BYJU'S All India Mock Board Exams: CBSE Grade XII - Term I Mock Papers Set - 1

BYJU’S All India Mock Board Exams
CBSE Grade XII - Term I

Mock Papers Set -1
TABLE OF CONTENTS
Mathematics Questions…………………………………………………………………………..2
Mathematics Answer Keys………………………………………………………….…….……..13
Mathematics Solutions…………………………………………………………………….……..14
Physics Questions………………………………………………………………………………..46
Physics Answer Keys………………………………………………………….………………...59
Physics Solutions………………………………………………………………………………...60
Chemistry Questions…………………………………………………………………………….92
Chemistry Answer Keys………………………………………………………….……………..106
Chemistry Solutions……………………………………………………………………………..107
Biology Questions……………………………………………………………………………….131
Biology Answer Keys………………………………………………………….………………..149
Biology Solutions………………………………………………………………………………..150
MATHEMATICS
Time Allowed: 90 minutes Maximum Marks: 40
General Instructions:
1. The question paper contains three section- A, B and C. Each part is compulsory.
2. Section A consists of 20 MCQs, attempt any 16 out of 20.
3. Section B consists of 20 MCQs, attempt any 16 out of 20.
4. Section C consists of 10 MCQs, attempt any 8 out of 10.
5. There is no negative marking.
6. All question carry equal marks.
SECTION – A
In this section, attempt any 16 questions out of Questions 1 – 20.
Each Question is of 1 mark weightage
Q1 Let 𝐴 = {1,2,3, ⋯ ,40} and 𝑅 be an equivalence relation on 𝐴 × 𝐴 defined by (1)

(𝑎, 𝑏)𝑅(𝑐, 𝑑) if and only if 𝑎𝑑 = 𝑏𝑐. If 𝑚 is the number of elements in the equivalence
class [(1,4)], then the value of 𝑚 is
A. 5
B. 10
C. 15
D. 20
Q2 The relation 𝑅 on ℝ defined as 𝑅 = {(𝑎, 𝑏): 𝑎 ≤ 𝑏}, is (1)
A. Reflexive relation, Transitive relation
B. Reflexive relation, Symmetric relation
C. Reflexive relation, Equivalence relation
D. Symmetric relation, Equivalence relation
2
Q3 Let R = {(3,3), (6,6), (9,9), (12,12), (6,12), (3,9), (3,12), (3,6)} be a relation on the set (1)
𝐴 = {3,6,9,12}. Then the relation is
A. Reflexive and transitive only
B. Reflexive only
C. Reflexive, symmetric and transitive
D. Reflexive but neither symmetric nor transitive
Q4 If the function f: B → [−5, ∞) defined by f(x) = x 2 − 4x + 5 is one-one function, then 𝐵 (1)
is
A. [2, ∞)
B. [0, ∞)
C. [−5, ∞)
D. [−1, ∞ )
Q5 x2 (1)
If the function 𝑓: ℝ → 𝐴 given by f(x) = 𝑥 2 +1 is surjection, then 𝐴 =
A. (0, 1)
B. (0, 1]
C. [0, 1)
D. [0, 1]
Q6 The range of the function 𝑓(𝑥) = 4 tan−1 𝑥 + 3 sin−1 𝑥 + sec −1 𝑥 is (1)
3𝜋 3𝜋
A. {− , }
2 2
5𝜋 5𝜋
B. {− , }
2 2
5𝜋 3𝜋
C. {− , }
2 2
3𝜋 5𝜋
D. {− , }
2 2
Q7 The domain of the function 𝑓(𝑥) = sin−1(−𝑥 2 ) is (1)

A. [0,1]
B. (0,1)
C. [−1,1]
D. (−∞, −1) ∪ (1, ∞)
Q8 1 𝜋 (1)
If sin−1 𝑥 + cot −1 2 = 2 , then the value of 𝑥 is
A. 0
1
B.
√5
2
C.
√5
√3
D.
2
3
Q9 The principal value of sin−1 (−
√3
) is (1)
2
𝜋
A. 3
𝜋
B. − 3
𝜋
C. 6
𝜋
D. − 6
Q10 ln(1+3𝑥)−ln(1−2𝑥) (1)

; 𝑥≠0
If 𝑓(𝑥) = { 𝑥 is continuous at 𝑥 = 0, then the value of 𝑎 is
𝑎 ; 𝑥=0
A. 5
B. 1
C. −1
D. −2
Q11 1−sin 𝑥 𝜋 𝜋 𝜋 (1)
If 𝑓(𝑥) = sin 2𝑥
,𝑥 ≠ 2 is continuous at 𝑥 = 2 , then the value of 𝑓 ( 2 ) is
A. 0
1
B. 2
C. 1
D. 2
Q12 𝑥3 ; 𝑥 < 1 (1)
If 𝑓(𝑥) = { , then 𝑓(𝑥) is
𝑥 ; 𝑥≥1
A. continuous but non-differentiable at 𝑥 = 1
B. continuous and differentiable at 𝑥 = 1
C. neither continuous nor differentiable at 𝑥 = 0
D. none of these
Q13 𝑥−1 ; 𝑥 ≤ 1 (1)
At 𝑥 = 1, the function 𝑓(𝑥) = { is
𝑥3 − 1 ; 𝑥 > 1
A. continuous and differentiable
B. continuous but non-differentiable
C. discontinuous and differentiable
D. discontinuous and non-differentiable
Q14 sin 𝑥 𝑑𝑦 (1)
If 𝑦 = ln (1+cos 𝑥) , then 𝑑𝑥 =
A. sin 𝑥
B. cosec 𝑥
C. cos 𝑥
D. sec 𝑥
4
Q15 𝑑𝑦 (1)
If 𝑦 = 𝑒 sin √𝑥 , then =
𝑑𝑥
𝑒 cos √𝑥 ⋅sin √𝑥
A. 2√𝑥
𝑒 sin √𝑥
B. 2√𝑥
𝑒 sin √𝑥 ⋅cos √𝑥
C.
2√𝑥
D.
√𝑥
Q16 𝑑𝑦 (1)
If sin 𝑦 = x sin(𝑎 + 𝑦), then 𝑑𝑥 =
sin 𝑎
A.
sin2(𝑎+𝑦)
sin2(𝑎+𝑦)
B.
sin 𝑎
C. sin2(𝑎 + 𝑦)
D. sin(𝑎 + 𝑦)
Q17 𝑒 𝑡 +𝑒 {−𝑡} et −e{−t} dx (1)
If x = 2
,y = 2
, then dy =
𝑥
A. − 𝑦
𝑦
B.
𝑥
𝑥
C. 𝑦
𝑦
D. −
𝑥
Q18 The derivative of ln 𝑥 with respect to cot 𝑥 is (1)

− sin2 𝑥
A.
𝑥
cos2 𝑥
B.
𝑥
− sin3 𝑥
C.
𝑥
− sin2 𝑥
D. 𝑥2
Q19 𝑑2 𝑦 𝑑𝑦 (1)
If 𝑦 = acos(ln 𝑥) + 𝑏 sin(ln 𝑥) ,then 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 =
A. 0
B. 𝑦
C. 2𝑦
D. −𝑦
5
Q20 𝑑𝑦 (1)
If 𝑥 = 2 cos 𝑡 − cos 2𝑡 and 𝑦 = 2 sin 𝑡 − sin 2𝑡, then is equal to
𝑑𝑥
A. tan 𝑡
𝑡
B. − tan 2
3𝑡
C. tan 2
3𝑡
D. − tan
2
SECTION – B
In this section, attempt any 16 questions out of Questions 21 – 40.
Each Question is of 1 mark weightage
Q21 Which of the following is true about the monotonicity of 𝑓(𝑥) = sin 𝑥 + 𝑥 2 + 3𝑥, (1)
at 𝑥 = 0 ?
A. Increasing
B. Decreasing
C. Non-monotonic
D. Cannot be determined
Q22 𝑥 2 −8 (1)
If 𝑓(𝑥) = 𝑥−3
is an increasing function, then
A. 𝑥 ∈ (−∞, 2] ∪ [4, ∞)
B. 𝑥 ∈ (2,4)
C. 𝑥 ∈ (−∞, 3)
D. 𝑥 ∈ (2, ∞)
Q23 If 𝑓(𝑥) = cos(cos 𝑥) is an increasing function, then a possible interval of 𝑥 is (1)
𝜋 𝜋
A. [− 4 , 4 ]
𝜋
B. [0, 2 ]
𝜋 𝜋
C. [− 2 , 2 ]
D. [0, 𝜋]
Q24 4 (1)
The minimum value of 𝑓(𝑥) = 𝑥 + 𝑥+2 , (𝑥 ≥ 0) is
A. −1
B. -2
C. 1
D. 2
6
Q25 The equation of normal to the parabola 𝑥 2 = 4𝑦 drawn at point (2,1) is (1)
A. 𝑥 + 𝑦 = 1
B. 𝑥 + 𝑦 = 2
C. 𝑥 + 𝑦 = 3
D. 𝑥 + 𝑦 = 4
Q26 Tangent to the curve 𝑦 = 𝑥 2 − 5𝑥 + 5, parallel to the line 2𝑦 = 4𝑥 + 1 also passes (1)
through the point
7 1
A. (2 , 4)
1 7
B. (4 , 2)
1
C. (− 8 , 7)
1
D. (8 , −7)
Q27 4 (1)
The equation of tangent to the curve 𝑦 = 𝑥 + 𝑥 2 which is parallel to 𝑥 −axis, is
A. 𝑦 = 8
B. 𝑦 = 3
C. 𝑦 = 0
D. 𝑦 = 2
Q28 A particle moves in a straight line according to the law 𝑣 2 = 4𝑎(𝑥 sin 𝑥 + cos 𝑥) where 𝑣 (1)
is the velocity of a particle at a distance 𝑥 from the fixed point. The acceleration of the
particle is
A. 2𝑎𝑥 sin 𝑥
B. 𝑎𝑥 sin 𝑥
C. 𝑎𝑥 cos 𝑥
D. 2𝑎𝑥 cos 𝑥
Q29 9 1 1 5 (1)
If 𝐴 = [ ] and 𝐵 = [ ] and 5𝐴 + 3𝐵 + 2𝐶 is a null matrix, then matrix 𝐶 is
7 8 7 12
−28 −10
A. [ ]
−4 −24
−18 −5
B. [ ]
−28 −38
−12 −10
C. [ ]
−10 −38
−24 −10
D. [ ]
−28 −38
7
𝑥
Q30 If 𝐴 = [𝑥 𝑦], 𝐵 = [𝑎 ℎ
] and 𝐶 = [𝑦], then 𝐴𝐵𝐶 = (1)
ℎ 𝑏
A. [𝑎𝑥 + ℎ𝑦 + 𝑏𝑥𝑦]
B. [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
C. [𝑎𝑥 2 − 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
D. [𝑏𝑥 2 − 2ℎ𝑥𝑦 + 𝑎𝑦 2 ]
Q31 3 −4 (1)
If 𝐴 = [ ], then 𝐴 − 𝐴𝑇 is
1 −1
A. Skew-symmetric matrix
B. Symmetric matrix
C. Neither skew-symmetric nor symmetric matrix
D. Upper triangular matrix
Q32 If 𝐴 and 𝐵 are square matrices of the same order and 𝐴𝐵 = 3𝐼, where 𝐴, 𝐵 are non- (1)
singular matrices and 𝐼 is identity matrix, then 𝐴−1 is equal to
A. 3𝐵
1
B. 𝐵
3
C. 3𝐵−1
1 −1
D. 3
𝐵
Q33 (𝑖+𝑗)+(𝑖.𝑗) (1)

If a matrix 𝐴 = [𝑎𝑖𝑗 ]2×2 is given by 𝑎𝑖𝑗 = 2
, then the matrix 𝐴 is
3 5
A. [23 2
]
4
2
3 5
B. [25 2
]
4
2
3 3
C. [23 2
]
2
4
3
3
D. [ 2 ]
3 8
Q34 The area of the triangle whose vertices are 𝐴(0,0), 𝐵(3,1) and 𝐶(2,4), is (1)
A. 0 square units
B. 25 square units
C. 5 square units
D. 15 square units
8
Q35 If 𝐴 is a 4 × 4 matrix such that |3 ⋅ 𝑎𝑑𝑗 𝐴| = 3, then |𝐴| is equal to (1)
1
A.
3
1
B. 9
1
C. 27
D. 1
Q36 1 0 −𝐾 (1)
The values of 𝐾 for which the matrix 𝐴 = [ 2 1 3 ] is invertible, is:
𝐾 0 1
A. All the values of 𝐾 ∈ ℝ
B. 𝐾 ∈ [−1,1]
C. No value of 𝐾 ∈ ℝ
D. 𝐾 ∈ {−1,1}
Q37 cos 𝜃 − sin 𝜃 (1)
The inverse of the matrix 𝐴 = [ ]
sin 𝜃 cos 𝜃
cos 𝜃 sin 𝜃
A. [ ]
−sin 𝜃 cos 𝜃
sin 𝜃 cos 𝜃
B. [ ]
cos 𝜃 sin 𝜃
sin 𝜃 − cos 𝜃
C. [ ]
cos 𝜃 sin 𝜃
cos 𝜃 sin 𝜃
D. [ ]
sin 𝜃 −cos 𝜃
Q38 3 4 𝑥 1 (1)
The number of solutions of [ ] [ ] = [ ] is
9 12 𝑦 3
A. No solution
B. Infinite solutions
C. Unique solutions
D. Two solutions
Q39 If the system of equations 𝑥 − 𝑘𝑦 − 𝑧 = 0, 𝑘𝑥 − 𝑦 − 𝑧 = 0 and 𝑥 + 𝑦 − 𝑧 = 0 has a non- (1)
trivial solution, then 𝑘 can be
A. 0,1
B. 1, −1
C. −1,2
D. 2, −2
9
Q40 For the LPP; maximize 𝑧 = 5𝑥 − 2𝑦 subject to the constraints 3𝑥 + 2𝑦 ≤ 6, 10𝑥 + 9𝑦 ≥ (1)
45, 𝑥, 𝑦 ≥ 0
A. 𝑧𝑚𝑖𝑛 = −15
B. 𝑧𝑚𝑖𝑛 = 0
C. 𝑧𝑚𝑖𝑛 = −6
D. The LPP has no feasible solution
SECTION – C
In this section, attempt any 8 questions.
Each Question is of 1 mark weightage.
Question 46-50 are based on a case study.
Q41 If 𝑍 = 𝑥 + 2𝑦, subject to the constraints: 2𝑥 + 𝑦 ≥ 3, 𝑥 + 2𝑦 ≥ 6, 𝑥, 𝑦 ≥ 0, then the (1)

minimum of 𝑍 occurs at
A. Only one point
B. Two points
C. More than two points
D. Minimum of 𝑍 is not possible
Q42 Sum of maximum and minimum value of the objective function 𝑧 = 10𝑥 + 7𝑦, subjected (1)
to the constriants 0 ≤ 𝑥 ≤ 60,0 ≤ 𝑦 ≤ 45, 5𝑥 + 6𝑦 ≤ 420 is
A. 740
B. 1025
C. 1055
D. 1100
Q43 The corner points of the feasible region determined by a system of linear constraints are (1)
(0,10), (5,5), (15,15), (0,20). Let z = px + qy where p, q > 0. If the maximum of 𝑧
occurs at both the points (15,15) and (0,20), then which of the following is true:
A. 𝑝 = 𝑞
B. 𝑝 = 2𝑞
C. 𝑝 = 3𝑞
D. 𝑞 = 3𝑝
Q44 For an LPP, maximize z = ax + by where 𝑎, 𝑏 ∈ ℝ subject to the constraints (1)
a1 x + b1 y ≤ 0
a2 x + b2 y ≤ 0
10
x, y ≥ 0,
Consider the following statements:
(I) The solution depends on the optimizing function.
(II) The solution depends on the constraints.
Which of the following statement(s) is/are correct?
A. Both (I) and (II)

B. (I) only
C. (II) only
D. The solution depends upon some more aspects.
Q45 The feasible region of a LPP is shown in the figure. If Z = 3x − y, then the maximum (1)
value of 𝑍 occur at
A. (0,0)
B. (3,0)
C. (12,20)
D. (0,15)
Case Study
The relation between the height of the plant (𝑦 in cm) with
respect to exposure to sunlight is governed by the following
1
equation 𝑦 = 4𝑥 − 2 𝑥 2 where 𝑥 is the number of days exposed to
sunlight.
Based on the given information, answer the following questions.
11
Q46 The rate of growth of the plant with respect to the number of days exposed to sunlight is (1)
1
A. 4𝑥 − 𝑥 2
2
B. 4 − 𝑥
C. 𝑥 − 4
1
D. 𝑥 − 2 𝑥 2
Q47 What is the number of days it will take for the plant to grow to the maximum height? (1)
A. 4
B. 6
C. 7
D. 10
Q48 What is the maximum height of the plant? (1)
A. 12 cm
B. 10 cm
C. 8 cm
D. 6 cm
Q49 What will be the height of the plant after 2 days? (1)
A. 4 cm
B. 6 cm
C. 8 cm
D. 10 cm
Q50 7 (1)
If the height of the plant is 2 cm, then the number of days it has been exposed to the
sunlight is
A. 2
B. 3
C. 4
D. 1
12
ANSWER KEYS
Q1 B. Q26 D.
Q2 A. Q27 B.
Q3 A. Q28 D.
Q4 A. Q29 D.
Q5 C. Q30 B.
Q6 D. Q31 A.
Q7 C. Q32 B.
Q8 B. Q33 B.
Q9 B. Q34 C.
Q10 A. Q35 A.
Q11 A. Q36 A.
Q12 A. Q37 A.
Q13 B. Q38 B.
Q14 B. Q39 B.
Q15 C. Q40 D.
Q16 B. Q41 C.
Q17 B. Q42 A.
Q18 A. Q43 D.
Q19 D. Q44 C.
Q20 C. Q45 C.
Q21 A. Q46 B.
Q22 A. Q47 A.
Q23 B. Q48 C.
Q24 D. Q49 B.
Q25 C. Q50 D.
13
SOLUTIONS
Q1 Let 𝐴 = {1,2,3, ⋯ ,40} and 𝑅 be an equivalence relation on 𝐴 × 𝐴 defined by
(𝑎, 𝑏)𝑅(𝑐, 𝑑) if and only if 𝑎𝑑 = 𝑏𝑐. If 𝑚 is the number of elements in the
equivalence class [(1,4)], then the value of 𝑚 is
A. 5
B. 10
C. 15
D. 20
Answer: (𝐵)10
Solution:
𝑅 is defined as (a, b)R(c, d) iff ad = bc
Here, (a, b)R(1,4) iff 4a = b × 1
𝑏
⇒ =4
𝑎
∴ [(1,4)] = { (1,4), (2,8), ⋯ , (10,40)}
𝑚 = 10
Q2 The relation 𝑅 on ℝ defined as 𝑅 = {(𝑎, 𝑏): 𝑎 ≤ 𝑏}, is
A. Reflexive relation, Transitive relation
B. Reflexive relation, Symmetric relation
C. Reflexive relation, Equivalence relation
D. Symmetric relation, Equivalence relation
Answer: (𝐴) Reflexive relation, Transitive relation
Solution:
Given 𝑅 = {(𝑎, 𝑏): 𝑎 ≤ 𝑏}
For reflexive:
Clearly 𝑅 is reflexive as 𝑎 = 𝑎 ∀𝑎 ∈ ℝ
For transitive:
Let (𝑎, 𝑏 ∈ ℝ) and (𝑏, 𝑐) ∈ ℝ
⇒ 𝑎 ≤ 𝑏 ⋯ (1) and 𝑏 ≤ 𝑐 ⋯ (2)
14
From (1) and (2), we get
𝑎≤𝑐
⇒ (𝑎, 𝑐) ∈ ℝ
So, ℝ is transitive.
For symmetric:
Let (𝑎, 𝑏) ∈ ℝ
⇒ 𝑎 ≤ 𝑏 ∀ 𝑎, 𝑏 ∈ ℝ
But 𝑏 ≤ 𝑎 is not always true.
So, ℝ is not symmetric.
Q3 Let R = {(3,3), (6,6), (9,9), (12,12), (6,12), (3,9), (3,12), (3,6)} be a relation on
the set 𝐴 = {3,6,9,12}. Then the relation is
A. Reflexive and transitive only

B. Reflexive only
C. Reflexive, symmetric and transitive
D. Reflexive but neither symmetric nor transitive
Answer: (𝐴) Reflexive and transitive only
Solution:
(1) It is reflexive relation as all ordered pairs {(3,3), (6,6), (9,9), (12,12)} are
present in 𝑅
(2)𝑅 is not symmetric as (6,12) ∈ 𝑅 but (12,6) ∉ 𝑅
(3)For transitive if (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
i.e., (3,6), (6,12) ∈ R ⇒ (3,12) ∈ R
As the given relation 𝑅 satisfies the above condition.

