CBSE Previous Year Question Papers Class 12 Maths 2016 Delhi

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

CBSE Previous Year Question Papers Class 12 Maths 2016 Delhi Set I

Section – A

Question 1.
Find the maximum value of [1]

Solution:

Question 2.
If A is a square matrix such that A² = I, then find the simplified value of (A – I)³ + (A + I)³ – 7A. [1]
Solution:
Given,
(A – I)³ + (A + I)³ – 7A
= A³ – I³ – 3A²I + 3AI² + A³ + I³ + 3A²I
= 2A³ + 6AI² – 7A
= 2A.A² + 6AI² – 7A
= 2AI + 6AI – 7A
= 8A – 7A = A

Question 3.
Matrix A = is given to be symmetric, find values of a and b. [1]
Solution:

Question 4.
Find the position vector of a point which divides the join of points with position vectors and externally in the ratio 2 : 1. [1]
Solution:
Let A and B be the given points with

Question 5.
The two vectors represent the two sides AB and AC, respectively of a ∆ABC. Find the length of the median through A. [1]
Solution:
In ∆ABC,

Question 6.
Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is . [1]
Solution:

Section – B

Question 7.
Prove that: [4]

Solution:
LHS =

OR
Solve for x : 2 tan^-1(cos x) = tan^-1(2 cosec x)
Solution:
Given, 2 tan^-1(cos x) = tan^-1(2 cosec x)

Question 8.
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value ? [4]
Solution:
Let the monthly incomes of Aryan and Babban be 3x and 4x respectively.
Suppose their monthly expenditures are 5y and 7y respectively.
Since each saves ₹ 15,000 per month.
Monthly saving of Aryan : 3x – 5y = 15,000
Monthly saving of Babban : 4 x – 7y = 15,000
The above system of equations can be written in the matrix form as follows:

⇒ x = 30,000 and y = 15,000
Therefore,
Monthly income of Aryan = 3 × 30,000 = ₹ 90,000
Monthly income of Babban = 4 × 30,000 = ₹ 1,20,000
Value : Saving in good time helps us to survive in bad times.

Question 9.
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), find the values of . [4]
Solution:
We have, x = a sin 2t (1 + cos 2t)


OR
If y = x^x prove that .
Solution:
Given, y = x^x
Taking log on both sides, we get
log y = log(x^x)

Question 10.
Find the values of p ans q for which [4]

Solution:

Question 11.
Show that the equation of normal at any point on the curve x = 3 cos t – cos³ t and y = 3 sin t- sin³ is 4(y cos³ t – x sin³t) = 3 sin 4 t. – sin³ t is [4]
Solution:
Given,

Question 12.
Fimd: [4]

Solution:


OR
Evaluate

Solution:

Question 13.
Find: [4]

Solution:

Question 14.
Evaluate : [4]

Solution:

Question 15.
Find the particular solution of the differential equation (1 – y²) (1 + logx) dx + 2xy dy=0, given that y = 0 when x = 1. [4]
Solution:
The given differential equation is,
(1 – y²) (1 + log x).dx + 2xy dy = 0

Question 16.
Find the general solution of the following differential equation: [4]

Solution:
The given differential equation is,

Question 17.
Show that the vectors are coplanar if are coplanar. [4]
Solution:

Question 18.
Find the vector and cartesian equations of the line through the point (1, 2, – 4) and perpendicular to the two lines.

Solution:
The equations of the given lines are

Question 19.
Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C. [4]
Solution:
Let E₁, E₂ and E₃ be the events denoting the selection of A, B and C as managers respectively.
P(E₁) = Probability of selection of A =
P(E₂) = Probability of selection of B =
P(E₃) = Probability of selection of C =
Let A be the event denoting the change not taking place.
P(A/E₁) = Probability that A does not introduce change = 0.2
P(A/E₂) = Probability that B does not introduce change = 0.5
P(A/E₃) = Probability that C does not introduce change 0.7 .
∴ Required probability = P(E₃/A)
By Bayes’ theorem, we have


OR
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.
Solution:
Total of 7 on the dice can be obtained in the following ways :
(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)

Section – C

Question 20.
Let f : N → N be a function defined as f(x) = 9x² + 6x – 5. Show that f : N → S, where S is the range off, is invertible. Find the inverse of f and hence find f^-1 (43) and f^-1 (163). [6]
Solution:
Given,

Question 21.
Prove that is visible by (x + y + z) and hence find the quotient. [6]
Solution:


OR
Using elementary transformations, find the inverse of the matrix A = and use it to solve the following system of linear equations:
8x + 4y + 3z = 19
2x + y + z = 5
x + 2y + 2z = 7
Solution:

Question 22.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is . Also find maximum volume in terms of volume of the sphere. [6]
Solution:
A sphere of fixed radius (r) is given. Let R and h be the radius and the height of the cone respectively.





OR
Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing.
Solution:
Consider the function
f(x) = sin 3x – cos 3x
⇒ f'(x) = 3 cos 3x + 3 sin 3x
= 3 (sin 3x + cos 3x)

Question 23.
Using integration find the area of the region {(x, y) : x² + y² ≤ 2ax, y² ≥ ax, x, y ≥ 0} [6]
Solution:
We have, {(x, y) : x² + y² ≤ 2ax, y² ≥ ax, x, y ≥ 0}
Consider x² + y² = 2ax
y² = ax …9ii)
x = 0, y = 0
Solving equation (i) and (ii), we get
x² + ax = 2ax
⇒ x² – ax = 0
⇒ x(x – a) = 0
∴ x = 0, a
So, points of intersections of (i) and (ii) are (0, 0) and (a, ± a). Also, equation (i) can be written as, (x – a)² + (y – 0)² = a² whose centre is at (a, 0) and radius is of ‘a’ units.

Question 24.
Find the coordinate of the point P where the line through A(3, -4, -5) and B (2, – 3, 1) crosses the plane passing through three points L (2, 2, 1), M (3, 0, 1) and N (4, – 1, 0). Also, find the ratio in which P divides the line segment AB. [6]
Solution:
The equation of the plane passing through three given points can be given by


At the point of intersection, these points satisfy the equation of the plane 2x + y + z – 7 = 0
Putting the values of x, y and z in the equation of the plane, we get the value of λ
2(-λ + 3) + (λ – 4) + (6λ – 5) – 7 = 0
⇒ -2λ + 6 + λ – 4 + 6λ – 5 – 7 = 0
⇒ 5λ =10
⇒ λ = 2
Thus, the point of intersection is P (1, – 2, 7). Now, let P divide the line AB in the ratio m : n.
By the section formula, we have

Hence, P divides externally the line segment AB in the ratio 2 : 1.

Question 25.
An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution. [6]
Solution:
Let X denote the total number of red balls when four balls are drawn one by one with replacement.

Question 26.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at ₹ 7 profit and B at a profit of ₹ 4. Find the production level per day for maximum profit graphically.
Solution:
Let the numbers of units of products A and B to be produced be x and y, respectively.


Therefore, the manufacturer has to produce 2 units of product A and 3 units of product B for the maximum profit of ₹ 26.

All questions are same in Outside Delhi Set II and Set III

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Maths 2016 Delhi appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Chemistry 2015 Outside Delhi

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

CBSE Previous Year Question Papers Class 12 Chemistry 2015 Outside Delhi Set I

Question 1.
Write the formulae of any two oxoacids of sulphur. [1]
Answer:
Oxoacids of sulphur are chemical compounds that contain sulphur, oxygen and hydrogen.
Two oxoacids of sulphur are:
1. Peroxodisulphuric acid: H₂S₂O₈
2. Dithionic acid: H₂S₂O₆.

Question 2.
Write the IUPAC name of the given compound: [1]

Answer:
Ethoxy-2-methyl propane.

Question 3.
A delta is formed at the meeting point of seawater and river water. Why? [1]
Answer:
River water is the negatively charged colloidal solution whereas seawater contains a number of electrolytes the meeting point of seawater and river water, the electrolytes present in seawater coagulate the colloidal solution of clay resulting in its deposition with the formation of the delta.

Question 4.
Which would undergo S_N1 reactions faster in the following pair: [1]

Answer:

Question 5.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 273rd of tetrahedral voids? [1]

Question 6.
Write one similarity and one difference between the chemistry of lanthanoids and that of actinoids. [2]
Answer:
Similarity
1. Both involve the filling of f-orbital (i.e. 4f and 5f)
2. Both show contraction i.e. Lanthanide contraction and actinoid contraction.
Differences between lanthanoids and actinoids

Question 7.
(i) Write down the IUPAC name of the following complex: [CO(NH₃)₅Cl]²⁺
(ii) Write the formula for the following complex: Potassium tetrachloridonickelate (II). [2]
Answer:
(i) [CO(NH₃)₅Cl]₂: Pentaamminechlorocobalt (III) ion
(ii) Potassium tetrachloridonickelate(II): K₂[NiCl₄]

Question 8.
Write the reagents required in the following reactions:

OR
Arrange the following compounds in increasing order of their property as indicated:
(i) CH₃COCH₃, C₆H₅COCH₃, CH₃CHO (reactivity towards nucleophilic addition reaction)
(ii) Cl-CH₂-COOH, F-CH₂-COOH, CH₃-COOH (acidic character).

Answer:
(i) CH₂=CH-CH₂OH

Nucleophiles are negatively charged entity, the positivity of carbon in the carbonyl group facilitates nucleophilic addition. Thus positivity of carbon in carbonyl group is in order HI > I > II. Due to +1 effect of -CH₃ group and benzene ring π-cloud, the positivity of carbonyl carbon is decreased so much than (I) and (II). Thus nucleophilic addition order III > I > II.

The stability order II > I > III. Due to +1 effect of -CH₃ group electron density of oxygen is more. But in (I) and (II), -I effect of -F is more than -Cl. Thus II is more stabilized than I. Thus acidity order is II > I > III.

Question 9.
(i) On mixing liquid X and liquid Y, the volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason. [2]
Answer:
(i) Negative deviation of Raoult’s law occurred. There is an elevation of boiling point i.e., the temperature of the solution increases.
(ii) When a blood cell is placed in water (hypotonic solution), water penetrate to blood cell and blood cell gets bulged and then disrupt.

Question 10.
Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO₄.
(Molar mass of Cu = 63.5 g mol^-1, 1F = 96500 C mol^-1). [2]
Answer:
Cu²⁺ + 2e^– → Cu
63.5 g of copper deposited by 2 × 96500 C
1.27 g of copper will be deposited by

Question 11.
A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308 K = 32 mm Hg). [3]
Answer:
According to Raoult’s law,

Where
P⁰ → Vapour pressure of pure water
P → Vapour pressure of the solution
w → Weight of solute
m → Molecular weight of solute
W → Weight of solvent
M → Molecular weight of solvent

or, 0.005 m = 0.9 or, m = 180.
Thus molar mass of the solute 180 gm./mol

Question 12.
(i) Name the method of refining to obtain silicon of high purity.
(ii) What is the role of SiO₂ in the extraction of copper?
(iii) What is the role of depressants in froth floatation process? [3]
Answer:
(i) Zone refining is the method of refining to obtain silicon of high purity.
(ii) SiO₂ combined with the iron in the copper ore to form iron(II) silicate slag which is easily removed. Thus it (SiO₂) acts as a flux to remove the impurity of Iron oxide.

(iii) In the froth flotation process, depressant prevents the formation of froth. It is used to separate two sulphide ore by preventing the formation of froth of one sulphide ore and allowing the other to form the froth e.g., NaCN a depressant selectively prevents ZnS from coming in froth but allows PbS to come with the froth.

Question 13.
(i) Which one of the following is a polysaccharide:
Starch, maltose, fructose, glucose.
(ii) Write one difference between α-helix and β-pleated sheet structures of the protein.
(iii) Write the name of the disease caused by the deficiency of vitamin B₁₂. [3]
Answer:
(i) Starch is a polysaccharide.
(ii)

(iii) Pernicious Anaemia caused by the deficiency of Vitamin B₁₂.

Question 14.
(i) What type of isomerism is shown by the complex [Cr(H₂O)₆] Cl₃?
(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ₀ > P.
(iii) Write the hybridization and shape of [CoF₆]^3-.
(Atomic number of Co = 27). [3]
Answer:
(i) Hydration isomerism is shown by the complex [Cr(H₂O)₆] Cl₃.
[Cr(H₂O)₅ Cl] Cl₂ (H₂O); [Cr(H₂O)₃ Cl₃] 3(H₂O).
(ii) Since Δ₀ > P it is on splitting t⁴_2g e_g⁰
(iii) [CoF₆]^3-: sp³d² hybridisation and the shape is octahedral.

[CoF₆]^3- has four unpaired electrons. thus it is paramagnetic.

Question 15.
How can the following conversion be carried out:
(i) Aniline to bromobenzene
(ii) Chlorobenzene to 2-chloroacetophenone
(iii) Chloroethane to butane. [3]
OR
What happens when
(i) Chlorobenzene is treated with Cl₂/FeCl₃
(ii) Ethyl chloride is treated with AgNO₂
(iii) 2-bromopentane is treated with alcoholic KOH?
Write the chemical equations in support of your answer.
Answer:

Question 16.
Examine the given defective crystal:

Answer the following questions:
(i) Is the above defect stoichiometric or non- stoichiometric?
(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.
(iii) How does this defect affect the density of the crystal? [3]

Question 17.
The conductivity of 2.5 × 10^-4 M methanoic acid is 5.25 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation.
Given: λ(H⁺) = 349.5 S cm² mol^-1 and λ⁰(HCOO^–) = 50.5 S cm² mol^-1. [3]
Answer:
We know molar conductivity

Question 18.
Write any three differences between Physisorption and Chemisorption. [3]
Answer:

Question 19.
Give reasons for the following:
(i) Phenol is more acidic than methanol.
(ii) The C – O – H bond angle in alcohols is slightly less than the tetrahedral angle (109⁰28′).
(iii) (CH₃)₃C – O – CH₃ on reaction with HI gives (CH₃)₃C – I and CH₃ – OH as the main products and not (CH₃)₃C – OH and CH₃ – I. [3]
Answer:
(i) Phenoxide ion is more stabilized than CH₃ – O- ion because of resonance:

Thus phenol is more acidic than CH₃ – OH.
(ii) Due to the presence of lone pair on oxygen which causes repulsion, the bond angle in alcohol is slightly less than tetrahedral angle (109⁰28′).
(iii) Since (CH₃)₃ C⁺ is more stabilized through +I effect of three methyl group thus (CH₃)₃ C – I and CH₃ – OH formed on the treatment of HI with (CH₃)₃C-O-CH₃.

Question 20.
Predict the products of the following reactions: [3]

Answer:

Question 21.
(a) Account for the following:
(i) Cu⁺ is unstable in an aqueous solution.
(ii) Transition metals form complex compounds.
(b) Complete the following equation:
Cr₂O^–₇ + 8H⁺ + 3NO^–₂ → [3]
Answer:
(a) (i) 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ ion in aqueous solution is due to its greater negative charge than Cu. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.
(ii) Since transition metals have unfilled or partially filled d-orbital thus to satisfy its octane forms complex compounds.

Question 22.
Write the names and structures of the monomers of the following polymers:
(i) Terylene
(ii) Buna-S
(iii) Neoprene [3]
Answer:

Question 23.
Seeing the growing cases of diabetes and depression among young children, Mr Chopra, the principal of one reputed school organized a seminar in which he invited parents and principals. They all resolved this issue by strictly banning junk food in schools and introducing healthy snacks and drinks like soup, lassi, milk, etc. in school canteens. They also decided to make compulsory half an hour of daily physical activities for the students in the morning assembly. After six months, Mr Chopra conducted the health survey in most of the schools and discovered a tremendous improvement in the health of the students.

After reading the above passage, answer the following questions:
(i) What are the values (at least two) displayed by Mr Chopra?
(ii) As a student, how can you spread awareness about this issue?
(iii) Why should antidepressant drugs not be taken without consulting a doctor?
(iv) Give two examples of artificial sweeteners. [4]
Answer:
(ii) Awareness can be spread by performing nukkad Natak in community, displaying posters, cartoons and slogans and by conducting seminars.
(iii) Antidepressant drugs have lots of side effects like indigestion, headache, stomach aches, drowsiness, weight gain. That is why it should not be taken without consulting with doctors.
(iv) Example of artificial sweeteners is Aspartame, Saccharin, Sucralose etc.

Question 24.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is a large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF₃
(ii) XeF₄ [5]
OR
(i) Which allotrope of phosphorus is reactive and why?
(ii) How are the supersonic jet aeroplanes responsible for the depletion of the ozone layer?
(iii) F₂ has a lower bond dissociation enthalpy than Cl₂. Why?
(iv) Which noble gas is used in filling balloons for meteorological observations?
(v) Complete the following equation:
XeF₂ + PF₅ → [5]
Answer:
(a) (i) Size of halide ions is of the order
F^– < Cl^– < Br^– < I^– with increase in size negative charge is dispersed throughout ions and ions gets stabilized. Thus acidity order
HF < HCl < HBr < HI.
(ii) Due to the presence of vacant d-orbital and combining forms of sulphur (S₈) which is not present in oxygen, the cohesive energy of sulphur is more than oxygen leading to large melting point and boiling point.

(ii) The exhaust emitted from supersonic jet aeroplane contains CO₂, NO and other particles which are the killers of stratospheric ozone layer along with the supersonic sound produced by these aeroplanes are destroy the ozone layer.
(iii) Due to the smaller size and high electro-negativity of fluorine, more energy is required to break the bond of F₂ than Cl₂. Thus F₂ has lower bond dissociation energy than Cl₂.
(iv) Helium (He) gas is used in filling balloons for meteorological observations because it is non-inflammable and light gas.
(v) XeF₂ + PF₅ → [XeF]⁺ [PF₆ ]^–.

Question 25.
An aromatic compound ‘A’ of molecular formula C₇H₆O₂ undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions: [5]

OR
(a) Write the structures of main products when benzene diazonium chloride reacts with the following reagents:
(i) H₃PO₂ + H₂O
(ii) CuCN/KCN
(iii) H₂O
(b) Arrange the following in the increasing order of their basic character in an aqueous solution:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N
(c) Give a simple chemical test to distinguish between the following pair of compounds:
C₆H₅-NH₂ and C₆H₅-NH-CH₃ [5]
Answer:
A: C₆H₅COOH (Benzoic acid)
B: C₆H₅NH₂ (Aniline)
C: C₆H₅NHCOCH₃ (N-phenyl ethanamide)
D: C₆H₅CH₂NH₂ (Benzylamine)
E: C₆H₄(Br)₃ (NH₂) (2,4, 6-Tribromoaniline)
OR


(b) C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
+I effect of three C2Hs group increases the enormous availability of lone pair of nitrogen atom than +I effect of 2 ethyl group.
Thus the order of basicity is I < III < II.
(c) C₆H₅ – NH₂ (Primary amine) and C₆H₅ – NH – CH₃ (Secondary amine) can be distinguished by Hinsberg’s test. In this test, amines are allowed to react with Hinsberg’s reagent, benzene sulphonyl chloride (C₆H₅SO₂Cl). Primary amines react with this reagent to form N-alkylbenzene sulphonyl amide which is soluble in alkali but secondary amines give sulphonamide which is insoluble in alkali.

Question 26.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

(a) Show that it follows pseudo-first-order reaction, as the concentration of water remains constant.
(b) Calculate the average rate of reaction between the time interval 10 to 20 seconds.
(Given: log 2 = 0.3010, log 4 = 0.6021). [5]
OR
(a) For a reaction A + B → P, the rate is given by Rate = k[A][B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first-order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
Answer:


OR
(a) (i) Since the given reaction has order two with respect to reactant B, thus if the concentration of B is doubled in the given reaction, then the rate of reaction will become four times.
(ii) It the concentration of B is doubled i.e.; [B]² the overall reaction will be two, because if A is present in large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B.
Order of reaction = 2.

Note: All questions are same in outside Delhi Set II and III.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Chemistry 2015 Outside Delhi appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Maths 2015 Delhi

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

CBSE Previous Year Question Papers Class 12 Maths 2015 Delhi Set I

Section – A

Question 1.
If , then find the projection of . [1]
Solution:
Given,

Question 2.
Find λ, if the vectors and are coplanar. [1]
Solution:
Since, the given vectors are coplanar.

∴ λ = 7

Question 3.
If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ. [1]
Solution:
Given,

Question 4.
Write the element a₂₃ of a 3 × 3 matrix A = (a_ij) whose elements a_ij are given be a_ij = . [1]
Solution:
Given,
a_ij =

Question 5.
Find the differential equation representing the family of curves v = + B, where A and B are arbitrary constants.  [1]
Solution:
We have,
v = + B …(i)
Since, the given equation contains two arbitrary constants, we shall differentiate it two times. Now, differentiating (i) w.r.t. r, we get

This is the required differential equation representing the family of the given curve.

Question 6.
Find the integrating factor of the differential equation . [1]
Solution:
We have,

Section – B

Question 7.
If A = , find A² – 5A + 4I and hence find a matrix X such that A² – 5A + 4I + X = 0. [4]
Solution:
We have,


OR

Solution:
Given,

Question 8.
If f(x) = , using properties of determinants find the value of f(2x) – f(x). [4]
Solution:

Question 9.
Find: [4]

Solution:



OR
Integrate the following w.r.t. x: .
Solution:
Given,

Question 10.
Evaluate: [4]

Solution:

Question 11.
A bag ‘A’ contains 4 black and 6 red balls and bag ‘B’ contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black. [4]
Solution:
Consider the following events :
E₁ = Getting 1 or 2 on die.
E₂ = Getting 3, 4, 5 or 6 on die.
E = One of the ball drawn is red and another is black.

