of photo electrons.
- Existence of threshold frequency for metal surface.
- K.E. of emitted electrons is independent of intensity of light and depends on frequency.
The concept of photon explained that energy is not only emitted and absorbed in discrete energy quanta, but also it propagates through space in definite quanta with the speed of light. It can explain all the above photoelectric effect, which wave theory cannot explain.
Question 12.
Plot a graph showing variation of de Broglie wavelength (λ) associated with a charged particle of a mass m, versus 1/√v, where V is the potential difference through which the particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle ? [2]
Answer:
Section-C
Question 13.
(a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole. [3]
Answer:
Question 14.
Using Kirchhoff’s rules, calculate the current through the 40 Q and 20 Q resistors in the following circuit: [3]
What is end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5Ω and S respectively.
When the resistance S is shunted with an equal resistance, the new balance length found to be 1.5 l1 where l1 is the initial balancing length. Calculate the value of S.
Question 15.
(a) Identify the part of the electromagnetic
spectrum used in
(i) radar and
(ii) eye surgery.
Write their frequency range.
(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. [3]
Answer:
(a)(i) Radar—Microwaves are used in radar. The frequency range is from 3xl0:i Hz to lxlO9 Hz.
(ii) Eye Surgery — Infrared waves are used. The frequency range is from 4xl014Hz to 3xlOu Hz.
(b) Energy density in oscillating electric field is
Question 16.
Define the term wave front. Using Huygens’s wave theory, verify the law of reflection. [3]
OR
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wave front is propagating from a denser to a rarer medium.
Answer:
A wave front is the continuous locus of vibrating particles which are in the same state of vibration or phase.
Laws of Reflection from Huygens principle
Consider a plane wave front AB incident on a plane reflecting surface PQ, Let v be the velocity of the wave. At time t – 0 one end of the wave front just touches the reflecting surface at B. Draw normal NB to PQ. When the wave front strikes the reflecting surface, then due to the presence of it, it cannot advance further. When wave front strikes at B, secondary wavelets starts emitting from B. The secondary wavelets will travel a distance AD = vt during the time the other end A of the wave front AB reaches the surface PQ at Q. To find the reflected wave front, B as a center and AD as radius draw an arc, which represent the secondary wavelets originating from B. As the incident wave front AB advances, the secondary wavelets will touch CD simultaneously.
According to Huygens’s principle CD represents the reflected wave front corresponding to incident wave front AB. BD is common triangles BAD and CBD and BC = AD = vt. Therefore, two triangles are congruent.
So, ∠ABD = ∠CDB
i. e., ∠i = ∠r
Thus, angle of incidence is equal to the angle of reflection. This is the second law of reflection. Also, the incident wave front AB, reflecting surface PQ and the reflected wave front CD are perpendicular to the plane of the paper. So, the incident ray, reflected ray and the normal at the point of incidence, all lie in same plane. This is first law of reflection.
Thus, Huygens’s principle explains both the law of reflection.
OR
The refractive index of a medium is defined as the ratio of speed of light in vacuum to that of the speed of the light in medium.
i.e n= c/v
Where, n = refractive index of medium when light ray passes from vaccum into a medium.
c = velocity of light in vacuum
v = velocity of light in the medium
Proof of Snell’s law of refraction
when a wave front travels from one medium to other, it deviates from its path. In travelling from one medium to other, the frequency of wave remains same and speed and wavelength changes. Let, XY be a surface separating two media T and ‘2’. Let the speed of waves of v1 and v2.
Suppose, a plane wave front AB in first medium is incident obliquely on the boundary surface XY and its end touches the surface at A at time t = 0, while the other end B reaches the surface at point B’ after time-interval ‘f Clearly, BB1 = v1t.
In the same time, wavelets starts from A and reaches A’ in time ‘f with velocity v2. Therefore, AA1 = v2t According to Huygen’s principle, A1B1 is the new position of the wave front in second medium. A1B1 is a refracted wave front.
Let, the incident wave front (AB) and reflected wave front (A1B1) makes angle i and r with surface XY.
In ΔAB1B,
Hence, ratio of sine of angle of incidence and the sine of angle of refraction for a given pair of media is constant. This is Snell’s law of refraction.
Question 17.
(a) Define mutual inductance and write its S.I. unit.