Hence, 𝑅 is transitive.
Q4 If the function f: B → [−5, ∞) defined by f(x) = x 2 − 4x + 5 is one-one function,
then 𝐵 is
A. [2, ∞)
B. [0, ∞)
C. [−5, ∞)
D. [−1, ∞ )
15
Answer: (𝐴) [2, ∞)
Solution:
f(x) = x 2 − 4x + 5
= (x − 2)2 + 1
Clearly, we can conclude from above graph and given options 𝑓(𝑥) will be one-
one function forx ∈ [2, ∞).
Q5 x2
If the function 𝑓: ℝ → 𝐴 given by f(x) = 𝑥 2 +1 is surjection, then 𝐴 =
A. (0, 1)
B. (0, 1]
C. [0, 1)
D. [0, 1]
Answer: (𝐶) [0, 1)
Solution:
The domain of 𝑓(𝑥) is all real numbers.
∵ f: ℝ → A is surjective, therefore 𝐴 must be the range of 𝑓(𝑥)
1
⇒ f(x) = 1 −
1 + 𝑥2
We know that,
x2 + 1 ≥ 1
1
⇒ 0< ≤ 1
𝑥2
+1
1
⇒ −1 ≤ − < 0
1 + 𝑥2
16
⇒ 0 ≤ f(x) < 1, ∀ x ∈ ℝ
Q6 The range of the function 𝑓(𝑥) = 4 tan−1 𝑥 + 3 sin−1 𝑥 + sec −1 𝑥 is
3𝜋 3𝜋
A. {− , }
2 2
5𝜋 5𝜋
B. {− , }
2 2
5𝜋 3𝜋
C. {− , }
2 2
3𝜋 5𝜋
D. {− , }
2 2
3𝜋 5𝜋
Answer: (D) {− , }
2 2
Solution:
𝑓(𝑥) = 4 tan−1 𝑥 + 3 sin−1 𝑥 + sec −1 𝑥
Domain of tan−1 𝑥 is ℝ ⋯ (1)
Domain of sin−1 𝑥 is [−1,1] ⋯ (2)
Domain of sec −1 𝑥 is (−∞, 1] ∪ [1, ∞) ⋯ (3)
Hence, domain of 𝑓(𝑥) is (1) ∪ (2) ∪ (3), we get
⇒ 𝑥 = {−1,1}
Now,
𝑓(−1) = 4 tan−1 (−1) + 3 sin−1(−1) + sec −1 (−1)
𝜋 𝜋
= 4 (− ) + 3 (− ) + 𝜋
4 2
3𝜋
∴ 𝑓(−1) = −
2
𝑓(1) = 4 tan−1 (1) + 3 sin−1(1) + sec −1(1)
𝜋 𝜋
= 4( ) + 3( )+ 𝜋
4 2
5𝜋
∴ 𝑓(1) =
2
3𝜋 5𝜋
Hence, the range of 𝑓(𝑥) is {− , }.
2 2
Q7 The domain of the function 𝑓(𝑥) = sin−1(−𝑥 2 ) is
A. [0,1]
B. (0,1)
C. [−1,1]
17
D. (−∞, −1) ∪ (1, ∞)
Answer: (C) [−1,1]
Solution:
𝑓(𝑥) = sin−1(−𝑥 2 )
∴ −1 ≤ −𝑥 2 ≤ 1
⇒ −1 ≤ 𝑥 2 ≤ 1
⇒ 0 ≤ 𝑥 2 ≤ 1 (∵ 𝑥 2 ≥ 0)
⇒ 𝑥 ∈ [−1,1]
Hence, the domain of 𝑓(𝑥) is [−1,1].
Q8 1 𝜋
If sin−1 𝑥 + cot −1 2 = 2 , then the value of 𝑥 is
A. 0
1
B.
√5
2
C.
√5
√3
D. 2
1
Answer: (B)
√5
Solution:
1 𝜋
Given: sin−1 𝑥 + cot −1 2 = 2
1
Let cot −1 2 = 𝛼, sin−1 𝑥 = 𝜃
1
⇒ cot 𝛼 = 2 , sin 𝜃 = 𝑥
Now,
𝜋
𝜃+𝛼 = ⇒ sin 𝜃 = cos 𝛼
2
1 1
⇒ sin 𝜃 = ⇒𝑥=
√5 √5
Q9 The principal value of sin−1 (−
√3
) is
2
𝜋
A. 3
𝜋
B. − 3
18
𝜋
C. 6
𝜋
D. − 6
𝜋
Answer: (B) − 3
Solution:
Let the principal value be 𝑦, we get
√3 𝜋 𝜋
𝑦 = sin−1 (− ) , 𝑦 ∈ [− , ]
2 2 2
√3
⇒ sin 𝑦 = −
2
𝜋
⇒ sin 𝑦 = sin (− )
3
𝜋
∴𝑦=−
3
Q10 ln(1+3𝑥)−ln(1−2𝑥)
; 𝑥≠0
If 𝑓(𝑥) = { 𝑥 is continuous at 𝑥 = 0, then the value of 𝑎 is
𝑎 ; 𝑥=0
A. 5
B. 1
C. −1
D. −2
Answer: (A) 5
Solution:
𝑓(𝑥) to be continuous at 𝑥 = 0, we must have
𝑓(0) = lim 𝑓(𝑥)
𝑥→0
ln(1 + 3𝑥) − ln(1 − 2𝑥)

⇒ 𝑎 = lim
𝑥→0 𝑥
(1 + 3𝑥) (1 − 2𝑥)
⇒ 𝑎 = lim ln ⋅ 𝑥 − ln ⋅ (−2)
𝑥→0 3𝑥 −2𝑥
∴𝑎 =3+2=5
19
Q11 1−sin 𝑥 𝜋 𝜋 𝜋
If 𝑓(𝑥) = sin 2𝑥
,𝑥 ≠ 2
is continuous at 𝑥 = 2 , then the value of 𝑓 ( 2 ) is
A. 0
1
B.
2
C. 1
D. 2
Answer: (A) 0
Solution:
𝜋
𝑓(𝑥) to be continuous at 𝑥 = 2 , we must have
𝜋
𝑓 ( ) = lim𝜋 𝑓(𝑥)
2 𝑥→
2
𝜋 1 − sin 𝑥
⇒ 𝑓 ( ) = lim𝜋
2 𝑥→ sin 2𝑥
2
𝜋 1 − sin 𝑥 1 + sin 𝑥
⇒ 𝑓 ( ) = lim𝜋 ×
2 𝑥→ sin 2𝑥 1 + sin 𝑥
2
𝜋 cos 2 𝑥 cos 𝑥
⇒ 𝑓 ( ) = lim𝜋 = lim𝜋
2 𝑥→ 2 sin 𝑥 cos 𝑥 𝑥→ 2 sin 𝑥
2 2
𝜋
∴ 𝑓( ) = 0
2
Q12 𝑥3 ; 𝑥 < 1
If 𝑓(𝑥) = { , then 𝑓(𝑥) is
𝑥 ; 𝑥≥1
A. continuous but non-differentiable at 𝑥 = 1

B. continuous and differentiable at 𝑥 = 1
C. neither continuous nor differentiable at 𝑥 = 0
D. none of these
Answer: (A) continuous but non-differentiable at 𝑥 = 1
Solution:
Given:
𝑥3 ; 𝑥 < 1
𝑓(𝑥) = {
𝑥 ; 𝑥≥1
Now,
20
R.H.L.
𝑓(1− ) = lim+(1 − ℎ)3 = 1
ℎ→0
L.H.L.
𝑓(1+ ) = lim+(1 + ℎ) = 1
ℎ→0
And 𝑓(1) = 1
⇒ 𝑓(1− ) = 𝑓(1+ ) = 𝑓(1) = 1
∴ 𝑓(𝑥) is continuous at 𝑥 = 1.
Now,
R.H.D.
1+ℎ−1
𝑓 ′ (1+ ) = lim =1
ℎ→0 ℎ
L.H.D.
′ (1− )
(1 − ℎ)3 − 1 1 − ℎ3 − 3ℎ + 3ℎ2 − 1
𝑓 = lim = lim
ℎ→0 −ℎ ℎ→0 −ℎ
𝑓 ′ (1− ) = lim+ ℎ2 + 3 − 3ℎ = 3
ℎ→0
⇒ 𝑓 ′ (1− ) ≠ 𝑓 ′ (1+ )
Therefore, 𝑓 is continuous but non-differentiable at 𝑥 = 1
Q13 𝑥−1 ; 𝑥 ≤1
At 𝑥 = 1, the function 𝑓(𝑥) = { is
𝑥3 − 1 ; 𝑥 > 1
A. continuous and differentiable

B. continuous but non-differentiable
C. discontinuous and differentiable
D. discontinuous and non-differentiable
Answer: (B) continuous but non-differentiable
Solution:
𝑥−1 ; 𝑥 ≤ 1
Given: 𝑓(𝑥) = {
𝑥3 − 1 ; 𝑥 > 1
R.H.L.
𝑓(1− ) = lim+ 1 − ℎ − 1 = 0
ℎ→0
L.H.L.
𝑓(1+ ) = lim+(1 + ℎ)3 − 1 = lim+ 1 + ℎ3 + 3ℎ + 3ℎ2 − 1
ℎ→0 ℎ→0
+)
𝑓(1 =0
And 𝑓(1) = 0
21
⇒ 𝑓(1− ) = 𝑓(1+ ) = 𝑓(1) = 0
∴ 𝑓(𝑥) is continuous at 𝑥 = 1.
Now,
R.H.D.
′ (1+ )
(1 + ℎ)2 − 1 − 0 ℎ3 + 3ℎ + 3ℎ2
𝑓 = lim = lim
ℎ→0 ℎ ℎ→0 ℎ
𝑓 ′ (1+ ) = 3
L.H.D.
1−ℎ−1−0 −ℎ
𝑓 ′ (1− ) = lim = lim
ℎ→0 −ℎ ℎ→0 −ℎ
𝑓 ′ (1+ ) = 1
⇒ 𝑓 ′ (1− ) ≠ 𝑓 ′ (1+ )
Therefore, 𝑓 is continuous but non-differentiable at 𝑥 = 1
Q14 sin 𝑥 𝑑𝑦
If 𝑦 = ln (1+cos 𝑥) , then 𝑑𝑥 =
A. sin 𝑥
B. cosec 𝑥
C. cos 𝑥
D. sec 𝑥
Answer: (B) cosec 𝑥
Solution:
sin 𝑥
Given: 𝑦 = ln (1+cos 𝑥)
𝑥 𝑥
2 sin 2 cos 2
⇒ 𝑦 = ln ( 𝑥 )
1 + 2cos 2 2 − 1
𝑥 𝑥
2 sin 2 cos 2
⇒ 𝑦 = ln ( 𝑥 )
2cos 2 2
𝑥
sin 2 𝑥
⇒ 𝑦 = ln ( 𝑥 ) = ln (tan 2)
cos 2
Differentiation w.r.t. 𝑥, we get

𝑑𝑦 1 𝑥 1
= 𝑥 × sec 2 ×
𝑑𝑥 tan 2 2
2
22
𝑑𝑦 1 1
⇒ = 𝑥 𝑥 =
𝑑𝑥 2 sin ⋅ cos sin 𝑥
2 2
𝑑𝑦
∴ = cosec 𝑥
𝑑𝑥
Q15 𝑑𝑦
If 𝑦 = 𝑒 sin √𝑥 , then 𝑑𝑥 =
𝑒 cos √𝑥 ⋅sin √𝑥
A. 2√𝑥
𝑒 sin √𝑥
B.
2√𝑥
C. 2√𝑥
D.
√𝑥
Answer: (C) 2√𝑥
Solution:
Given:
𝑦 = 𝑒 sin √𝑥
Differentiation w.r.t. 𝑥, we get
𝑑𝑦 1
= 𝑒 sin √𝑥 × cos √𝑥 ×
𝑑𝑥 2√𝑥
𝑑𝑦 𝑒 sin √𝑥 ⋅ cos √𝑥
∴ =
𝑑𝑥 2√𝑥
Q16 𝑑𝑦
If sin 𝑦 = x sin(𝑎 + 𝑦), then 𝑑𝑥 =
sin 𝑎
A. sin2(𝑎+𝑦)
sin2(𝑎+𝑦)
B. sin 𝑎
C. sin2(𝑎 + 𝑦)
D. sin(𝑎 + 𝑦)
sin2 (𝑎+𝑦)
Answer: (𝐵) sin 𝑎
Solution:
23
Given,
sin 𝑦 = x sin(𝑎 + 𝑦),
sin 𝑦
⇒ x=
sin(𝑎 + 𝑦)
dx sin(𝑎 + 𝑦) ⋅ cos 𝑦 − sin 𝑦 ⋅ cos(𝑎 + 𝑦)
=
dy sin2(𝑎 + 𝑦)
dx sin(𝑎 + 𝑦 − 𝑦) sin 𝑎
⇒ = 2
= 2
dy sin (𝑎 + 𝑦) sin (𝑎 + 𝑦)
dy sin2 (𝑎 + 𝑦)
∴ =
dx sin 𝑎
Q17 𝑒 𝑡 +𝑒 {−𝑡} et −e{−t} dx