OR
An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained.
Solution:
Let X denote the number of heads in the four tosses of the coin, then X is a random variable that can have values 0, 1, 2, 3, 4.
P(X = 0) = Probability of getting no head (TTTT)

Question 12.
If , find + xy. [4]
Solution:
Given,

Question 13.
Find the distance between the point (- 1, – 5, – 10) and the point of intersection of the line and the plane x – y + z = 5. [4]
Solution:

Question 14.
If sin [cot^-1 (x + 1)] = cos(tan^-1 x), then find x. [4]
Solution:
Given, sin [cot^-1 (x +1)] = cos(tan^-1 x)

OR
If (tan^-1 x)² + (cot^-1 x)² = , then find x.
Solution:
Given, (tan^-1 x)² + (cot^-1 x)² =

Question 15.
If y = tan^-1, x² ≤ 1, then find . [4]
Solution:
Given,

Question 16.
IF x = a cos θ + b sin θ, y = a sin θ – b cos θ, show that . [4]
Solution:
We have,
x = a cos θ + b sin θ …(i)
y = a sin θ – b cos θ …(ii)

Question 17.
The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of the triangle is 20 cm ? [4]
Solution:
We know that, Area of an equilateral triangle,

Question 18.
Find . [4]
Solution:

Question 19.
Three schools A, B and C organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats and plates from recycled material at a cost of ₹ 25, ₹ 100 and ₹ 50 each. The number of articles sold are given below:

Find the fund collected by each school separately by selling the above articles. Also find the total funds collected for the purpose. Write one value generated by the above situation. [4]
Solution:
The number of articles sold by each school can be represented by the 3 × 3 matrix

Hence, the funds collected by schools A, B and C are ₹ 7,000, ₹ 6,125 and ₹ 7,875 respectively.
The total funds collected for flood victims
= ₹ (7,000 + 6,125 + 7,875)
= ₹ 21,000
The above situation exhibits the helping nature of students.

Section – C

Question 20.
Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) if ad(b + c) = bc(a + d). Show that R is an equivalence relation. [6]
Solution:
We know that relation R will be an equivalence relation, if we prove it as a reflexive, symmetric and transitive relation.
(i) Reflexivity:
Let (a, b), be an arbitrary element of N × N
then, (a, b) ϵ N × N
a, b ϵ N
⇒ ab(b + a) = ba(a + b)
⇒ (a, b) R (a, b)
(a, b) R (a, b) ∀ (a, b) ϵ N × N.
(ii) Symmetry:’
Let (a, b), (c, d) be ah arbitrary element of N × N such that (a, b) R (c, d)
⇒ ad(b + c) = bc(a + d)
⇒ cb(d + a) = da(c + b)
⇒ (c, d) R (a, b)
∴ (a, b) R (c, d) ⇒ (c, d) R (a, b) ∀ (a, b), (c, d) ϵ N × N So, R is symmetric on N × N.
(iii) Transitivity:
Let (a, b), (c, d), (e, f) be an arbitrary element of N × N such that (a, b) R (c, d) and (c, d) R then (a, b) R (c, d)

Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f). ∀ (a, b) (c, d) (e, f) ϵ N × N
So, R is transitive on N × N.
Hence, R being reflexive, symmetric and transitive, is an equivalence relation on N × N.
Hence Proved.

Question 21.
Using integration find the area of the triangle formed by positive x-axis and tangent and normal to the circle x² + y² = 4 at (1, ). [6]
Solution:
The equation of the given circle is x² + y² = 4. The equation of the normal to the circle at (1, ) is same as the line joining the points (1, ) and (0, 0) which is given by




OR
Evaluate dx as a limit of a sum.
Solution:

Question 22.
Solve the differential equation: (tan^-1 y – x) dy = (1 + y²) dx. [6]
Solution:
Same as solution Q. 23 (OR) Set 1 (Outside Delhi) upto eq. x = tan^-1 y – 1 + c e^{tan-1 y}
OR
Find the particular solution of the differential equation given that y = 1, when x = 0.
Solution:

Question 23.
If lines intersect, then find the value of k and hence find the equation of the plane containing these lines. [6]
Solution:
The coordinates of any point on first line are

Question 24.
If A and B are two independent events such that P( ∩ B) = and P(A ∩ ) = then find P(A) and P(B). [6]
Solution:

Question 25.
Find the local maxima and local minima of the function f(x) = sin x – cos x, 0 < x < 2π. Also find the local maximum and local minimum values. [6]
Solution:
We have,/(x) = sin x – cos x, 0 < x < 2π.
⇒ f'(x) – cos x + sin x
For local maximum or minimum, we have
⇒ f'(x) = o
⇒ cos x + sin x = 0
⇒ cos x = – sin x
⇒ tan x = -1

Question 26.
Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below: [6]
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≤ 0
Solution:
We first convert the inequalities into equations to obtain lines
2x + 4y = 8 ….(i)
3x + y = 6 …..(ii)
x + y = 4 —(iii)
x = 0
and y = 0.
We need to maximize the objective function z = 2x + 5y
These lines are drawn and the feasible region of the L.P.P. is the shaded region :

The point of intersection of (i) and (ii) is B (1.6, 1.2)
The coordinates of the corner points of the feasible region are 0(0, 0), A(0, 2), B (1.6, 1.2) and C(2, 0).
The value of the objective function at these points are given in the following table:

Out of these values of z, the maximum value of z is 10 which is attained at the point (0, 2). Thus the maximum value of z is 10.

All questions are same in Delhi Set II and Set III

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Maths 2015 Delhi appeared first on Learn CBSE.

NTSE Punjab 2019-20: State Council of Educational Research and Training (SCERT) will conduct the NTSE Punjab. The NTSE Punjab will be held on 3rd November 2019. The Punjab application process will start in the last week of August 2019. Class 10th students can apply for NTSE Punjab through their schools till the last week of September 2019. NTSE Punjab admits card will be released online in the third week of October 2019. Students preparing for NTSE Punjab can also find the study materials, test papers and mock test papers provided by SCERT.

The NTSE Exam Stage 1 qualified candidates will only be eligible to attend the stage 2 exam conducted by NCERT. Finally, shortlisted ones will be awarded scholarships till the Ph.D. level of their academics. The talented students will get financial assistance from the scholarship sponsored by NCERT. Students should read the article below to get information about NTSE Punjab 2019-20. Information including dates, eligibility, application, admit card, answer key and question pattern, etc.

NTSE Punjab Important Dates

State-level National Talent Search Exam

NTSE Punjab Eligibility Criteria

Refer to the below eligibility conditions to appear for the NTSE Punjab 2019-20.

• Students studying in Navodaya Vidyalaya, Kendriya Vidyalaya or PSEB affiliated school are eligible.
• Class 10th regular students of Punjab who are studying in the academic year of 2019-20.
• General category students who have scored at least 70% marks in Class 9th can apply for the exam.
• Reserved category students who have scored at least 55% marks in Class 9th can apply for the exam.
• Students who are employed or getting a scholarship from any other source are not eligible for this year.
• Students from open schooling or distance learning within the age limit are eligible.
• NMMS scholarship holders are also eligible to apply for NTSE Punjab.

NTSE Punjab Application Form

The application form for NTSE Punjab will be released online by SCERT. Students can submit the application starting from the last week of August till the last week of September 2019. Students can apply for NTSE Punjab through their schools on the official portal of the education department. The below-mentioned information indicates how to apply for NTSE Punjab 2019-20.

• The school authority has to visit the official website epunjabschool.gov.in and login with their credentials.
• Click on new NTSE application form appeared on the Dashboard of the portal.
• Make sure to fill all the required information correctly to avoid rejection of the form by SCERT.
• Students must attach their passport size photo with signature and photocopies of 9th class mark sheets.
• Students also upload the photocopies of caste certificates, and disability certificates if required.
• Submit the application along with the attached documents to the state liaison officer.

NTSE Punjab Application Fee

Applicants no need to pay any application fee to appear for the NTSE Punjab Stage 1.

NTSE Punjab Admit Card

• The admit card will be released online on the official website epunjabschool.gov.in by SCERT.
• The admit card can only be downloaded using the school login credentials.
• Students can collect the admit card from the schools in the third week of October 2019.
• Admit card contains the student and exam center information.
• Students must bring the admit card while going to attend the exam.
• Without the admit card students will not be allowed to sit in the exam.

NTSE Punjab Question Pattern

Students must know the NTSE Punjab exam pattern while preparing for the exam. The NTSE Punjab contains two papers i.e, the Mental Ability Test (MAT) and Scholastic Ability Test (SAT). Refer to the table below to know the exam pattern for NTSE Punjab Stage 1.

• There will be given 1 mark for each right answer whereas no negative marking for wrong answers.
• MAT questions test the candidates thinking abilities, problem-solving and reasoning skills.
• SAT questions test the candidate’s knowledge on each subject of their academics.
• Total of 200 questions will be asked in MAT and SAT with a duration of 240 minutes.
• Students can expect the questions asked in the exam will be from the class 9th and 10th syllabus.

NTSE Punjab Stage 1 Mock Test

• NTSE Punjab Stage 1 also conducts mock tests to help students in preparing well for the exam.
• The mock test papers can be downloaded from the official website to practice before the exam.
• The answer key will be released by SCERT immediately on the next day. Download the answer key and compare it with the solved ones to determine the performance.
• Students can also download the previous year mock test papers to practice for the exam.
• In this way, students can determine their level of preparation and can do some workaround to improve their performance.

NTSE Punjab Answer Key

The official answer key will be published online by SCERT in the first week of December 2019. Candidates can also check the answer keys in some coaching institutes websites after the exam. It helps the candidates to estimate their probable scores before the result declaration and start preparing for stage 2. Candidates can also compare their answers with the answer keys and determine their performance in NTSE Punjab. Using answer keys candidates can find out the correct answers to all the questions asked in MAT and SAT. The following points are given to check the NTSE Punjab answer keys:

1. Visit the official website ssapunjab.org to download the answer key.
2. Students can compare their answers with the ones available in the answer keys to calculate the scores.
3. Students must remember the exam pattern and cut-off scores while calculating the marks.

NTSE Punjab Exam Result

NTSE Punjab result will be declared along with the cut-off by SCERT. The result will be released on the official website in the last week of January 2020. The exam results can be downloaded in pdf format. It contains the details of all the shortlisted candidates who appeared in the exam. The exam results will be released as a merit list which includes whole information of the shortlisted candidates. The merit list contains the names of those candidates who are eligible for the NTSE stage 2 to be held on May 10, 2019. Refer to the steps below to know the NTSE Punjab exam results:

• Visit the official website www.ssapunjab.org.
• Click on the SCERT circular and download the pdf file of results.
• Check the roll number in the downloaded pdf file.
• The resulting file consists of the details of the selected candidates for the stage 2 exam.
• Take a photocopy of the result and keep it safely for future purpose.

Details Mentioned on NTSE Punjab Exam Result

Punjab candidates can check the merit list on the official website of SCERT. The NTSE Punjab merit list contains the below-mentioned information.

• Candidates Roll Number, School, and Name
• Marks obtained in MAT, SAT and Total scores of the candidate
• Gender, Category, Date of Birth and Disability Status of the candidate

NTSE Punjab Stage 1 Cut-Off

The cut-off score is the minimum score must be required to get selected for the stage 2 exam. Punjab cut-off will be declared along with the result by SCERT. The NTSE Punjab cut-off will be released in the last week of January 2020. Students can check the previous year’s cut-off scores and get an idea of this year’s cut-off. The NTSE Punjab cut-off is published for all category candidates depending on various factors as given below:

• Total number of candidates belongs to each category
• Highest marks obtained by the candidate in the exam
• Minimum marks obtained by the candidate in the exam
• The exam difficulty level
• Previous year’s cut-off

The NTSE Punjab cut-off is different for different categories. Refer to the table below to find the previous year’s cut-off for NTSE Punjab.

Candidates can check the qualifying cut-off for NTSE Punjab category candidates

NTSE Punjab Reservation Criteria

The reservation criteria for scholarships to all the states/UTs have been determined by NCERT. Total 183 scholarships are reserved for Punjab state. Based on the below reservation percentages students will be selected by SCERT.

NTSE Punjab Stage 1 Syllabus

SCERT, Punjab has not specified any syllabus for stage 1 exam. However, Punjab candidates can practice and expect the questions from the class 9th and 10th syllabus. Refer to the table below to know the important topics for NTSE Punjab.

NTSE Scholarship Information 2019

Punjab students who qualified in NTSE Stage 2 will get the scholarship amount from NCERT every month. The table below shows information about scholarship amounts distributed to the qualified candidates.

FAQ’s on NTSE Punjab 2019

Question 1.
Is there another way to check the NTSE Punjab result other than checking online?

Answer:
Yes, you can contact your school authorities to know whether you are qualified in stage 1 or not.

Question 2.
What is the reservation criteria in NTSE Punjab 2019?

Answer:
Total 183 scholarships are reserved for Punjab state. All states reservation criteria for scholarships has been determined by NCERT.

Question 3.
How do I check my NTSE Punjab result?

Answer:
NTSE Punjab result will be released online in the form of merit list by SCERT. Students can visit the official website www.ssapunjab.org to check their result.

Question 4.
What happens after the NTSE Punjab result declaration?

Answer:
Students whose names and roll number are available in the stage 1 result can only be eligible for stage 2 exam. The stage 2 exam will be held on May 10, 2020, conducted by NCERT. The deserved student who qualified stage 2 exam will be awarded a scholarship for higher studies.

Question 5.
What will be the result date of stage 1 for the candidates who belong to Punjab state?

Answer:
The stage 1 result will be released on the official website of SCERT in the last week of January 2020.

Question 6.
Do I know whether NTSE Punjab cut-off is same as NTSE stage 2 cut-off?

Answer:
Yes, you can know that both the cut-offs are not the same. Stage 1 cut-off is totally different from stage 2 cut-off. Since, stage 1 is released by SCERT, Punjab whereas stage 2 is released by NCERT.

Question 7.
Is NTSE Punjab is tougher than NTSE stage 2?

Answer:
No, NTSE stage 2 is tougher than NTSE Punjab. Since NTSE stage 2 is conducted at the national level by NCERT.

Hope this article will help you to get information about NTSE Punjab 2019-20. For any queries in NTSE Punjab, leave your queries in the comment box to get in touch with us.

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

CBSE Previous Year Question Papers Class 12 Physics 2016 Outside Delhi

Section — A

Question 1.
A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process. [1]
Answer:

Work done, W = q(V_B – V_A ) = q × 0 = 0

Question 2.
In what way is the behavior of a diamagnetic material different from that of a para magnetic, when kept in an external magnetic field ? [1]
Answer:
The magnetic field lines pass through the para magnetic material while the magnetic field lines move away from the diamagnetic material, or para magnetic material get aligned along B and dianegnetic aliqied perpendicular to B.

Question 3.
Name the essential components of a communication system. [1]
Answer :
The essential components are : Transmitter, communication channel and receiver.

Question 4.
Why does sun appear red at sunrise and sunset? [1]
Answer :
Sun appears red at sunrise and sunset due to the least scattering of red light as it has the longest wavelength.

Question 5.
The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell ? [1]

Answer:

Section—B

Question 6.
Define modulation index. Why is it kept low ? What is the role of a band pass filter ? [2]

Question 7.
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 emerge ? Justify your answer. [2]


Thus, the light  ray PQ will emergy out from face AC.
Question 8.
Calculate the de-Broglie wavelength of the electron orbiting in the n = 2 stage of hydrogen atom. [2]

Question 9.
Define ionization energy.
How would the ionization energy change when electron in hydrogen atom replaced by a particle of mass 200 times that of the electron but having the same charge ? [2]
OR
Calculate the shortest wavelength of the spectral lines emitted in Balmer series. [Given Rydberg constant, R= 10⁷ m^-1]
Answer :
Ionization energy is defined as the amount of energy needed to remove the valence electron of an isolated gaseous atom.
The ionization energy of hydrogen atom is

Question 10.
A battery of emf 12 V and internal resistance 2 Ω is connected to a 4 Ω resistor as shown in the figure. [2]
(a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.
(b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit ?


So, a voltmeter when placed across the cell and across the resistor, gives the same reading,
(b) An ammeter is connected in series because it has very low resistance. So, when, an ammeter is connected in series, then there is not much increase in the resistance of the circuit and hence the current through the circuit unchanged.
A voltmeter is connected in parallel because it has very high resistance. So, it draws a very small current from the circuit.

Section—C

Question 11.
Define an equipotential surface. Draw equipo- tential surfaces: [3]
(i) in the case of a single point charge and
(ii) in a constant electric field in Z-direction. Why the equipotential surfaces about a single charge are not equidistant ?
(iii) Can electric field exist tangential to an equipotential surface ? Give reason.
Answer: An equipotential surface is the surface which has same potential at its every point.
(i)

(ii)

The electric field due to single charge is not constant, this is the reason why the equipotential surfaces about a single charge are not equidistant and potential vary
inversely with radius i.e., V × 1/r

(iii) No, electric field cannot exist tangential to an equipotential surface. If it happen then a charged particle will experience a force along the tangential line and can move along it. As a charged particle can move only due to the potential difference i.e., along the direction of charge of potential, this contradicts the concept of an equipotential surface.

Question 12.
(i) State law of Malus.
(ii) Draw a graph showing the variation of intensity (I) of polarised light transmitted by an analyser with angle (θ) between polariser and analyser.
(iii) What is the value of refractive index of a medium of polarising angle 60°? [3]
Answer :
(i) Malus discovered that when a beam of completely plane polarized light is passed through the analyser, the intensity T of transmitted light changes directly as the square of the cosine of the angle 0 between the transmission directions of polarizer and analyzer. This is known as the law of Malus.
I ∝ cos² θ
I = I₀ cos² θ
Where, I₀ is the maximum intensity of the transmitted light.

Question 13.
Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies V_A > V_B. [3]
(i) In which case is the stopping potential more and why ?
(ii) Does the slope of the graph depend on the nature of the material used ? Explain.
Answer:

Given, V_A > V_B
Therefore, threshold frequency for metal B is less than the threshold frequency for metal A. Hence, stopping potential is more for metal B.
(ii) No, the slope of the graph tells us the value of h/e which is same for both the materials. So, it does not depend on the nature of the materials.

Question 14.
Write the basic nuclear process involved in the emission of β⁺ in a symbolic form, by a radioactive nucleus.
(b) In the reactions given below:


Find the  value  of  x,y,z and a,b,c.  [3]
Answer:
(a) In β⁺ -decay, the atomic number of the radioactive nucleus decreases by one and its mass number remains same. In this process, a positron (e⁺) and a new particle neutrino (v) are emitted from the nucleus.

The corresponding  values of a,b,c are 10,2 and 4, respectively.

Question 15.
(i) Derive an expression for drift velocity of free electrons.
(ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature ? Explain. [3]
Answer :
(i) Consider a conductor in which an electric field E is produced. Let a free electron experience a force (-eE) in this electric field. So, the acceleration of free electron is



This is the required relation.
(ii) The drift velocity of free electrons in a metallic conductor decreases with increase in temperature, because, if we increase the temperature of the metallic conductor the collision between the electrons and ions increases, which decreases relaxation time. Hence, drift velocity decreases.

Question 16.
(i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor ? Explain. [3]

Answer:
(i) The average power supplied by the source over a complete cycle is

When the circuit contains an ideal inductor, then the phase difference between the current and voltage is π/2.
So,
Hence 
So, when an ac source is connected to an ideal inductor, the average power supplied by the source over a complete cycle is zero.

(ii) The brightness of the lamp will decrease. When the key is plugged in and the iron rod is inserted inside the inductor, it increases the inductance. Hence, the reactance of the inductor (X_L = ωL) increases. So, the impedance of the circuit (Z = R + jωL) increases, which decreases the current in the circuit.

Question 17.
(i) Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction.
(ii) Draw the circuit diagram of a half wave rectifier and explain its working. [3]
Answer :
(i) During the formation of p-n junction, the holes diffuse from p-type semiconductor to the n-type, and electrons diffuse from n-type to p-type. This is because of the concentration gradient across p-side and n-side.

When a hole diffuse from p to n type, it leaves an unmovable negative charge. Similarly, when an electron diffuses from n to p type, it leaves an unmovable positive charge. When the diffusion of holes and electrons takes place continuously across the junction, a layer of unmovable positive and negative charges are developed on either side of the junction. This layer is called the depletion layer or the depletion region and the potential difference across the region is called barrier potential.

(ii) Half wave rectifier:

Working: When an input ac voltage is applied across the primary coil, a potential difference is developed across the ends of the secondary coil. Consider that in half cycle of input ac signal, the end A acts as the +ve end and B acts as the -ve end of the battery. So, the diode is in forward bias and we get output across the ends of the load resistance R_L.

In the second half cycle, ends A and B reverse in polarity. Now, A acts as the -ve end and B acts as the +ve end. So, the diode D is in reverse bias and no output is obtained due to the high resistance offered by the diode.

So, in this process, we get output alternately, and hence the diode is called the half wave rectifier.

Question 18.
(i) Which mode of propagation is used by short wave broadcast service having frequency range from a few MHz up to 30 MHz ? Explain diagrammatically how long distance communication can be achieved, by this mode.
(ii) Why is there an upper limit to frequency of waves used in this mode ? [3]

Question 19.
(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 electron.
(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. [3]
Answer :
(i) (a) Microwaves,
(b) X-rays
(ii) During the charging and discharging of a capacitor, a flow of charges take place from the battery to the plates of the capacitor. This produces a conduction current in the circuit and a displacement current between plates. Hence the galvanometer shows a momentary deflection.

Where

Question 20.
For a CE – transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ . [3]

Question 21.
Define the term wave front. State Huygens’s principle. Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front traverses through the lens and after refraction focuses on the focal point of the lens, giving the shape of the emergent wave front. [3]
OR
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 ?
Answer :
Wave front: A wave front is the locus of all the points in space that reach a particular distance by a propagating wave in same phase at any instant.
Huygens’s principle : It is based on two assumptions :
(a) Each point of the wave front behaves like a source of secondary disturbances and secondary wavelets from there points spread out in all directions with the same speed as that of the original wave front.
(b) When we draw an envelope in the forward direction of the secondary disturbances at any instant, And this envelope tells the new position of the wave front at that instant.

(i) Both the reflection and refraction takes place due to the interaction of light with the atoms at the surface of the separation. Light incident on these atoms, force them to vibrate with the frequency of light. But, the light emitted by these charged atoms is equal to their own frequency of oscillation. So, both the reflected and refracted lights have same frequency, hence frequency remains changed.
(ii) The energy carried by a wave depends on the amplitude of the wave. It does not depend on the speed of the wave propagation. Hence the energy of the wave remains same and does not decrease.
(iii) The intensity of light is determined by the number of photons incident per unit area around the point at which intensity is to be determined.

Question 22.
Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.
Draw the magnetic field lines due to a circular wire carrying current I. [3]
Answer:
Imagine a circular coil of radius R with center O. Let the current flowing through the circular loop be I. Suppose P is any point on the axis at a distance of r from the center O. Let the circular coil be made up of a large number of small elements of current, each having a length of dl.