(b) A square loop of side carrying a current I2 is kept at distance x from an infinitely long straight wire carrying a current I1 as shown in the figure. Obtain the expression for the resultant force acting on the loop. [3]
Answer:
(a) Mutual inductance : It is a property of two coils due to which each coil opposes any change of current flowing in the other. The mutually induced e.m.f, in one coil produces the opposition in other coil. Its SI unit is henry (H).
(b)
Question 18.
(a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.
(b) Explain the significance of a radial magnetic field when a current carrying coil is kept in it. [3]
Answer : Torque on current carrying loop in magnetic field :
Consider a coil PQRS placed in a magnetic field. Let 0 be the angle between the plane of the coil and direction of B. When current flow through coil each side experiences a force. The forces on vertical side will constitute a couple.
Moment of torque = One of forces × Perpendicular distance between lines of action of force
Where, is magnetic moment of loop.
(b)A magnetic field, in which the plane of the coil in all positions remains parallel to the direction of magnetic field is called radial magnetic field. In a radial magnetic field, magnetic torque remains maximum for all the positions of the coil.
Question 19.
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.8 x 106 m and the radius of lunar orbit is 3.8 x 106 m. [3]
Answer:
Astronomical telescope in near points adjustment:
Question 20.
(a) State Gauss’ law for magnetism. Explain its significance.
(b) Write the four important properties of the magnetic field lines due to a bar magnet. [3]
OR
Write three points of differences between para, dia, and ferromagnetic materials, giving one example for each.
Answer:
(a) Gauss law for magnetism : If a closed surface is imagined in a magnetic field, the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero.
Gauss law signifies that magnetic mono poles does not exist.
(b)
1. In a bar magnet, each lines of force, starts from a north pole and reaches the south pole externally and then goes from south pole to a north pole internally. Thus, magnetic line of force forms a closed loop.
2. No two lines of force will never intersect each other.
3. In a uniform field, the lines are parallel and equidistant from each other.
4. The lines of force are crowded near the poles.
OR
Properties | Ferro-magnetic Materials | Para-magnetic Materials | Dia-magnetic Materials |
State | They are solid | They can be solid, liquid or gas. | They can be solid, liquid or gas. |
Effect of Magnet | Strongly attracted by a magnet | Weakly attracted by a magnet. | Weakly repelled by a magnet. |
Effect of Temperature | Above curie point, it becomes a paramagnetic. | With the rise of temperature, it becomes a diamagnetic. | No effect. |
Examples | Iron, Nickel, Cobalt | Lithium, Molybdenum, magnesium | Copper, Silver, Gold |
Note : Any three difference can be written in exam.
Question 21.
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10000 disintegrations and 5,000 disintegrations after 20 hr. and 30 hr. respectively from start. Calculate the half life and initial number of nuclei at t = 0. [3]
Answer:
Decay constant’ of a radioactive sample is defined as the ratio of its instantaneous rate of disintegration to the number of atoms present at that time.
Question 22.
(a) Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm ?
(b) Why photo diodes are required to operate in reverse bias? Explain. [3]
Answer:
(a) Energy corresponding to wavelength 600 nm
The photon energy (E = 2.06 eV) is greater than the band gap for diode D2 only. Hence, diode D1 and D3 will not be able to detect the given wavelength.
(b)A photo diode is operated reverse bias because in reverse bias it is easier to observe change in current with change in light intensity,
Question 23.
(a) Describe briefly the functions of the three segments of n-p-n transistor.
(b) Draw the circuit arrangement for studying the output characteristics of n-p-n transistor in CE configuration. Explain how the output characteristics is obtained. [3]
OR
Draw the circuit diagram of a full wave rectifier and explain its working. Also, give the input and output wave forms.
Answer:
For full wave rectifier, we use two junction diodes as shown in figure.
During first half cycle of input a.c. signal the terminal Si is positive relative to S and S2 is negative, then diode Di is in forward biased and diode D2 is reverse biased. Therefore, current flows in D1 not in D2. In next half cycle, Si is negative and S2 is positive relative to S. Then D1 is in reverse biased and D2 is in forward biased. Therefore, current flows in D2 not in Di. Thus, for input a.c. signal the output current is a continuous series of unidirectional pulse.
The input and output wave forms are shown in the figure.
Question 24.
(a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of A & B.
(b) A message signal of frequency 20 kHz and peak voltage 10 V is used to modulate a carrier of frequency 2 MHz and peak voltage of 15 V. Calculate the modulation index. Why the modulation index is generally kept less than one ?