If x = ,y = , then =
2 2 dy
𝑥
A. −
𝑦
𝑦
B.
𝑥
𝑥
C.
𝑦
𝑦
D. − 𝑥
𝑦
Answer: (𝐵) 𝑥
Solution:
We have,
et + e{−t}
x= ,
2
dx et − e{−t}
⇒ = = y,
dt 2
et −e{−t}
and, y =
2
dy et + e{−t}
⇒ = =x
dt 2
dx
dx dt
=
dy dy
dt
1
=y×
x
𝑦
=
𝑥
24
Q18 The derivative of ln 𝑥 with respect to cot 𝑥 is
− sin2 𝑥
A. 𝑥
cos2 𝑥
B. 𝑥
− sin3 𝑥
C. 𝑥
− sin2 𝑥
D. 𝑥2
− sin2 𝑥
Answer: (𝐴) 𝑥
Solution:
Let 𝑦 = ln 𝑥 and 𝑧 = cot 𝑥
Now,
𝑑𝑦 1
=
𝑑𝑥 𝑥
𝑑𝑧
= − csc 2 𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑦 𝑑𝑥 −1
⇒ = =
𝑑𝑧 𝑑𝑧 𝑥 ⋅ csc 2 𝑥
𝑑𝑥
𝑑𝑦 − sin2 𝑥
∴ 𝑑𝑧
= 𝑥
Q19 𝑑2 𝑦 𝑑𝑦
If 𝑦 = acos(ln 𝑥) + 𝑏 sin(ln 𝑥) ,then 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 =
A. 0
B. 𝑦
C. 2𝑦
D. – 𝑦
Answer: (𝐷) − 𝑦
Solution:
𝑦 = a cos(ln 𝑥) + 𝑏 sin(ln 𝑥) ,
Differentiating w.r.t 𝑥, we get
𝑑𝑦 1 1
= −𝑎 sin(ln 𝑥) ⋅ + 𝑏 cos(ln 𝑥) ⋅
𝑑𝑥 𝑥 𝑥
25
𝑑𝑦
⇒𝑥 = −𝑎 sin(ln 𝑥) + 𝑏 cos(ln 𝑥)
𝑑𝑥
Again differentiating w.r.t 𝑥,we get
𝑑2 𝑦 𝑑𝑦 𝑦
⇒𝑥 2
+ =−
𝑑𝑥 𝑑𝑥 𝑥
𝑑2 𝑦 𝑑𝑦
∴ 𝑥2 +𝑥 = −𝑦
𝑑𝑥 2 𝑑𝑥
Q20 𝑑𝑦
If 𝑥 = 2 cos 𝑡 − cos 2𝑡 and 𝑦 = 2 sin 𝑡 − sin 2𝑡, then 𝑑𝑥 is equal to
A. tan 𝑡
𝑡
B. − tan
2
3𝑡
C. tan 2
3𝑡
D. − tan 2
3𝑡
Answer: (𝐶) tan
2
Solution:
𝑥 = 2 cos 𝑡 − cos 2𝑡
𝑑𝑥
⇒ = −2 sin 𝑡 + 2 sin 2𝑡
𝑑𝑡
𝑑𝑥
⇒ = 2 (sin 2𝑡 − sin 𝑡)
𝑑𝑡
𝑦 = 2 sin 𝑡 − sin 2𝑡
𝑑𝑦
⇒ = 2(cos 𝑡 − cos 2𝑡)
𝑑𝑡
Now,
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
𝑑𝑥 𝑑𝑥
𝑑𝑡
cos 𝑡 − cos 2𝑡
=
sin 2𝑡 − sin 𝑡
3𝑡 𝑡
2 sin 2 ⋅ sin 2
=
3𝑡 𝑡
2 cos 2 ⋅ sin 2
3𝑡
= tan
2
26
Q21 Which of the following is true about the monotonicity of 𝑓(𝑥) = sin 𝑥 + 𝑥 2 + 3𝑥,
at 𝑥 = 0 ?
A. Increasing
B. Decreasing
C. Non-monotonic
Answer: (A) Increasing
Solution:
Given, 𝑓(𝑥) = sin 𝑥 + 𝑥 2 + 3𝑥
⇒f'(x)=cos x+2x+3
⇒ 𝑓 ′ (0) = 1 + 0 + 3 > 0
So, 𝑓(𝑥) is increasing at 𝑥 = 0.
Q22 𝑥 2 −8
If 𝑓(𝑥) = 𝑥−3
is an increasing function, then
A. 𝑥 ∈ (−∞, 2] ∪ [4, ∞)
B. 𝑥 ∈ (2,4)
C. 𝑥 ∈ (−∞, 3)
D. 𝑥 ∈ (2, ∞)
Answer: (A) 𝑥 ∈ (−∞, 2] ∪ [4, ∞)
Solution:
𝑥 2 −8
Given, 𝑓(𝑥) = 𝑥−3
(𝑥 − 3)(2𝑥) − (𝑥 2 − 8) 𝑥 2 − 6𝑥 − 8
⇒ 𝑓 ′ (𝑥) = =
(𝑥 − 3)2 (𝑥 − 3)2
(𝑥 − 2)(𝑥 − 4)
⇒ 𝑓 ′ (𝑥) =
(𝑥 − 3)2
Plotting on number line:
27
For function to be increasing, 𝑓 ′ (𝑥) ≥ 0
⇒ 𝑥 ∈ (−∞, 2] ∪ [4, ∞)
Q23 If 𝑓(𝑥) = cos(cos 𝑥) is an increasing function, then a possible interval of 𝑥 is
𝜋 𝜋
A. [− 4 , 4 ]
𝜋
B. [0, 2 ]
𝜋 𝜋
C. [− 2 , 2 ]
D. [0, 𝜋]
Answer: (𝐵) [0, 𝜋/2]
Solution:
Given, 𝑓(𝑥) = cos(cos 𝑥)
⇒ 𝑓 ′ (𝑥) = − sin(cos 𝑥) ⋅ (− sin 𝑥)
⇒ 𝑓 ′ (𝑥) = sin(cos 𝑥) ⋅ sin 𝑥
For 𝑓(𝑥) to be increasing 𝑓 ′ (𝑥) ≥ 0
⇒ sin(cos 𝑥) ⋅ sin 𝑥 ≥ 0
So, option (B) is correct.
Q24 4
The minimum value of 𝑓(𝑥) = 𝑥 + 𝑥+2 , (𝑥 ≥ 0) is
A. −1
B. -2
C. 1
D. 2
Answer: (D) 2
Solution:
4
Given, 𝑓(𝑥) = 𝑥 + 𝑥+2 , (𝑥 ≥ 0)
28
4
⇒ 𝑓 ′ (𝑥) = 1 −
(𝑥 + 2)2
𝑥(𝑥 + 6)
⇒ 𝑓 ′ (𝑥) = ≥ 0, ∀𝑥 ≥ 0
(𝑥 + 2)2
So, 𝑓(𝑥) is increasing
⇒ 𝑓(𝑥) ≥ 𝑓(0) for 𝑥 ≥ 0
⇒ 𝑓(𝑥) ≥ 2
So, minimum value of 𝑓(𝑥) is 2.
Q25 The equation of normal to the parabola 𝑥 2 = 4𝑦 drawn at point (2,1) is
A. 𝑥 + 𝑦 = 1
B. 𝑥 + 𝑦 = 2
C. 𝑥 + 𝑦 = 3
D. 𝑥 + 𝑦 = 4
Answer: (C) 𝑥 + 𝑦 = 3
Solution:
Given curve, 𝑥 2 = 4𝑦
Differentiating both sides w.r.t. 𝑥,
4𝑑𝑦
⇒ 2𝑥 =
𝑑𝑥
𝑑𝑦 𝑥
⇒ =
𝑑𝑥 2
𝑑𝑥 2
We know, slope of normal = − 𝑑𝑦 = − 𝑥
At point (2,1), 𝑚𝑛𝑜𝑟𝑚𝑎𝑙 = −1

So, equation of normal is (𝑦 − 1) = −1(𝑥 − 2)
⇒𝑥+𝑦 =3
Q26 Tangent to the curve 𝑦 = 𝑥 2 − 5𝑥 + 5, parallel to the line 2𝑦 = 4𝑥 + 1 also
passes through the point
7 1
A. (2 , 4)
1 7
B. (4 , 2)
1
C. (− 8 , 7)
1
D. ( , −7)
8
29
1
Answer: (D)(8 , −7)
Solution:
Given curve:
𝑦 = 𝑥 2 − 5𝑥 + 5 ⋯ (1)
𝑑𝑦
⇒ = 2𝑥 − 5
𝑑𝑥
Slope of the line 2𝑦 = 4𝑥 + 1 is 2.
7
∴ 2𝑥 − 5 = 2 ⇒ 𝑥 =
2
Putting this value in the equation (1), we ger
49 35 1
𝑦= − +5=−
4 2 4
Now, equation of tangent
1 7
𝑦+ = 2 (𝑥 − )
4 2
⇒ 8𝑥 − 28 = 4𝑦 + 1
⇒ 8𝑥 − 4𝑦 − 29 = 0
1
∴ (8 , −7) satisfies above equation.
Q27 4
The equation of tangent to the curve 𝑦 = 𝑥 + 𝑥 2 , which is parallel to 𝑥 −axis, is
A. 𝑦 = 8
B. 𝑦 = 3
C. 𝑦 = 0
D. 𝑦 = 2
Answer: (B)𝑦 = 3
Solution:
If a tangent(line) is parallel to 𝑥 − axis, then slope of the tangent is 0.
𝑑𝑦
i.e., 𝑑𝑥 = 0
8
⇒1− =0
𝑥3
⇒𝑥=2
At 𝑥 = 2 the value of 𝑦 is given by
4
𝑦 =2+
22
30
⇒𝑦=3
Equation of the tangent at (2,3) is
𝑦 − 3 = 0(𝑥 − 2)
∴𝑦=3
Q28 A particle moves in a straight line according to the law 𝑣 2 = 4𝑎(𝑥 sin 𝑥 + cos 𝑥)
where 𝑣 is the velocity of a particle at a distance 𝑥 from the fixed point. The
acceleration of the particle is
A. 2𝑎𝑥 sin 𝑥
B. 𝑎𝑥 sin 𝑥
C. 𝑎𝑥 cos 𝑥
D. 2𝑎𝑥 cos 𝑥
Answer: (D) 2𝑎𝑥 cos 𝑥
Solution:
Given: 𝑣 2 = 4𝑎(𝑥 sin 𝑥 + cos 𝑥)
𝑑𝑣
We know, acceleration = 𝑑𝑡
𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥
⇒ 2𝑣 = 4𝑎 (𝑥 cos 𝑥 + sin 𝑥 − sin 𝑥 )
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑣 𝑑𝑥
∴ = 2𝑎𝑥 cos 𝑥 (∵ = 𝑣)
𝑑𝑡 𝑑𝑡
Q29 9 1 1 5
If 𝐴 = [ ] and 𝐵 = [ ] and 5𝐴 + 3𝐵 + 2𝐶 is a null matrix, then matrix 𝐶
7 8 7 12
is
−28 −10
A. [ ]
−4 −24
−18 −5
B. [ ]
−28 −38
−12 −10
C. [ ]
−10 −38
−24 −10
D. [ ]
−28 −38
−24 −10
Answer: (D) [ ]
−28 −38
Solution:
𝑥 𝑦
Let 𝐶 = [ ]
𝑧 𝑤
31
Given: 5𝐴 + 3𝐵 + 2𝐶 = 𝑂
9 1 1 5 𝑥 𝑦 0 0
⇒ 5[ ]+ 3[ ] +2[ ]=[ ]
7 8 7 12 𝑧 𝑤 0 0
45 5 3 15 2𝑥 2𝑦 0 0
⇒[ ]+[ ]+[ ]=[ ]
35 40 21 36 2𝑧 2𝑤 0 0
45 + 3 + 2𝑥 5 + 15 + 2𝑦 0 0
⇒[ ]=[ ]
35 + 21 + 2𝑧 40 + 36 + 2𝑤 0 0
⇒ 2𝑥 + 48 = 0 ⇒ 𝑥 = −24
⇒ 2𝑦 + 20 = 0 ⇒ 𝑥 = −10
⇒ 2𝑧 + 56 = 0 ⇒ 𝑧 = −28
⇒ 2𝑤 + 76 = 0 ⇒ 𝑤 = −38
−24 −10
Hence, 𝐶 = [ ].
−28 −38
𝑥
Q30 If 𝐴 = [𝑥 𝑦], 𝐵 = [𝑎 ℎ] and 𝐶 = [ ], then 𝐴𝐵𝐶 =
ℎ 𝑏 𝑦
A. [𝑎𝑥 + ℎ𝑦 + 𝑏𝑥𝑦]
B. [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
C. [𝑎𝑥 2 − 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
D. [𝑏𝑥 2 − 2ℎ𝑥𝑦 + 𝑎𝑦 2 ]
Answer: (B) [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
Solution:
Given:
𝑥
𝐴 = [𝑥 𝑦], 𝐵 = [𝑎 ℎ
] and 𝐶 = [𝑦]
ℎ 𝑏
Now,
𝑦] [𝑎 ℎ 𝑥
𝐴𝐵𝐶 = [𝑥 ][ ]
ℎ 𝑏 𝑦
𝑥
⇒ 𝐴𝐵𝐶 = [𝑎𝑥 + ℎ𝑦 𝑥ℎ + 𝑏𝑦] [𝑦]
⇒ 𝐴𝐵𝐶 = [𝑎𝑥 2 + ℎ𝑥𝑦 + ℎ𝑥𝑦 + 𝑏𝑦 2 ]

∴ 𝐴𝐵𝐶 = [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]
Q31 3 −4
If 𝐴 = [ ], then 𝐴 − 𝐴𝑇 is
1 −1
A. Skew-symmetric matrix
B. Symmetric matrix
C. Neither skew-symmetric nor symmetric matrix
D. Upper triangular matrix
32
Answer: (A) Skew-symmetric matrix
Solution:
Let 𝐵 = 𝐴 − 𝐴𝑇
3 −4 3 1
Then, 𝐵 = [ ]−[ ]
1 −1 −4 −1
0 −5
𝐵=[ ]
5 0
0 5
Now, 𝐵𝑇 = [ ]
−5 0
∴ 𝐵𝑇 = −𝐵
Hence, 𝐵 is a Skew-symmetric matrix.
Q32 If 𝐴 and 𝐵 are square matrices of the same order and 𝐴𝐵 = 3𝐼, where 𝐴, 𝐵 are
non-singular matrices and 𝐼 is identity matrix, then 𝐴−1 is equal to
A. 3𝐵
1
B. 3
𝐵
C. 3𝐵−1
1 −1
D. 𝐵
3
1
Answer: (B) 3
𝐵
Solution:
Given: 𝐴𝐵 = 3𝐼
Multiplying both the sides by 𝐴−1 , we get
𝐴−1 𝐴𝐵 = 3𝐴−1 𝐼
⇒ 𝐼𝐵 = 3𝐴−1
1
∴ 𝐴−1 = 𝐵
3
Q33 (𝑖+𝑗)+(𝑖.𝑗)
If a matrix 𝐴 = [𝑎𝑖𝑗 ]2×2 is given by 𝑎𝑖𝑗 = 2
, then the matrix 𝐴 is
3 5
A. [23 2
]
2
4
3 5
B. [25 2
]
2
4
33
3 3
C. [23 2
]
4
2
3
3
D. [ 2 ]
3 8
3 5
Answer: (B) [25 2

]
4
2
Solution:
𝑎11 𝑎12
Let 𝐴 = [𝑎𝑖𝑗 ] = [𝑎 𝑎22 ]
2×2 21
(𝑖+𝑗)+(𝑖.𝑗)
Given, 𝑎𝑖𝑗 =
2
(1 + 1) + (1.1) 3
⇒ 𝑎11 = =
2 2
(1 + 2) + (1.2) 5
⇒ 𝑎12 = =
2 2
(2 + 1) + (2.1) 5
⇒ 𝑎21 = =
2 2
(2 + 2) + (2.2)
⇒ 𝑎22 = =4
2
3 5
∴ 𝐴 = [ 2 2]
5
4
2
Q34 The area of the triangle whose vertices are 𝐴(0,0), 𝐵(3,1) and 𝐶(2,4), is
A. 0 square units
B. 25 square units
C. 5 square units
D. 15 square units
Answer: (C) 5 square units
Solution:
Area of triangle with the given vertices can be obtained as
1 0 0 0
𝐴 = |3 1 1|
2
2 4 1
Expanding the above determinant by using expansion techniques of
34
determinant, we get
1
⇒ 𝐴 = [0(1 − 4) − 0(3 − 2) + 1(12 − 2)]
2
∴ 𝐴 = 5 square units
Q35 If 𝐴 is a 4 × 4 matrix such that |3 ⋅ 𝑎𝑑𝑗 𝐴| = 3, then |𝐴| is equal to
1
A.
3
1
B. 9
1
C. 27
D. 1
1
Answer: (𝐴) 3
Solution:
|3 ⋅ 𝑎𝑑𝑗 𝐴| = 3
34 |𝑎𝑑𝑗 𝐴| = 3
1
|𝑎𝑑𝑗 𝐴| =
27
1
|𝐴|4−1 =
27
1
|𝐴| =
3
Q36 1 0 −𝐾
The values of 𝐾 for which the matrix 𝐴 = [ 2 1 3 ] is invertible, is:
𝐾 0 1
A. All the values of 𝐾 ∈ ℝ

B. 𝐾 ∈ [−1,1]
C. No value of 𝐾 ∈ ℝ
D. 𝐾 ∈ {−1,1}
Answer: (𝐴) All the values of 𝐾 ∈ ℝ
Solution:
Matrix 𝐴 is invertible if |𝐴| ≠ 0
35
1 0 −𝐾
i.e | 2 1 3 |≠0
𝐾 0 1
⇒ 1(1) − 𝐾(−𝐾) ≠ 0
⇒ |𝐴| = 𝐾 2 + 1 ≠ 0 which is true for all real values of 𝐾.
Hence, 𝐴 is invertible for all real values of 𝐾.
Q37 cos 𝜃 − sin 𝜃
The inverse of the matrix 𝐴 = [ ]
sin 𝜃 cos 𝜃
cos 𝜃 sin 𝜃
A. [ ]
sin 𝜃 cos 𝜃
B. [ ]
cos 𝜃 sin 𝜃
sin 𝜃 − cos 𝜃
C. [ ]
cos 𝜃 sin 𝜃
cos 𝜃 sin 𝜃
D. [ ]
sin 𝜃 −cos 𝜃
cos 𝜃 sin 𝜃
Answer: (𝐴) [ ]
Solution:
cos 𝜃 − sin 𝜃
𝐴=[ ]
sin 𝜃 cos 𝜃
|𝐴| = cos 2 𝜃 + sin2 𝜃 = 1
cos 𝜃 − sin 𝜃 𝑇
𝑎𝑑𝑗 𝐴 = [ ]
sin 𝜃 cos 𝜃
cos 𝜃 sin 𝜃
=[ ]
𝑎𝑑𝑗 𝐴 cos 𝜃 sin 𝜃
𝐴−1 = =[ ]
|𝐴| −sin 𝜃 cos 𝜃
Q38 3 4 𝑥 1
The number of solutions of [ ] [ ] = [ ] is
9 12 𝑦 3
A. No solution
B. Infinite solutions
C. Unique solutions
D. Two solutions
Answer: (𝐵) Infinite solutions
36
Solution:
3 4 𝑥 1
Given [ ] [ ] = [ ],
9 12 𝑦 3
We know that for
𝑎1 𝑏1 𝑥 𝑐1
[ ] [𝑦] = [𝑐 ], if
𝑎2 𝑏2 2
𝑎1 𝑏 𝑐
𝑎2
= 𝑏1 = 𝑐1 , there are infinite solutions.
2 2
In the given equation, we have

3 4 1
= =
9 12 3
Thus, we have infinite solutions.
Q39 If the system of equations 𝑥 − 𝑘𝑦 − 𝑧 = 0, 𝑘𝑥 − 𝑦 − 𝑧 = 0 and 𝑥 + 𝑦 − 𝑧 = 0 has

a non-trivial solution, then 𝑘 can be
A. 0,1
B. 1, −1
C. −1,2
D. 2, −2
Answer: (𝐵) 1, −1
Solution:
The given system of equation has a non-trivial solution if Δ = 0.
For the given system of equations, we have
1 −𝑘 −1
Δ = |𝑘 −1 −1|
1 1 −1
𝐶1 → 𝐶1 + 𝐶3
0 −𝑘 −1
Now, Δ = |𝑘 − 1 −1 −1| = 0
0 1 −1
On solving, we get
⇒ (𝑘 − 1)(𝑘 + 1) = 0 ⇒ 𝑘 = −1,1
Q40 For the LPP; maximize 𝑧 = 5𝑥 − 2𝑦 subject to the constraints 3𝑥 + 2𝑦 ≤
6, 10𝑥 + 9𝑦 ≥ 45, 𝑥, 𝑦 ≥ 0
A. 𝑧𝑚𝑖𝑛 = −15
B. 𝑧𝑚𝑖𝑛 = 0
37
C. 𝑧𝑚𝑖𝑛 = −6
D. The LPP has no feasible solution
Answer: (𝐷) The LPP has no feasible solution
Solution:
From the graph, we can conclude that the given LPP has no feasible solution.
Q41 If 𝑍 = 𝑥 + 2𝑦, subject to the constraints: 2𝑥 + 𝑦 ≥ 3, 𝑥 + 2𝑦 ≥ 6, 𝑥, 𝑦 ≥ 0, then
the minimum of 𝑍 occurs at
A. Only one point