According  to biot-savart’s law magnetic field at point P will be



Section – D

Question 23.
Ram is a student of class x in a village school. His uncle gifted him a bicycle with a dynamo fitted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. He asked this question to his teacher. The teacher considered it an opportunity to explain the working to the whole class.
Answer the following questions : [4]
(a) State the principle and working of a dynamo.
(b) Write two values each displayed by Ram and his school teacher.
Answer:
(a) A dynamo works on the principle of electromagnetic induction. A dynamo includes a coil attached to a small turbine fitted with a plastic cap. The coil is placed in a magnetic field. When the plastic cap comes in contact with moving tyres of the bicycle, the coil placed between the poles of a magnet rotates, thus, the flux through the coil changes continuously. This induces a current in the coil which is connected to a bulb which lights up. As long as the bicycle is moving, the coil keeps on rotating, and hence, the flux keeps on changing. At a steady rate, we get a steady current and hence a light of steady intensity is obtained.

Section—E

Question 24.
(i) Draw a labelled diagram of a step-down trans former. State the principle of its working.
(ii) Express the turns ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turns ratio in an ideal transformer.
(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V— 550 W refrigerator ? [5]
OR
(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius and the other of radius r₂ (r₁ < r₂) placed co axially with centers coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
(b) A rectangular coil of area A, having number of turns N is rotated at ‘f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2 πf/NBA.
Answer:
(i)

Principle : A transformer works on the principle of mutual induction. Whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighboring coil.

Working: When an alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. According to Faraday’s laws the e.m.f. induced in the primary coil,


(a) Mutual inductance: It is the property of a pair of coils due to which an e.m.f. is induced in one coil due to the change in the flux or current in the other coil. Let a current I₂ flow through the outer circular coil. The magnetic field at the center of the coil is  …… (i)

Question 25.
(a) Derive the mathematical relation between refractive indices n₁ and n₂ of two media 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.
(b) 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 ? [5]
OR
(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 ?



In deriving Lens maker’s formula, we adopt the coordinate geometry sign convention and make the assumptions:
(i) The lens is thin so that the distances measured from the poles of its two surfaces can be taken as equal to the distances from its optical center.
(ii) The aperture of the lens is small.
(iii) The object is a point-object placed on the principal axis of the lens.
(iv) The incident and the refracted rays make small angles with the principal axis.




Magnifying power : The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.
(b) (i) We know that,

Therefore, the lens of 0.5 D should be used as objective and the lens of 10 D should be used as eye-piece in order to achieve higher magnification.
(ii) The aperture of the objective lens is made larger so, that it receives as much light as coming from the distant object and the resolving power of the telescope increases.

Question 26.
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities ?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitance’s in the ratio 1:2 so that the energy stored in the two cases becomes the same. [5]
OR
(i) If two similar large plates, each of area A having surface charge densities +σ and – σ are separated by a distance in air, find the expressions for
(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.
(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density σ. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?
Answer: (i)

Consider a thin infinite uniformly charged plane sheet having the surface charge density of a. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction. Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A. The cylinder is made from three surfaces A, S₂, and A and the electric flux linked with S₂ is 0. So, the total electric flux linked through the Gaussian surface is

The direction of field for positive charge density is in outward direction away from sheet and perpendicular to the plane infinite sheet whereas for the negative charge density the direction becomes inward i.e., towards the sheet and perpendicular to the sheet.

Equating equation (i) and (ii), since energy stored in both cases are same we get,

OR
(i) (a) Consider a parallel plate capacitor with ‘ two identical plates X and Y, each having an area of A, and separated by a distance d. Let the space between the plates be filled by a dielectric medium with its dielectric constant as K and σ be the surface charge density on each of the plates.
Surface charge

CBSE Previous Year Question Papers

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Important Questions for Class 12 Physics Chapter 15 Communication Systems Class 12 Important Questions

Communication Systems Class 12 Important Questions Very Short Answer Type

Question 1.
What is sky wave propagation? (Delhi 2009)
Answer:
Propagation of frequencies less than 40 MHz using the reflecting property of ionosphere is called Sky wave propagation.

Question 2.
What is ground wave propagation? (Delhi 2009)
Answer:
A radiowave that can travel directly from one point to another following the surface of the earth is called a ground wave propagation. Ground wave propagation is possible only when the transmitting and receiving antenna are close to the surface of the earth.

Question 3.
What is space wave propagation? (Delhi 2009)
Answer:
When the signal travels in a straight line from the transmitting antenna to the receiving antenna with frequencies more than 40 MHz, it is called space-wave propagation. Space waves are used for line of sight (LOS) communication as well as satellite communication.

Question 4.
What is the function of a repeater used in communication system? (Comptt. Delhi 2012)
Answer:
The function of a repeater in communication system is to extend the range of communication.

Question 5.
What does the term ‘attenuation’ used in communication system mean? (Comptt. Delhi 2012)
Answer:
Attenuation used in communication system means loss of strength of a signal during its propagation through the communication channel.

Question 6.
What is the function of a transducer used in a communication system? (Comptt. Delhi 2012)
Answer:
Transducer : Any device/arrangement that converts one form of energy into another is called a transducer.

Question 7.
The carrier wave is given by C(t) = 2 sin (8πt) volt

The modulating signal is a square wave as shown. Find modulation index. (Delhi 2012)
Answer:

[Hint: Comparing the general expression for a wave C(f) = a sin ωt with the expression given here, we get the value of amplitude of carrier wave (A_c) = 2 volts.
In the diagram shown here for modulating signal, amplitude of modulating signal (A_m) is 1 volt]

Question 8.
The given figure shows the block diagram of a generalized communication system. Identify the element labelled ‘X’ and write its function. (Delhi 2012)

Answer:
X = Channel

Function: The physical path between the transmitter and reciver is known as the communication channel. It comprises of the wire links, wireless and optic fibres.

Question 9.
The carrier wave is represented by C(t) = 5 sin (10πt) volt.
A modulating signal is a square wave as shown. Determine modulation index. (Delhi 2012)

Answer:
Similar to Q. 7, Page 335 = 0.4

Question 10.
Tire carrier wave of a signal is given by C(t) = 3 sin (8πt) volt.
The modulating signal is a square wave as shown. Find its modulation index. (Delhi 2012)

Answer:
Similar to Q. 7, Page 335

Question 11.
How does the effective power radiated from a linear antenna depend on the wavelength of the signal to be transmitted? (Comptt. Delhi 2012)
Answer:
Effective power radiated decreases with an increase in wavelength, i.e.,

Question 12.
Draw a block diagram of a detector for amplitude modulated signal. (Comptt. Delhi 2012)
Answer:

Question 13.
What is the meaning of the term ‘attenuation’ used in communication system? (Comptt. Delhi 2012)
Answer:
‘Attenuation’ means the loss of strength of a signal while propagating through the medium.

Question 14.
Give one example of point-to-point communication mode. (Comptt. All India 2012)
Answer:
Example of point-to-point communication mode : Telephone

Question 15.
Give one example of broadcast mode of communication. (Comptt. All India 2012)
Answer:

1. Radio
2. Television (any one)

Question 16.
Define the term ‘modulation index’ in communication system. (Comptt. All India 2012)
Answer:
Modulation index is defined as the ratio of amplitude of modulating signal to the amplitude of carrier wave, i.e.,

Question 17.
What does the term ‘demodulation’ in communication system mean? (Comptt. All India 2012)
Answer:
Demodulation is the process of retrieval of information from the carrrier wave at the receiver end.

Question 18.
Distinguish between ‘point-to-point’ and ‘broadcast’ modes of communication. (Comptt. All India 2012)
Answer:
In a point-to-point communication, the communication takes place over a single link between transmitter and receiver, whereas, in the broadcast mode, there are a large number of receivers corresponding to a single transmitter.

Question 19.
How are side bands produced? (Delhi 2012)
Answer:
Side bands are produced due to superposition of carrier waves of frequencey cof over modulating/ Audio signal of frequency com; with the frequencies (ω_c ± ω_m).

Question 20.
Which basic mode of communication is used for telephonic communication? (All India 2012)
Answer:
Point to point communication mode is used for telephonic communication.

Question 21.
Why is the frequency of outgoing and incoming signals different in a mobile phone? (Comptt. Delhi 2012)
Answer:
To avoid overlapping of signals.

Question 22.
Distinguish between amplitude modulation and frequency modulation. (Comptt. All India 2012)
Answer:
The amplitude modulation provides a larger coverage area, while frequency modulation provides a better quality transmission.

Question 23.
Write two factors which justify the need of modulating a low frequency signal into high frequencies before transmission. (All India 2012)
Answer:
Need of modulating a low frequency signal :

(ii) Transmission of audio frequency electrical signals need long impracticable antenna.
(iii) To avoid mixing-up of signals from different transmitters.

Question 24.
Name the essential, components of a communication system. (All India 2016)
Answer:
Transmitter, medium (channel) and receiver are three essential components of communication system.

Question 25.
Write the full forms of the terms :
(i) LAN
(ii) WWW (Comptt. Delhi 2017)
Answer:
(i) Local Area Networking
(ii) World Wide Web

Question 26.
Name the two basic modes of communication system. (Comptt. All India 2017)
Answer:

1. Point to Point Communication and
2. Broadcast are the two basic modes of communication.

Question 27.
Define modulation index. Why is it generally kept less than one? (Comptt. All India 2017)
Answer:
Modulation index is defined as the ratio of amplitude of modulating signal (A_m) to amplitude of carrier wave (A_c)

It is kept less than one to avoid distortion

Question 28.
Why is sky wave propagation of signals restricted to a frequency of 30 MHz? (Comptt. All India 2017)
Answer:
Sky wave propagation of signals is restricted to a frequency of 30 MHz, because waves of frequency greater than 30 MHz get penetrated through the ionosphere and thus they do not get reflected by it.

Question 29.
State two reasons why high frequency carrier waves are needed in transmitting a message signal. (Comptt. All India 2017)
Answer:
For the following reasons, high frequency carrier waves are needed in transmitting a message signal :

1. Length of transmitting antenna is short.
2. Power radiated is more.
3. Mixing of signals can be avoided.

Communication Systems Class 12 Important Questions Short Answer Type SA-I

Question 30.
A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is maximum distance between them, for satisfactory communication in the LOS mode? (Radius of earth = 6400 km) (Delhi 2017)
Answer:
d_M = Maximum line of sight distance between the transmitting and receiving antennas

Question 31.
Draw a block diagram of a simple amplitude modulation. Explain briefly how amplitude modulation is achieved. (All India 2017)
Answer:
Production of Amplitude Modulated wave : A conceptually simple method to produce Am wave is shown in the following block diagram :

This signal is passsed through a band pass filter which rejects dc. The output of the band pass filter is therefore, A_m wave

Question 32.
By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21%? (Delhi 2009)
Answer:
Range of a TV tower of height,

Answer:
Transmitter, medium (channel) and receiver are three essential components of communication system.

Question 25.
Write the full forms of the terms :
(i) LAN
(ii) WWW (Comptt. Delhi 2009)
Answer:
(i) Local Area Networking
(ii) World Wide Web ,

Question 26.
Name the two basic modes of communication system. (Comptt. All India 2009)
Answer:

1. Point to Point Communication and
2. Broadcast are the two basic modes of communication.

Question 27.
Define modulation index. Why is it generally kept less than one? (Comptt. All India 2017)
Answer:
Modulation index is defined as the ratio of amplitude of modulating signal (A_m) to amplitude of carrier wave (A_C)

It is kept less than one to avoid distortion

Question 28.
Why is sky wave propagation of signals restricted to a frequency of 30 MHz? (Comptt. All India 2017)
Answer:
Sky wave propagation of signals is restricted to a frequency of 30 MHz, because waves of frequency greater than 30 MHz get penetrated through the ionosphere and thus they do not get reflected by it.

Question 29.
State two reasons why high frequency carrier waves are needed in transmitting a message signal. (Comptt. All India 2017)
Answer:
For the following reasons, high frequency carrier waves are needed in transmitting a message signal :

1. Length of transmitting antenna is short.
2. Power radiated is more.
3. Mixing of signals can be avoided.

Communication Systems Class 12 Important Questions Short Answer Type SA-II

Question 30.
A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is maximum distance between them, for satisfactory communication in the LOS mode? (Radius of earth = 6400 km) (Delhi 2008)
Answer:
d_M = Maximum line of sight distance between the transmitting and receiving antennas

Question 31.
Draw a block diagram of a simple amplitude modulation. Explain briefly how amplitude modulation is achieved. (All India 2008)
Answer:
Production of Amplitude Modulated wave : A conceptually simple method to produce Am wave is shown in the following block diagram :

This signal is passed through a band pass filter which rejects dc. The output of the band pass filter is therefore, an A_m wave.

Question 32.
By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21% ? (Delhi 2008)
Answer:

Question 33.
Why are high frequency carrier waves used for transmission? (Delhi 2008)
Answer:

1. Radiation loss at low frequencies is more.
2. Length of receiver antenna becomes very high for low frequency. So we use high frequency carrier waves for transmission.

Question 34.
What is meant by term ‘modulation’? Draw a block diagram of a simple modulator for obtaining an AM signal. (Delhi 2008)
Answer:
The process of placement or mounting of a low frequency signal over the high frequency signal is known as modulation.

Question 35.
Answer the following questions :
(a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?
(b) The small ozone layer on top of the
stratosphere is crucial for human survival. Why? (All India 2017)
Answer:
(a) The earth’s atmosphere is transparent to visible light and radio-waves but it absorbs X-rays. X-ray astronomy is possible only from satellites orbitting the earth. These satellites orbit at a height of 36,000 km, where the atmosphere is very thin and X- rays are not absorbed.

(b) Ozone layer absorbs ultraviolet radiation from the Sun and prevents it from reaching the earth and thus avoids causing damage to life.

Question 36.
Write the function of
(i) Transducer and
(ii) Repeater in the context of communication system. (All India 2017)
Answer:
(i) Transducer : A transducer is a device which converts one form of energy into another.
(ii) Repeater : A repeater is a combination of receiver and transmitter used for extending the range of communication system.

Question 37.
Write two factors justifying the need of modulation for transmission of a signal. (All India 2017)
Answer:

1. Length of receiving antenna (a wavelength) becomes practicable with larger frequency carrier wave used in the process of modulation.
2. With larger frequency or reduced wavelength, the power radiated will be larger from the transmitter and the loss will be less.

Question 38.
Explain the function of a repeater in a communication system. (Delhi 2017)
Answer:
A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency. Repeaters are used to extend the range of a communication system.

Question 39.
Explain the function of a repeater in a communication system. (Delhi 2017)
Answer:
A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency. Repeaters are used to extend the range of a communication system.

Question 40.
What is the range of frequencies used for TV transmission? What is common between these waves and light waves? (Delhi 2017)
Answer:
Television frequencies lie in the range of 54-890 MHz which cannot be reflected by ionosphere. Both waves are electromagnetic waves and can travel through vaccum with same speed.

Question 41.
What is the range of frequencies used in satellite communication? What is common between these waves and light waves? (Delhi 2017)
Answer:
Range of frequencies used in satellite communication is :
uplink = 5.925 to 6.425 GHz and downlink = 3.7 to 4.2 GHz.

Both the waves uplink and downlink used in satellite communication and light waves are electromagnetic waves and can travel through vaccum with the same speed.

Question 42.
Write two factors justifying the need of modulating a signal.
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? (All India 2017)
Answer:
(a) A signal requires modulation because :
(i) a modulated signal being of high frequency can be transmitted with the help of an antenna of reasonable size.
(ii) a modulated signal can be transmitted with more power because power radiated is proportional to

Question 43.
(i) Define modulation index.
(ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave? (Delhi 2017)
Answer:
(i) Modulation index : The modulation index is defined as the ratio of the change in the amplitude of the carrier wave to the amplitude of the original carrier wave.

(ii) The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid distortion.

Question 44.
What is sky wave communication? Why is this mode of propagation restricted to the frequencies only upto few MHz? (All India 2017)
Answer:
Sky wave communication : A radiowave directed towards the sky and reflected by the ionosphere . towards the desired location of the earth is called a sky wave. The ionospheric layers act as a reflector for a certain range of frequencies. Radio waves of frequencies between 3 MHz to 30 MHz can be reflected by the ionosphere. This mode of propagation is used by short wave broadcast service.

The propagation is restricted to the electro-magnetic waves of frequencies greater than 30 MHz because they penetrate the ionosphere and escape.

Question 45.
What is ground wave communication? On what factors does the maximum range of propagation in this mode depend? (All India 2017)
Answer:
Ground wave propagation : A radio wave that can travel directly from one point to another following the surface of the earth is called a ground wave. Groud wave propagation is possible only when the transmitting and receiving antenna are close to the surface of the earth.
The maximum range of ground wave propagation depends on two factors :

1. The frequency of the transmitted wave, and
2. The power of the transmitter.

Question 46.
What is space wave communication? Write the range of frequencies suitable for space wave communication? (All India 2017)
Answer:
When the signal travels directly from the transmitting antenna to the receiving antenna with frequencies more than 30 MHz, is called space wave communitaion.
High frequencies (above 40 MHz) can be transmitted through space wave propagation.

Question 47.
Distinguish between ‘Analog and Digital signals’. (Delhi 2017)
Answer:
Analog signal: A signal v- in which current or voltage varies continuously with time is called analog signal.
Example : Sinosidal wave

Digital signal : A signal in which current or voltage can take only two discrete values, is called a digital signal.
A digital signal can take only two values 1 and 0 which are labelled as high and low values.

Question 48.
Mention the function of any two of the following used in communication system:
(i) Transducer
(ii) Repeater
(iii) Transmitter
(iv) Bandpass Filter (Delhi 2017)
Answer:
(i) Transducer. It converts energy from one form to another.
(ii) Repeater. It picks up a signal from the transmitter, amplifiers and retransmits it to the receiver, sometimes with a change of carrier frequency.

(iii) Transmitter. It is a device which processes a message signal into a form suitable for transmission and then transmits it to the receiving end through a transmission channel.
(iv) Bandpass filter. A bandpass filter blocks lower and higher frequencies and allows only a band of frequencies to pass through.

Question 49.
In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their functions. (All India 2017)

Answer:
X ➝ IF (Intermediate frequency) stage
Y ➝ Amplifier/Power amplifier
Function of IF Stage : IF stage represents intermediate frequency stage preceding the detection. The carrier frequency is usually changed to a lower frequency by the IF stage.
Function of Amplifier : It is to amplify the signal i.e. to increase the strength of the input signal.

Question 50.
A carrier wave of peak voltage 12 V is used to transmit a message signal. Calculate the peak voltage of the modulating signal in order to have a modulation index of 75%.(Comptt. Delhi 2017)
Answer:

Question 51.
(a) Identify the boxes, ‘P’ and ‘Q’ in the block diagram of a receiver shown in the figure.

(b) Write the functions of the blocks ‘P’ and ‘Q’. (Comptt. Delhi 2017)
Answer:
(a) ‘P’ is IF (Intermediate frequency) Stage.
‘Q’ is Detector.

(b) IF stage represents intermediate frequency stage preceding the detection, the carrier frequency is usually changed to a lower frequency by the IF stage. Detection is the process of recovering the modulating signal from the modulated carrier wave.

Question 52.
A carrier wave of peak voltage 18 V is used to transmit a message signal. Calculate the peak voltage of the modulating signal in order to have a modulation index of 50%.(Comptt. Delhi 2017)
Answer:

Question 53.
A carrier wave of peak voltage 15 V is used to transmit a message signal. Calculate the peak voltage of the modulating signal in order to have a modulation index of 60%.(Comptt. Delhi 2017)
Answer:

Question 54.
Which mode of wave propagation is suitable for television broadcast and satellite communication, and why? Draw a suitable diagram depicting this mode of propagation of wave. (Comptt. All India 2017)
Answer:
Space wave propagation is suitable for television broadcast and satellite communication.
Reason: These waves, having frequency > 40 MHz, are not likely to be reflected back by the ionosphere.

Question 55.
Block diagram of a receiver is shown in the figure :

(a) Identify ‘X’ and ‘Y’.
(b) Write their functions (Delhi 2013)
Answer:
X is IF stage (intermediate frequency stage).
Function: The carrier frequency is changed to a lower frequency by intermediate frequency (IF) stage preceding the detection. .
Y is Amplifier.
Function : To strengthen the detected signal, it is fed to an amplifier.

Question 56.
In the block diagram of a simple modulator for obtaining an AM signal, shown in the figure, identify the boxes A and B. Write their functions. (All India 2013)

Answer:
Box A is square law device while Box B is band pass filter centred at ω_c.
Function of (A) square law device :
It is a non-linear device which gives an output in

Question 57.
A message signal of frequency 10 kHz and peak voltage 10 V is used to modulate a carrier of frequency 1 MHz and peak voltage 20 V. Determine
(i) the modulation index,
(ii) the side bands produced. (Comptt. Delhi 2013)
Answer:

Question 58.
Explain briefly how ground waves are propagated? Why can’t this mode be used for long distance using high frequency?
(Comptt. All India 2013)
Answer:
When the radiowaves from the transmission antenna propagate along surface of the earth, so as to reach the receiving antena, these are known as ground waves propagation.
Ground waves are rapidly attenuated due to scattering by the curved surface of earth. It is why these cannot be used for long distances.

Question 59.
Explain briefly how sky waves are propagated. Why are sky waves not used for transmission of TV signals? (Comptt. All India 2013)
Answer:
Sky wave propagation. When the radio-waves from the transmitting antenna reach the receiving antenna after reflection in the ionosphere, the wave propaga-tion is called sky wave propagation.

Sky waves are not used in transmitting TV signals because of high frequency (> 80 MHz). These waves are not reflected back by the ionosphere and . penetrate into outer space.

Question 60.
Describe briefly the space wave mode of propagation. Name one transmission which uses space wave mode. (Comptt. All India 2013)
Answer:
Space wave propagation. When the radio-waves from the transmitting antenna reach the receiving antenna either directly or after reflection from the ground or in the troposphere, the wave propagation is called space wave propagation.

In practice, direct wave mode is more dominant. However, it is limited to the so called ‘line-of-sight’ transmission distances and curvature of earth as well as height of antenna restrict the extent of coverage.
TV transmission uses the space wave mode.

Question 61.
Write the functions of the following in communication systems :
(i) Transducer
(ii) Repeater (All India 2014)
Answer:
(i) Transducer : A transducer is a device which converts one form of energy into another.
(ii) Repeater : A repeater is a combination of receiver and transmitter used for extending the range of communication system.