Section-D
Question 25.
(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.
(b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit ?
(c) When an inductor is connected to 200 V dc voltage, a current of 1 A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why ? Also, calculate the self inductance of the inductor. [5]
OR
(a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
(b) A small town with a demand of 1200 kW of electric power at 220V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Q per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.
Answer:
(a) Consider an alternating e.m.f. is connected in series with an inductor, resistance R and capacitance C. Let, E and I be the instantaneous values of e.m.f. and current in the LCR circuit VL, VC and VR be the instantaneous values of voltage across inductor, capacitor and resistor respectively. Then,
(b) At resonance
XL = Xc
iXL = iXc
VL= VC
The voltage across inductance and capacitance are equal and have a phase difference of 180° at resonance.
(c) Since the reactance of an inductor is zero for d.c. circuit. But the inductor offers resistance to an a.c. circuit. Therefore, the current decreases for the same inductor when it is connected with an a.c. source.
When inductor is connected in a.c. circuit:
OR
(a) Step-down transformer:
It is a device used for converting high alternating voltage at low current into low alternating voltage at high current and vice-versa.
The device works on the principle of mutual induction i.e., if the current or magnetic flux linked with a coil changes then an e.m.f. is induced in the other coil.
In step-down transfomer Np > Ns and transformation ratio is less than 1.
1. Copper losses: Due to resistance of winding’s in primary and secondary coils, some electrical energy is converted into heat energy.
2. Flux losses : Some of the flux produced in primary coil is not linked up with secondary coils.
3. Hysteresis losses : When the iron core is subjected to a cycle of magnetization the core
gets heated up due to hysteresis known as hysteresis loss.
4. Iron losses : The varying magnetic flux produces eddy current in the iron core, which leads to the wastage of energy in the form of heat.
(b) Length of wire line = 20 x 2 = 40 km
Resistance of wire line, r = 40 x 0.5 = 20 Ω
Power to be supplied = 1200 kW = 1200 × 103 W
Voltage at which power supplied = 4000 V
Question 26.
(a) Describe any two characteristic features
which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double slit experiment.
(b)In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. [5]
OR
(a) Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal length +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.
Answer:
(a)
Interference | Diffraction |
1. It is the result of interaction of light coming from two different wave fronts originating from two coherent sources. | 1. It is the result of interaction of light come from different parts of same waterfronts |
2. All the bright fringes are of same intensity. | 2. The bright fringes are of varying intensity (Intensity of bright fringes decreases from central bright fringe on either sides.) |
A sources of monochromatic light illuminates two narrow slits S1 and S2.The two illuminated slits act as the two coherent sources. The two slits is very close to each other and at equal distance from source. The wave front S1 and S2 spread in all direction and superpose and produces dark and bright fringe on screen. Let the displacement of waves from Si and S2 at point P on screen at time t is
OR
(a) Conditions for total internal refraction :
1. The ray must travel from a denser medium into a rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for the pair of media.
Relation between critical angle and refractive index:
When a ray of light travels from denser to rarer medium, the ray bends away from the normal. When the angle of incidence is equal to the critical angle then the refracted ray grazes the surface of separation represented by ray and
Question 27.
(a) Describe briefly the process of transferring
the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. [5]
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(b) The identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
Answer:
(a) When the plates of the parallel plate capacitor is connected to a battery. Then the first insulated metal plate gets, the positive charge till its potential become maximum. Then, the charge will leak to surroundings. So, the negative charge will be induced on the nearer face of the second plate and the positive charge will be induced on its farther plate.
Consider a capacitor of capacitance C. Initial charge and potential difference be zero. Let, a charge Q be given in small steps. Let at any instant when charge on capacitor be q, the potential difference between its plates,
Now work done in giving an additional charge dq is,
OR
(a) Consider an electric dipole of charges -q and +q separated by a distance 2a and placed in a free space. Let P be a point on equitorial line of dipole at a distance r from the centre of a dipole.
This is the required expression.
(b) Let the two charges of + q each placed at point A and B at a distance 2 m apart in air.
Suppose, the third charge Q (unknown magnitude and charge) is placed at a point O, on the line joining the other two charges, such that OA= x and OB 2-x.
For the system to be in equilibrium, net force on each 3 charges must be zero.