B. Two points
C. More than two points
D. Minimum of 𝑍 is not possible
Answer: (𝐶) More than two points
Solution:
The feasible region under the given constraints is
38
The coordinates of corner points and values of 𝑍 are
Corner points 𝑍 = 𝑥 + 2𝑦
(0,3) 6
(6,0) 6
From the above table minimum value of 𝑍 = 6

Now, compare the constraints and objective function.
We observe that the slope of the constraint (x + 2y ≥ 6) and objective
function is same, so 𝑍 will be minimum at all the points which lie on the
constraint line (x + 2y = 6) between the points (0,3) and(6,0).
Q42 Sum of maximum and minimum value of the objective function 𝑧 = 10𝑥 + 7𝑦,
subjected to the constriants 0 ≤ 𝑥 ≤ 60,0 ≤ 𝑦 ≤ 45, 5𝑥 + 6𝑦 ≤ 420 is
A. 740
B. 1025
C. 1055
D. 1100
Answer: (𝐴) 740
Solution:
Given, 𝑧 = 10𝑥 + 7𝑦 and constraints are 0 ≤ 𝑥 ≤ 60,0 ≤ 𝑦 ≤ 45, 5𝑥 + 6𝑦 ≤ 420
𝑥 𝑦
⇒ 0 ≤ x ≤ 60,0 ≤ y ≤ 45 and 84 + 70 ≤ 1
Now, we get common area as marked in figure
39
Finally,
Corner points z = 10x + 7y
(0,45) 315
(60,20) 740 (Maximum)
(60,0) 600
(30,45) 615
(0,0) 0 (Minimum)
Thus, zmax + zmin = 740 + 0 = 740
Q43 The corner points of the feasible region determined by a system of linear
constraints are (0,10), (5,5), (15,15), (0,20). Let z = px + qy where p, q > 0. If
the maximum of 𝑧 occurs at both the points (15,15) and (0,20), then which of
the following is true:
A. 𝑝 = 𝑞
B. 𝑝 = 2𝑞
C. 𝑝 = 3𝑞
D. 𝑞 = 3𝑝
Answer: (𝐷) 𝑞 = 3𝑝
Solution:
Let 𝑧0 be the maximum value of 𝑧 in the feasible region. Since maximum
occurs at both (15,15) and(0,20), the value 𝑧0 is attained at both (15,15) and
40
(0,20).
⇒ z0 = p(15) + q(15) ⋯ (1)
and z0 = p(0) + q(20) ⋯ (2)
From (1) and (2), we get:
15p + 15q = 20q
⇒ 15p = 5q
⇒ 3p = q
Q44 For an LPP,
maximize z = ax + by where 𝑎, 𝑏 ∈ ℝ subject to the constraints
a1 x + b1 y ≤ 0
a2 x + b2 y ≤ 0
x, y ≥ 0,
consider the following statements:
(I) The solution depends on the optimizing function.
(II) The solution depends on the constraints.
Which of the following statement(s) is/are correct?
A. Both (I) and (II)

B. (I) only
C. (II) only
D. The solution depends upon some more aspects.
Answer: (𝐶) (II) only
Solution:
For an arbitrary LPP, the solution depends upon the values at the corner points
of the feasible region (if there is any). So, the solution depends only on the
constraints.
41
Q45. The feasible region of a LPP is shown in the figure. If Z = 3x − y, then the
maximum value of 𝑍 occurs at
A. (0,0)
B. (3,0)
C. (12,20)
D. (0,15)
Answer: (𝐶) (12,20)
Solution:
The table of values at corner points for subjective function Z = 3x − y is given
below
Corner points: (𝑥, 𝑦) Value: Z = 3x − y
𝑂(0,0) 0
𝐴(0,15) −15
𝐵(3,0) 9
𝐶(12,20) 16
So, 𝑍𝑚𝑎𝑥 occurs at C(12,20).
42
The relation between the height of the plant (𝑦 in cm) with
respect to exposure to sunlight is governed by the following
1
equation 𝑦 = 4𝑥 − 2 𝑥 2 where 𝑥 is the number of days
exposed to sunlight.
Based on the given information, answer the following questions.

Q46 The rate of growth of the plant with respect to sunlight is
1
A. 4𝑥 − 2 𝑥 2
B. 4 − 𝑥
C. 𝑥 − 4
1
D. 𝑥 − 2 𝑥 2
Answer: (B) 4 − 𝑥
Solution:
1
Given: 𝑦 = 4𝑥 − 𝑥 2
2
Differentiating both sides w.r.t. 𝑥, we get

𝑑𝑦 1
= 4 − ⋅ 2𝑥
𝑑𝑥 2
𝑑𝑦
∴ = 4−𝑥
𝑑𝑥
Hence, the rate of growth of the plant with respect to sunlight is 4 − 𝑥.
Q47 What is the number of days it will take for the plant to grow to the maximum height?
A. 4
B. 6
C. 7
D. 10
Answer: (A) 4
Solution:
For maximum/minimum
𝑑𝑦
=0
𝑑𝑥
43
⇒4−𝑥 =0
∴𝑥=4
Hence, it will take 4 days to grow to the maximum height.
Q48 What is the maximum height of the plant?
A. 12 cm
B. 10 cm
C. 8 cm
D. 6 cm
Answer: (C) 8 cm
Solution:
As we know plant will take 4 days to grow to the maximum height.
i.e., 𝑥 = 4
Then the value of maximum height is
1
𝑦 = 4(4) − (4)2 = 16 − 8
2
∴ 𝑦 = 8 cm
Hence, maximum height of the plant is 8 cm
Q49 What will be the height of the plant after 2 days?
A. 4 cm
B. 6 cm
C. 8 cm
D. 10 cm
Answer: (B) 6 cm
Solution:
1
Given: 𝑦 = 4𝑥 − 2 𝑥 2
At 𝑥 = 2, we get
1
𝑦 = 4(2) − (2)2 = 8 − 2
2
∴ 𝑦 = 6 cm
44
Hence, the height of the plant after 2 days is 6 cm
Q50 7
If the height of the plant is 2 c. m., then the number of days it has been exposed to
the sunlight is
A. 2
B. 3
C. 4
D. 1
Answer: (D)1
Solution:
1
Given: 𝑦 = 4𝑥 − 𝑥 2
2
7
At 𝑦 = 2 , we get
7 1
= 4𝑥 − 𝑥 2
2 2
⇒ 𝑥 2 − 8𝑥 + 7 = 0
⇒ 𝑥 2 − 7𝑥 − 𝑥 + 7 = 0
⇒ 𝑥(𝑥 − 7) − 1(𝑥 − 7) = 0
⇒ 𝑥 = 1 or 7
Hence, the number of days it has been exposed to the sunlight is 1 or 7.
45
PHYSICS
Time: 90 minutes Maximum Marks: 35
1. The Question Paper contains three sections
2. Section A has 25 questions. Attempt any 20 questions.
3. Section B has 24 questions. Attempt any 20 questions.
4. Section C has 6 questions. Attempt any 5 questions.
5. All questions carry equal marks.
SECTION A
This section consists of 25 multiple choice questions with overall choice to attempt any 20
questions. In case more than desirable number of questions are attempted, ONLY first 20 will be
considered for evaluation.
Q11. When a body is charged by induction, then the body (0.77)

A. Becomes neutral
B. Does not lose any charge
C. Loses whole of the charge on it
D. Loses part of the charge on it
Q22. Which of the following is a wrong statement? (0.77)

A. The charge on a body is always quantized.
B. Charge on proton has a positive sign and charge on electron has a negative
sign
C. The net charge of an isolated system is always conserved
D. The net charge of an object connected to ground through earthing, is
always constant
Q33. Consider three charged bodies P, Q and R. If P and Q repel each other, (0.77)
while P and R attract, what is the nature of the force between Q and R?
A. Repulsive force
B. Attractive force
C. No force
D. None of these
Q44. In a process, when two bodies are charged by rubbing against each other, one (0.77)
becomes positively charged while the other becomes negatively charged. Then
A. Mass of each body remains unchanged
B. Mass of positively charged body slightly increases
C. Mass of negatively charged body slightly decreases
D. Mass of each body changes slightly but the total mass of system remains
the same
46
Q55. Two points charges 𝑄1 and 𝑄2 are placed at separation d in vacuum and the force (0.77)
𝑑
acting between them is F. Now a dielectric slab of thickness 2 and dielectric
constant K = 4 is placed between them. The new force between the charges will
be
4𝐹
A. 9
2𝐹
B. 9
𝐹
C. 9
5𝐹
D. 9
Q66. Five-point charges, each of +q, are placed at five vertices of a regular hexagon of (0.77)
side ‘𝑎’ unit. What is the magnitude of force experienced by a charge −q placed
at the centre of the hexagon?
A. Zero
𝑞2
B. 4 𝜋𝜖0 𝑎2
𝑞2
C.
√2 𝜋𝜖0 𝑎2
𝑞2
D. 8 𝜋𝜖0 𝑎2
Q77. Two equally charged identical metallic spheres A and B, repel each other with a (0.77)
force of 2 × 10−5 𝑁, when placed in air (neglect the dimension of sphere as they
are very small). Another identical uncharged sphere C is touched to B and then
placed at the midpoint of line joining A and B. What is the net electrostatic force
on C?
A. 1 × 10−5 𝑁, toward BA
B. 2 × 10−5 𝑁, towards AB
C. 4 × 10−5 𝑁, towards BA
D. 0.5 × 10−5 𝑁, towards AB
Q88. The magnitude of force due to an electric dipole, of dipole moment 3.6 × (0.77)
10−29 C-m, on an electron at a distance of 25 nm from the centre of the dipole and
along the dipole axis, is
A. 6.6 × 10−15 𝑁
B. 7.8 × 10−15 𝑁
C. 9.4 × 10−15 𝑁
D. 12 × 10−15 𝑁
Q99. A charge of 1 C is located at the centre of a sphere of radius 10 cm whose centre (0.77)
coincides with that of a cube of side 20 cm. The ratio of, the outgoing flux from
the sphere, to the outgoing flux from the cube, will be
A. More than one
B. Less than one
C. One
47
D. Nothing can be said
10. Three particles, each having a charge of 10 𝜇𝐶, are placed at the vertices of an (0.77)
Q10
equilateral triangle of side 10 cm. Find the work done by a person in pulling them
apart, to infinite separation.
A. 0J
B. 27J
C. -54J
D. -27J
11. A parallel plate capacitor has rectangular plates of area 400 𝑐𝑚2 and are separated (0.77)
Q11
by 2 mm, with air as a dielectric medium. What is the magnitude of charge on
each plate, if a potential difference of 200 volts is applied across them?
(Take ϵo=8.85 × 10−12 𝐹/𝑚)
A. 3.54 × 10−6 𝐶
B. 3.54 × 10−8 𝐶
C. 1770.8 × 10−9 𝐶
D. 1770.8 × 10−13 𝐶
12. A parallel plate capacitor of capacitance 100 𝜇𝐹 is connected to a power supply (0.77)
Q12
of 200 V. A dielectric slab of dielectric constant 5 is now inserted into the gap
between the plates. The work done by power supply will be:
A. -12 J
B. 16 J
C. -24 J
D. 8 J
Q13
13. In a balanced Wheatstone’s bridge, current in the galvanometer is zero. It remains (0.77)
zero when
(1) The emf of is increased
14. (2) All resistances are increased by 10 Ω
15. (3) All resistances are made five times.
16. (4) The battery and the galvanometer are interchanged.
A. Only (1) is correct
B. (1), (2) and (3) are correct
C. (1), (3) and (4) are correct
D. (1) and (3) are correct
Q14
17. When a metal conductor connected to the left gap of a meter bridge is heated, the (0.77)
balancing point,
A. Shifts towards right
B. Shifts towards left
C. Remains unchanged
D. Remains zero
48
Q15
18. In a meter bridge for measurement of resistance, the known and unknown (0.77)
resistances are interchanged and mean reading is calculated for null point
measurement. The error thus removed is
A. End Correction
B. Index error
C. Temperature effect
D. Random error
19. The Kirchhoff’s first law (∑𝑖 = 0) and second law (∑𝑖 𝑅 = 0 = ∑𝐸), where the (0.77)
Q16
symbols have their usual meanings, are respectively based on
A. Conservation of charge and conservation of energy
B. Conservation of charge and conservation of momentum
C. Conservation of energy and conservation of charge
D. Conservation of momentum and conservation of charge
Q17
20. In a wire of uniform cross-section and radius r, free electrons travel with drift (0.77)
velocity v when a current I flow through the wire. What is the current in another
wire of half the radius and of the same material when the drift velocity is 2v?
A. 2I
B. I
𝐼
C. 2
𝐼
D. 4
21. A current of 1.34 A exists in a copper wire of cross – section 1 𝑚𝑚2 . Assuming (0.77)
Q18
that each copper atom contributes one free electron, the drift speed of the free
electron in the wire will be:
(Density of copper is 8990 𝑘𝑔/𝑚3, atomic mass is 63.50 and charge on electron
is 1.6 × 10−19 𝐶)
A. 0.5 mm/s
B. 0.3 mm/s
C. 0.1 mm/s
D. 0.5 mm/s
Q19
22. Current through wire XY in the circuit shown is (0.77)
A. 1A
B. 4A
C. 2A
D. 3A
49
23. Two wires, of same dimensions but resistivities 𝜌1 and 𝜌2 , are connected in series. (0.77)
Q20
The equivalent resistivity of the combination is
A. √𝜌1 𝜌2
B. 𝜌1 + 𝜌2
𝜌 +𝜌
C. 1 2 2
D. None of the above
Q21
24. The storage battery of a car has an emf of 12 V. If the internal resistance of the (0.77)
battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
A. 10 A
B. 20 A
C. 30 A
D. 40 A
25. In the figure shown, battery 1 has emf=6 V and internal resistance = 1 𝛺. (0.77)
Q22
Battery 2 has emf=2 V and internal resistance = 3 𝛺. The wires have negligible
resistance. What is the potential difference across the terminals of battery 2?
A. 4V
B. 1.5 V
C. 5V
D. 0.5 V
Q23
26. A battery consists of n number of identical cells having internal resistance (0.77)
connected in series. The terminals of the battery are short circuited and the
current I is measured.
27. Which of the following graphs shows correct relationship between I and n?
A.
50
B.
C.
D.
28. In a meter bridge experiment, the value of unknown resistance is 2𝛺. To get the (0.77)
Q24
balancing point at 40 cm distance from the same end, then what will be the
resistance in the resistance box?
A. 3 Ω
B. 6 Ω
C. 8 Ω
51
D. 9 Ω
Q25
29. In Oersted's experiment, deflection in magnetic needle (0.77)
A. Decreases if current is increased in the wire
B. Reverses if current is reversed in the wire
C. Remains constant if current is increased
D. All above statement are true
SECTION B
questions. In case more than desirable number of questions are attempted, ONLY first 20 will be
considered for evaluation
Q26
30. Which of the following statements is true in case of a solenoid? (0.77)
A. A solenoid is a type of permanent magnet
B. It generates a controlled magnetic field through a coil wound into a tightly
packed helix.
C. A solenoid has a very weak magnetic field.
D. Multiple turns of more than two metallic wires are used to make a solenoid.
Q27
31. A wire of fixed length is bent to form a coil of one turn. It is again bent to form a (0.77)
coil of three turns. If in both, cases same amount of current is passed, then the
ratio of the intensities of magnetic field produced at the centre of the coil, of one
turn and three turns respectively, will be,
A. 2 : 1
B. 9 : 1
C. 1 : 9
D. 1 : 3
Q28
32. What can we say about the solenoid more accurately? (0.77)
A. A solenoid is a combination of multiple circular loops.
B. A solenoid is a combination of multiple rectangular shape loops.
C. The circular loops at the ends represent poles of the bar magnet.
D. Both a and c.
Q29
33. A solenoid has 1000 turns and 20 cm long. Find the magnetic induction produced (0.77)
at the centre of the solenoid by the current of 2 A
A. 1.25 × 10−3 𝑇
B. 3.75 × 10−2 𝑇
C. 2.50 × 10−2 𝑇
D. 1.25 × 10−2 𝑇
52
Q30
34. A proton of energy 8 eV is moving in a circular path in a uniform magnetic field. (0.77)
The energy of an 𝛼 - particle moving in the same magnetic field and along the
same path will be
A. 2eV
B. 4 eV
C. 6 eV
D. 8 eV
Q31
35. A magnetic needle is kept in a non-uniform magnetic field. It experiences (0.77)
A. A force and a torque
B. A force but not a torque
C. A torque but not a force
D. Neither a torque nor a force
Q32
36. What is the consequence of placing a core of a material made up of something (0.77)
other than a soft iron?
A. The strength of a net magnetic field decreases due to a soft iron core.
B. The soft iron core is used due to its high value of permeability.
C. Overall strength of an electromagnet increases due to the soft iron core.
D. Both b and c are correct
Q33
37. If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic (0.77)
field of strength B, the work done in rotating the magnet through an angle 𝜃 is
A. 𝑀𝐵(1 − sin 𝜃)
B. 𝑀𝐵 sin 𝜃
C. 𝑀𝐵 cos 𝜃
D. 𝑀𝐵(1 − cos 𝜃)
Q34
38. A long magnetic needle of length 2L, magnetic moment M and pole (0.77)
strength m units is broken into two pieces at the middle point. The magnetic
moment and pole strength of each piece will be
𝑀 𝑚
A. 2 , 2
𝑚
B. 𝑀, 2
𝑀
C. 2 , 𝑚
D. 𝑀, 𝑚
39. A domain in a magnet is in the form of a cube of side length 1 𝜇𝑚. If it (0.77)
Q35
contains 8 × 1010 atoms and each atomic dipole has a dipole moment
of 9 × 10−24 𝐴 − 𝑚2 and all of these dipoles are aligned in the same direction,
then the magnetisation of the domain is
A. 7.2 × 105 𝐴𝑚−1
B. 7.2 × 103 𝐴𝑚−1
C. 3,6 × 105 𝐴𝑚−1
D. 1.8 × 1063 𝐴𝑚−1
53
Q36
40. A magnet oscillating in a horizontal plane has a time period of 2 sec at a place (0.77)
where the angle of dip is 300 and 3 sec at another place where the angle of dip
is 600 . The ratio of resultant magnetic fields at the two places is:
4 √3
A. 7
4
B. 9 √3
9
C. 4 √3
9
D.
√3
Q37
41. In electromagnetic induction, the induced e.m.f. in a coil is independent of (0.77)
A. Change in the magnetic flux
B. Time interval
C. Resistance of the circuit
D. Number of turns in the coil
Q38
42. The dimension of √𝐿𝐶 is (0.77)
A. [L]
B. [M]
C. [T]
D. [I]
43. Magnetic flux 𝜙 (in Webers) linked with a closed circuit of resistance 10 Ω varies (0.77)
Q39
with time t (in seconds) as 𝜙 = 5 𝑡 2 − 4𝑡 + 1. The induced emf in the circuit at t
= 0.2 sec is
A. 5 mA
B. 10 mA
C. 20 mA
D. 1 A
Q40
44. Two solenoids A and B spaced close to each other and sharing the same cylindrical (0.77)
axis have 400 and 700 turns respectively. A current of 3.5A in coil A produce an
arrange flux of 300 𝜇𝑇 − 𝑚2 through each turn of A and a flux of 90 𝜇𝑇 − 𝑚2
through each turn B. The mutual inductance of the two solenoids is
A. 1.8 × 10−2 𝐻
B. 2.8 × 10−2 𝐻
C. 3.8 × 10−2 𝐻
D. 4.8 × 10−2 𝐻
45. Power dissipated in an LCR series circuit connected to an a.c. source of emf ϵ is
Q41 (0.77)
ϵ2 𝑅
A. 2
√𝑅 2 +(𝐿 𝜔− 1 )
𝐶𝜔
ϵ2 𝑅
B. 1 2
[𝑅 2 +( 𝐿 𝜔− ) ]
𝐶𝜔
54
1 2
√[𝑅 2 +( 𝐿 𝜔− ) ]
𝐶𝜔
C. ϵ2 𝑅
1 2
ϵ2 [𝑅 2 +(𝐿 𝜔− ) ]
𝐶𝜔
D. 𝑅
Q42
46. In an AC circuit, when an AC ammeter is connected, it reads i. If a student uses a (0.77)
DC ammeter in place of the ac ammeter, the reading in the DC ammeter will be:
𝑖
A.
√2
B. √2𝑖
C. 0.637 𝑖
D. Zero
47. An alternating current is given as 𝑖 = 𝑖1 cos 𝜔𝑡 − 𝑖2 sin 𝜔𝑡. The rms current is (0.77)
Q43
given by
𝑖 +𝑖
A. 1 2
√2
𝑖1 + 𝑖2 |
B.
√2
𝑖 +𝑖 2 2
C. √ 1 2 2
𝑖 2 +𝑖22
D. √ 1
√2
Q44
48. An alternating current having peak value 14 A is used to heat a metal wire. To (0.77)
produce the same heating effect, a constant current i can be used where i is
A. 14 A
B. About 20 A
C. 7A
D. About 10 A
Q45
49. Given below are two statements labelled Assertion (A) and Reason (R). (0.77)
Select the most appropriate answer from the options given below.
1. Given below are two statements labelled Assertion (A) and Reason (R).
Assertion: If three capacitors of capacitance 𝐶1 < 𝐶2 < 𝐶3 are connected in
parallel then their equivalent capacitance 𝐶𝑝 < 𝐶𝑠
1 1 1 1
Reason: 𝐶 = 𝐶 + 𝐶 + 𝐶
𝑝 1 2 3
A. Both assertion and reason are true and the reason is the correct explanation
of the assertion.
B. Both assertion and reason are true but reason is not the correct explanation
of the assertion.
C. Assertion is true but reason is false.
D. Assertion and reason both are false.
55
Q46
Assertion: A potentiometer of longer length is used for accurate measurement.
Reason: The potential gradient for a potentiometer of longer length, with a given
source of e.m.f., becomes small.
of the assertion.
of the assertion.
D. Assertion and reason both false.
Q47 Given below are two statements labelled Assertion (A) and Reason (R). (0.77)
Assertion : If a proton and an -particle enter a uniform magnetic field
perpendicularly, with the same speed, then the time period of revolution of the -
particle is double than that of proton.
Reason: In a magnetic field, the time period of revolution of a charged particle is
directly proportional to mass.
of the assertion.
of the assertion.
Q48
51. Given below are two statement labelled Assertion (A) and Reason (R). (0.77)
Assertion: Electromagnets are made of soft iron.
Reason: coercivity of soft iron is small
of the assertion.
of the assertion
Q49
Assertion: When ac circuit contain resistor only, its power utilization is minimum.
Reason: Power utilized in an AC circuit is independent of phase angle.
of the assertion.
56
of the assertion
C. Assertion is true but reason is false
SECTION C
This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In
case more than desirable number of questions are attempted, ONLY first 5 will be considered for
evaluation.
Q50 A transformer is used to light a 140 W, 24 V lamp from 240 AC mains. The current (0.77)
in mains cable is 0.7 A, find the efficiency of transformer.
A. 50%
B. 76.2%
C. 83.33%
D. 91.25%
Q51
53. In a series CR circuit shown in figure, the applied voltage is 10 V and the voltage (0.77)
across capacitor is found to be 8 V. Then, the voltage across R, and the phase
difference between current and the applied voltage, will respectively be
4
A. 6𝑉, tan−1 (3)
3
B. 3𝑉, tan−1 ( )
4
−1 5
C. 6𝑉, tan (3)
D. None
54.
55. Case study:
Every petrol engine needs separate arrangement for the ignition of the air-fuel mixture. This
arrangement is called its ignition system. The ignition system starts with the engine coil. It is a
stationary coil mounted around the engine shaft while a magnet is mounted on the engine shaft,
57
such that the magnet is revolving around the coil as the engine is running. The rotation of the
magnet induces an EMF (of relatively smaller magnitude) in the engine coil. This induced EMF
is applied across the ignition coil. The output of ignition coil is given to the spark plug, which is
placed inside the combustion chamber of the engine. The spark plug has a small gap between its
electrodes. When a high voltage is applied across these electrodes, there is an electric spark
between them, due to dielectric breakdown of air between them. At the appropriate time instant,
when the air fuel mixture is to be ignited, the switch 'S' is closed so that there is an electric spark
across the gap, in the spark plug.
Q52 The engine coil works on the principle of (0.77)