Question 62.
Write the functions of the following in communication systems :
(i) Transmitter
(ii) Modulator (All India 2014)
Answer:
(i) Transmitter. A set-up that transmits the message to the receiver through a communication channel.
(ii) Modulator. A set-up which makes necessary modification of message signal to make it suitable for transmission to long distances through carrier waves.

Question 63.
Write the functions of the following in communication systems :
(i) Receiver
(ii) Demodulator (All India 2014)
Answer:
(i) Receiver. It extracts the desired message signals from the received signals at the channel output.
(ii) Demodulator. It is the set-up which makes the process of retrival of information from the carrier wave at the receiver. It is the reverse of ‘modulator’

Question 64.
Explain the terms
(i) Attenuation and
(ii) Demodulation used in communication system. (Delhi 2016)
Answer:
(i) The loss of strength of a signal while propagating through a medium is called ‘Attenuation’.
(ii) The process of separation of information signal from modulated wave at the receving end is called ‘Demodulation’.

Question 65.
Define modulation index. Why is it kept low? What is the role of a bandpass filter? (All India 2016)
Answer:
(i) Modulation index is the ratio of the amplitude of modulating signal to that of carrier wave and is given by
(ii) It is kept low to avoid distortion.
(iii) Role of bandpass filter is that it rejects low and high frequencies and allows a band of frequencies to pass through.

Question 66.
Distinguish between ‘Sky wave’ and ‘Space wave’ modes of propagation in a communication system. (Comptt. Delhi 2016)
Answer:
Distinction between ‘Sky Wave’ and ‘Space Wave’

Question 67.
Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving reason, the frequency range used in this mode of propagation. (Delhi 2017)
Answer:

• Broadcast/point to point is the mode of communication used in satellite communication.
• Space wave propagation is used in this mode.
• The frequency range is above 40 MHz, because e.m. waves of frequency above 40 MHz, are not reflected back by the ionosphere and penetrate through the ionosphere.

Question 68.
Distinguish between a transducer and a repeater. (Delhi 2017)
Answer:
Transducer : A device which converts one form of energy into another.
Repeater : A combination of receiver and transmitter. It picks signals from a transmitter; amplifies and retransmits them.

Question 69.
(i) What is the line of sight communication?
(ii) Why is it not possible to use sky waves for transmission of TV signals? tJpto what distance can a signal be transmitted using an antenna of height ‘h’ (Delhi 2017)
Answer:
(i) Communication, using waves which travels in straight line from transmitting antenna to receiving antenna, constitutes line of sight communication.
(ii) (a) We cannot use sky waves because T.V. . signal waves are not reflected back by the ionosphere.
(b) d = (where ‘R’ is the radius of Earth)

Question 70.
State the two points to distinguish between sky wave and space wave modes of propagation. (Comptt. All India 2017)
Answer:
Distinction between ‘Sky Wave’ and ‘Space Wave’

Question 71.
Distinguish between point-to-point and broadcast modes of communication. Give one example for each. (Comptt. All India 2017)
Answer:
Point to point communication takes place between a single transmitter and a receiver; While in broadcast mode, a large number of receivers can receive signal from a single transmitter.
Example of point to point mode : telephony Example of broadcast mode : Radio/TV

Communication Systems Class 12 Important Questions Short Answer Type SA-II

Question 72.
Draw a plot of the variation of amplitude versus ω for an amplitude modulated wave. Define modulation index. State its importance for effective amplitude modulation. (Delhi 2017)
Answer:
(i) Plot of ‘amplitude’ versus ‘ω’ for an amplitude modulated signal


(ii) Modulation index : It is the ratio of amplitude of message or modulating signal to the amplitude of the carrier wave

(iii) Importance :

• Modulation index determines the strength and quality of the transmitted signal.
• Distortions are avoided by keeping p < 1.

Question 73.
Explain, why high frequency carrier waves are needed for effective transmission of signals.
A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the
(i) modulation index
(ii) side-band frequencies. (All India 2017)
Answer:
(a) With high frequency carrier waves

• transmission can take place with reasonable antenna length.
• as power radiated is proportional to 1/λ², power radiation increases.

Question 74.
Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated. (All India 2017)
Answer:
Sky wave propagation involves frequencies in the range 30 to 40 MHz and ionospheric reflection. Space wave propagation facilitates line of sight communication at frequencies more than 40 MHz. For the space wave to be received beyond the horizon, the receiving antenna of higher lengths are used.

Question 75.
What is space wave propagation? Give two examples of communication system which use space wave mode.
A TV tower is 80m tall. Calculate the maximum distance upto which the signal transmitted from the tower can be received. (Delhi 2017)
Answer:
If a radiowave transmitted from an antenna, travelling in a straight line, directly reaches the receiving antenna, it is called a space wave and the wave propagation is called space wave propagation.
Space waves used for line of sight (LOS) communication as well as satellite communication travels in straight line from transmitting antenna to the receiving antenna.

The figure shows the communication by LOS. If height of transmitting and receiving antenna are h_T and h_R, the maximum distance d_M between two antenna is given by

Question 76.
(i) Why is communication using line of sight mode limited to frequencies above 40 MHz?
(ii) A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? (Delhi 2017)
Answer:
(i) It is evident that above 40 MHz frequencies, the size of transmitting and receiving antenna reduces and has to be placed at a sufficient . height from the ground. Whereas below this
frequency, the waves from transmitting antenna get interrupted and blocked at many points by curvature of earth.
(ii) Given : h_T = 32 m, h_R = 50 m,
R = 6400 = 64 × 10⁵ m
Maximum line of sight distance dm between the transmitting antenna and receiving antenna is given by

Question 77.
Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz upto 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. Why is there an upper limit to frequency of waves used in this mode? (All India 2017)
Answer:
(a) Sky wave propagation is used by Broadcast services.
(b) As signals get reflected back to the earth (due to internal reflection) from 65 km to 100 km above its surface range of transmission g increases from ionosphere.

(c) Electromagnetic waves of frequencies higher than 30 MHz, penetrate the ionosphere and escape.

Question 78.
Draw a schematic diagram showing the
(i) ground wave
(ii) sky wave and
(iii) space wave propagation modes for em waves.
Write the frequency range for each of the following :
(i) Standard AM broadcast
(ii) Television
(iii) Satellite communication (Delhi 2011)
Answer:


Frequency range is 100-220 MHz Frequency range for
(i) Standard AM broadcast 540-1600 kHz
(ii) Televisiuon 54-890 MHz
(iii) Satellite communication
5.925-6.425 GHz Uplink
3.7-4.2 GHz Downlink

Question 79.
Write briefly any two factors which demonstrate the need for modulating a signal.
Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal. (All India 2011)
Answer:
Need for modulation :
(i) Audio frequencies below 20 kHz are poor to radiate. They die out after covering small distance in air. Hence they cannot travel large distance from the point of transmission.
(ii) To transmit audio waves, a very small antenna is needed. For example, if a 15 kHz signal is to be broadcasted then it requires a vertical antenna of heights 5 km which is impossible to think even.
Diagram showing amplitued modulation:

Question 80.
Write any three factors which justify the need for modulating a signal.
Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave. (Delhi 2012)
Answer:
Need for modulation:
(i) Size of an tenna: For transmitting a signal, an antennaa or an aerial of a practical size comparable to the wavelength of the signal is required. If 15 kHz signal is to be broadcasted then it requires a vertical antenna of height 5 km which is impossible to even think off.

(ii) Effective power radiated by an antenna: Audio frequencies are below 20 kHz. The low frequencies are poor to radiate. They die out after covering small distance in air. For good transmission, high powers and high frequency transmission is needed.

(iii) Mixing up of signals from different transmitters : If many transmitters are transmitting message signals simultaneously, all these signals will get mixed up and it will be difficult to distinguish between them.

Question 81.
Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode of propagation used in the frequency range above 40 MHz. (Delhi 2012)
Answer:
The modes of propagation of electromagnetic waves are :
(i) Ground waves
(ii) Sky waves
(iii) Space waves
Above 40 MHz, the mode of propagation used is via space waves. A space wave travels in a straight line from the transmitting antenna to the receiving antenna. Space waves are used for line of sight (LOS) communications as well as satellite communication.

Question 82.
Mention three different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves. (All India 2012)
Answer:
Three different modes of propagation used in communication are :
(i) Space wave propagation.
(ii) Ground wave propagation
(iii) Sky wave or ionospheric propagation
The radio waves of the high frequency band having frequency range 3-30 MHz cannot penetrate through the ionosphere. They are reflected back towards the earth. This region of the AM band is called short wave band. Above 40 MHz, the ionosphere bends the electromagnetic waves and does not reflect them back towards the earth.

Question 83.
Explain briefly the following terms used in communication system :
(i) Transducer
(ii) Repeater
(iii) Amplification (All India 2012)
Answer:
(i) Transducer: Any device which converts energy from one form to another is called a transducer. We may define an electrical transducer as a device which converts variation in physical quantity such as pressure, displacement, force, temperature etc., into corresponding variations in the electrical signal at its output.

(ii) Repeater: As signal while passing through the transmission medium may get attenuated due to the various energy losses along its path. So a signal booster or amplifying repeater is placed at suitable distance along its path. A repeater is a combination of a transmitter, an amplifier and a receiver which picks up a signal from the transmitter, amplifiers and retransmits it to the receiver sometimes with a change of carrier frequency.

(iii) Amplification : It is the process of increasing the amplitude and hence the strength of an electrical signal by using a suitable electric circuit (consisting of atleast one transistor) is called the amplifier.

Question 84.
(a) Distinguish between sinusoidal and pulse-shaped signals.
(b) Explain, showing graphically, how a sinu-soidal carrier wave is superimposed on a modulating signal to obtain the resultant amplitude modulated (AM) wave. (Comptt. All India 2012)
Answer:

When a modulating signal (Figure b) is superimposed on sinusoidal carrier wave (Figure a), and as a result, the amplitude of the carrier wave varies in accordance with the modulating signal, the resultant wave (Figure c) is known as an amplitude modulated wave.

Question 85.
Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulated wave is obtained when a modulating signal is superimposed on a carrier wave. (Delhi 2013)
Answer:
Need for modulation:
(i) Size of an tenna: For transmitting a signal, an antennaa or an aerial of a practical size comparable to the wavelength of the signal is required. If 15 kHz signal is to be broadcasted then it requires a vertical antenna of height 5 km which is impossible to even think off.

(ii) Effective power radiated by an antenna: Audio frequencies are below 20 kHz. The low frequencies are poor to radiate. They die out after covering small distance in air. For good transmission, high powers and high frequency transmission is needed.

(iii) Mixing up of signals from different transmitters : If many transmitters are transmitting message signals simultaneously, all these signals will get mixed up and it will be difficult to distinguish between them.

Question 86.
Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communication system.
(a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz?
(b) Give two examples where space wave mode of propagation is used. (Delhi 2013)
Answer:
Difference between skywave and space wave modes :

(a) The radio waves of frequencies more than 40 MHz penetrate into the ionosphere.
(b) Television broadcast, microwave link and satellite communication.

Question 87.
Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 × 10⁶ m) (All India 2013)
Answer:
(i) Space waves
(ii) More than 50 Hz

Question 88.
Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a tower has a height of 45 m and the receiving antenna is on the ground. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 ∞ 10⁶ m) (All India 2013)
Answer:
• Space waves
• Frequency range above 40 MHz

Question 89.
What is meant by ‘detection of a modulated signal’? Draw block diagram of a detector for AM waves and state briefly, showing the waveforms, how the original message signal is obtained. (Comptt. Delhi 2013)
Answer:
Detection of a modulated signal: Detection is the process of recovering the modulated signal from the modulated carrier wave. The modulated carrier wave contains the frequencies <oc, eoc+ com and In order to obtain the original message signal m (+) of angular frequency com following simple method is shown in the form of a block diagram.

Question 90.
Write the function of each of the following used in communication sytem :
(i) Transducer
(ii) Repeater
(iii) Transmitter (Comptt. Delhi 2013)
Answer:
(i) Transducer. It converts energy from one form to another.

(ii) Repeater. It picks up a signal from the transmitter, amplifiers and retransmits it to the receiver, sometimes with a change of carrier frequency.

(iii) Transmitter. It is a device which processes a message signal into a form suitable for transmission and then transmits it to the receiving end through a transmission channel.

(iv) Bandpass filter. A bandpass filter blocks lower and higher frequencies and allows only a band of frequencies to pass through.

Question 91.
(a) Describe briefly the three factors which justify the need for translating a low frequency signal into high frequencies before transmission.
(b) Figure shows a block diagram of a detector for AM signal

Draw the waveforms for the
(i) input AM wave at A,
(ii) output B at the rectifier, and
(iii) output signal at C. (Comptt. All India 2013)
Answer:
(a) Modulation is needed

• to transmit a low frequency signal to a distant place.
• to keep the height of antenna small.
• not to allow the signals from different stations getting mixed up.

(b)

Question 92.
Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave. (All India 2013)
Answer:
The two basic modes of communication are :
(1) point-to-point communication
(2) broadcast communication
Amplitude modulation. It is produced by varying the amplitude of carrier waves in accordance with the amplitude of the modulating wave.

Production of amplitude modulated wave. A conceptually simple method to produce AM wave is shown in block diagram.

Question 93.
Draw a block diagram of a detector for AM signal and show, using necessary processes and the waveforms, how the original message signal is detected from the input AM wave. (Delhi 2015)
Answer:
Detection of a modulated signal: Detection is the process of recovering the modulated signal from the modulated carrier wave. The modulated carrier wave contains the frequencies ω_c, (ω_c + ω_m) and (ω_c – ω_m). In order to obtain the original message signal m (+) of angular frequency ω_m following simple method is shown in the form of a block diagram.

Question 94.
Define modulation index. Why is its value kept, in practice, less than one?
A carrier wave of frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz producing 50% amplitude modulation. Calculate the amplitude of the AM wave and frequencies of the side bands produced. (All India 2015)
Answer:
(i) Definition of Modulation Index. It is defined as the ratio of the amplitude of modulating signal (A_m) to the amplitude of carrier wave (A_c)

= (1500 kHz ± 10 kHz)
= 1510 and 1490 kHz respectivley.

Question 95.
Answer the following questions:
(i) Why is the thin ozone layer on top of the stratosphere crucial for human survival? Identify to which a part of electromagnetic spectrum does this radiation belong and write one important application of the radiation.
(ii) Why are infrared waves referred to as heat waves? How are they produced? What role do they play in maintaining the earth’s warmth through the greenhouse effect? (Comptt. Delhi 2015)
Answer:
(i) The thin ozone layer on the top of stratosphere is crucial for human survival, because it absorbs ultraviolet radiations from the Sun and thus prevents them from reaching the earth’s surface causing damage to life.
Identification : ultraviolet radiations. Its correct application is Sanitization.

(ii) (a) Water molecules present in most materials readily absorb infra red waves. Hence, their thermal motion * increases. Therefore, they heat their surroundings and are hence referred to as heat waves.
(b) They are produced by hot bodies and molecules.
(c) Incoming visible light is absorbed by earth’s surface and radiated as infra red radiations. These radiations are trapped by green house gases.

Question 96.
(a) Given a block diagram of a generalized communication system.

Identify the boxes ‘X’ and ‘Y’ and write their functions.
(b) Distinguish between “Point to Point” and “Broadcast” modes of communication. (Comptt. Delhi 2015)
Answer:
(a) X : Transmitter Y : Channel Their function:
Transmitter : To convert the message signal into suitable form for transmission through channel.
Channel : It sends the signal to the receiver.

(b) In point to point mode, communication takes place between a single transmitter and receiver while in broadcast mode, large number of receivers are connected to a single transmitter.

Question 97.
Explain the following terms in relation to the use of internet :
(i) Internet surfing
(ii) Social networking
(iii) E-mail (Comptt. Delhi 2015)
Answer:
(i) Internet surfing : “It is the exploration of me world wide web (www) by following one interesting link to another one usually without a definitive object or search strategy, through the internet.”

(ii) Social networking : It is a platform to build social networks or social relations among people who share similar interests, activities, backgrounds or real life connections. A social network service consists of a representation of each user (often a profile), his or her social links, and a variety of services.

(iii) E-mail : E-maii (Electronic mail) is a method of exchanging digital messages from an author to one or more recipients. E-mail operates across the internet or other complex networks. E-mail servers accept, forward, deliver and store messages.

An internet Email message consists of these components :
(a) message envelope,
(b) message header and
(c) message body. Thus e-mail is an information and communications technology.

Question 98.
(a) Explain any two factors which justify the need of modulating a low frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation. (Delhi 2016)
Answer:
(a) Need of modulation : A low frequency signal is modulated for the following purposes :
(i) It reduces the wavelength of transmitted signal, and the minimum height of antenna for effective communication is .
Therefore height of antenna becomes practically achievable.

(ii) Power radiated into the space by an antenna is inversely proportional to λ². Therefore, the power radiated into the space increases and signal can travel larger distance.

(b) Advantages of frequency modulation :

1. High efficiency
2. Less noise .
3. Maximum use of transmitted power.

Question 99.
(i) Which mode of propagation is used by shortwave broadcast services having frequency range from a few MHz upto 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode.
(ii) Why is there an upper limit to frequency of waves used in this mode? (All India 2015)
Answer:
(i) (a) Sky wave propagation
(b)
Three different modes of propagation used in communication are :

1. Space wave propagation.
2. Ground wave propagation
3. Sky wave or ionospheric propagation

The radio waves of the high frequency band having frequency range 3-30 MHz cannot penetrate through the ionosphere. They are reflected back towards the earth. This region of the AM band is called shortwave band. Above 40 MHz, the ionosphere bends the electromagnetic waves and does not reflect them back towards the earth.

(ii) Electromagnetic waves of frequencies higher than 30 MHz, penetrate the ionosphere and escape whereas the waves less than 30 MHz are reflected back to the earth by the ionosphere.

Question 100.
What does the term ‘Modulation’, used in communication system, mean?

Identify the two types of modulation shown here. Give two advantages of any one of these over the other. (Comptt. Delhi 2015)
Answer:
(i) Modulation is a process in which one of the characteristics (amplitude, frequency, phase) of a high frequency carrier wave is made to change in accordance with a given low frequency message signal.
(ii) (a) Modulated wave 1 : Frequency Modulation (FM)
(b) Modulated wave 2 : Amplitude Modulation (AM)

Two advantages of FM over AM :

1. Lower noise, better power efficiency.
2. Higher operating range.
3. Higher fidelity reception.

Two advantages of AM over FM :

1. Simple circuits are required.
2. Lower frequency space for transmission.

Question 101.
Give (brief) reasons for the following :
(a) We use the ‘sky wave’ mode of propagation, of electromagnetic waves, only for frequencies up to 30 to 40 MHz.
(b) The LOS communication, via space waves, has a (fairly) limited range.
(c) A mobile phone user gets an ‘uninterrupted link to talk’ while walking. (Comptt. All India 2015)
Answer:
(a) The ionosphere can act as a ‘reflector’ only for e.m. waves of frequencies upto 30 to 40 MHz. Higher frequency e.m. waves penetrate into the atmosphere and escape.

(b) The range is (fairly) limited because the e.m. waves lose energy (fairly rapidly) when they glide over the surface of the earth.

(c) This is because of the presence of a network of base stations’/cells’ which keep on passing the signals from one base station/cell to the other.

Question 102.
Define the term ‘amplitude modulation’.
Explain any two factors which justify the need for modulating a low frequency base-band signal. (Delhi 2017)
Answer:
‘Amplitude modulation’ is the process of superposition of information/message signal over a carrier wave in such a way that the amplitude of carrier wave is varied according to the information signal/message signal.
Direct transmission, of the low frequency base band information signal, is not possible due to the following reasons :
(i) Size of Antenna: For transmitting a signal, minimum height of antenna should be . With the help of modulation, wavelength of signal decreases, hence height of antenna becomes manageable.

(ii) Effective power radiated by an antenna :
Effective power radiated by an antenna varies inversely as λ², hence effective power radiated into the space, by the antenna, increases.

(iii) To avoid mixing up of signals from different transmitters.

Question 103.
(a) How is amplitude modulation achieved?
(b) The frequencies of two side bands in an AM wave are 640 kHz and 660 kHz respectively. Find the frequencies of . carrier and modulating signal. What is the bandwidth required for amplitude modulation? (All India 2017)
Answer:
(a) Amplitude modulation is achieved by applying the message signal, and the carrier wave, to a non-linear (square law device) followed by a band pass filter as shown below in the block diagram:

Question 104.
Draw a block diagram of generalized communication system. Write the functions of each of the following:
(a) Transmitter
(b) Channel
(c) Receiver (All India 2017)
Answer:
Block diagram of generalized communication system :

Functions:
(a) Transmitter : A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception.
(b) Channel: It carries the message signal from a transmitter to a receiver.
(c) Receiver : A receiver extracts the desired message signals from the received signals at the channel output.

Question 105.
Explain the term, ‘amplitude modulation’ of a signal. For an amplitude modulated wave, the maximum amplitude is 10 V and the minimum amplitude is 2 V. Calculate the modulation index. (Comptt. Delhi 2017)
Answer:
Amplitude modulation is the process of superposition of a message signal over a carrier, wave in which amplitude of the carrier wave is varied in accordance with the message/ information signal.

Question 106.
Distinguish between sky wave and space modes of communication. What is the main limitation of space wave mode? Write the expression for the optimum separation between the transmitting and receiving antenna for effective reception of signals in this mode of communication. (Comptt. Delhi 2017)
Answer:
(i) In sky wave mode of communication, waves reach from transmitting antenna to receiving antenna through reflections from ionosphere; while in space wave mode of communication, waves travel either directly from the transmitter to receiver or through satellites.

(ii) Direct waves get blocked at some point due to the curvature of earth.

(iii) Optimum distance between transmitting and receiving antenna

Question 107.
Explain the meaning of terms : Attenuation and Demodulation. For an amplitude modulated wave, the maximum amplitude is 12 V and the minimum amplitude is 2 V. Calculate the modulation index. (Comptt. Delhi 2017)
Answer:
Attenuation : It is the loss of strength of signal, while propagating through a medium.
Demodulation : It is the process of recovery of audio signal from the modulated wave.