If we assume that charge Q placed at O is positive, the force on it at O may be zero. But the force on charge q at point A or B will not be zero. It is because, the forces on a charge q due to the other two charges will act in same direction. If charge Q is negative, then the forces on q due to other two charges will act in opposite direction.
Hence, Q will be negative in nature. For charge (-Q) to be in equilibrium Force on charge (-q) due to charge (+q) at point A should be equal and opposite to charge (+Q) at B
Therefore, for the system to be in equilibrium a charge – Q is placed at a mid point between the two charges of + q each.
CBSE Previous Year Question Papers Class 12 physics 2019 Outside Delhi Set-II
Note : Except for the following questions, all the remaining questions have been asked in previous set.
Section-A
Question 1.
When unpolarised light is incident on the interface separating the rarer medium and the denser medium. Brewster angle is found to be 60°. Determine the refractive index of the denser medium. [1]
Answer:
Question 2.
When a potential difference is applied across the ends of a conductor, how is the drift velocity of the electrons related to the relaxation time? [1]
Answer:
Question 3.
Draw the equipotential surfaces due to an isolated point charge. [1]
Answer:
Section-B
Question 4.
Explain with the help of Einstein’s photoelectric equation any two observed features in photo-electric effect which cannot be explained by wave theory. [2]
Answer:
Features of photoelectric equation which can not be explained by wave theory :
(a) The wave theory could not explain the instantaneous process of photoelectric effect.
(b) ‘Maximum kinetic energy’ of the emitted photo electrons is independent of intensity of incident light.
Question 5.
A deuteron and an alpha particle having same momentum are in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of the radii of the circular paths described by them. [2]
Answer:
Radius of circular path
Question 6.
(a) Plot a graph showing variation of de Broglie wavelength (X) associated with a charged particle of mass m, versus √v, where V is the accelerating potential.
(b) An electron, a proton and an alpha particle
have the kinetic energy. Which one has the shortest wavelength ? [2]
Answer:
Section-C
Question 7.
(a) State the underlying principle of a moving
coil galvanometer.
(b) Give two reasons to explain why a galvanometer cannot as such be used to measure the value of the current in a given circuit
(c) Define the terms:
(i) voltage sensitivity and
(ii) current sensitivity of a galvanometer. [3]
Answer:
(a) The Principle : When a current flows through the conductor coil, a torque acts on it due to the external radial magnetic field. Counter torque due to suspension balances coil after appropriate deflection due to current in the circuit.
(b) A galvanometer can be used as such to measure current due to following two reasons.
- A galvanometer has a finite large resistance and is connected in series in the circuit, so it will increase the resistance of circuit and hence change the value of current in the circuit.
- A galvanometer is a very sensitive device, it gives a full scale deflection for the current of the order of micro ampere, hence if connected as such it will not measure current of the order of ampere.
(c) (i) Voltage sensitivity : It is defined, as the deflection produced in the galvanometer when a unit voltage is applied across it.
(ii) Current Sensitivity: The ratio of deflection produced by the coil Φ to the current in the coil is called the current sensitivity. It is the deflection of the meter per unit current.
Question 8.
(a) Draw equipotential surfaces corresponding to the electric field that uniformly increases in magnitude along with the z-directions.
(b) Two charges -q and + q are located at point (0, 0, – a) and (0, 0, a). What is the electrostatic potential at the points (0, 0, ± z) and (x, y, 0) ? [3]
Answer:
(a)
Question 9.
(a) Write the relation between half life and average life of a radioactive nucleus.
(b) In a given sample two isotopes A and B are initially present in the ratio of 1 : 2. Their half lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio of 2 :1 ? [3]
Question 10.
(a) Define the term ‘self inductance’ of a coil. Write its S.I. unit.
(b) A rectangular loop of sides a and b carrying current I2 is kept at a distance ‘a’ from an infinitely long straight wire carrying current I1 as shown in the figure. Obtain an expression for the resultant force acting on the loop. [3]
Answer:
(a) Self-inductance : Self-inductance of a coil is numerically equal to the amount of magnetic flux linked with the coil when and current flows through the coil. The S.I. unit of Self-inductance is henry (H) or weber per Ampere lH = 1 wb/A
(b) The force on the side AB of rectangle is attractive as current is flowing in the same direction and on side CD will be repulsive the current is flowing in opposite direction with respect to straight conductor. The resultant magnetic force. On sides AD and BC is zero. The side AB is the straight wire. So, the net force will be attractive and rectangular loop will move towards the straight wire.