A. Ampere’s law
B. Faraday’s law
C. Gauss’ law
D. Both (a) and (b)
Q53
56. The ignition coil is necessarily: (0.77)
A. A solenoid with soft iron core
B. An R-L-C series circuit
C. Step down transformer
D. Step up transformer
Q54
57. As the speed of rotation of the engine increases (0.77)
A. The electric spark will get brighter and stronger
B. The electric spark will get dimmer and weaker
C. The electric spark will be unaffected
D. There will be no spark above a certain speed
Q55
58. Thick and multi-stranded copper wire will be needed (0.77)
A. For the engine coil
B. Between engine coil and ignition coil
C. Between ignition coil and spark plug
D. For all connections
58
ANSWER KEYS
PHYSICS
Q1 B. Q20 C. Q39 C.
Q2 D. Q21 C. Q40 A.
Q3 B. Q22 C. Q41 B.
Q4 D. Q23 D. Q42 D.
Q5 A. Q24 A. Q43 C.
Q6 B. Q25 B. Q44 D.
Q7 B. Q26 B. Q45 C.
Q8 A. Q27 C. Q46 B.
Q9 C. Q28 D. Q47 C.
Q10 B. Q29 D. Q48 A.
Q11 B. Q30 D. Q49 D.
Q12 B. Q31 A. Q50 C.
Q13 C. Q32 D. Q51 A.
Q14 A. Q33 D. Q52 B.
Q15 A. Q34 C. Q53 D.
Q16 A. Q35 A. Q54 A.
Q17 C. Q36 C. Q55 C.
Q18 C. Q37 C.
Q19 C. Q38 C.
59
SOLUTIONS
PHYSICS
Q1 When a body is charged by induction, then the body

A. Becomes neutral
B. Does not lose any charge
C. Loses whole of the charge on it
D. Loses part of the charge on it
Answer: B.
Solution:
Electrostatic induction is a redistribution of charges in an object, caused by influence of
nearby charges. Charging by induction involves transfer of charges from one part to the
other of the body. So, no charge is lost in the process.
Q259. Which of the following is a wrong statement?

A. The charges on a body is quantized always
B. Charge on proton has a positive sign and charge on electron has a negative sign
C. The net charge of an isolated system is always conserved
D. The net charge of an object connected to ground through earthling is always
constant
Answer: D.
Solution:
Law of conservation of charge is always applicable for an isolated system.
Isolated system means system are not connected to any external source of charge like a
battery or connected to ground.
The system shown in figure (a) and (b) are not isolated one.
60
Q3 Consider three charged bodies P, Q and R. If P and Q repel each other,
while P and R attract. What is the nature of force between Q and R?
A. Repulsive force
B. Attractive force
C. No force
D. None of these
Answer: B.
Solution:
P and Q repel each other.
⇒ 𝑃 and Q are similarly charged, thus either P and Q both are +ve or –ve charged body.
Since, P and R attract each other.
⇒ 𝑃 and R are oppositely charged.
Case (I):
If P is +ve charged body, then
⇒ 𝑄 → +𝑣𝑒 & 𝑅 → −𝑣𝑒
∴ 𝑄 and R will attract each other.
Case (II) :
If P is – ve charged body, then
⇒ 𝑄 → −𝑣𝑒 & 𝑅 → +𝑣𝑒
∴ 𝑄 and R will attract each other.
Hence, attractive force will act between Q and R. So, option B. is correct.
Q460. In a process, when two bodies are charged by rubbing against each other, one becomes
positively charged while the other becomes negatively charged. Then
A. mass of each body remains unchanged.
B. mass of positively charged body slightly increases.
C. mass of negatively charged body slightly decreases.
D. mass of each body changes slightly but the total mass of system remains the same.
Answer: D.
Solution:
The transfer of electrons from one body to the other produces charge on the body.
Hence, number of electrons given by one body = number of electrons obtained by the
other body
So, the mass of the negatively charged body slightly increases while the mass of the
positively charged body slightly decreases. However, the total mass of the system
remains the same.
Q561. Two points charges 𝑄1 and 𝑄2 are placed at separation d in vacuum and force acting
𝑑
between them is F. Now a dielectric slab of thickness 2 and dielectric constant K = 4 is
placed between them. The new force between the charges will be
4𝐹
A. 9
61
2𝐹
B. 9
𝐹
C. 9
5𝐹
D. 9
Answer: A.
Solution:
We know that, the force between charges separated by distance r in a dielectric medium
is equivalent to that of charges placed at distance 𝑟 √𝐾 in vacuum (here K is dielectric
constant of medium)
1
Now, from coulombs law 𝐹 ∝ 𝑑2 Ratio of force when charges are placed in dielectric to
when charges are placed in vaccum is
𝐹𝑑 𝑑2 4
= 2 =9
𝐹 𝑑 𝑑 √4
( + )
2 2
4𝐹
𝐹𝑑 = 9
Q662. Five-point charges each +q is placed on five vertices of a regular hexagon of side a unit.
What is the magnitude of force experienced by a charge −q placed at the centre of the
hexagon?
A. Zero
𝑞2
B. 4 𝜋𝜖0 𝑎2
𝑞2
C.
√2 𝜋𝜖0 𝑎2
𝑞2
D. 8 𝜋𝜖0 𝑎2
Answer: B.
Solution:
62
The regular hexagon shown in figure the distance between centre and vertex = a
Thus, each pair of interaction force will be same i.e., F
𝑘𝑞 2
|𝐹| = 2
𝑎
If there had been a sixth charge at the remaining vertex B of the hexagon, then at the
charge placed at center
𝐹⃗𝑛𝑒𝑡 = 0
i.e, the sixth charge will exert force F along OB.
∴ unbalanced force is the net force on the central charge
𝑘𝑞 2
⇒ 𝐹𝑛𝑒𝑡 = 𝐹 = 𝑎2
Direction of 𝐹𝑛𝑒𝑡 is along BO
1 𝑞2
∴ |𝐹𝑛𝑒𝑡 | = 4 𝜋𝜖
0 𝑎2
Tip: When identical charges are placed symmetrically at the vertices of a regular
polygon, then net force, 𝐹⃗𝑛𝑒𝑡 = 0 at geometric centre of polygon.
Q763. Two equally charged identical metallic spheres A and B repel each other with a
force 2 × 10−5 𝑁, when placed in air (neglect the dimension of sphere as they are very
small). Another identical uncharged sphere C is touched to B and then placed at the mid-
point of line joining A and B. What is the net electrostatic force on C?
A. 1 × 10−5 𝑁, toward BA
B. 2 × 10−5 𝑁, towards AB
C. 4 × 10−5 𝑁, towards BA
D. 0.5 × 10−5 𝑁, towards AB
Answer: B.
Solution:
Let the initial charge on the sphere A and B be +q and separated by r. The force of
repulsion,
𝑘𝑞 2
𝐹= = 2 × 10−5 𝑁
𝑟2
63
When the sphere C is touched with B; the charge of B will get distributed equally on B
and C due to their identical nature.
𝑞
∴ 𝑞𝐵 = 𝑞𝐶 = 2
Force on sphere C by A,
𝑘𝑞 2 2𝑘𝑞 2
𝐹𝐶𝐴 = 𝑟 2
= = 2𝐹
2( ) 𝑟2
2
𝐹𝐶𝐴 is directed from A to B
Similarly, force on the sphere C by sphere B
𝑞 𝑞
𝑘( )( ) 𝑘𝑞 2
2 2
𝐹𝐶𝐵 = 𝑟 2
=
( ) 𝑟2
2
⇒ 𝐹𝐶𝐵 = 𝐹
And 𝐹𝐶𝐵 = 𝐹 is directed from B to A.
Thus, net force on C is,
𝐹𝑐 = 2𝐹 − 𝐹 = 𝐹
⇒ 𝐹𝑐 = 2 × 10−5 𝑁 from A to B
Q864. The magnitude of force due to an electric dipole of dipole moment 3.6 × 10−29 C-m on
an electron 25 nm from the centre of the dipole along the dipole axis is
A. 6.6 × 10−15 𝑁
B. 7.8 × 10−15 𝑁
C. 9.4 × 10−15 𝑁
D. 12 × 10−15 𝑁
Answer: A.
Solution:
Given:
𝑝 = 3.6 × 10−29 𝐶 − 𝑚; 𝑒 = 1.6 × 10−19 𝐶
𝑥 = 25 × 10−9 𝑚
We know that electric field due to electric dipole on axial line is given by
2𝑘𝑝𝑥
𝐸 = (𝑥 2−𝑙2)2
In this case, x >> 1
2𝑘𝑝
So, 𝐸 = 𝑥 3
64
2×9×109 ×3.6×10−29
⇒𝐸= (25×10−9 )3
4
⇒ 𝐸 = 4.15 × 10 𝑁/𝐶
Due to electric field E, force experience by the electron will be
F = eE
⇒ 𝐹 = 1.6 × 10−19 × 4.15 × 104
⇒ 𝐹 = 6.6 × 10−15 𝑁
Q965. A charge of 1 C is located at the centre of a sphere of radius 10 cm whose centre coincides
with that of a cube of side 20 cm. The ratio of the outgoing flux from the sphere to the
outgoing flux from the cube will be
A. More than one
B. Less than one
C. One
D. Nothing can be said
Answer: C.
Solution:
The charge q = 1C is placed at the center of a sphere of radius R = 10 cm, As the side
of the square is 20 cm, so the cube will contain the sphere as shown.
As we can see in this figure, the field lines originating from the charge will first intersect
the sphere followed by the cube. This tells us that the same number of field lines will
intersect the sphere and the cube.
∴ 𝜙𝑐𝑢𝑏𝑒 = 𝜙𝑠𝑝ℎ𝑒𝑟𝑒
𝜙𝑠𝑝ℎ𝑒𝑟𝑒
Or, =1
𝜙𝑐𝑢𝑏𝑒
66. Three particles, each having a charge of 10 𝜇𝐶 are placed at the vertices of an equilateral
Q10
triangle of side 10 cm. Find the work done by a person in pulling them apart to infinite
separation.
A. 0J
B. 27J
C. -54J
D. -27J
Answer: B.
Solution:
Given that,
Length of each side of the triangle
65
R = 10 cm = 0.1 m
Charge, 𝑞 = 10 𝜇𝐶
The potential energy (U) of a system of two charges (𝑞1 𝑎𝑛𝑑 𝑞2 ) is given by
𝑘𝑞 𝑞
𝑢 = 𝑟1 2
In the case of system of multiple charges, the potential energy is
𝑘𝑞 𝑞 𝑘𝑞 𝑞 𝑘𝑞 𝑞
𝑈 = 𝑟1 3 + 𝑟2 3 + 𝑟1 2
2 1 3
Here, 𝑞1 = 𝑞2 = 𝑞3 = 𝑞 = 10 × 10−6 𝐶
𝑟1 = 𝑟2 = 𝑟3 = 𝑟 = 0.1 𝑚
Substituting the values,
9×109 ×10×10−6 ×10×10−6 ×3
𝑈=
0.1
⇒ 𝑈 = 27 𝐽
This is the initial potential energy. So,
𝑈𝑖 = 27 𝐽
Now, when charges are infinitely separated, the potential energy is reduced to zero. So,
𝑈𝑓 = 0
Work done by electric force is
𝑊 = 𝑈 − 𝑓 − 𝑈𝑖 = −27 𝐽
Thus, work done by a person on the system is 27 J.
67. A parallel plate capacitor has rectangular plates of area 400 𝑐𝑚2 and separated
Q11
by 2 mm with air as a medium. What is the magnitude of charge on each plate if the
applied potential difference is 200 Volts? (Take 𝜖𝑜 =8.85 × 10−12 𝐹/𝑚)
A. 3.54 × 10−6 𝐶
B. 3.54 × 10−8 𝐶
C. 1770.8 × 10−9 𝐶
D. 1770.8 × 10−13 𝐶
Answer: B.
Solution:
66
We can solve this problem in two simple steps.
Step 1:
𝐴𝜖
First, let’s find the capacitance of the capacitor by using, 𝐶 = 𝑑 0 we get,
400×10−4 ×8.85×10−12
⇒𝐶= 2×10−3
⇒ 𝐶 = 17.7 × 10−11 𝐹
Step 2: Now Let’s use Q = CV
⇒ 𝑄 = 17.7 × 10−11 × 200 = 3.54 × 10−8 𝐶
68. A parallel plate capacitor of capacitance 100 𝜇𝐹 is connected to a power supply of 200 V.
Q12
A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates.
The work done by power supply will be:
A. -12 J
B. 16 J
C. -24 J
D. 8 J
Answer: B.
Solution:
Before insertion of dielectric slab,
𝐶𝑖 = 100 𝜇𝐹; 𝑉 = 200 𝑉
𝑄𝑖 = 100 𝜇 𝐹 × 200 𝑉 = 20000 𝜇𝐶
∴ 𝑄𝑖 = 20 𝑚𝐶
After the dielectric introduced:
𝐶𝑓 = 𝐾𝐶𝑖 = 5 × 100 = 500 𝜇𝐹
𝑉 ′ = 𝑉 = 200 𝑉𝑜𝑙𝑡
(∵ battery remains connected)
So,
𝑄𝑓 = 500 𝜇𝐹 × 200 𝑉 = 100000 𝜇𝐶
∴ 𝑄1 = 100 𝑚𝐶
Therefore, the charge supplied by battery is,
Δ𝑄 = 𝑄𝑓 − 𝑄𝑖 = 100 − 20 = 80 𝑚𝐶
So, work done by battery will be
𝑊𝑏 = Δ𝑄 × 𝑉
⇒ 𝑊𝑏 = (80 𝑚𝐶 × 200 𝑉) = 16000 𝑚𝐽
∴ 𝑊𝑏 = 16 𝐽
67
Q13
69. In a balanced Wheatstone bridge, current in the galvanometer is zero. It remains zero
when
(1) The emf of is increased
70. (2) All resistances are increased by 10 Ω
71. (3) All resistances are made five times.
72. (4) The battery and the galvanometer are interchanged.
A. Only (1) is correct
B. (1), (2) and (3) are correct
C. (1), (3) and (4) are correct
D. (1) and (3) are correct
Answer: C.
Solution:
The necessary condition for a balanced Wheatstone bridge:

𝑅1 𝑅
= 𝑅4
𝑅2 3
⇒ 𝑅1 𝑅3 = 𝑅2 𝑅4 … … … . (𝑖)
We can observe that the emf of battery has nothing to do with the condition for a
balanced Wheatstone bridge.
∴ A balanced bridge remains unaffected from the change in emf of battery.
Hence statement (1) is correct
When each resistance is increased by 10 Ω, then equation (i) will no longer hold. Thus,
balanced condition of Wheatstone bridge will get altered.
⇒ (𝑅1 + 10)(𝑅3 + 10) ≠ (𝑅2 + 10)(𝑅4 + 10)
Hence statement (2) is incorrect
If all resistances are increased five times, then the multiplication factor gets cancelled
from the two sides of Eq. (i), thus the balancing condition is maintained.
⇒ (5𝑅1 )5𝑅3 ) = (5𝑅2 )(5𝑅4 )
Or, 𝑅1 𝑅3 = 𝑅2 𝑅4
Hence statement (3) is correct
When the position of battery and galvanometer are interchanged in a balanced
Wheatstone bridge, it will again lead to equal potential across the diagonal arm.
Since, 𝑅1 𝑅3 = 𝑅2 𝑅4
68
Hence this is also a balanced Wheatstone bridge.
Therefore statement (4) is correct
Q14
73. When a metal conductor connected to the left gap of a meter bridge is heated, the
balancing point,
A. Shifts towards right
B. Shifts towards left
C. Remains unchanged
D. Remains zero
Answer: A.
Solution:
When the metal conductor A connected to the left gap is heated then its resistance will
increase (∵ 𝑅 ∝ 𝑇).
Now from the balanced bridge condition,
𝑅𝐴 𝑙1 𝑙1
= = … … (1)
𝑅𝐵 𝑙
2 100−𝑙1
(𝑙1 being the distance of null point on meter bridge wire from left end)
From equation (1), if 𝑅𝐴 ↑ then 𝑙1 ↑.

Thus, the balance point or null point will shift towards right.
Q15
74. In a meter bridge for measurement of resistance, the known and unknown resistance are
interchanged and mean reading is calculated for null point measurement. The error hence
removed is
A. End Correction
B. Index error
C. Temperature effect
D. Random error
69
Answer: A.
Solution:
In a meter bridge experiment, it is assumed that the resistance of the L shaped plate is
negligible, but actually it is not so. The error created due to this is called end error. To
remove this, the resistance box and the unknown resistance must be interchanged and
then the mean reading must be taken.
75. The Kirchhoff’s first law (∑𝑖 = 0) and second law (∑𝑖 𝑅 = 0 = ∑𝐸), where the symbols
Q16
have their usual meanings, are respectively based on
A. Conservation of charge and conservation of energy
B. Conservation of charge and conservation of momentum
C. Conservation of energy and conservation of charge
D. Conservation of momentum and conservation of charge
Answer: A.
Solution:
According to Kirchhoff's first law, a junction can act neither as a source of charge nor
as a sink of charge. This supports the law of conservation of charge.
According to Kirchhoff's second law, the energy per unit charge transferred to the
moving charges is equal to the energy per unit charge transferred from them. This
supports the law of conservation of energy.
Q17
76. In a wire of cross-section radius r, free electrons travel with drift velocity v when a
current I flow through the wire. What is the current in another wire of half the radius and
of the same material when the drift velocity is 2v?
A. 2I
B. I
𝐼
C. 2
𝐼
D. 4
Answer: C.
Solution:
From current and drift velocity relation, we know that
𝐼 = 𝑛𝑒 𝐴 𝑣𝑑
For first wire:
𝐴 = 𝜋𝑟 2 ; 𝑣𝑑 = 𝑣
∴ 𝐼 = 𝑛𝑒(𝜋𝑟 2 )𝑣 … … . . (1)
For second wire:
𝑟 2
𝐴 = 𝜋 (2) ; 𝑣𝑑 = 2𝑣
𝑟 2
∴ 𝐼 ′ = 𝑛𝑒 [𝜋 (2) ] 2𝑣 … … . (2)
70
From eq. (1) & (2)
𝐼
𝐼′ = 2
77. A current of 1.34 A exists in a copper wire of cross – section 1 𝑚𝑚2 . Assuming that each
Q18
copper atom contributes one free electron, the drift speed of the free electron in the wire
will be:
(Density of copper is 8990 𝑘𝑔/𝑚3, atomic mass is 63.50 and charge on electron is
1.6 × 10−19 𝐶)
A. 0.5 mm/s
B. 0.3 mm/s
C. 0.1 mm/s
D. 0.5 mm/s
Answer: C.
Solution:
The density of copper is 8990 𝑘𝑔/𝑚3.
Mass of 1 𝑚3 volume of the copper is 8990 𝑘𝑔
i.e. 𝑚 = 8990 × 103 𝑔𝑚
number of moles of copper in 1 𝑚3 is
8990×103 𝑔
𝑛′ = 63.5 𝑔/𝑚𝑜𝑙𝑒 = 1.4 × 105 moles
Since each mole will contain 𝑁𝐴 atoms, where 𝑁𝐴 = 6.023 × 1023
The number of atoms in 1𝑚3 copper will be,
𝑛 = 1.4 × 105 × 6.023 × 1023
⇒ 𝑛 ≈ 8.4 × 1028
Using current drift velocity relation,
𝐼 = 𝑛𝑒 𝐴 𝑣𝑑
𝐼
⇒ 𝑣𝑑 = 𝑛𝑒𝐴
1.34
⇒ 𝑣𝑑 = 8.4×1028×1.6×10−19 ×(1×(10−3)2)
1.34
⇒ 𝑣𝑑 = 8.4×1.6×103 = 0.997 × 10−4
𝑚
∴ 𝑣𝑑 ≈ 10−4 = 0.1 𝑚𝑚/𝑠
𝑠
Q19
78. Current through wire XY of circuit shown is
A. 1A
B. 4A
71
C. 2A
D. 3A
Answer: C.
Solution:
Let current through XY is 𝑖3 .

Applying Kirchhoff’s law in loop (1),
−(𝑖1 × 1) + (0 × 𝑖3 ) + (3 × 𝑖2 ) = 0
∴ 𝑖1 = 3𝑖2 … … … … (1)
Again, applying Kirchhoff’s law in loop (2),
+ 2 (𝑖1 − 𝑖3 ) − 4 (𝑖2 + 𝑖3 ) = 0
⇒ 2𝑖1 − 4𝑖2 = 6 𝑖3 … … (2)
Applying Kirchhoff’s law of outer loop,
50 = 1 𝑖1 + 2 (𝑖1 − 𝑖3 )
⇒ 50 = 𝑖1 + 2𝑖2 − 2𝑖3
∴ 3 𝑖1 − 2𝑖3 = 50 … … . . (3)
From eqs. (1), (2) and (3), we get
𝑖3 = 2𝐴
79. Two wires of same dimensions but resistivities 𝜌1 and 𝜌2 are connected in series. The
Q20
equivalent resistivity of the combination is
A. √𝜌1 𝜌2
B. 𝜌1 + 𝜌2
𝜌 +𝜌
C. 1 2 2
Answer: C.
Solution:
Let the length of each wire will be l and A be the area of cross-section of each wire.
Let us consider a single rod of same dimension (i.e. length 2l, Area A as of composite
rod. If resistance of the single rod is the same as that of the composite rod will become
equivalent resistivity of the composite rod.
Therefore,
𝑅 = 𝑅1 + 𝑅2
𝜌𝑒𝑞 (2𝑙) 𝜌1 𝑙 𝜌2 𝑙
⇒ = +
𝐴 𝐴 𝐴
72
𝜌1 +𝜌2
⇒𝜌= 2
Q21
80. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery
is 0.4 Ω, what is the maximum current that can be drawn from the battery?
A. 10 A
B. 20 A
C. 30 A
D. 40 A
Answer: C.
Solution:
Given:
EMF, E = 12 V;
Internal resistance, 𝑟 = 0.4 Ω
The maximum current is drawn from the battery when the external resistance in the
circuit is zero. i.e., R = 0.
𝐸 12
Thus, 𝐼𝑚𝑎𝑥 = 𝑟 = 0.4 = 30 𝐴
81. In the figure shown, battery 1 has emf=6 V and internal resistance = 1 𝛺.
Q22
Battery 2 has emf=2 V and internal resistance = 3 𝛺. The wires have negligible
resistance. What is the potential difference across the terminals of battery 2?
A. 4V
B. 1.5 V
C. 5V
D. 0.5 V
Answer: C.
Solution:
Let I be the current in the circuit.
73
Applying KVL in loop:
6 − 3𝑖 − 2 − (1) = 0
∴ 𝑖 = 1𝐴
Now potential difference across terminals of battery 2 is
𝑉𝐴 − 𝑉𝐵
Potential difference across battery 2 can be calculated as:
𝑉𝐴 − 𝑖(3) − 2 = 𝑉𝐵
Putting i = 1
𝑉𝐴 − 𝑉𝐵 = 5 𝑉
Q23
82. A battery consists of n number of identical cells having internal resistance connected in
series. The terminals of the battery are short circuited and the current I is measured.
Which of the following graphs shows correct relationship between I and n ?
a.
b.
c.
74
d.
Answer: D.
Solution:
Equivalent emf = nE
Equivalent resistance = nr
𝑛𝐸 𝐸
∴ 𝑖 = 𝑛𝑟 = 𝑟
Current is independent of n.
Hence the correct graph is given in option D.
83. In a meter bridge experiment, the value of unknown resistance is 2𝛺. To get the balancing
Q24
point at 40 cm distance from the same end, then what will be the resistance in the
resistance box?
A. 3 Ω
B. 6 Ω
C. 8 Ω
D. 9 Ω
Answer: A.
Solution:
75
From above diagram, given data is 𝑆 = 2 Ω; 100 − 𝑙 = 40 𝑐𝑚
Apply condition for balanced meter bridge.
𝑅 𝑆
=
𝑙 100−𝑙
𝑅 2
= 40
60
∴𝑅 =3Ω
Q25
84. In Oersted's experiment, deflection in magnetic needle
A. Decreases if current is increased in the wire
B. Reverses if current is reversed in the wire
C. Remains constant if current is increased
D. All above statement are true
Answer: B.
Solution:
In Oersted’s experiment
1. Deflection in the magnetic needle decreases when the current through the wire is
decreased and vice versa.
2. If the direction of current is reversed in the wire, the deflection in magnetic needle
also reverses.
Based on these observations, Oersted concluded that moving charges or current produce
a magnetic field in surrounding space.
So, option B. is correct.
Q26
85. Which of the following statements is true in case of a solenoid?
A. A solenoid is a type of permanent magnet
B. It generates a controlled magnetic field through a coil wound into a tightly packed
helix.
C. A solenoid has a very weak magnetic field.
D. Multiple turns of more than two metallic wires are used to make a solenoid.
Answer: B.
Solution:
76
A solenoid is an electromagnet i.e. a temporary magnet. The turns are normally of
copper wire wound closely in the form of a helix or shape of a spring. The solenoid has
a soft iron core as a base which increases its strength.
Q27
86. A wire of fixed length is bent to form a coil of one turn. It is again bent to form a coil of
three turns. If in both cases same amount of current is passed, then the ratio of the
intensities of magnetic field produced at the centre of the coil of one turn and three turns
respectively will be,
A. 2 : 1
B. 9 : 1
C. 1 : 9
D. 1 : 3
Answer: C.
Solution:
Magnetic field at the center of a coil having n and carrying a current I is given by.
𝑛𝜇 𝑖
𝐵 = 2𝑟0
𝜇 𝑖
For, 𝑛 = 1, 𝐵 = 2𝑟0
If the same wire is again turned three times, then the radius of the new coil be f’
2 𝜋𝑟 = 3 (2 𝜋𝑟 ′ )
𝑟
Or 𝑟 ′ = 3
Now, n =3, Hence the magnetic field at center of the new coil will be
𝑛(𝜇 𝑖) 3(𝜇 𝑖) 9𝜇 𝑖
𝐵 ′ = 2𝑟′0 = 𝑟0 = 2𝑟0
2( )
3
′
⇒ 𝐵 = 9𝐵
𝐵 1
′ =
𝐵 9
Q28
87. What can we say about the solenoid more accurately?
A. A solenoid is a combination of multiple circular loops.
B. A solenoid is a combination of multiple rectangular shape loops.
C. The circular loops at the ends represent poles of the bar magnet.
D. Both a and c.
Answer: D.
Solution:
77
From the above two figures, it is clear that the solenoid is formed with close circular
loops and ends circular loops represents N – S pole of a bar magnet depending upon the
direction of current.
Q29
88. A solenoid has 1000 turns and 20 cm long. Find the magnetic induction produced at the
centre of the solenoid by the current of 2 A
A. 1.25 × 10−3 𝑇
B. 3.75 × 10−2 𝑇
C. 2.50 × 10−2 𝑇
D. 1.25 × 10−2 𝑇
Answer: D.
Solution:
Given:
N = 1000; l = 20 cm = 0.2 m; I = 2 A
Magnetic field at the center of long solenoid is given by
𝐵 = 𝜇0 𝑛𝑖
𝑁
𝐵 = 𝜇0 ( 𝑙 ) 𝑖
1000
𝐵 = 4𝜋 × 10−7 × 0.2 × 2
𝐵 = 4 × 3.14 × 10−2 = 1.25 × 10−2 𝑇
Q30
89. A proton of energy 8 eV is moving in a circular path in a uniform magnetic field. The
energy of an 𝛼 - particle moving in the same magnetic field and along the same path will
be
A. 2eV
B. 4 eV
C. 6 eV
D. 8 eV
Answer: D.
Solution:
We know that,
𝑞𝑎 = 2𝑞𝑝 , 𝑚𝑎 = 4 𝑚𝑝
1 𝑃2
𝐾𝐸 = 2 𝑚𝑣 2 = 2𝑚 … … . . (1)
Force due to magnetic field.
𝑚𝑣 2
𝐹𝑚 = 𝑞𝐵𝑣 = 𝑟
𝑚𝑣 𝑃
𝐵𝑞 = 𝑟 = 𝑟
𝑃 = 𝐵𝑞𝑟 … … … . (2)
Using equation (1) and (2)
𝐵2 𝑞 2 𝑟 2
𝐾𝐸 = 2𝑚
78
𝑞2
𝐾𝐸 𝛼 𝑚 [∵ 𝐵 and r are constants]
2
𝐾𝐸𝑎 𝑞 𝑚𝑝
= (𝑞𝑎 ) × (𝑚 )
𝐾𝐸𝑃 𝑝 𝑎
2 2 1 4
= (1) × (4) = 4
𝐾𝐸𝑎 = 𝐾𝐸𝑝 = 8𝑒𝑉
Q31
90. A magnetic needle is kept in a non-uniform magnetic field. It experiences
A. A force and a torque
B. A force but not a torque
C. A torque but not a force
D. Neither a torque nor a force
Answer: A.
Solution:
For a non-uniform magnetic field, the force as well as torque acting on the needle is
non-zero.
𝐹𝑛𝑒𝑡 ≠ 0
𝜏𝑛𝑒𝑡 ≠ 0
The needle will experience a torque that will make it align with the region of magnetic
field of higher strength in the total non-uniform magnetic field region. And the direction
of net force on the needle will be in the direction of stronger magnetic field.
Q32
91. What is the consequence of placing a core of a material made up of something other than
a soft iron?
A. The strength of a net magnetic field decreases due to a soft iron core.
B. The soft iron core is used due to its high value of permeability.
C. Overall strength of an electromagnet increases due to the soft iron core.
D. Both b and c are correct
Answer: D.
Solution:
Soft iron core has greater permeability than any other material like hard iron. It means
it increases magnetic field lines pass through it easily. This amplifies the net magnetic
strength of an electromagnetic.
Hence, option D. is the correct answer.
Q33
92. If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic field
of strength B, the work done in rotating the magnet through an angle 𝜃 is
A. 𝑀𝐵(1 − sin 𝜃)
B. 𝑚𝐵 sin 𝜃
C. 𝑀𝐵 cos 𝜃
D. 𝑀𝐵(1 − cos 𝜃)
79
Answer: D.
Solution:
We know that,
𝑊𝑓𝑖𝑒𝑙𝑑 = −Δ𝑈
𝑊𝑓𝑖𝑒𝑙𝑑 = 𝑈𝑖 − 𝑈𝑓
U = Magnetic potential energy of dipole in external field, which is given by,
𝑈 = −𝑀 ⃗⃗⃗. 𝐵
⃗⃗
Initially the magnet is aligned to field, so that angle 𝜃 between
𝑀⃗⃗⃗ and 𝐵 ⃗⃗ = 00
𝑈𝑖 = 𝑀𝐵 cos 00 = −𝑀𝐵
Finally, 𝜃𝑓 = 𝜃 ⇒ 𝑈𝑓 = −𝑀𝐵 cos 𝜃
So 𝑊𝑓𝑖𝑒𝑙𝑑 = −𝑀𝐵 − (−𝑀𝐵 cos 𝜃)
= 𝑀𝐵 (cos 𝜃 − 1)
Since, there is no change in kinetic energy,
𝑊𝑒𝑥𝑡 = −𝑊𝑓𝑖𝑒𝑙𝑑 = 𝑀𝐵(1 − cos 𝜃)
Q34
93. A long magnetic needle of length 2L, magnetic moment M and pole strength m units is
broken into two pieces at the middle point. The magnetic moment and pole strength of
each piece will be
𝑀 𝑚
A. 2 , 2
𝑚
B. 𝑀,
2
𝑀
C. 2 , 𝑚
D. 𝑀, 𝑚
Answer: C.
Solution:
Initially,
M = mL
When magnet is broken, new pair of dipoles with same pole strength m but half the
magnetic length are generated so now,
𝐿 𝑀
M’ for each part = 𝑚 2 = ( 2 )
80
So, the magnetic moment becomes half but the pole strength remains same.
94. A domain in a magnet is in the form of a cube of side length 1 𝜇𝑚. If it