Question 108.
Write the functions of
(i) Repeater and
(ii) Receiver. For an amplitude modulated wave, the maximum amplitude is 15 V and the minimum amplitude is 3 V. Calculate the modulation index. (Comptt. Delhi 2017)
Answer:
Repeater is a combination of a receiver and a transmitter. It is used to increase the range of communication of signals. A repeater picks up the signal from the transmitter, amplifies it and retransmits it to the receiver.

Receiver extracts the desired message signals from the received signals at the channel output. It is the combination of receiving antenna, amplifier, intermediate frequency converter, demodulator and amplifier of audio signal.

Question 109.
Briefly explain the three factors which justify the need of modulating low frequency signal into high frequencies. (Comptt. All India 2017)
Answer:
(a) Modulation is needed

• to transmit a low frequency signal to a distant place.
• to keep the height of antenna small.
• not to allow the signals from different stations getting mixed up.

Important Questions for Class 12 Physics

The post Important Questions for Class 12 Physics Chapter 15 Communication Systems Class 12 Important Questions appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Chemistry 2014 Outside Delhi

Time allowed: 3 hours
Maximum Marks: 70

CBSE Previous Year Question Papers Class 12 Chemistry 2014 Outside Delhi Set I

Question 1.
What is the effect of temperature on chemisorption? [1]
Answer:
Chemisorption initially increases then decreases with rising in temperature. The initial increase is due to the high energy of activation and the decrease afterwards is due to the exothermic nature of adsorption equilibrium.

Question 2.
What is the role of zinc metal in the extraction of silver? [1]
Answer:
Zinc is used as a reducing agent to recover silver from its cyanide complex. It reduces Ag⁺ to Ag and itself get oxidised to Zn²⁺.

Question 3.
What is the basicity of H₃PO₃? [1]

Question 4.
Identify the chiral molecule in the following pair: [1]

Answer:

Question 5.
Which of the following is a natural polymer? [1]
Buna-S, Proteins, PVC Economics type
Answer:
Protein is a natural polymer having amino acid as a monomer.

Question 6.
The conversion of primary aromatic amines into diazonium salts is known as ____ [1]
Answer:
Diazotisation reaction.

Question 7.
What are the products of hydrolysis of sucrose? [1]
Answer:
The products of Hydrolysis of sucrose are: Glucose and Fructose

Question 8.
Write the structure of p-methylbenzaldehyde.
Answer:

Question 9.
An element with density 2.8 g cm^-3 forms an f.c.c. unit cell with edge length 4 × 10^-8 cm. Calculate the molar mass of the element. (Given N_A = 6.022 × 10²³ mol^-1) [2]

Question 10.

1. What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
2. What type of stoichiometric defect is shown by NaCl? [2]

OR
How will you distinguish between the following pairs of terms?

1. Tetrahedral and octahedral voids
2. Crystal lattice and unit cell

Question 11.
State the Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? [2]
Answer:
It states that the limiting molar / Conductivity of an electrolyte can be expressed as a sum of the individual contribution of anion and cation of the electrolyte.
With dilution, the number of ions per unit volume of electrolytes decreases and hence, conductivity decreases.

Question 12.
For a chemical reaction R → P, the variation in the concentration (R) Vs. time (t) plot is given as [2] (i)

1. Predict the order of the reaction.
2. What is the slope of the curve?

Answer:

1. Zero-order reaction
2. The slope of the curve is (-K). i.e., a negative slope.

Question 13.
Explain the principle of the method of electrolytic refining of metals. Give one example. [2]
Answer:
In the electrolytic method, the impure metal is made anode. A strip of the same metal in pure form is used as the cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.

For e.g.: Copper is refined using an electrolytic method, the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode.
Anode: Cu → Cu²⁺ + 2e^–
Cathode: Cu²⁺ + 2e^– → Cu

Question 14.
Complete the following equations: [2]

1. P₄ + H₂O →
2. XeF₄ + O₂F₂ →

Answer:

1. P₄ + H₂O → H₃PO₄ + H₂
2. XeF₄ + O₂F₂ → XeF₆ + O₂

Question 15.
Draw the structures of the following:
(i) XeF₂
(ii) BrF₃ [2]
Answer:

Question 16.
Write the equations involved in the following reactions: [2]
(i) Reimer-Tiemann reaction
(ii) Williamson synthesis
Answer:
(i) Reimer Tiemann reaction:

Question 17.
Write the mechanism of the following reaction:

Answer:

Question 18.
Write the name of monomers used for getting the following polymers: [2]

1. Bakelite
2. Neoprene

Answer:

1. Phenol and formaldehyde are the monomers used for the formation of Bakelite.
2. Chloroprene is the monomer used to prepare Neoprene.

Question 19.
(a) Calculate Δ_rG⁰ for the reaction
Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Given: E⁰_cell = 2.71 V, 1F = 96500 C mol^-1
(b) Name the type of cell that was used in Apollo space program for providing electrical power. [3]
Answer:
(a) Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Δ_rG° = – nFE°_cell
= -2 × 96500 × 2.71
= -523.03 kj mol^-1
(b) Hydrogen-oxygen fuel cells and solar cells.

Question 20.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume:
SO₂Cl₂ (g) → SO₂(g) + Cl₂ (g)

Calculate the rate constant. (Given: log 4 = 0.6021, log 2 = 0.3010) [3]
Answer:
SO₂Cl₂ (g) → SO₂(g) + Cl₂ (g)
dt1 t = 0, P₀ ….. 0
At t = t, P₀ – P …….. P
The total pressure of the thermal decomposition of SO₂Cl₂ time t.
P_t = (P₀ – P) + P + P
P_t = P₀ + P
Hence, P = P_t – P₀
P₀ – P = P₀ – (P_t – Po) = 2P₀ – Pt

Question 21.
What are emulsions? What are their different types? Give one example of each type. [3]
Answer:
Emulsions are colloids in which both the dispersed phase and dispersing medium are liquid. It is a mixture of two or more liquids that are normally immiscible. They are of two types:
(i) Oil in water type emulsions (O/W): In such emulsions, oil is the dispersed phase and water is the dispersed medium.
e.g., Milk, Vanishing cream.

(ii) Water in oil type emulsions (W/O): In such emulsions, water is the dispersed phase and oil is the dispersed medium.
e.g., Butter, Cod liver oil.

Question 22.
Given the reasons for the following:
(i) (CH₃)₃ P = O exists but (CH₃)₃N = O does not.
(ii) Oxygen has less electron gain enthalpy with a negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃. [3]
Answer:
(ii) Due to a small size and high electronegativity of oxygen compared to sulphur, oxygen has less electron gain enthalpy.

Question 23.
(i) Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂]Cl.
(ii) What type of isomerism is exhibited by the complex [Co(en)₃]³⁺? (en = ethane-1,2-diamine)
(iii) Why is [NiCl₄]^2- paramagnetic but [Ni(CO)₄] is diamagnetic [3]
(At. nos.: Cr = 24, Co = 27, Ni = 28)
Answer:
(i) Tetraammine dichloridochromium(III) chloride

(ii) Optical isomerism is shown by the complex [CO(en)₃]³⁺

(iii) [Ni(Cl)₄]^2-, Ni is in +2 oxidation state with the electronic configuration 3d⁸4s⁰. As Cl^– is a weak ligand it cannot pair up the electrons in 3d orbitals therefore, [NiCl4]^2- is paramagnetic. In [Ni(CO)₄], Ni is in 0 oxidation state with the electronic configuration 3d⁸4s². Co is a strong ligand it causes 4s electrons to shift to 3d and pair up 3d electrons. Therefore, Ni(CO)₄ is diamagnetic.

Question 24.
(a) Draw the structure of major moon halo products in each of the following reactions:

(b) which halogen compound in each of the following pairs will react faster in SN2 reaction:
(i) CH₃Br or CH₃I
(ii) (CH₃)₃C-Cl or CH₃-Cl
Answer:

(b) (i) CH₃-I will react faster because Iodine (I) is a better leaving group.
(ii) CH₃-CI will react faster because it is a primary halide and it undergoes S_N2 reaction faster.

Question 25.
Account for the following:
(i) Primary amines (R-NH₂) have a higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. [3]
OR
Give the structures of A, B and C in the following reactions:

Answer:
(i) Due to intermolecular hydrogen bonding in primary amines (presence of more number of H-atoms). They have a high boiling point in comparison to tertiary amines.

(ii) Aniline does not undergo Friedel-Crafts reaction because Aniline is basic in nature and reacts with AlCl₃ to form a salt. The positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated.

Question 26.
Define the following terms related to proteins: [3]

1. Peptide linkage
2. Primary structure
3. Denaturation

Answer:

1. Peptide linkage is the amide linkage formed by -COOH group of one a-amino acid and -NH₂ group of other α-amino acids by loss of a water molecule.
2. The sequence in which various amino acids are arranged in a linear structure with no intermediate bonding is called the primary structure of a protein.
3. When a protein in its native form is subjected to a change, such as a temperature, pressure etc. Due to this protein loses its biological activity and this is called denaturation of a protein e.g., curdling of milk.

Question 27.
On the occasion of World Health Day.
Dr Satpal organized a ‘health camp’ for the poor fanners living in a nearby village. After check¬up, he was shocked to see that most of the farmers suffered from cancer due to regular exposure to pesticides and many were diabetic. They distributed free medicines to them. Dr Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care, financial assistance, setting up of super-speciality hospitals for treatment and prevention of the deadly disease in the affected villages all over India.
(i) Write the values shown by (a) Dr Satpal (b) NHRC?
(ii) What type of analgesics are chiefly used for the relief of pains of terminal cancer?
(iii) Give an example of artificial sweetener that could have been recommended to diabetic patients. [3]
Answer:
(ii) Narcotic analgesics like morphine and heroin.
(iii) Aspartame.

Question 28.
(a) Define the following terms:
(i) Molarity
(ii) Molal elevation constant (k_b)
(b) A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of a solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 190 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. [2, 3]
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL^-1)
Answer:
(a) (i) The number of moles of a solute present in one litre of solution is known as its molarity.
(ii) The elevation in the boiling point of a solution when one mole of non-volatile solute is dissolved in one kilogram of a volatile solvent is known as molal elevation constant (K_b).

W_B = Mass of solute
M_B = Molar mass of solute
W_A = Mass of solvent.

(b) For isotonic solution,
π₁ = π₂
C₁ = C₂ (at same temp.)
or n₁ = n₂ (at same Vol.)
=
x = 45 g, mass of glucose per litre of solution.
OR
(a) The mixture of ethanol and acetone shows positive deviation from Raoult’s law. In pure ethanol, hydrogen bond exists between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of ethanol thus breaking some of the hydrogen bonds and weakening the molecular interactions this leads to an increase in vapour pressure resulting in a positive deviation from Raoult’s law.

(b) Mass of glucose = 10 g
Mass of solution = 100 g
Mass of water = 100 – 10 = 90 g

Question 29.
(a) Complete the following equations:
(i) Cr₂O₇^2- + 2OH^– →
(ii) MnO₄^– + 4H⁺ + 3e^– →
(b) Account for the following:
(i) Zn is not considered as a transition element.
(ii) Transition metals form a larger number of complexes.
(iii) The E value for the Mn³⁺/Mn²⁺ couple is much more positive than that for C³⁺/Cr²⁺ couple. [2, 3]
OR
(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation :
MnO₄^– + 8H⁺ + 5e^– →
(iv) Out of Mn³⁺ and Cr³⁺, which is more paramagnetic and why? (atomic nos : Mn = 25, Cr = 24) [5]
Answer:
(a) (i) Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O
(ii) MnO₄^– + 4H⁺ + 3e^– → MnO₂ + 2H₂O

(b) (i) Zinc has 3d¹⁰4s² configuration with no unpaired d-orbital electron and hence it is not considered as a transition element.
(ii) Transition metals have vacant orbitals to accommodate lone pairs of electrons for bond formation and have high charge density, therefore, they form complexes.
(iii) Due to the high stability of Mn²⁺ (due to its half-filled 3d subshell) than Mn³⁺ while Cr³⁺ is more stable than Cr²⁺.
OR
(i)

(ii) Cerium (Ce) is a lanthanoid element, which is well known to exhibit a +4 oxidation state.
(iii) MnO^–₄ + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O
(iv) Mn³⁺ is more paramagnetic as it has four impaired electrons while Cr³⁺ has only three.
Mn²⁺ = [Ar] 4s⁰3d⁴
Cr³⁺ = [Ar] 4s⁰3d³

Question 30.
(a) Write the products formed when CH₃CHO reacts with the following reagents:
(i) HCN
(ii) H₂N-OH
(iii) CH₃CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Benzoic acid and Phenol
(ii) Propanal and Propanone. [3, 2]
OR
(a) Account for the following: (2, 2, 1)
(i) Cl^–CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of the carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro’s reaction
(c) Out of CH₃CH₂-CO-CH₃ and CH₃CH₂-CH₂-CO-CH₃, which gives iodoform test?
Answer:
(a) (i) On the reaction of acetaldehyde with hydrogen cyanide, it gives 2-Hydroxypropanenitrile as the product.

(iii) When 2 molecules of acetaldehyde react with each other in presence of dil. NaOH, 3-Hydroxybutanal is obtained. This is an aldol reaction and further proceeds the reaction when heated.

(b) (i) Phenol reacts with FeCl₃ to give violet coloured precipitate while benzoic acid gives a buff coloured precipitate.

OR
(a) (i) Because of -I effect of Cl atom in ClCH₂COOH and +I effect of CH₃ group in CH₃COOH the electron density in the O-H bond in ClCH₂COOH is much lower than CH₃COOH. Hence ClCH₂COOH acid is a stronger acid than CH₃COOH.
(ii) In carboxylic acids, the carboxyl group is not free as it is involved in resonance

(c) CH₃CH₂CH₂COCH₃, being a methyl ketone gives iodoform test.

CBSE Previous Year Question Papers Class 12 Chemistry 2014 Outside Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in the previous set.

Question 1.
Why is adsorption always exothermic? [1]
Answer:
Adsorption being a surface phenomenon leads to a decrease in surface energy and hence is exothermic in nature.

Question 2.
Name the method used for refining of Nickel. [1]
Answer:
Mond’s process is used for refining of Nickel.

Question 3.
Why does NO₂ dimerise? [1]

Question 4.
Based on molecular forces, what type of polymer is neoprene? [1]
Answer:
Elastomer.

Question 5.
What are the products of hydrolysis of maltose? [1]
Answer:
Two molecules of α-D-glucose

Question 6.
Write the structure of 4-chloropentan-2 one. [1]
Answer:

Question 9.
Write the name of monomers used for getting the following polymers: [2]

1. Terelyne
2. Nylon-6,6

Answer:

1. Ethylene glycol and Terephthalic acid.
2. Adipic acid and Hexamethylene diamine.

Question 10.
Describe the role of the following: [2]

1. SiO₂ in the extraction of copper from copper matte.
2. NaCN in froth floatation process.

Answer:

1. SiO₂ (Silica) acts as a flux in the extraction of copper from copper matte to remove ferrous oxide as ferrous silicate slag.
   FeO + SiO₂ → FeSiO₃ (Slag)
2. NaCN is used as a depressant as it forms zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS thereby preventing it from forming froth.

Question 11.
Complete the following equations: [2]
(i) Ag + PCl₅
(ii) CaF₂ + H₂SO₄
Answer:
(ii) CaF₂ + H₂SO4 → CaSO₄ + 2HF

Question 12.
Draw the structures of the following: [2]
(i) XeF₄
(ii) HClO₄
Answer:

Question 13.
(i) Write the type of magnetism observed when the magnetic moment is oppositely aligned and cancel out each other.
(ii) Which stoichiometric defect does not change the density of the crystal? [2]

Question 14.
Define the following terms: [2]

1. Fuel cell
2. Limiting molar conductivity (∧_m°)

Answer:

1. A fuel cell is a device that converts chemical energy from the combustion of a fuel into electric energy through a chemical reaction.
2. Molar conductivity of electrolyte at infinite dilution or when concentration approaches zero is called limiting molar conductivity. It is expressed as ∧_m°

Question 19.
Define the following terms: [3]
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Oligo saccharides
Answer:
(i) The linkage between two monosaccharide units through oxygen atom is known as glycosidic linkage.

(ii) Sucrose is called invert sugar as it produces equimolar quantities of glucose and fructose on hydrolysis it gives an equimolar mixture of D – (+)-glucose, and D – (-)-fructose, which is laevorotatory. This change of specific rotation from dextrorotation to laevorotation is called inversion of sugar and the mixture so obtained is called invert sugar.

(iii) Carbohydrate which on hydrolysis give two to ten molecules of monosaccharides is called oligosaccharide e.g. sucrose.

CBSE Previous Year Question Papers Class 12 Chemistry 2014 Outside Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question 1.
What are the dispersed phase and dispersion medium in milk? [1]
Answer:
The dispersed phase is oil or fat and the dispersion medium is water.

Question 2.
Name the method used for refining of copper metal. [1]
Answer:
Electrolytic refining.

Question 3.
Why does NH₃ act as a Lewis base? [1]
Answer:
Due to the presence of lone pair on nitrogen in NH₃, It can donate its lone pair of electrons and it forms coordinate bonds with Lewis acids and acts as Lewis base.

Question 5.
Which of the following is a fibre? [1]
Nylon, Neoprene, PVC
Answer:
Nylon

Question 6.
Write the products of hydrolysis of lactose. [1]
Answer:

Question 8.
Write the structure of 2-hydroxybenzoic acid. [1]
Answer:

Question 9.
Complete the following equations: [2]

1. Cu + 2H2SO4 (Cone.) →
2. XeF2 + H2O →

Answer:

1. Cu + conc. 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
2. 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

Question 10.
Draw the structure of the following: [2]
(i) XeO₃
(ii) H₂SO₄
Answer:

Question 11.
Write the name of monomers used for getting the following polymers: [2]

1. Teflon
2. Buna-N

Answer:

1. Tetrafluoroethylene
2. 1, 3-Butadiene and Acrylonitrile

Question 13.
(i) Write the type of magnetism observed when the magnetic moment is aligned in parallel and anti-parallel directions in unequal numbers. [2]
(ii) Which stoichiometric defect decreases the density of the crystal?

Question 14.
Define the following terms:

1. Molar conductivity (∧_m)
2. Secondary batteries [2]

Answer:

1. Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.
   ∧_m =
2. Secondary batteries are those batteries which can be recharged by passing an electric current through them and hence can be used over again e.g. Lead storage battery.

Question 17.
Write the principle behind the froth floatation process. What is the role of collectors in this process? [2]
Answer:
Froth floatation method has been in use for removing gangue from sulphide ores. In this process, a suspension of the powdered ore is made with water.
Collectors (e.g. pine oils, fatty acids, etc.) enhance non-wettability of the mineral particles and used to skim the froth off the surface.

Question 23.
Define the following terms : [3]
(i) Nucleotide
(ii) Anomers
(iii) Essential amino acids
Answer:
(i) A Nucleotide contains all the three basic components of nucleic acid i.e, a pentose sugar, a nitrogenous base and a phosphoric add. When nucleoside is linked to phosphoric add at 5′ position of the sugar moiety, we get a nucleotide.

(ii) The carbohydrate which differs in configuration at the glycosidic carbon (i.e., C₁ in aldoses and C₂ in ketoses) are called anomers. e.g. α-D-(+) glucose and β-D-(+) glucose.

(iii) Essential amino acids are those amino acids which cannot be synthesised by the body and need to be consumed through diet. eg. Valine.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Chemistry 2014 Outside Delhi appeared first on Learn CBSE.

Important Questions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Class 12 Important Questions

Alcohols, Phenols and Ethers Class 12 Important Questions Very Short Answer Type

Question 1.
Give the IUPAC name of the following compound : (Delhi 2009)

Answer:

IUPAC name : 2-Bromo-3-methyl-but-2-ene-1-ol

Question 2.
Give the IUPAC name of the following (All India 2009)

Answer:

Question 3.
Write the structure of the molecule of a compound whose IUPAC name is 1-phenylpropan-2-ol. (All India 2010)
Answer:
1-phenylpropan-2-ol

Question 4.
How would you convert ethanol to ethene? (All India 2011)
Answer:

Question 5.
Draw the structure of 2, 6-Dimethylphenol. (All India 2011)
Answer:

Question 6.
Draw the structural formula of 2-methylpropan- 2-ol molecule. (Delhi 2012)
Answer:

Question 7.
Draw the structure of hex-l-en-3-ol compound. (Delhi 2012)
Answer:
CH₂ = CH – CH(OH) – CH₂ – CH₂ – CH₃

Question 8.
Ortho nitrophenol has lower boiling point than p-nitrophenol. Why ? (Comptt. Delhi 2012)
Answer:
Ortho-nitrophenol has lower boiling point due to formation of intramolecular H-bonding whereas p-nitrophenol forms intermoleeular H-bonding.

Question 9.
Ortho-nitrophenol is more acidic than ortho-methoxyphenol. Why? (Comptt. Delhi 2012)
Answer:
NO₂ group is an electron withdrawing group while methoxy group is electron donating in nature. The release of H⁺ is easier from O-nitrophenol while it is difficult from O-methoxyphenol.

Question 10.
The C-O bond is much shorter in phenol than in ethanol. Give reason. (Comptt. Delhi 2012)
Answer:
Carbon of C-O bond of phenol is Sp² hybridised, so it acquires a partial double bond character but in ethanol it is Sp³ hybridised and a single bond. Double bond is shorter than a single bond.

Question 11.
Write the IUPAC name of the following : (Comptt. All India 2012)

Answer:
IUPAC name : 2-Bromo-3-methyl but-2-en-l-ol.

Question 12.
Of the two hydroxy organic compounds ROH and R’OH, the first one is basic and other is acidic in behaviour. How is R different from R’? (Comptt. Delhi 2013)
Answer:
When R = alkyl, ROH behaves as a bronsted base and when R’ = aryl, R’OH behaves as a bronsted acid.

Question 13.
Give a chemical test to distinguish between 2-Pentanol and 3-Pentanol. (Comptt. Delhi 2013)
Answer:
2-pentanol gives Iodoform test with yellow ppt. of Iodoform while 3-pentanol does not give this test.