Now, force between AB and straight wire
Force between CD and straight wire
CBSE Previous Year Question Papers Class 12 Physics 2019 Outside Delhi Set-III
Note : Except for the following questions, all the remaining questions have been asked in previous set.
Section-A
Question 1.
Distinguish between unpolarized and linearly polarized light. [1]
Answer:
Unpolarized light : The light having vibration of electric field vector in all possible directions perpendicular to the direction of wave propagation the light is known as unpolarized light.
Linearly polarized light : The light having vibrations of electric field vector in only one direction perpendicular to the direction of propagation of light is as plane or linearly polarized light
Question 2.
How is the drift velocity in a conductor affected with the rise in temperature ? [1]
Answer:
With the rise in temperature, the collision of electrons occurs more frequently, so relaxation time decreases and hence drift velocity increases.
Question 3.
Draw the pattern of electric field lines when a point charge + q is kept near an uncharged conducting plate. [1]
Answer :
The lines of force start from + Q and terminates at metal plate inducing negative charge on it.
Section -B
Question 4.
(a) Define the terms,
(i) threshold frequency and
(ii) stopping potential in photoelectric effect. [2]
Answer:
(a) (i) Threshold frequency : The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. It is denoted by V0
(ii) Stopping potential : The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is denoted by V0 (or Vs).
Question 5.
Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom. [2]
Answer:
According to Bohr’s postulate,
Question 6.
A charged particle q is moving in the presence of a magnetic field B which is inclined to an angle 30° with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path. [2]
Answer:
When a charged particle enter in a magnetic field making angle at 30°. Then velocity component is resolved into 2 components, v cos θ (along the magnetic field) and v sin θ (normal to the magnetic field). As the charged particle moves along XY- plane due to velocity component v sin θ, it also advances linearly due to the velocity component v cos θ. As a result, the charged particle will move in a helical path as shown in figure.
Section-C
Question 7.
(a) Explain briefly how Rutherford scattering of a-particle by a target nucleus can provide information on the size of the nucleus.
(b) Show that density of nucleus is independent of its mass number A. [3]
Answer:
(a) In Rutherford’s scattering experiment of a-particle, it was observed that the fast and heavy a-particles could be deflected through 180°. But only very small number of particles i.e., 1 in about 8,000 a-particles are deflected through 180° that too from centre only. So by this it was assumed that the size of central part i.e., nucleus is about of
the size of the atom and whole positive charge is concentrated in it.
(b) Consider an atom whose mass number is A and R be the radius of the nucleus. If we neglect the mass of orbital electrons, then mass of the nucleus of the atom of mass number A = A a.m.u.
Question 8.
State the underlying principle of a cyclotron. Explain its working with the help of a schematic diagram. Obtain the expression for cyclotron frequency. [3]
Answer:
(a) Principle : It is based on a principle that a positive ion can acquire sufficiently large energy with a comparatively smaller alternating potential difference by making it to cross the same electric field again and again by making use of a strong magnetic field.
Working: The positive ions are produced from the source at the centre are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of a.c source are so chosen that as the ions comes out of a dee, it changes its polarity and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. This phenomenon is continued till the ion reaches at the periphery of the dees where an deflecting plate deflects the accelerated ion on the target to be bombarded. Expression for cyclotron frequency Suppose a position ion with charge q moving with a velocity v, then
Question 9.
Two infinitely long straight wire A1 and A2 carrying currents I and 21 flowing in the same direction are kept ‘d’ distance apart. Where should a third straight wire A3 carrying current 1.5 I be placed between A1 and A2 so that it experiences no net force due to A1 and A2 ? Does the net force acting on A3 depend on the current flowing throught it ? [3]
Answer:
No, at a same distance the force on the wire A3 is independent of the direction of the current. As if current is in opposite direction then F1 and F2 will be in opposite direction, but will be in equilibrium.
Question 10.
(a) Draw the equipotential surfaces due to an electric dipole.
(b) Derive an expression for the electric field due to a dipole of dipole moment if at a point on its perpendicular bisector. [3]
Answer:
(a)
The Magnitudes of the electric field due to the two charges +q and -q given by.
The directions of E+q and E-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
.’. Total electric field
CBSE Previous Year Question Papers
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