Q35
contains 8 × 1010 atoms and each atomic dipole has a dipole moment of 9 × 10−24 𝐴 −
𝑚2 and all of these dipoles are aligned in the same direction, then the magnetisation of
the domain is
A. 7.2 × 105 𝐴𝑚−1
B. 7.2 × 103 𝐴𝑚−1
C. 3,6 × 105 𝐴𝑚−1
D. 1.8 × 1063 𝐴𝑚−1
Answer: A.
Solution:
We know that magnetization is given by,
𝑀𝑛𝑒𝑡
𝐼 = 𝑣𝑜𝑙𝑢𝑚𝑒
Here,
𝑀𝑛𝑒𝑡 = 8 × 1010 × 9 × 10−24 𝐴 − 𝑚2
𝑀𝑛𝑒𝑡 = 72 × 10−14 𝐴 − 𝑚2
Volume of the cube.
𝑉 = (10−6 )3 = 10−18 𝑚3
72×10−14
𝐼= = 7.2 × 105 𝐴𝑚−1
10−18
Q36
95. A magnet oscillating in a horizontal plane has a time period of 2 sec at a place where the
angle of dip is 300 and 3 sec at another place where the angle of dip is 600 . The ratio of
resultant magnetic fields at the two places is:
4 √3
A. 7
4
B. 9 √3
9
C. 4 √3
9
D.
√3
Answer: C.
Solution:
The time period of the magnet is,
𝐼 𝐼
𝑇 = 2 𝜋√𝑀𝐵 = 2 𝜋√𝑀𝐵𝑐𝑜𝑠 𝛿
𝐻
1
Here, 𝑇 𝛼
√𝐵𝐻
1
⇒𝑇∝
√𝐵 cos 𝛿
𝑇1 𝐵2 cos 𝛿2
∴ 𝑇 = √𝐵
2 1 cos 𝛿1
81
𝐵 𝑇 2 cos 𝛿
So that, 𝐵1 = 𝑇22 cos 𝛿2
2 1 1
𝐵1 (3)2 cos 600

= (2)2 cos 300
𝐵2
𝐵 9
∴ 𝐵1 = 4
2 √3
Q37
96. In electromagnetic induction, the induced e.m.f. in a coil is independent of
A. Change in the magnetic flux
B. Time interval
C. Resistance of the circuit
D. Number of turns in the coil
Answer: C.
Solution:
We know that the emf induced in a coil is given by,
𝑑𝜙
𝜖 = 𝑁 𝑑𝑡
And, 𝜙 = 𝐵𝐴 cos 𝜃
So, the induced emf depends upon the number of turns in the coil, magnitude and
direction of magnetic field, area of the coil, and angle between magnetic field and area
vector.
So, we can say that it is independent of the resistance of the coil,
Q38
97. The dimension of √𝐿𝐶 is
A. [L]
B. [M]
C. [T]
D. [I]
Answer: C.
Solution:
We know that, frequency of oscillation,
1
𝑓 = 2 𝜋√𝐿𝐶
1
Time period, ⇒ [𝑇0 ] = 𝑓 = 2 𝜋√𝐿𝐶
⇒ [𝑇0 ] = [√𝐿𝐶]
[∴ 2 𝜋 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡]
⇒ [√𝐿𝐶] = [𝑇]
Therefore, √𝐿𝐶 has dimension [T]
98. Magnetic flux 𝜙 (in Webers) linked with a closed circuit of resistance 10 Ω varies with
Q39
time t 9in seconds) as 𝜙 = 5 𝑡 2 − 4𝑡 + 1. The induced emf in the circuit at t = 0.2 sec is
82
A. 5 mA
B. 10 mA
C. 20 mA
D. 1 A
Answer: C.
Solution:
Given,
𝑅 = 10 Ω
𝜙 = 5𝑡 2 = 4𝑡 + 1
𝑡 = 0.2𝑠𝑒𝑐
We know that emf induced in a circuit is given by,
𝑑𝜙
𝜖𝑖𝑛𝑑 = | |
𝑑𝑡
𝑑
⇒ 𝜖𝑖𝑛𝑑 = |𝑑𝑡 (5𝑡 2 − 4𝑡 + 1)|
𝜖𝑖𝑛𝑑 = |10 𝑡 − 4|
𝜖 |10𝑡−4|
∴ 𝑖𝑖𝑛𝑑 = 𝑖𝑛𝑑 =
𝑅 100
|10×0.2−4|
(𝑖𝑖𝑛𝑑 )𝑡=0.2 =
100
∴ 𝑖𝑖𝑛𝑑 = 20 𝑚𝐴
Q40
99. Two solenoids A and B spaced close to each other and sharing the same cylindrical axis
have 400 and 700 turns respectively. A current of 3.5A in coil A produce an arrange flux
of 300 𝜇𝑇 − 𝑚2 through each turn of A and a flux of 90 𝜇𝑇 − 𝑚2 through each turn B.
The mutual inductance of the two solenoids is
A. 1.8 × 10−2 𝐻
B. 2.8 × 10−2 𝐻
C. 3.8 × 10−2 𝐻
D. 4.8 × 10−2 𝐻
Answer: A.
Solution:
Given,
Number of turns in solenoid 𝐴, 𝑁𝐴 = 400
Number of turns in solenoid 𝐴, 𝑁𝐵 = 700
Current in coil 𝐴, 𝑖𝐴 = 3.5 𝐴
Magnetic flex through each turn of coil 𝐴, 𝜙𝐴 = 300 𝜇𝑡 − 𝑚−2
We know mutual inductance,
𝑁 𝜙
𝑀 = 𝑀𝐴𝐵 = 𝑀𝐵𝐴 = 𝐵𝑖 𝐵
𝐴
700×90×10−6
= 3.5
= 1.8 × 10−2 𝐻
Q41 Power dissipated in an LCR series circuit connected to an a.c. source of emf ϵ is
83
ϵ2 𝑅
A. 2
√𝑅 2 +(𝐿 𝜔− 1 )
𝐶𝜔
ϵ2 𝑅
B. 1 2
[𝑅 2 +( 𝐿 𝜔− ) ]
𝐶𝜔
1 2
√[𝑅 2 +( 𝐿 𝜔− ) ]
𝐶𝜔
C. ϵ2 𝑅
1 2
ϵ2 [𝑅 2 +(𝐿 𝜔− ) ]
𝐶𝜔
D. 𝑅
Answer: B.
Solution:
Power dissipation for A – C circuit is given by < 𝑃 > = 𝐸𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 . cos 𝜙 where
𝑅
cos 𝜙 = Power factor = 𝑍
𝑅
So, < 𝑝 > = 𝑉𝑟𝑚𝑠 . 𝑖𝑟𝑚𝑠 . 𝑍
𝑉𝑟𝑚𝑠
Now, 𝐼𝑟𝑚𝑠 = 𝑍
2
𝑉𝑟𝑚𝑠 .𝑅
<𝑝>= and
𝑍2
1 2
𝑍 2 = 𝑅 2 + (𝑤𝐿 − 𝑤𝑐)
ϵ2 𝑅
So, < 𝑝 > = 1 2
𝑅 2 +(𝑤 𝐿− )
𝑤𝑐
Q42 In ac circuit when ac ammeter is connected it reads i current if a student uses dc ammeter
in place of ac ammeter the reading in the dc ammeter will be:
𝑖
A.
√2
B. √2𝑖
C. 0.637 𝑖
D. Zero
Answer: D.
Solution:
DC ammeter doesn’t will have np deflection for AC current so, 𝑖𝐴𝐶 read by DC ammeter
= Zero.
Q43 An alternating current is given as 𝑖 = 𝑖1 cos 𝜔𝑡 − 𝑖2 sin 𝜔𝑡. The rms current is given by
𝑖 +𝑖
A. 1 2
√2
𝑖1 + 𝑖2 |
B.
√2
𝑖 +𝑖 2 2
C. √ 1 2 2
84
𝑖 2 +𝑖22
D. √ 1
√2
Answer: C.
Solution:
𝑖 = 𝑖1 cos 𝜔𝑡 − 𝑖2 sin 𝜔𝑡
In RHS multiply & divide by √𝑖12 + 𝑖22
𝑖1 𝑡2
𝑖 = √𝑖12 + 𝑖22 ( cos 𝜔𝑡 + sin 𝜔𝑡 )
√𝑖12 +𝑖22 √𝑖12 + 𝑖22
Now consider a right-angle triangle with sides 𝑖1 & 𝑖2
𝑖1
sin 𝜃 =
√𝑖12 +𝑖22
𝑖2
cos 𝜃 =
√𝑖12 +𝑖22
So, 𝑖 = √𝑖12 + 𝑖22 {sin 𝜃 cos 𝜔𝑡 + cos 𝜃 sin 𝜔𝑡}

So, 𝑖 = 𝑖0 sin(𝜔𝑡 + 0) where 𝑖0 = √𝑖12 + 𝑖22
Comparing the equation from general equation of current 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙)
We get peak current 𝑖0 = √𝑖12 + 𝑖12
𝑖0 𝑖 2 +𝑖22
So 𝑖𝑟𝑚𝑠 = = √ 12
√2
Q44 An alternating current having peak value 14 A is used to heat a metal wire. To produce
the same heating effect, a constant current i can be used where i is
A. 14 A
B. About 20 A
C. 7A
D. About 10 A
Answer: D.
Solution:
For a sinusoidal AC,
𝑖𝑝𝑒𝑎𝑘 14
𝑖𝑟𝑚𝑠 = = ≈ 10 Amp
√2 √2
Since for heat calculation 𝑖𝑟𝑚𝑠 is equivalent d.c current
85
So 𝑖𝑟𝑒𝑓 = 10 Amp
Q45 Given below are two statements labelled Assertion (A) and Reason (R).
Assertion: If three capacitors of capacitance 𝐶1 < 𝐶2 < 𝐶3 are connected in parallel then
their equivalent capacitance then 𝐶𝑝 > 𝐶𝑠
1 1 1 1
Reason: 𝐶 = 𝐶 + 𝐶 + 𝐶
𝑝 1 2 3
A. Both assertion and reason are true and the reason is the correct explanation of the
assertion.
B. Both assertion and reason are true but reason is not the correct explanation of the
assertion.
Answer: C.
Solution:
1
For the parallel combination of three capacitors, 𝐶𝑃 = 𝐶1 + 𝐶2 + 𝐶3 and for series 𝐶 =
𝑆
1 1 1
+𝐶 +𝐶
𝐶1 2 3
It is clear that, 𝐶𝑃 > 𝐶𝑆 . This means that assertion is true but Reason is incorrect.
Assertion: A potentiometer of longer length is used for accurate measurement.
Reason: The potential gradient for a potentiometer of longer length, with a given source
of e.m.f., becomes small.
assertion.
assertion.
D. Assertion and reason both false.
Answer: B.
Solution:
The accuracy of a potentiometer is larger if the potential gradient across its length is of
smaller magnitude. If we use a potentiometer wire of longer length the potential gradient
decreases.
Here, assertion and reason both are correct but the reason does not justify the assertion.
86
Assertion : If a proton and an -particle enter a uniform magnetic field
perpendicularly, with the same speed, then the time period of revolution
of the -particle is double than that of proton.
Reason : In a magnetic field, the time period of revolution of a charged particle is
directly proportional to mass.
assertion.
assertion.
Answer: C.
Solution:
2 𝜋𝑚
The time period of a charged particle in external magnetic field is given by, 𝑇 = 𝐵𝑞 .
The time period of a charged particle is directly proportional to its mass and inversely
proportional to the charge. Hence, the assertion is correct i.e., radius of 𝛼 – particle will
be double that of proton. But the reason is wrong.
Q48 Given below are two statement labelled Assertion (A) and Reason (R).
Assertion: Electromagnets are made of soft Iron.
Reason: coercivity of soft iron is small
assertion.
assertion
Answer: A.
Solution:
Electromagnets are expected to get demagnetized after switching off the current through
their coil. So, the core of the electromagnet should have a tendency to lose its magnetic
moment, when the external field is removed. Such materials have smaller coercivity i.e.
they get demagnetized easily. Also, soft iron has smaller coercivity.
Assertion: When ac circuit contain resistor only, its power utilization is minimum.
Reason: Power utilized in an AC circuit is independent of phase angle.
87
assertion.
assertion
C. Assertion is true but reason is false
Answer: D.
Solution:
Both the statements are false. When ac circuit contain resistor only, its power utilization
is maximum. And, power utilization in an AC circuit depends upon phase angle. So, both
the statements are false.
Q50 A transformer is used to light a 140 W, 24 V lamp from 240 AC mains. The current in
mains cable is 0.7 A, find the efficiency of transformer.
A. 50%
B. 76.2%
C. 83.33%
D. 91.25%
Answer: C.
Solution:
Voltage primary and secondary coil of transfer are given as
𝑒1 = 240 𝑉 𝑎𝑛𝑑𝑒2 = 24 𝑉
Resistance of lamps given as
𝑉 (24)2 144
𝑅 = 𝑃2 = 140 = 35 Ω
Current in primary coil is given as
𝑒 24 35
𝑖2 = 𝑅2 = 144 × 35 = 6 𝐴
Power at secondary coil is given as
𝑃1 = 𝑒1 𝑖1 = 240 × 0.7 = 168 𝑊
Power at secondary coil is given as
35
𝑃2 = 𝑒2 𝑖2 = 24 × 6 = 140 𝑊
Transformer efficiency is given as
𝑒 𝑖
𝑛 = 𝑒2𝑖2
1 1
148
Times 100 = 168 × 100
⇒ 𝑛 = 83.33
Q51 In a series CR circuit shown in figure, the applied voltage is 10 V and the voltage across
capacitor is found to be 8 V. Then the voltage across R, and the phase difference between
current and the applied voltage will respectively be
88
4
A. 6𝑉, tan−1 (3)
3
B. 3𝑉, tan−1 (4)
5
C. 6𝑉, tan−1 (3)
D. None
Answer: A.
Solution:
For rms/ peak values of voltage from phasor Analysis we can wirte
2
⇒ 𝑉𝑛𝑒𝑡 = 𝑉𝑅2 + 𝑉𝐶2
This relation is valid forms and peak values both.
Rms values are given.
So, 𝑉𝑛𝑒𝑡 = 10 𝑣𝑜𝑙𝑡
⇒ 𝑉𝑅2 = 102 − 82 = 62 [𝑉𝑅 = 6 𝑉𝑜𝑙𝑡]
𝑉𝐶 = 8 𝑉𝑜𝑙𝑡
𝑉 8 4
|tan 𝜃| = | 𝐶 | = =
𝑉 𝑅 6 3
4
Phase different between current and 𝑉𝑛𝑒𝑡 = tan−1 (3)
89
Case study:
Every petrol engine needs separate arrangement for the ignition of the air-fuel mixture. This
arrangement is called its ignition system. The ignition system starts with the engine coil. It is a
stationary coil mounted around the engine shaft while a magnet is mounted on the engine shaft,
such that the magnet is revolving around the coil as the engine is running. The rotation of the
magnet induces an EMF (of relatively smaller magnitude) in the engine coil. This induced EMF
is applied across the ignition coil. The output of ignition coil is given to the spark plug, which is
placed inside the combustion chamber of the engine. The spark plug has a small gap between its
electrodes. When a high voltage is applied across these electrodes, there is an electric spark
between them, due to dielectric breakdown of air between them. At the appropriate time instant,
when the air fuel mixture is to be ignited, the switch 'S' is closed so that there is an electric spark
across the gap, in the spark plug.
Q52 The engine oil works on the principle of

A. Ampere’s law
B. Faraday’s law
C. Gauss’ law
D. Both (a) and B.
Answer: B.
Solution:
Due to rotation of magnet around the coil, the magnetic flux is changing and we get
induced emf in the engine coil. Hence, the engine coil works on the principle of
Faraday’s law of electromagnetic induction
Q53 The ignition coil is necessarily:

A. A solenoid with soft iron core
B. An R-L-C series circuit
C. Step down transformer
D. Step up transformer
Answer: D.
90
Solution:
The spark plug (connected to the output side of the ignition coil) requires a high voltage
(more than 10,000 volts) to function but the engine coil (connected to the input side of
the ignition coil) produces a smaller voltage. Hence, ignition coil is a necessarily a step-
up transformer
Q54 As the speed of rotation of the engine increases

A. The electric spark will get brighter and stronger
B. The electric spark will get dimmer and weaker
C. The electric spark will be unaffected
D. There will be no spark above a certain speed
Answer: A.
Solution:
As the speed of rotation of the engine increases, the magnitude of induced EMF in the
engine coil also increases. This ultimately, increases the voltage across the electrodes
of the spark plug and, therefore the electric Spark produced becomes brighter and
stronger.
Q55 Thick and multi-stranded copper wire will be needed