Question 14.
Write the chemical reaction to explain Kolbe’s reaction. (Comptt. Delhi 2013)
Answer:
Kolbe’s reaction : Phenol reacts with CO₂ in presence of sodium hydroxide (NaOH) at 4 – 7 Atm and 390 – 410 K giving salicylic acid

Question 15.
How would you obtain ethane-1, 2-diol from ethanol? (Comptt. All India 2013)
Answer:

Question 16.
How would you obtain acetophenone from phenol? (Comptt. All India 2013)
Answer:

Question 17.
Write IUPAC name of the following : (Comptt. All India 2013)

Answer:
IUPAC name : 2~Bromo-3-methylbut -2-ene-l-ol

Question 18.
How would you obtain phenol from benzene? (Comptt. All India 2013)
Answer:

Question 19.
Write IUPAC name of the following (Comptt. All India 2013)

Answer:
IUPAC name : 2-Methoxy-5-methyl phenol

Question 20.
Which of the following isomers is more volatile : o-nitrophenol or p-nitrophenol? (Delhi 2014)
Answer:
o-nitrophenol is more volatile than p-nitrophenol due to intramolecular hydrogen bonding.

Question 21.
Name the alcohol that is used to make the following ester : (Comptt All India 2014)

Answer:
Alcohol used : Propan-2-ol

Question 22.
Write the IUPAC name of the given compound (Delhi 2015)

Answer:
2, 5-dinitrophenol.

Question 23.
Write the IUPAC name of given compound: (All India 2015)

Answer:
IUPAC name : 1-Ethoxy-2-methylpropane

Question 24.
Write the IUPAC name of the given compound: (All India 2016)

Answer:
2-Phenylethanol

Question 25.
Write equation of the nitration of anisole. (Comptt. Delhi 2016)
Answer:

Question 26.
Out of CH₃OH and
, which one is more acidic? (Comptt. All India 2016)
Answer:
is more acidic, as Phenoxide formed is more stabilized by Resonance.

Question 27.
Write the IUPAC name of the following compound: (All India 2017)

Answer:
2-Bromo-3-methylbut-2-enol-1 -ol

Question 28.
Write the IUPAC name of the following compound: (All India 2017)

Answer:
3-phenylprop-2-en-1-ol

Question 29.
Write IUPAC name of the following compound : (Comptt. Delhi 2017)

Answer:
2, 3-Dinitro phenol.

Question 30.
What happens when phenol is oxidized by Na₂Cr₂O₇/H₂SO₄?
Answer:
Phenol forms benzoquinone on oxidation with Na₂Cr₂O₇ / H₂SO₄,

Question 31.
What happens when phenol is heated with zinc dust? (Comptt. All India 2017)
Answer:
Benzene is formed when phenol is heated with zinc dust.

Question 32.
What happens when phenol is treated with bromine water? (Comptt. All India 2017)
Answer:
2, 4, 6-tribromophenol is formed when phenol is treated with bromine water.

Question 33.
Give simple chemical tests to distinguish between the following pairs of compounds: Benzoic acid and Phenol (All India 2017)
Answer:
Ferric chloride test. Add neutral FeCl₃ in both the solutions, phenol reacts with neutral FeCl₃ to form an iron-phenol complex giving violet colour but benzoic acid does not.

Alcohols, Phenols and Ethers Class 12 Important Questions Short Answer Type -I [SA – I]

Question 34.
Complete the following reaction equations : (Delhi 2009)

Answer:

Question 35.
Illustrate the following reactions giving a chemical equation for each :
(i) Kolbe’s reaction
(ii) Williamsons synthesis of an ether (Delhi 2010)
Answer:
(i) Kolbe’s reaction : Phenol reacts with CO₂ in presence of sodium hydroxide (NaOH) at 4 – 7 Atm and 390 – 410 K giving salicylic acid

(ii) Williamsons synthesis of an ether : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

Question 36.
Explain the following reactions with an example for each :
(i) Reimer-Tiemann reaction
(ii) Friedel-Crafts reaction. (Delhi 2010)
Answer:
(i) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(ii) Friedel-Crafts reaction : This reaction is used for introducing an alkyl or an acyl group into an aromatic compound in presence of Lewis acid catalyst (AlCl₃)
Example:

Question 37.
How are the following conversions carried out?
(i) Propene to propan-2-ol
(ii) Ethylmagnesium chloride to propan-1-ol. (Delhi 2010
Answer:
(i) Propene to propan-2-ol

(ii) Ethylmagnesium chloride to propan-1-ol

Question 38.
How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol,
(ii) Methyl magnesium bromide to 2-methylpropan-2-ol. (All India 2010)
Answer:
(i) Benzyl chloride to benzyl alcohol

Question 39.
Explain the following giving one example for each :
(i) Reimer-Tiemann reaction.
(ii) Friedel-Craft’s acetylation of anisole. (Delhi 2011)
Answer:
(i) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(ii) Friedel-Craft’s acetylation of anisole :

Question 40.
How would you obtain
(i) Picric acid (2, 4, 6-trinitrophenol) from phenol,
(ii) 2-Methylpropene from 2-methylpropanol? (Delhi 2011)
Answer:
(i) Picric acid from phenol :

Question 41.
Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol. (All India 2012)
Answer:
Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols

Mechanism : It involves three steps :
(i) Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺

Question 42.
Explain the following behaviours :
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol. (All India 2012)
Answer:
(i) Alcohols can form H-bonds with water and break the H-bonds already existing between water molecules. So they are soluble in water.

On the other hand, hydrocarbons cannot form H-bonds with water and hence are insoluble in water.

(ii) Due to strong – R and -1 effect of the – NO₂ group, electron density in the – OH bond decreases and hence the loss of a proton becomes easier. Moreover O-nitrophenoxide ion is stabilized by resonance, 1 thereby making O-nitrophenol a stronger acid.
In O-methoxyphenol, due to + R effect of the – OCH₃ group the electron density in the O – H bond increases thereby making the loss of proton difficult. Furthermore, the Q-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance because the two negative charges repel each other. So O-methoxyphenol is a weaker acid.

Question 43.
Explain the mechanism of following reaction: (Delhi 2013)

Answer:

Question 44.
How will you convert:
(i) Propene to propan-2-ol?
(ii) Phenol to 2, 4, 6-trinitrophenol? (Delhi 2013)
Answer:
(i) Propene to propan-2-ol:

(ii) Phenol to 2, 4, 6-trinitrophenol?

Question 45.
How will you convert the following?
(i) Propan-2-ol to propanone
(ii) Phenol to 2, 4, 6-tribromophenol (Delhi 2013)
Answer:
(i) Propan-2-ol to propane

(ii) Phenol to 2, 4, 6-tribromophenol: Phenol reacts with bromine in presence of polar solvent H₂O to form 2, 4, 6-tribromophenol (white ppt.)

Question 46.
How will you convert:
(i) Propene to Propane-1-ol?
(ii) Ehtanal to Propan-2-ol (Delhi 2013)
Answer:
(i) Propene to Propane-1-ol

Question 47.
Explain the mechanism of the following reaction: (All India 2013)

Answer:

Question 48.
Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
(ii) Williamson’s ether Synthesis (All India 2013)
Answer:
(i) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(ii) Williamson’s ether synthesis : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

Question 49.
Write the equations involved in the following reactions :
(i) Reimer – Tiemann reaction
(ii) Williamson synthesis (All India 2014)
Answer:
(i) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(ii) Williamson’s ether synthesis : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

Question 50.
Write the mechanism of the following reaction : (All India 2014)

Answer:

Question 51.
Write the equations involved in the following reactions:
(i) Williamson ether synthesis
(ii) Kolbe’s reaction (Comptt. Delhi 2014)
Answer:
(i) Williamson ether synthesis : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

(ii) Kolbe’s reaction : Phenoxide ion generated by treating phenol with sodium hydroxide is more reactive than phenol and undergoes electrophilic substitution with carbon dioxide. Ortho hydroxybenzoic acid is formed as the main reaction product.

Question 52.
How are the following conversions carried out?
(i) Propene to Propan-2-ol
(ii) Ethyl chloride to Ethanal (Comptt. Delhi 2014)
Answer:
(i) Propene to propan-2-ol

Question 53.
Explain the mechanism of dehydration steps of ethanol : (Comptt. Delhi 2015)

Answer:

Question 54.
Write the mechanism of acid dehydration of ethanol to yield ethene. (Comptt. All India 2015)
Answer:
The mechanism of dehydration of ethanol involves the following steps :
Step 1 : Formation of protonated alcohol

Question 55.
Write the mechanism of the following reaction: (Delhi 2016)

Answer:
Mechanism:

Question 56.
(a) Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenol
(b) Write the mechanism (using curved arrow notation) of the following reaction: (All India 2016)

Answer:
(a) Order of acidic strength:
p-cresol < phenol < p-nitrophenol
(b) Mechanism of acid catalysed hydration of alkene:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.

Question 57.
Write the structures of the products when Butan-2-ol reacts with the following:
(a) CrO₃
(b) SOCl₂ (All India 2016)
Answer:

Alcohols, Phenols and Ethers Class 12 Important Questions Short Answer Type -II [SA – II]

Question 58.
Explain the mechanism of the following reactions :
(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(ii) Acid catalysed dehydration of an alcohol forming an alkene.
(iii) Acid catalysed hydration of an alkene forming an alcohol. (Delhi 2009)
Answer:
(i) Carbonyl group undergoes nucleophillic addition reaction with Grignard reagent to form an adduct which undergoes hydrolysis to give alcohol in the following manner:

(ii) The mechanism of dehydration of ethanol involves the following steps :
Mechanism : It involves the following three steps :
Step 1 : Formation of protonated alcohol

(iii) Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols. Mechanism : It involves the following three steps :
Step 1 : Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺

Question 59.
Explain the following observations :
(i) The boiling point of ethanol is higher than that of methoxymethane.
(ii) Phenol is more acidic than ethanol.
(iii) o- and p-nitrophenols are more acidic than phenol. (All India 2009)
Answer:
(i) Due to presence of intermolecular H-bonding, associated molecules are formed, hence ethanol has high boiling point while methoxymethane does not have intermolecular H-bonding.
(ii) Phenol on losing H⁺ ion forms phenoxide ion, and ethanol on losing H⁺ ion forms ethoxide ion. Phenoxide ion is more stable than ethoxide ion as phenoxide ion exists in resonance structure. Due to this phenol is more acidic than ethanol.
(iii) Both o- and p-nitrophenols contain the NO₂ group which is an electron withdrawing group. Due to -R and -I effect of the -NO₂ group, electron density in the OH bond of substituted phenol decreases and hence the loss of proton becomes easy and therefore more acidic.

Question 60.
How would you convert the following :
(i) Phenol to benzoquinone
(ii) Propanone to 2-methylpropan-2-ol
(iii) Propene to propan-2-ol (All India 2010)
Answer:
(i) Phenol to benzoquinone

Question 61.
How would you obtain the following ;
(i) Benzoquinone from phenol
(ii) 2-Methylpropan-2-ol from ethylmagnesium chloride
(iii) Propan-2-ol from propene (All India 2010)
Answer:
(i) Benzoquinone from phenol

(ii) Ethylmagnesium chloride to propan-1-ol

(iii) Propene to propan-2-ol

Question 62.
Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidising agent is used.
(i) CH₃CH₂CH₂CH₂OH
(ii) 2-butanol
(iii) 2-methyl-l-propanol (Delhi 2012)
Answer:

Question 63.
(a) Illustrate the following name reactions :
(i) Reimer-Tiemann Reaction (ii) Williamson Synthesis.
(b) Give a chemical test to distinguish between 2-propanol and 2-methyl-2-propanol. (Comptt. Delhi 2012)
Answer:
(a)  (i) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(ii) Williamson’s ether synthesis : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

(b) 2-Propanol is a secondary alcohol. When it reacts with I₂ in NaOH, it forms a yellow ppt of iodoform but 2-methyl-2 propanol does not respond to this test.

Question 64.
(a) Give a seperate chemical test to distinguish between the following pairs of compounds:
(i) Ethanol and Phenol (ii) 2-Pentanol and 3-Pentanol
(b) Explain Kolbe’s reaction with the help of suitable example. (Comptt. Delhi 2012)
Answer:
(a) (i) Ethanol on reacting with I₂ in NaOH gives yellow ppt of iodoform whereas phenol does not respond to this test.
(ii) 2-Pentanol on reacting with I₂ in NaOH gives yellow ppt of iodoform whereas 3-pentanol does not respond to this test.

(b) Kolbe’s reaction : Phenol reacts with NaOH to give sodium phenoxide which on reaction with CO₂ in acid gives salicylic acid.

Question 65.
(a) How would you obtain the following :
(i) 2-methylpentan-2-ol from 2-methyl-1-pentene
(ii) Acetophenone from phenol
(b) Write IUPAC name of the following : (Comptt. All India 2012)

Answer:
(a) (i) 2-Methylpentan-2-ol from 2-methyl-1-pentene

(b) IUPAC name : 1 -ethoxy-2-nitrocyclohexane.

Question 66.
(a) Give mechanism of preparation of ethoxy ethane from ethanol.
(b) How is toluene obtained from phenol? (Comptt. Delhi 2013)
Answer:
(a) Mechanism of formation of ethoxy ethane
Step 1 : Protonation of ethanol

Question 67.
(a) Give chemical tests to distinguish between the following pairs of compounds :
(i) Pentan-2-ol and Pentan-3-ol (ii) Methanol and Phenol
(b) o-nitro phenol is more acidic than o-methoxy phenol. Explain why. (Comptt. All India 2013)
Answer:
(a) (i) 2-pentanol gives Iodoform test with yellow ppt. of Iodoform while 3-pentanone does not give this test.
(ii) Distinction between Methanol and Phenol :
By FeCl₃ test : Phenol gives violet coloured solution with FeCl₃ while methanol does not.

(b) O-nitro phenol is more acidic than o-methoxy phenol due to presence of NO₂ group which has -I effect. It weakens the O-H bond of phenol by withdrawing their electrons and thus releases H+ ion easily while due to +I, +R effect of OCH₃, o-methoxy phenol is less acidic.

Question 68.
(a) Write the mechanism of the following reaction :

(b) Write the equation involved in Reimer-Tiemann reaction. (Delhi 2014)
Answer:


(b) Ortho-nitrophenol has lower boiling point due to formation of intramolecular H-bonding whereas p-nitrophenol forms intermoleeular H-bonding.

Question 69.
Explain the following with an example for each :
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis (Comptt. All India 2014)
Answer:
(i) Kolbe’s reaction: Phenoxide ion generated by treating phenol with sodium hydroxide is more reactive than phenol and undergoes electrophilic substitution with carbon dioxide. Ortho hydroxybenzoic acid is formed as the main reaction product.

(ii) Reimer-Tiemann reaction : Treatment of phenol with CHCl₃ in presence of aqueous NaOH at 340K followed by hydrolysis gives salicylaldehyde.

(iii) Williamson’s ether synthesis : The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism.
Example :

Question 70.
How are the following conversions carried out?
(i) Propene → 4 Propan-2-ol
(ii) Ethylmagnesium chloride → 4 Propan-l-ol
(iii) Benzyl chloride → Benzyl alcohol (Comptt All India 2014)
Answer:
(i) Propene to propan-2-ol

(ii) Ethylmagnesium chloride to propan-1-ol

(iii) Benzyl chloride → Benzyl alcohol

Question 71.
How do you convert the following :
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol (Delhi 2015)
Answer:

Question 72.
(a) Write the mechanism of the following reaction :

(b) Write the equation involved in the acetylation of Salicylic acid (Delhi 2015)
Answer:

Question 73.
Give reasons for the following :
(i) Phenol is more acidic than methanol.
(ii) The C—O—H bond angle in alcohols is slightly less than the tetrahedral angle (190°28′).
(iii) (CH₃)₃C—O—CH₃ on reaction with HI gives (CH₃)₃C—I and CH₃—OH as the main products and not (CH₃)₃C—OH and CH₃—I. (All India 2015)
Answer:
(i) Phenol is more acidic than methanol because in phenol, phenoxide ion formed is more stabilized by resonance than phenol. There is no resonance in methanol.
(ii) The C—O—H bond angle in alcohols is slightly less than tetrahedral angle due to repulsion between the lone pairs of electrons of oxygen.
(iii) (CH₃)₃C+ is 3° carbo-cation which is more stable than for S_N1 reaction.

Question 74.
How are the following conversions carried out?
(i) Propene to propan-2-ol
(ii) Benzyl chloride to Benzyl alcohol
(iii) Anisole to p-Bromoanisole (Comptt. Delhi 2015)
Answer:
(i) Propene to propan-2-ol

Question 75.
Write the major product in the following equations : (Comptt. All India 2015)

Answer:

Question 76.
Write the main product(s) in each of the following reactions: (Delhi 2016)

Answer:

Question 77.
Write the final product(s) in each of the following reactions: (All India 2016)

Answer:

Question 78.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol (Comptt. Delhi 2016)
Answer:

Question 79.
Explain the following with an example in each :
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis (Comptt. All India 2016)
Answer:
(i) Kolbe’s reaction : Phenol reacts with CO₂ at 390-400 K under pressure 4-7 giving Selicyle Acid

Question 80.
(a) What happens when CH₃—O—CH<sub3 is heated with HI?
(b) Explain mechanism for hydration of acid catalyzed ethene :
CH2 = CH2 + HzO CH3—CH,—OH (Comptt. Delhi 2017)
Answer:
(a) Methyl Iodide (CH₃I) and Methanol (CH₃OH) are formed when CH₃—O—CH₃ is heated with HI.

(b) Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols.
Mechanism : It involves the following three steps :
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.

Question 81.
(a) Why phenol is more acidic than ethanol?
(b) Write the mechanism of acid dehydration of ethanol to yield ether:  (Comptt. All India 2017)
Answer:
(a) Phenol on losing H+ ion forms phenoxide ion, and ethanol on losing H+ ion forms ethoxide ion. Phenoxide ion is more stable than ethoxide ion as phenoxide ion exists in resonance structure. Due to this phenol is more acidic than ethanol.

Question 82.
(i) Write Reimer-Timann reaction.
(ii) Write the mechanism of acid dehydration of ethanol to yield ethene :  (Comptt. All India 2017)
Answer:
(i) Reimer-Timman reaction—

Alcohols, Phenols and Ethers Class 12 Important Questions long Answer Type (LA)

Question 83.
(a) Write the product(s) in the following reactions: (Delhi 2017)

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanol and Phenol
(ii) Propanol and 2-methylpropan-2-ol
Answer:

Question 84.
Write the formula of reagents used in the following reactions:
(i) Bromination of phenol to 2,4,6-tribromophenol
(ii) Hydroboration of propene and then oxidation to propanol.
(b) Arrange the following compound groups in the increasing order of their property indicated:
(i) p-nitrophenol, ethanol, phenol (acidic character)
(ii) Propanol, Propane, Propanal (boiling point)
(c) Write the mechanism (using curved arrow notation) of the following reaction: (Delhi 2017)

Answer:

Important Questions for Class 12 Chemistry

The post Important Questions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Class 12 Important Questions appeared first on Learn CBSE.

DCE Scholarship: DCE scholarship which is also known as Kerala scholarship is being run by the Kerala government and it’s various departments. Every year the Kerala government calls for applications for scholarships to ensure that the finances are not a hindrance in student’s path in the pursuit of their academic career. The DCE scholarship is applicable for all the students coming from different strata of life. The main objective behind these scholarships is to provide financial help to the deserving and meritorious students of Kerala state.

Students from college as well a school can all apply for this scholarship. A total of 16 scholarships are available by the Kerala government under the central as well as state government. The main departments for Kerala government are the department of minority welfare and department of collegiate education.

DCE Scholarship – Detailed List

Department of Minority Welfare, Kerala government provides different type of scholarships for the students of Kerala state.

DCE Scholarship Eligibility Criteria

The eligibility criteria are not common for all the DCE scholarships. To be eligible for Kerala scholarship, one needs to be either a permanent resident or domicile of Kerala state. There are other conditions as well that needs to be fulfilled for the eligibility of DCE scholarship. Below are the eligibility conditions for DCE scholarship:

DCE Scholarship Application Process

All the scholarships provided by the Kerala government can be filled online through two main portals of the Kerala government. These are the department of collegiate education and minority welfare scholarship portal. Also, the Kerala scholarship that is run by the central government invites the applications through the NSP portal.

DCE Scholarship Rewards

The amount for the scholarship for each DCE scholarship varies from scholarship to scholarship to the students. Furthermore, it depends on the current standard in which you are studying. Every Kerala scholarship offers a substantial amount of rewards to the students through direct benefit transfer. Below is the table that contains a detailed description of all the scholarships. Also, you will get to know about the duration for which these scholarships are available in some of the cases.

Students can also Check

• NTSE Scholarship – Click Here
• Central Sector Scholarship – CSSS

The post DCE Scholarship | Complete List, Eligibility, Application Process, Rewards appeared first on Learn CBSE.

Important Questions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life Class 12 Important Questions

Chemistry in Everyday Life Class 12 Important Questions Very Short Answer Type

Question 1.
Differentiate between disinfectants and antiseptics. (Delhi 2012)
Answer:

Question 2.
What are limited spectrum antibiotics? Give one example. (Comptt. Delhi 2012)
Answer:
Those antibiotics which are specific for certain diseases are called limited spectrum antibiotics. Example: Streptomycin for tuberculosis.

Question 3.
Name the important by-products of soap industry. (Comptt. Delhi 2012)
Answer:
Glycerol is the important by-product of soap industry.

Question 4.
Why do we require artificial sweetening agents? (Comptt. All India 2012)
Answer:
To reduce calorie intake and to protect teeth from decaying, we need artificial sweetners.

Chemistry in Everyday Life Class 12 Important Questions Short Answer Type SA-I

Question 5.
What are food preservatives? Name two such substances. (All India 2012)
Answer:
Food preservatives : Food preservatives are the compounds which prevent spoilage of food due to microbial growth.
Two substances : Example : Sodium benzoate, vinegar.

Question 6.
Explain the cleaning action of soap. Why do soaps not work in hard water? (All India 2012)
Answer:
Cleaning action of soap : The cleansing action of soap is due to the fact that soap molecules form micelles around the oil droplets in such a way that hydrophobic part of stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats.

Reason : Hard water contains calcium and magnesium ions. These ions form insoluble Ca and Mg salts. These salts act as scum. The insoluble scum sticks on the clothes, so the cleaning capacity of soap is reduced when Na or K soaps are dissolved in hard water.

Question 7.
Explain the following terms with suitable examples :
(a) Cationic detergents
(b) Anionic detergents (Comptt. Delhi 2013)
Answer:
(a) Cationic detergents :
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(b) Anionic detergents : Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H2S04 followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

Chemistry in Everyday Life Class 12 Important Questions Short Answer Type SA-II

Question 8.
Explain the following types of substances with one suitable example, for each case :
(i) Cationic detergents.
(ii) Food preservatives.
(iii) Analgesics. (Delhi 2009)
Answer:
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Analgesics : Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, in coordination or paralysis or some other disturbance of nervous system.
They are of two types :
(a) Non-narcotic analgesics Example : Aspirin
(b) Narcotic analgesics Example : Morphine

Question 9.
How do antiseptics differ from disinfectants? Give one example of each type. (Delhi 2008)
Antiseptics Disinfectants
Answer:

Question 10.
What are the following substances? Give one example of each type.
(i) Antacid
(ii) Non-ionic detergents
(iii) Antiseptics (All India 2008)
Answer:
(i) Antacid : Those substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example : Sodium bicarbonate, Mg(OH)₂

(ii) Non-ionic detergents : These are esters of high molecular mass alcohols obtained by reaction between polyethylene glycol and steric acid.
Example :

(iii) Antiseptics : These are chemical substances which prevent the growth of micro-organisms and may even kill them and safe to be applied on living tissues.
Example : Furacin, soframycin etc.

Question 11.
Describe the following substances with one suitable example of each type : (All India 2009)
(i) Non-ionic detergents
(ii) Food preservatives
(iii) Disinfectants
Answer:
(i) Non-ionic detergents : These are esters of high molecular mass alcohols obtained by reaction between polyethylene glycol and steric acid.
Example :

(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Disinfectants :
Disinfectants are chemical compounds which kill microorganisms but are not safe when applied on living organisms.
Example : Phenol, chlorine.

Question 12.
What are the following substances? Give one example of each of them. (All India 2009)
(i) Cationic detergents
(ii) Enzymes
(iii) Sweetening agents
Answer:
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(ii) Enzymes : Enzymes are biological catalysts which are chemically globular proteins having high molecular mass and highly specific in their actions due to presence of active sites of definite shape and size on their surfaces so that only specific substrate can fit in them.
Example : Pepsin, amylase

(iii) Sweetening agents : Those chemical substances which are sweet in taste but do not add any calories to our body are called artificial sweetening agents. These are excreted as such through urine.
Example : Saccharin, aspartame etc.

Question 13.
What are analgesic medicines? How are they classified and when are they commonly recommended for use? (Delhi 2010)
Answer:
Analgesic medicine : Drugs which reduce or abolish pain without causing reduction of consciousness, mental confusion, incoordination or paralysis or some other disorder of the nervous system are called analgesic medicines.
They are classified into the following two categories :
(i) Non-narcotic (non-addictive) drugs Example : Aspirin, Ibuprofen
(ii) Narcotic (addictive) drugs Example : Morphine, Heroin.
They are recommended with proper prescription because they are habit forming drugs.

Question 14.
Explain the following terms with one suitable example in each case.
(i) Cationic detergents
(ii) Enzymes
(iii) Antifertility drugs (Delhi 2010)
Answer:
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(ii) Enzymes :
The enzymes may be defined as bio-catalysts which catalyse the bio-chemical reactions in the living organisms.
Example : Pepsin and Amylase.

(iii) Antifertility drugs : Chemical substances, which are used to check pregnancy in women, are called antifertility drugs or birth control pills or oral contraceptives. These control the female menstrual cycle and ovulation.
Example : Norethindrone, Ethinyl estradol, Mestranol.

Question 15.
Explain the following terms with one example in each case :
(i) Food preservatives
(ii) Enzymes
(iii) Detergents (Delhi 2010)
Answer:
(i) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(ii) Enzymes : The enzymes may be defined as bio-catalysts which catalyse the bio-chemical reactions in the living organisms.
Example : Pepsin and Amylase.

(iii) Detergents : They may be defined as ammonium, sulphonate or sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms.
Example : Sodium lauryl sulphate.

Unlike soaps they are non-biodegradable and hence cause water pollution but they can be conveniently used even in hard water.

Question 16.
What are analgesic drugs? How are they classified and when are they usually recommended for use? (Delhi 2010)
Answer:
Analgesic drugs : These drugs are the chemical substances which are given to relieve body pains. These act on the central nervous system.

These are classified as Narcotics i.e. habit forming and non-narcotics i.e. not habit forming.
Examples of Narcotics : Opium which contains alkaloids such as Codeine and Morphine.
Examples of Non-Narcotics : Aspirin and Ibuprofen.

Question 17.
Explain the following terms with an example for each :
(i) Antibiotics
(ii) Antiseptics
(iii) Analgesics (Delhi 2010)
Answer:
(i) Antibiotics : Those chemical substances which are produced completely or partially by chemical synthesis in low concentration and either kill or inhibit the growth of microorganisms by intervening in their metabolic processes, are known as antibiotics.
Examples:
Tetracycline ➝ Bacteriostatic antibiotics, Penicillin Bactericidal antibiotics

(ii) Antiseptics :
These are chemical substances which prevent the growth of micro-organisms and may even kill them and safe to be applied on living tissues.
Example : Furacin, soframycin etc.

(iii) Analgesics :
Analgesic drugs : These drugs are the chemical substances which are given to relieve body pains. These act on the central nervous system.

These are classified as Narcotics i.e. habit forming and non-narcotics i.e. not habit forming.

Examples of Narcotics : Opium which contains alkaloids such as Codeine and Morphine.
Examples of Non-Narcotics : Aspirin and Ibuprofen.

Question 18.
Describe the following giving one example for each :
(i) Detergents
(ii) Food preservatives
(iii) Antacids (Delhi 2010)
Answer:
(i) Detergents :
They may be defined as ammonium, sulphonate or sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms.
Example : Sodium lauryl sulphate.
Unlike soaps they are non-biodegradable and hence cause water pollution but they can be conveniently used even in hard water.

(ii) Food preservatives: They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Antacids : The substances which neutralize the excess acid and raise the pH to an appropriate level in the stomach are called antacids.
Example: Sodium bicarbonate, Ranitidine etc.

Question 19.
Explain the following terms with one suitable example for each :
(i) A sweetening agent for diabetic patients
(ii) Enzymes
(iii) Analgesics (Delhi 2010)
Answer:
(i) The sweetening agent used in the preparation of sweets for a diabetic patient is Saccharin.

(ii) Enzymes :
The enzymes may be defined as bio-catalysts which catalyse the bio-chemical reactions in the living organisms.
Example : Pepsin and Amylase.

(iii) Analgesics :

Analgesic drugs : These drugs are the chemical substances which are given to relieve body pains. These act on the central nervous system.

These are classified as Narcotics i.e. habit forming and non-narcotics i.e. not habit forming.
Examples of Narcotics : Opium which contains alkaloids such as Codeine and Morphine.
Examples of Non-Narcotics : Aspirin and Ibuprofen.

Question 20.
Answer the following questions :
(i) Why do soaps not work in hard water?
(ii) What are the main constituents of dettol?
(iii) How do antiseptics differ from disinfectants? (Delhi 2010)
Answer:
(i) Hard water contains insoluble calcium and magnesium chlorides which forms insoluble precipitate (scum) with soap and thus cannot be rinsed off easily.
(ii) Dettol is mixture of chloroxylenol and α-terpineol in a suitable solvent.
(iii)

Question 21.
What are the following substances? Give one example of each.
(i) Food preservatives
(ii) Synthetic detergents
(iii) Antacids (All India 2010)
Answer:
(i) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(ii) Synthetic detergents : Synthetic detergents are cleansing agents which have all the properties of soap but which actually do not contain any soap.
Example : Sodium lauryl sulphate (or any one other)

(iii) Antacids :
The substances which neutralize the excess acid and raise the pH to an appropriate level in the stomach are called antacids.
Example: Sodium bicarbonate, Ranitidine etc.

Question 22.
(a) Differentiate between a disinfectant and an antiseptic. Give one example of each.
(b) What is tincture of iodine and what is it used for? (All India 2010)
Answer:
(a)

(b) Tincture of iodine is 2-3% solution of iodine in alcohol and water.

• Use : It is used as a powerful antiseptic and applied on wounds to kill and prevent growth of micro-organisms.

Question 23.
What are the following substances? Give one example of each one of them. (Delhi 2012)
(i) Tranquilizers
(ii) Food preservatives
(iii) Synthetic detergents
Answer:
(i) Tranquilizers : Tranquilizers are chemical compounds used for the treatment of stress and mild or even severe mental diseases. Example : Equanil, meprobamate, veronal. (any one)

(ii) Food preservatives : Food preservatives are the compounds which prevent spoilage of food due to microbial growth.
Example : Sodium benzoate, table salt, vegetable oils etc.

(iii) Synthetic detergents : Synthetic detergents are cleansing agents which have all the properties of soap but which actually do not contain any soap.
Example : Sodium lauryl sulphate.

Question 24.
Explain the following terms giving one example of each type : (Delhi 2012)
(i) Antacids
(ii) Disinfectants
(iii) Enzymes
Answer:
(i) Antacids: Those substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example : Sodium bicarbonate, Mg(OH)₂

(ii) Disinfectants : Disinfectants are chemical compounds which kill microorganisms but are not safe when applied on living organisms.
Example : Phenol, chlorine.

(iii) Enzymes : Enzymes are biological catalysts which are chemically globular proteins having high molecular mass and highly specific in their actions due to presence of active sites of definite shape and size on their surfaces so that only specific substrate can fit in them. Example : Pepsin, amylase

Question 25.
What are the following substances ? Give one example of each.
(i) Antihistamines
(ii) Tranquilizers
(iii) Broad spectrum antibiotics (Comptt. Delhi 2012)
Answer:
(i) Antihistamines : Antihistamines are amines which are used as drugs to control the allergy effects produced by histamines. Example : Terfenadine

(ii) Tranquilizers : Tranquilizers are a class of chemical compounds used for the treatment of stress, and mild or even severe mental disease.
Example : Equanil.

(iii) Broad spectrum antibiotics : Antibiotics which kill or inhibit a wide range of Gram¬positive and Gram-negative bacteria are said to be broad spectrum antibiotics.
Example : Chloroamphenicol.

Question 26.
(a) How do antiseptics differ from disinfectants? Give one example of each. (Give two differences)
(b) Why do soaps not work in hard water? (Comptt. Delhi 2012)
Answer:
(a)

(b) Hard water contains Ca⁺² and Mg⁺² ions. These ions form insoluble calcium and magnesium salts respectively, when Na or K soaps are dissolved in hard water. These insoluble salt, separate as scum which adheres to fabric, thereby making soap ineffective for cleansing action.

Question 27.
What are the following substances ? Give one example of each.
(i) Analgesics
(ii) Antibiotics
(iii) Tranquilizers (Comptt. All India 2012)
Answer:
(i) Analgesics : Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, incoordination or paralysis or some other disturbances of nervous system.
Example : Aspirin.

(ii) Antibiotics : Antibiotic refers to a substance produced wholly or partly by chemical synthesis which in low concentration inhibits the growth or destroys microorganisms by intervening in their metabolic processes.
Example : Penicillin.

(iii) Tranquilizers : Tranquilizers are a class of chemical compounds used for the treatment of stress, and mild or even severe mental disease.
Example : Equanil.

Question 28.
What are the following substances? Give one example of each.
(i) Broad Spectrum antibiotics
(ii) Narcotic analgesics
(iii) Synthetic detergents (Comptt. All India 2012)
Answer:
(i) Broad spectrum antibiotics : Antibiotics which kill or inhibit a wide range of Gram¬positive and Gram-negative bacteria are said to be broad spectrum antibiotics.
Example : Chloroamphenicol.

(ii) Narcotic analgesics : Narcotic analgesics are administered in medicinal doses, relieve pain and produce sleep,
Example : Morphine and many of its homologues.

(iii) Synthetic detergents : Synthetic detergents are cleansing agents which have all the properties of soap but which actually do not contain any soap.
Example : Sodium lauryl sulphate.

Question 29.
(a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills? (Delhi 2013)
Answer:
(a) Sodium Benzoate: It is a food preservative.
(b) Bithional acts as deodorant in soaps, hence it works as an antiseptic agent and reduces the odours produced by bacterial decomposition of organic matter on the skin.
(c) Tranquilizers like barbiturates are used in sleeping pills.

Question 30.
(i) What class of drug is Ranitidine?
(ii) If water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
(iii) Which of the following is an antiseptic? 0.2% phenol, 1% phenol (All India 2013)
Answer:
(i) Ranitidine is an Antacid.
(ii) We will use synthetic detergents because they can produce lather with the hard water containing Ca²⁺ ions.
(iii) 0.2% phenol acts as an antiseptic.

Question 31.
(a) How do antiseptics differ from disinfectants? Give one example of each.
(b) What are tranquilizers? Give one example. (Comptt. All India 2013)
Answer:
(a) AntisepticsDisinfectants1. They are chemical substances which prevent the growth of micro-organisms and may even kill them.1. They are chemical substances which kill micro-organisms.2. They are safe to be applied to the living tissues.2. They are not safe to be applied to the living tissues.3. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Example : Furacin, soframycin, dettol and savlon, 0.2% solution of phenol.3. They are used to kill micro-organisms present in the drains, toilets, floors etc.
Example: Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm).

(b) Tranquilizers : Drugs which are used for the treatment of stress, fatigues, mild and severe mental diseases are called tranquilizers.
Example : Iproniazid, Phenelzine etc.

Question 32.
Explain the following and give one example for each :
(i) Broad spectrum antibiotics
(ii) Antipyretics
(iii) Anti-oxidants (Comptt. All India 2013)
Answer:
(i) Broad spectrum antibiotics : Antibiotics which kill or inhibit a wide range of Gram¬positive and Gram-negative bacteria are called broad spectrum antibiotics.
Example : Chloro-amphical

(ii) Antipyretics : Chemicals, which are used to bring down the body temperature during high fever, are called antipyretics Example : Paracetamol, Aspirin etc.

(iii) Anti-oxidants : Those molecules, which inhibit the oxidation of other molecules, are called anti-oxidants
Example : Thiols, Ascorbic acid etc.

Question 33.
(i) Give two examples of macromolecules that are chosen as drug targets.
(ii) What are antiseptics? Give an example.
(iii) Why is use of aspartame limited to cold foods and soft drinks? (Delhi 2014)
Answer:
(i) Carbohydrates and proteins

(ii)  Antiseptics : These are chemical substances which prevent the growth of micro-organisms and may even kill them and safe to be applied on living tissues.
Example : Furacin, soframycin etc.

(iii) Because it decomposes at baking or cooking temperature.

Question 34.
(i) Name the sweetening agent used in the preparation of sweets for a diabetic patient.
(ii) What are antibiotics? Give an example.
(iii) Give two examples of macromolecules that are chosen as drug targets. (Delhi 2014)
Answer:
(i) Saccharin is used for a diabetic patient for preparation of sweets.

(ii) Antibiotics :
Those chemical substances which are produced completely or partially by chemical synthesis in low concentration and either kill or inhibit the growth of microorganisms by intervening in their metabolic processes, are known as antibiotics.
Examples:
Tetracycline ➝ Bacteriostatic antibiotics, Penicillin Bactericidal antibiotics

(iii) Carbohydrates, proteins, nucleic acid etc.

Question 35.
(i) What are disinfectants? Give an example.
(ii) Give two examples of macromolecules that are chosen as drug targets.
(iii) What are anionic detergents? Give an example. (Delhi 2014)
Answer:
(i) Disinfectants : Disinfectants are chemical compounds which kill microorganisms but are not safe when applied on living organisms.
Example : Phenol, chlorine.

(ii) Macromolecules used as drug targets are carbohydrates, proteins, nucleic acid and lipids.

(iii) Anionic detergents.
Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H₂SO₄ followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

Question 36.
Explain the following terms with a suitable example for each:
(i) Disinfectants
(ii) Antacids
(iii) Food preservatives (Comptt. Delhi 2014)
Answer:
(i) Disinfectants : These are the chemical substances which are used for killing or
preventing the growth of micro-organisms but they are not safe for living tissues.

(ii) Antacid : Those substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example : Sodium bicarbonate, Mg(OH)₂

(iii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

Question 37.
What are the following? Give one example of each.
(i) Sweetening agents
(ii) Food preservatives
(iii) Antibiotics (Comptt. Delhi 2014)
Answer:
(i) Sweetening agent : Those chemical substances which are sweet in taste but do not add any calories to our body are called artificial sweetening agents. These are excreted as such through urine.
Example ; Saccharin, aspartame etc.

(ii) Food preservatives: They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Antibiotics : Those chemical substances which are produced completely or partially ’by chemical synthesis in low concentration and either kill or inhibit the growth of micro-organisms by intervening in their metabolic processes, are known as antibiotics.
Example : Tetracycline, Vancomycin

Question 38.
What are biodegradable and non-biodegradable detergents? Give one example of each. (Comptt. Delhi 2014)
Answer:
Biodegradable detergents : Detergents, having straight hydrocarbon chains are easily degraded or decomposed by micro-organism, are known as biodegradable detergents.
Example : Sodium lauryl sulphate.

Non-biodegradable detergents : Detergents containing branched hydrocarbon chains and are not easily decomposed by the micro-organisms, are known as non-biodegradable detergents.
Example : Sodium 4 – (1, 3, 5, 7-tetramethyloctyl) benzene sulphonate.

Question 39.
What is meant by the following terms? Explain with an example for each.
(i) Target molecules as used in medicinal chemistry
(ii) Food preservative
(iii) Non-ionic detergents (Comptt. All India 2014)
Answer:
(i) Drugs interact with macromolecules like proteins, carbohydrates, lipids etc. and are called as target molecules.

(ii) Food preservatives: They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Non-ionic detergents : These are esters of high molecular mass alcohols obtained by reaction between polyethylene glycol and steric acid.
Example :

Question 40.
Answer the following questions :
(i) Why should medicines not be taken without consulting a doctor?
(ii) What is meant by ‘broad spectrum antibiotics’?
(iii) What are the main constituents of Dettol? (Comptt. All India 2014)
Answer:
(i) Because medicines can cause harm to human body if a person does not know its physiological function on body.

(ii) Antibiotics which kill or inhibit a wide range of gram positive and gram negative bacteria, are called broad spectrum antibiotics.

(iii) Dettol is mixture of chloroxylenol and a-terpineol in a suitable solvent.

Question 41.
Answer the following :
(i) Why is the use of aspartame limited to cold foods and drinks?
(ii) How do antiseptics differ from disinfectants?
(iii) Why do soaps not work in hard water? (Comptt. All India 2014)
Answer:
(i) Use of aspartame is limited to cold foods and soft drinks because it is unstable at cooking temperature.

(ii)

(iii) Hard water contains insoluble calcium and magnesium chlorides which forms insoluble ppt (scum) with soap and thus cannot be rinsed off easily.

Question 42.
Define the following:
(i) Anionic detergents
(ii) Broad spectrum antibiotics
(iii) Antiseptic (Delhi 2017)
Answer:
(i) Anionic detergents : Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H2S04 followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

(ii) Broad spectrum antibiotics : Antibiotics which kill or inhibit a wide range of Gram¬positive and Gram-negative bacteria are said to be broad spectrum antibiotics.
Example : Chloroamphenicol.

(iii) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues.

Question 43.
Define the following:
(i) Cationic detergents
(ii) Narrow spectrum antibiotics
(iii) Disinfectants (Delhi 2017)
Answer:
(i) Cationic detergents. They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example: Cetyltrimethyl ammonium bromide —

(ii) Narrow spectrum antibiotics. Narrow spectrum antibiotics are those antibiotics which are mainly effective against gram positive or gram negative bacteria.

(iii) Disinfectants. Disinfectants kill or prevent growth of microbes and are applied on inanimate/non living objects.
Example: Phenol

Question 44.
Define the following:
(i) Anionic detergents
(ii) Limited spectrum antibiotics
(iii) Tranquilizers (Delhi 2017)
Answer:
(i) Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example: Sodium alkyl sulphates
(ii) Limited spectrum antibiotics are those which are effective against a single organism or disease.
(iii) Tranquilizers are class of chemicals used for treatment of stress or mild or severe mental diseases.

Question 45.
Define the following
(a) Anionic detergents
(b) Limited spectrum antibiotics
(c) Antiseptics (All India 2017)
Answer:
(a) Anionic detergents : Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H2S04 followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

(b) Limited spectrum antibiotics are effective against a single organism or disease, e.g., Streptomycin.
(c) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues, e.g., Penicillin.

Question 46.
Define the following:
(a) Anionic detergents
(b) Narrow spectrum antibiotics
(c) Antacids (All India 2017)
Answer:
(a) Anionic detergents : Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H2S04 followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

(b) Narrow spectrum antibiotics are those which are effective against either gram positive or gram negative bacteria.

(c) Antacids are chemical compounds which are used for the treatment of excess acid produced in the stomach.

Question 47.
Define the following:
(a) Cationic detergents
(b) Broad spectrum antibiotics
(c) Tranquilizers (All India 2017)
Answer:
(a)
(i) Cationic detergents. They are quaternary
ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example: Cetyltrimethyl ammonium bromide —

(ii) Narrow spectrum antibiotics. Narrow spectrum antibiotics are those antibiotics which are mainly effective against gram positive or gram negative bacteria.

(iii) Disinfectants. Disinfectants kill or prevent growth of microbes and are applied on inanimate/non living objects.
Example: Phenol

(b) Broad spectrum antibiotics. Antibiotics which kill or inhibit a wide range of gram-positive and gram-negative bacteria.

(c) Tranquilizers. Chemical compounds used for the treatment of stress and mild or severe mental diseases.

Question 48.
Write the therapeutic action of following on human body and mention the class of drugs to which each of these belong:
(i) Ranitidine
(ii) Morphine
(iii) Aspirin (Comptt. Delhi 2017)
Answer:
(i) Ranitidine belongs to antacids and it neutralizes the excess acid and raises the pH to an appropriate level in stomach.

(ii) Morphine belongs to narcotic analgesics and it relieves pain and produces sleep even when taken in small dose.

(iii) Aspirin belongs to non-narcotic analgesics and it inhibits the synthesis of compounds which stimulate inflammation in the tissues and cause pain. Aspirin relieves pain and reduces fever.

Question 49.
Write the therapeutic action of following on human body and mention the class of drugs to which each of the these belong:
(i) Equanil
(ii) Aspirin
(iii) Chloramphenicol (Comptt. Delhi 2017)
Answer:
(i) Equanil belongs to the class of tranquilizers and it is used in controlling depression and hypertension.

(ii) Aspirin belongs to non-narcotic analgesics and it inhibits the synthesis of compounds which stimulate inflammation in the tissues and cause pain. Aspirin reduces pain and fever.

(iii) Chloramphenicol belongs to antibiotics and it is used for treatment of typhoid. It kills or inhibits the growth of micro-organisms.

Question 50.
(i) Name a substance which can be used as an antiseptic as well as disinfectant.
(ii) Name an artificial sweetener whose use is limited to cold foods and drinks.
(iii) What are cationic detergents? (Comptt. All India 2017)
Answer:
(i) Phenol
(ii) Aspartame
(iii) Cationic detergents. They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example: Cetyltrimethyl ammonium bromide —

Important Questions for Class 12 Chemistry

The post Important Questions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life Class 12 Important Questions appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Maths 2014 Outside Delhi

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

CBSE Previous Year Question Papers Class 12 Maths 2014 Outside Delhi Set I

Section – A

Question 1.
If R = [(x, y): x + 2y = 8] is a relation on N, write the range of R. [1]
Solution:
The given relation on N is
R = [(x, y): x + 2y = 8]
Since both x, y ϵ N
x can take values 2, 4, 6 for other values of y ϵ N.
For x = 2, 2 + 2y = 8
⇒ y = 3
For x = 4, 4 + 2y = 8
⇒ y = 2
For x = 6, 6 + 2y = 8
⇒ y = 1
∴ R = {(2, 3), (4, 2), (6, 1)}
∴ The range of R = Set of second element’s
= {1, 2, 3}.

Question 2.
If tan^-1 x + tan^-1 y =, xy < 1, then write the value of x + y + xy. [1]
Solution:
Given,

Question 3.
If A is a square matrix such that A² = A, then write the value of 7A – (I + A)³, where I is an identity matrix. [1]
Solution:
7A – (I + A)³ = 7A – (I³ + A³ + 3I²A + 3IA²)
[ ∵ (a + b)³ = a³ + b³ + 3a²b + 3ab²]
= 7A – (I + A². A + 3IA + 3IA²) [ v Iⁿ = I ∀ n ϵ N]
= 7A – (I + A² + 3A + 3IA) [∵ A² = A, IA = A]
= 7A – (I + A + 3A + 3A)
= 7A – 1 – 7A = -1.

Question 4.
If , find the value of x + y. [1]
Solution:
Given,

Comparing the corresponding elements, we get
x – y = -1, z = 4
2x – y = 0, w = 5
Solving these equations, we get
x = 1, y = 2
x + y = 1 + 2 = 3.

Question 5.
If , find the value of x. [1]
Solution:
Given,

Question 6.
If f(x) = , then write the value of f'(x). [1]
Solution:

Question 7.
Evaluate: [1]

Solution:

Question 8.
Find the value of ‘p’ for which the vectors are parallel. [1]
Solution:

Question 9.

Solution:
Given,

Question 10.
If the Cartesian equations of a line are write the vector equation for the line. [1]
Solution:
The Cartesian equations of a line are

Section – B

Question 11.
If the function f : R→R be given by f(x) = x² + 2, and : R→R be given by g(x) = , x ≠ 1, find fog ang gof and hence find fog(2) and gof(-3). [4]
Solution:
Given, f : R→R

Question 12.
Prove that:

Solution:
L. H. S


OR
If , find the value of x.
Solution:
Given,

Question 13.
Using properties of determinants, prove that: [4]

Solution:
Taking L.H.S

Question 14.
Find the value of at θ = if x = ae^θ(sin θ – cos θ) and y = ae^θ(sin θ + cos θ). [4]
Solution:
Given,

Question 15.
If y = Pe^ax + Qe^bx, Show that [4]

Solution:
Given,

Question 16.
Find the value (s) of x for which y = [x(x – 2)]² is an increasing function.
Solution:
Given
y=[x(x- 2)]²
=> y = x² (x- 2)²
= f(x) (Let)
Differentiating w.r.t. x, we get
f'(x) =
2x(x – 2)² + 2x²(x- 2)
= 2x(x – 2) (x – 2 + x₁)
= 4x(x – 1)(x – 2).


OR
Find the equations of the tangent and normal to the curve at the point ( a, b).
Solution:
The given curve is

Question 17
Evaluate: [4]

Solution:


OR
Evaluate:

Solution:

Question 18.
Find the particular solution of the differential equation = 1 + x + y + xy, given that y = 0 when x = 1. [4]
Solution:
The given differential equation is

Question 19.
Solve the differential equation (1 + x²) + y = e^tan-1x. [4]
Solution:
The given differential equation is

Question 20.
Show that the four points A, B, C and D with position vectors respectively are coplanar. [4]
Solution:
Given the position vectors are

The given points A, B, C and D are coplanar.
Hence Proved
OR
The scalar product of the vector with a unit vector along the sum of vectors is equal to one. Find the value of λ and hence find the unit vector along .
Solution:

Question 21.
A line passes through (2, -1, 3) and is perpendicular to the lines . Obtain its equation in vector and Cartesian form. [4]
Solution:
The given lines are

Question 22.
An experiment succeeds thrice as often as it fails. Find the probability that in the next five trials, there will be at least 3 successes. [4]
Solution:
Let, p = Probability of success

Section – C

Question 23.
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to award its 4,1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹ 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award. [6]
Solution:
Given the awards for sincerity, truthfulness and helpfulness are ₹ x, ₹ y and ₹ z respectively.
3x + 2y + z = 1,600
4x + y + 3z = 2,300
x + y + z = 900
The given equation can be written in matrix form,
AX = B …(i)



Apart from the three values, sincerity, truthfulness and helpfulness, another value for award should be discipline

Question 24.
Show that the altitude of the right circular cone of maximum volume that can be described in a sphere of radius r is . Also show that the maximum volume of the cone is of the volume of the sphere. [6]
Solution:
From the figure OA = OC = r (Radius of the sphere)
From right angled ∆ OBC,
BC = r sin θ,
OB = r cos θ

Question 25.
Evaluate: [6]

Solution:

Question 26.
Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and(3, 4). [6]
Solution:
Let A (-1, 2); B (1, 5) and C (3, 4) Equation of AB is

Question 27.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Also find the distance of the plane obtained above, from the origin. [6]
Solution:
Equation of any plane through the line of intersection of the planes.
x + y + z – 1 = 0
and 2x + 3y + 4z – 5=0
x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0
⇒ (1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z – (1 + 5λ) = 0 ……(i)
This plane is perpendicular to the plane
x – y + z = 0
∴ (1 + 2λ). 1 + (1 + 3λ)(- 1) + (1 + 4λ). 1 = 0
⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0

OR
Find the distance of the point (2, 12, 5) from the point of intersection of the line and the plane .
Solution:
The given line is

Question 28.
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ‘ 80 on each piece of type A and ₹ 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit ? Make it as an LPP and solve graphically. What is the maximum profit per week ? [6]
Solution:
Let x and y be the number of teaching aids of type A and B respectively. Then the LPP is
Maximize Z = 80x + 120y Subject to constraints:
9x + 12y ≤ 180
x + 3y ≤ 30
and x ≥ 0, y ≥ 0,
First we draw the lines AB and CD whose equations are
9x + 12y = 180
⇒ 3x + 4y = 60 …(i)

The feasible region is OAPDO which is shaded in the figure.
The vertices of the feasible region are O (0, 0), A (20, 0), P(12, 6) and D (0, 10)
The value of objective function Z = 80x + 120y as follows :

∴ The profit is maximum at P(12, 6) i.e., when the teaching aids of types A and B are 12 and 6 respectively.
Also maximum profit = ₹ 1680 per week.

Question 29.
There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin ? [6]
Solution:
Let A be the two headed coin, B be the biased coin showing up heads 75% of the times and C be the biased coin showing up tails 40% (i.e., showing up heads 60%) of the times.
Let E₁, E₂ and E₃ be the events of choosing coins of the type A, B, C respectively. Let S be the event of getting a head. Then,


OR
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution.
Solution:
Sample space
S = {1, 2, 3, 4, 5, 6}
Number of ways of selecting any two members of S is

CBSE Previous Year Question Papers Class 12 Maths 2014 Outside Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in previous set.

Section – A

Question 9.
Evaluate: [1]

Solution:

Question 10.
Find a vector of magnitude , making an angle of with x-axis, with y-axis and an acute angle θ wirh z-axis. [1]
Solution:
Here,

Section – B

Question 19.
Using properties of determinants, prove that [4]

Solution:
Taking L. H. S

Hence Proved.

Question 20.
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t = . [4]
Solution:
Here,
x = a sin 2t (1 + cos 2t) …(i)
y = b cos 2t (1 – cos 2t) …(ii)
Differentiating (i) w. r. t. ‘t’, we get

Question 21.
Find the particular solution of the differential equation x(1 + y²) dx – y( 1 + x²) dy = 0, given that y = 1 when x = 0. [4]
Solution:
The given differential equation is
x(1 + y²)dx – y(1 + x²) dy = 0 …(i)
Separate the given differential equation, we get

Question 22.
Find the vector and Cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines and . [4]
Solution:
Let the equation of any line passing through (2, 1, 3) and perpendicular to the lines

Section – C

Question 28.
Evaluate: [6]

Solution:

Question 29.
Prove that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also find the maximum volume. [6]
Solution:
From the figure,

CBSE Previous Year Question Papers Class 12 Maths 2014 Outside Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – A

Question 9.
If , find the value of a. [1]
Solution:
Given,

Question 10.
If are perpendicular vectors, find the value of . [1]
Solution:

Section – B

Question 19.
Using properties of determinants, prove that: [4]

Solution:
Taking L. H. S

Question 20.
If x = cos t (3 – 2 cos² t) and y = sin t (3 – 2 sin²t), find the value of . [4]
Solution:

Question 21.
Find the particular solution of the differential equation log = 3x + 4y, given that y = 0 when x = 0. [4]
Solution:
The given differential equation is

Question 22.
Find the value of p, so that the lines l₁ :

are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l₁. [4]
Solution:
The given lines are

Section – C

Question 28.
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle of between them is 60°. [6]
Solution:
Let ∆ ABC be right angled with side a and hypotenuse l be

Question 29.
Evaluate: [6]

Solution:
Let

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CBSE Previous Year Question Papers Class 12 Physics 2016 Delhi

Section – A

Question 1.
A point charge +Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero ? [1]

Answer : Potential at a distance r from a given point charge Q is given by


Question 2.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ? [1]
Answer: According to Gauss’s law

Flux depends only on the charge enclosed and not on the radius. Hence, the electric flux remains constant.

Question 3.
Write the underlying principle of a moving coil galvanometer. [1]
Answer :
When a current carrying coil is placed in magnetic field then it experiences a torque.

Question 4.
Why are microwaves considered suitable for radar systems used in aircraft navigation ? [1]
Answer:
Microwaves are considered suitable because they have a short wavelength range they are suitable for radar system used in aircraft navigation.

Question 5.
Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ? [1]
Answer:
The Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the impressed voltage, which is the voltage across R.

It is dimensionless quantity. Hence, it has no units.

Section-B

Question 6.
Explain the terms
(i) Attenuation and
(ii) Demodulation used in Communication System. [2]
Answer:
(i) Attenuation : The loss of strength of a signal while propagating through a medium is known as attenuation.
(ii) Demodulation : The process of retrieval of information from the carrier wave at the receiver end is termed as demodulation. This is the reverse process of modulation.

Question 7.
Plot a graph showing variation of de-Broglie wavelength λ versus 1/√v, where V is accelerating potential for two particles A and B carrying same charge but of masses m₁, m₂ (m₁ > m₂). Which one of the two represents a particle of smaller mass and why ? [2]
Answer:
We know that,

Hence, the particle with lower mass (m₂) will have greater slope.

Question 8.
A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy. [2]
OR
Calculate the energy in fusion reaction:

Question 9.
Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω, and 0.3 Ω, respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. [2]


Question 10.
State Brewster’s law.
The value of Brewster angle for a transparent medium is different for light of different colors. Give reason. [2]
Answer:
Brewster’s law: The law states that the tangent of the polarizing angle of incidence for a given medium is equal to the refractive index of the medium. The light incident at this angle when reflects back is perfectly polarized,
i.e.  μ = tan i_p
The refractive index of a material depends on the color or wavelength of light. As the polarizing angle depends on refractive index (p = tan i_p), so it also depends on wavelength of light.

Section-C

Question 11.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. [3]
Answer:
Suppose we have a ring of radius a that carries a uniformly distributed positive charge q.

The electric field dE has two components.
(i) The axial components dE cos θ and
(ii) The perpendicular component dE sin θ.
Since the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field.


This expression  is similar to electric field due to a point charge

Question 12.
Write three characteristic features in photo electric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. [3]
Answer :

• Existence of threshold frequency : According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of threshold frequency.
• Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it dependence on frequency and not on intensity of the incident light.
• Instantaneous emission of electrons : According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.

Question 13.
(a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer. [3]

Answer :
(a) A charge particle having charge q is moving with velocity V in a magnetic field of field strength ‘B’ then the force acting on it is given by the formula and F = qv B sin θ (where θ is the angle between velocity vector and magnetic field).
Direction of force is given by the cross product of velocity and magnetic field.
(b)

α particle will trace circular path in anticlockwise direction as it’s deviation will be in the direction
of 
Neutron will pass without any deviation as magnetic field does not exert and force on neutral particle. Electron will trace circular path in clockwise direction as its deviation will be in the direction opposite to with a smaller radius due to large  charge/mass ratio as

Question 14.
(a) Define mutual inductance.
(b) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ? [3]
Answer:
(a) Mutual induction is the phenomenon of production of induced emf in one coil due to change of current or flux in the neighboring coil. The coil in which the current changes is called primary coil and the coil in which emf is induced is called the secondary coil.

Question 15.
Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of ∈_r = 4.   [3]

(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 μF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y.
Answer :
(i) Let capacitance of X be C1 and capacitance of Y be C2.


Question 16.
Two long straight parallel conductors carry steady current I₁ and I₁ separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up if one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. [3]
Answer :
Magnetic field produced on the wire (carrying current I₂) due to I₁ will be

So one ampere is defined as the current, which when maintained in two parallel infinite length conductors, held at a separation of one meter will produce a force of 2 x 10^-7 N per meter on each conductor.

Question 17.
How are e.m. waves produced by oscillating charges ?
Draw a sketch of linearly polarized e.m. waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields. [3]
OR
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
where Φ_E is the electric flux produced during charging of the capacitor plates.
Answer :
A charge oscillating with some frequency, produces an oscillating electric field in space, which in turn produces an oscillating magnetic field perpendicular to the electric field. This process goes on repeating, producing e.m. waves in space perpendicular to both the fields.

The direction of electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.
OR
Correction in Ampere’s circuital law (Modified Ampere’s law) : Maxwell removed the problem of current continuity and inconsistency observed in Ampere’s circuital law by introducing the concept of displacement current. Displacement current arises due to change in electric flux with

Conduction current is due to the flow of charges but displacement current is not because of the flow of charges but it is due to the change in electric flux.

Question 18.
(a) Explain any two factors which justify the need of modulating a low frequency signal,
(b) Write two advantages of frequency modulation over amplitude modulation. [3]
Answer:
(a) 1. Size of Antenna : The size of antenna required will be of order of λ /4. When frequency is small, the height of antenna should be large. So audio frequency signal should be modulated over a high frequency carrier wave to increase its frequency.
2. Effective power radiated by an Antenna : As power radiated hence when frequency is As increased then the power radiated will be more,
(b) Advantages of frequency modulation over amplitude modulation:
1. Noise can be reduced.
2. Transmission efficiency is more because the amplitude of an FM wave is constant.

Question 19.
(a) Write the functions of three segments of a transistor.
(b) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.  [3]

Question 20.
(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. [3]



Hence, it will form  virtual and erect image.

Question 21.
(a) State Bohr’s quantization condition for defining stationary orbits. How does de- Broglie hypothesis explain the stationary orbits ?
(b) Find the relation between the three wavelengths λ₁ λ₂ and λ₃ from the energy level diagram shown below. [3]

Answer :
(a) Quantization condition : Of all possible circular orbits allowed by the classical theory, the electrons are permitted to circulate only in those orbits in which the angular momentum of an electron is an integral multiple of h/2π ; h being planck constant.
therefore, for any permitted  orbit,

Where L, m, and v are the angular momentum, mass and speed of the electron respectively, r is the radius of the permitted orbit and n is positive integer called principle quantum number.

The above equation is Bohr’s famous quantum condition. When an electron of mass m is confined to move in a line of length l with velocity v, the de-Broglie wavelength λ , associated with electron is:

multiple of h/2π, which is Bohr’s quantisation of angular momentum.

Question 22.
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. [3]
Answer :
Reflecting Telescope : The reflecting telescope make use of a concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflector). After reflection the rays of light meet at a point where another convex mirror is placed. This mirror focuses light inside the telescope tube. The final image is seen through the eye¬piece. The images produced by the reflecting telescope is very bright and its resolving power is high.

Advantages:
(i) The resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.
(ii) There is no chromatic aberration as the object is a mirror.

Section – D

Question 23.
Meeta’s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this. Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father?
(ii) What answer did Meeta’s father give ?
(iii) What are the tiny lights in traffic signals called and how do these operate ? [4]
Answer:
(ii) Meeta’s father said that these are LED light which consume less power and have high reliability.
(iii) The tiny lights in traffic signals are Light Emitting Diode. These are operated by connecting the p-n junction diode in forward biased condition

Section-E

Question 24.
(a) An a.c. source of voltage V = V₀ sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ?
(b) In a series LR circuit X_L = R and power factor of the circuit is P₁ When capacitor with capacitance C such that X_L = Xc, is put in series, the power factor becomes P₂. Calculate P₁/P₂. [5]
OR
(a) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.
(b) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220 V and 1100 W.
Calculate:
(i) number of turns in secondary.
(ii) current in primary.
(iii) voltage across secondary.
(iv) current in secondary
(v) power in secondary.
Answer:
(a) Let a series LCR circuit is connected to an a.c. source V (Fig). We take the voltage of the source to be V = V₀ sin ωt

The a.c. current in each element is the same at any time, having the same amplitude and phase. It is given by,
I = I₀ sin (ωt + Φ)




OR
(a) A transformer is an electrical device for converting an alternating current at low voltage into high voltage or vice-versa.
1. If it increases the input a.c. voltage and decreases the, it is called step up transformer.
2. If it decreases the input a.c. voltage and increases the current, it is called step down transformer.
Principle : It works on the principle of mutual induction i.e., when a changing current or flux is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.

Working Theory : As the a.c. flows through the primary, it generate an alternating magnetic flux in the core which passes through the secondary coil

Let N₁ = No. of turns in primary coils
N₂ = No. of turns in secondary coils
This changing flux set up an induced emf in the secondary, also a self induced emf in the primary.

If there is no leakage of magnetic flux, then flux linked with each turn of the primary will be equal to that linked with each of the secondary. According to Faraday’s law of induction

Induced emf in the primary coil,
Induced emf in the secondary coil,
Where, = Rate of change of magnetic flux associated with each turn.
Φ = Magnetic flux linked with each turn of the primary or secondary at any instant.

Energy losses in. transformer :

• Copper loss : Some energy is lost due to the heating of copper wires used in the primary and secondary winding’s. This power loss (P = I²R) can be minimized by using thick copper wires of low resistance.
• Eddy current loss or Iron loss: The alternating , magnetic flux induces eddy current in their on core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.
• Hysteresis loss : When the iron core is subjected to a cycle of magnetization the core gets heated up due to hysteresis, having low hysteresis loop.
• Flux leakage : The magnetic flux produced by the primary may not fully pass through the secondary. Some of the flux may leak into air. This loss can be minimized by winding the primary and secondary coils over one another.

Question 25.
(a) In Young’s double slit experiment, deduce the condition for
(i) constructive, and
(ii) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against positionon ‘x’ the screen.
(b) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features. [5]
OR
(a) 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.
(b) What is dispersion of light ? What is its cause ?
(c) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.

Answer :
(a) Let the two waves arising from the slits A and B have the amplitudes a and b and the phase difference Φ. Such that y₁=a sin ωt and y₂ = b sin (ωt + Φ).

The resultant displacement is given as :



b) Comparison of interference pattern observed in Young’s double slits and the single slit diffraction:

OR


Dispersion. of light : Dispersion is often observed as light passes through a triangular prism. Upon passing through the prism, the white light is separated into its component colours : red, orange, yellow, green, blue, and voilet. The separation of visible light into its different colours is known as dispersion. Dispersion occurs because for different colour of light a transparent medium will have different refractive indices (μ). as different colours have different speed in transparent medium

Question 26.
(a) Define the term drift velocity.
(b) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend ?
(c) Why alloys like constantan and manganin are used for making standard resistors ? [5]
OR
(a) State the principle of working of a potentiometer.
(b) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (= l).

Answer :
(a) Drift velocity is defined as the average velocity with which the free electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.



Where ρ is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.
(c) They are used to make standard resistors because:
1. They have high value of resistivity.
2. Temperature coefficient of resistance is less.
3. They are least affected by temperature.
OR
(a) Principle of potentiometer : The basic principle of potentiometer is that when a constant current flows through a wire of uniform cross-section area then the potential drop across any length of the wire is directly proportional to that length.
A potentiometer is a device used to measure an unknown emf or potential difference and internal resistance of a cell accuratly.

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A.DELHI

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CBSE Previous Year Question Papers Class 10 Science With Solutions: The first step towards the preparation of board exam is to solve CBSE Previous Year Papers for Class Class 10 Science. Solving CBSE Previous Year Question Papers for Class Class 10 Science will help candidates to analyze the paper pattern and marking scheme of the examination. Also based on the previous year trends it is analyzed that the present question paper will consist of 1 to 25% questions which are asked in the previous year exam. So candidates who are clear with the CBSE Previous Year Papers for Class Class 10 Science can easily score good marks in the exam.

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• CBSE Previous Year Question Papers Class 10 Science 2016 Outside Delhi Term 2
• CBSE Previous Year Question Papers Class 10 Science 2015 Outside Delhi Term 2
• CBSE Previous Year Question Papers Class 10 Science 2014 Outside Delhi Term 2
• CBSE Previous Year Question Papers Class 10 Science 2013 Outside Delhi Term 2
• CBSE Previous Year Question Papers Class 10 Science 2012 Outside Delhi Term 2
• CBSE Previous Year Question Papers Class 10 Science 2011 Outside Delhi Term 2

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CBSE Topper Answer Sheet Class 10 Science

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