A. For the engine coil
B. Between engine coil and ignition coil
C. Between ignition coil and spark plug
D. For all connections
Answer: C.
Solution:
Thick and multi-stranded copper wire will be needed between the ignition coil and the
spark plug. Because the ignition coil produces high voltage output, which is applied
across the electrodes of the spark plug.
91
CHEMISTRY
Time: 90 Minutes Max Marks: 35
1. The Question Paper contains three sections.
Section – A
This section consists of 25multiple choice questions with overall choice to attempt any 20
questions. In case more than desirable number of questions are attempted, ONLY first 20
will be considered for evaluation.
Q1 A brown ring is formed in the ring test for NO3- ion. It is due to the formation (0.77)
of:
.
A. [Fe (H2O)5(NO)]2+
B. FeSO4.NO2
C. [Fe(H2O)4(NO)2]2+
D. FeSO4.HNO3
Q2 A compound is formed by two elements M and N. The element N forms ccp (0.77)
and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the
compound?
A. MN2
B. M2N3
C. M3N2
D. M2N2
(0.77)
Q3 The plots of 1/XA vs 1/YA (XA mole fraction of A in liquid and YA in vapour) is
linear whose slope and intercept respectively
A. AP0B /P0A, (P0B- P0A)/P0B

B. (P0A- P0B), (P0A- P0B)/P0B
C. P0A /P0B, (P0B- P0A)/P0B
D. (P0B- P0A), (P0A- P0B)/P0B
92
Q4 The flame colours of metal ions are due to (0.77)
A. Frenkel defect
B. Schottky defect
C. Metal deficiency defect
D. Metal excess defect
Q5 The correct IUPAC name of the following compound is: (0.77)
A. 1-bromo-1-ethyl-2-fluoro-2-iodo-1-nitroethane
B. 3-bromo-4-fluoro-4-iodo-3-nitrobutane
C. 2-bromo-1-fluoro-1-iodo-2-nitrobutane
D. 1-fluoro-1-iodo-2-bromo-2-ethyl-2-nitroethane
Q6 1-propanol and 2-propanol can be best distinguished by (0.77)
A. oxidation with alkaline KMnO4 followed by reaction with Fehling’s

solution.
B. oxidation with acidic dichromate followed by reaction with Fehling’s
solution
C. oxidation by heating copper followed by reaction with Fehling’s
solution.
D. oxidation with concentrated H2SO4 followed by reaction with
Fehling’s solution.
Q7 What is the correct IUPAC name of : (0.77)
A. Dioxybutane
B. 2,6−dioxybutane
C. 1,2−dimethoxyethane
D. Dimethylethylether
93
Q8 Which of the following statements are true? (0.77)
A. Only one type of interactions between particles of noble gases are
due to weak dispersion forces
B. Ionisation enthalpy of molecular oxygen is very close to that of
xenon
C. Hydrolysis of XeF6 is a redox reaction
D. Xenon fluorides are not reactive.
Q9 Glucose molecule reacts with the ‘X’ number of molecules of (0.77)

phenylhydrazine to yield osazone. The value of ‘X’ is
A. Four
B. One
C. Two
D. Three
Q10 Which stoichiometric defect does not change the density of the crystal? (0.77)
A. Frenkel defect
B. Schottky defect
C. Interstitial defect
D. F-centres
Q11 The reaction conditions leading to the best yields of C2H5Cl are (0.77)
A. C2H5Cl (Excess) + Cl2 →

B. C2H5Cl + Cl2 + dark, room temperature →
C. C2H5Cl + Cl2 (Excess) →
D. C2H5Cl + Cl2 →
Q12 Considering the formation, breaking and strength of hydrogen bonds, (0.77)
predict which of the following mixture will show a positive deviation from
Raoult's law?
A. Methanol and acetone
B. Chloroform and acetone
C. Nitric acid and water
D. Phenol and aniline
Q13 Which of the following has least polar C−X bond? (0.77)
A. CH3−CH=CH −Cl
B. HO−CH=CH −Cl
C. O2N−CH=CH −Cl
D. HS−CH=CH −Cl
94
Q14 On heating with concentrated NaOH solution in an inert atmosphere of CO, (0.77)
white phosphorus gives a gas. Which of the following statements is
incorrect about the gas?
A. It is highly poisonous and has smell like rotten fish

B. Its solution in water decomposes in the presence of light
C. It is more basic than NH3
D. It is less basic than NH3
Q15 What is the correct name of the following compound? (0.77)
A. D- (+)-fructose
B. D- (-)-fructose
C. L- (+)-fructose
D. L- (-)-fructose
Q16 Saccharic acid is obtained by oxidation of glucose. Sum of the number of C, (0.77)
H and O present in saccharic acid is:
A. 24
B. 18
C. 16
D. 32
Q17 In Finkelstein Reaction which reactants are used (0.77)
A. NaI + C2H5OH
B. NaF + acetone
C. NaBr + CH3OH
D. NaI + C2H5Br
Q18 Which of the following elements can be involved in pπ–dπ bonding? (0.77)
A. Carbon
B. Nitrogen
C. Phosphorus
D. Boron
95
Q19 Which of the following aqueous solutions should have the highest boiling (0.77)
point?
A. 1.0 M NaOH
B. 1.0 M 𝑁𝑎2 𝑆𝑂4
C. 1.0 M 𝑁𝐻4 𝑁𝑂3
D. 1.0 M 𝐾𝑁𝑂3
Q20 Hydrogen fluoride is liquid unlike other hydrogen halides because (0.77)
A. fluorine atom is smaller in size

B. HF is the weakest acid
C. fluorine is highly reactive
D. HF molecules form intermolecular H-bonds
Q21 Which of the following alcohol has highest solubility in water? (0.77)
A. Tertiary butyl alcohol

B. Secondary butyl alcohol
C. Ethylene glycol
D. Glycerol
Q22 Main source of helium is (0.77)
A. Air
B. Radium
C. Monazite
D. Water
Q23 Which of the following statements are correct? (0.77)
A. Among halogens, radius ratio between iodine and fluorine is maximum.

B. Leaving F - F bond, all halogens have weaker X-X bond than X-X’ bond
in inter-halogens.
C. Among inter-halogen compounds maximum. number of atoms ate
present in iodine fluoride.
D. Inter-halogen compounds are more reactive than halogen compounds.
Q24 For a dilute solution, Raoult's law states that (0.77)
A. The lowering of vapour pressure is equal to the mole fraction of

solute
B. The relative lowering of vapour pressure is equal to the mole fraction
of solute
C. The relative lowering of vapour pressure is proportional to the
amount of solute in solution
D. The vapour pressure of the solution is equal to the mole fraction of
the solute
96
Q25 (0.77)
Which is the best reaction for preparation of t-butyl ethyl ether?
A.
B.
C.
D.
97
Section - B
questions. In case more than desirable number of questions are attempted, ONLY first 20
will be considered for evaluation.
Q26 Value of Henry's constant 𝐾𝐻 is _______. (0.77)
A. Increases with increase in temperature.

B. Decreases with increase in temperature
C. Remains constant
D. First increases then decreases.
Q27 Which one of the following compounds is most acidic? (0.77)
A.
B.
C.
D.
Q28 When 𝑋𝑒𝐹6 is partially hydrolysed, it yields: (0.77)
A. 𝑋𝑒𝑆𝑂3
B. 𝑋𝑒𝑂3
C. 𝑋𝑒𝑂𝐹4
D. 𝑋𝑒𝐹2
98
Q29 If the basic formula of an α-amino acid is R-CH(NH2)-COOH, where R is the (0.77)
side chain, what is the primary point of distinction between any two proteins?
A. Number of amino groups

B. Number of carboxyl groups
C. The side chain R
D. Relative positions of amino, carboxyl groups and R
Q30 Which of the following bases is not present in DNA? (0.77)
A. Adenine
B. Guanine
C. Thymine
D. Uracil
Q31 Which of the following statements are correct? (0.77)
A. S – S bond is present in𝐻2 𝑆2 𝑂6

B. In peroxosulphuric acid (𝐻2 𝑆𝑂5 )sulphur is in +6 oxidation state
C. Iron powder along with 𝐴𝑙2 𝑂3 and 𝐾2 𝑂is used as a catalyst in the
preparation of 𝑁𝐻3by Haber’s process
D. Change in enthalpy is positive for the preparation of 𝑆𝑂3 by catalytic
oxidation of 𝑆𝑂2
Q32 Silver crystallises in f.c.c. Lattice. It’s edge length of the unit cells is 4.07× 10−8 (0.77)
cm and density is 10.5 g𝑐𝑚−3. Calculate the atomic mass of silver.
A. 144 g/mol
B. 125 g/mol
C. 106.6 g/mol
D. 213 g/mol
Q33 The solubility of haloalkanes/haloarenes is highest in: (0.77)
A. Water
B. Non-polar solvents
C. Polar Solvents
D. Acetic acid
Q34 What are the products obtained when ammonia is reacted with excess (0.77)
chlorine?
A. 𝑁2 and 𝑁𝐶𝑙3
B. 𝑁2 and HCl
C. 𝑁2 and 𝑁𝐻4 𝐶𝑙
D. 𝑁𝐶𝑙3 and HCl
99
Q35 A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ (0.77)
takes place when a small amount of ‘A’ is added to the solution. The solution is
_________.
A. Saturated
B. Supersaturated
C. Unsaturated
D. Concentrated
Q36 (0.77)
Which of the following can not be the product of this reaction?
A. 1
B. 2
C. 3
D. 4
Q37 If 𝐴𝑙 3+ ions replace 𝑁𝑎+ ions at the edge centres of NaCl lattice, then the (0.77)
number of vacancies in one mole of NaCl will be
A. 3.01 × 1023
B. 6.02 × 1023
C. 9.03 × 1023
D. 12.04 × 1023
Q38 Ratio of the total volume of bcc to simple cubic structure is (0.77)
A. 3√3: 8
B. 8: 3√3
C. 24√3: 1
D. 1: 24√3
100
Q39 Which of the following are peroxoacids of sulphur? (0.77)
A. 𝐻2 𝑆𝑂5 and 𝐻2 𝑆2 𝑂8
B. 𝐻2 𝑆𝑂5 and 𝐻2 𝑆2 𝑂7
C. 𝐻2 𝑆2 𝑂7and 𝐻2 𝑆2 𝑂8
D. 𝐻2 𝑆2 𝑂6and 𝐻2 𝑆2 𝑂7
Q40 Fibrous and globular shapes come under which structure of proteins? (0.77)
A. Primary
B. Secondary
C. Tertiary
D. Quaternary
Q41 The following tripeptide can be represented as: (0.77)
A. Tyr - Val - Thr

B. Phe - Ala - Ser
C. Phe - Leu - Cys
D. Lys - Ala - Ser
Q42 Which one of the following oxides of nitrogen is blue solid? (0.77)
A. NO
B. 𝑁2 𝑂3
C. 𝑁2 𝑂
D. 𝑁2 𝑂5
101
Q43 Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene (0.77)
Reason : Alkyl halides are less reactive than acyl halides.
A. If both assertion and reason are correct and the reason explains the
Assertion.
B. If both Assertion and reason are correct but reason is not the correct
explanation of the Assertion.
C. If the assertion is correct but the reason is incorrect.
D. If both the Assertion and reason are incorrect.
Q44 Assertion: The bond angle in alcohol is slightly less than the tetrahedral angle. (0.77)
3
Reason: In alcohols, the oxygen of the -OH group is attached to sp hybridized
carbon
Assertion.
explanation of the Assertion
C. If the assertion is correct but the reason is incorrect.
Q45 Assertion(A): F2 is a strong oxidizing agent. (0.77)

Reason (R): Electron gain enthalpy of fluorine is less negative
A. if both (A) and (R) are correct and the (R) explains the (A).
B. if both (A) and (R) are correct but (R) is not the correct explanation of
(A).
C. (A) is correct but the (R) is incorrect.
D. (A) is false and (R) is True
102
Q46 Assertion: SN2 reaction of an optically active aryl halide with an aqueous (0.77)
solution of KOH always gives an alcohol with opposite sign of rotation.
Reason: SN2 reactions always proceed with inversion configuration
Assertion.
C. If the assertion is correct but the reason is incorrect
Q47 Assertion(A): 0.1 M Solution of KCl has greater osmotic pressure than 0.1 M (0.77)
solution
of glucose at the same temperature.
Reason(R): in solution KCl dissociates to produce more number of particles
B. (A) and (R) both are correct but (R) is not the correct explanation of (A)
C. A is correct but the R is incorrect.
D. A is incorrect but the R is correct.
Q48 Assertion (A): Vapour pressure increases with increase in temperature. (0.77)
Reason (R): With increase in temperature, more molecules of the liquid can go
into the vapour phase.
103
Q49 Assertion (A): N2 is less reactive than P4 (0.77)
Reason (R): nitrogen gas has more electron gain enthalpy than phosphorus
A. if both (A) and (R) are correct and the (R) explains the (A)
Section – C
This section consists of 6multiple choice questions with an overall choice to attempt any5. In
case more than desirable number of questions are attempted, ONLY first 5 will be
considered for evaluation.
Q50 Match the orbital overlap figures shown in List-I with the description given in List- (0.77)
II and select the correct combination using the code provided in the options.
A. A-R; B-P; C-S; D-Q

B. A-R; B-P; C-Q; D-S
C. A-P; B-R; C-Q; D-S
D. A-P; B-R; C-S; D-Q
104
Q51 Which of the following analogies is correct: (0.77)
A. F2 > Cl2 > Br2 > I2: Bond dissociation energy

B. F2 > Cl2 > Br2 > I2: Electronegativity
C. F2 > Cl2 > Br2 > I2: Oxidising power
D. F2 < Cl2 < Br2 > I2: Electron gain enthalpy
Q52 Which of the following analogies is correct: (0.77)

A. XeF4
B. XeOF2
C. XeO2F2
D. XeO4
Q53 I. Read the passage given below and answer the following question: All real (0.77)
structures are three dimensional structures. They can be obtained by stacking
two dimensional layers one above the other while placing the second square
closed packed layer above the first we follow the same rule that was followed
when one row was placed adjacent to the other. The second layer is placed over
the first layer such that the spheres of the upper layer are exactly above there of
the first layer. In this arrangement spheres of both the layers are perfectly
aligned horizontally as well as vertically. A metallic element crystallises into a
lattice having a ABC ABC pattern and packing and packing of spheres leaves
out voids in the lattice.
What types of structures are formed by this arrangement?

A. Ccp
B. hcp
C. ccp/hcp
Q54 Name the non-stoichiometric point defect responsible for colour in alkali metal (0.77)
halides.
A. Frenkel defect
B. Interstitial defect
C. Schottky defect
D. F-centres
Q55 What is the total volume of the atom in a fcc centred cubic unit cell of metal ? (0.77)
A. 16/3 𝝅r3
B. 𝝅r3
C. 24/3 𝝅r3
D. 12/3 𝝅r3
105
ANSWER KEYS
Chemistry
Q1 A. Q29 C.
Q2 B. Q30 D.
Q3 C. Q31 B.
Q4 D. Q32 C.
Q5 C. Q33 B.
Q6 C. Q34 D.
Q7 C. Q35 B.
Q8 A. Q36 D.
Q9 D. Q37 A.
Q10 A. Q38 B.
Q11 A. Q39 A.
Q12 A. Q40 C.
Q13 C. Q41 B.
Q14 C. Q42 B.
Q15 C. Q43 C.
Q16 A. Q44 A.
Q17 D. Q45 B.
Q18 C. Q46 D.
Q19 B. Q47 A.
Q20 D. Q48 B.
Q21 D. Q49 C.
Q22 C. Q50 B.
Q23 B. Q51 A.
Q24 A. Q52 A.
Q25 B. Q53 A.
Q26 A. Q54 D.
Q27 C. Q55 A.
Q28 C.
106
SOLUTIONS
Q1 A brown ring is formed in the ring test for NO3- ion. It is due to the formation of:
.
A. [Fe (H2O)5(NO)]2+
B. FeSO4.NO2
C. [Fe (H2O)4(NO)2]2+
D. FeSO4.HNO3
Answer: (A) [Fe (H2O)5(NO)]2+
Solution: When freshly prepared solution of ferrous sulphate (FeSO4) is added in a

solution containing NO3- ion, formation of brown-coloured complex will take place. This
is called as brown ring test of nitrate. Hence, two moles of ammonia will produce two
moles of NO.
NO3− + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O
[Fe (H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O

Brown ring
Q2 A compound is formed by two elements M and N. The element N forms ccp and atoms
of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
.
A. MN2
B. M2N3
C. M3N2
D. M2N2
Answer: (B) M2N3
Solution: Suppose the atoms N in ccp = a

∴ No. of tetrahedral voids = 3a
No. of atoms M = 2a/3: a = 2:3
Hence, compound as M2N3.
Q3 The plots of 1/XA vs 1/YA (XA mole fraction of A in liquid and YA in vapour) is linear
whose slope and intercept respectively
A. P0B /P0A, (P0B- P0A)/P0B

B. (P0A- P0B), (P0A- P0B)/P0B
C. P0A /P0B, (P0B- P0A)/P0B
D. (P0B- P0A), (P0A- P0B)/P0B
Answer: (C) P0A /P0B, (P0B- P0A)/P0B
Solution:
PA = P0Ax XA = Total pressure x YA
PA = P0Bx XB = Total pressure x YB
107
(P0B XB)÷(P0A XA) = YB / YA
P0B (1-XA)÷(P0A XA) = (1-YA)÷ YA
1/XA− 1 = P0A / P0B(1/YA− 1) = (P0A / P0B) (1/YA)−(P0A / P0B)
Or, 1/XA = P0A / P0B(1/YA) + (1- P0A / P0B) = P0A / P0B(1/YA) + (P0B- P0A)/P0B
This is the equation of a straight line.
Slope = P0A /P0B, Intercept = (P0B- P0A)/P0B
Q4 The flame colours of metal ions are due to
A. Frenkel defect
B. Schottky defect
C. Metal deficiency defect
D. Metal excess defect
Answer: (D ) Metal excess defect
Solution: The flame colors of metal ions are due to metal excess defects. These types
of crystals are generally coloured. This is due to the presence of free electrons. These
electrons get excited easily to higher energy levels by absorption of certain
wavelengths from the visible light and therefore, the compounds appear coloured.
Q5 The correct IUPAC name of the following compound is:
A. 1-bromo-1-ethyl-2-fluoro-2-iodo-1-nitroethane
B. 3-bromo-4-fluoro-4-iodo-3-nitrobutane
C. 2-bromo-1-fluoro-1-iodo-2-nitrobutane
D. 1-fluoro-1-iodo-2-bromo-2-ethyl-2-nitroethane
Answer: (C ) 2-bromo-1-fluoro-1-iodo-2-nitrobutane
Solution: NO2 does not have any priority order and is considered as alkyl substituents
while numbering.
108
Q6 1-propanol and 2-propanol can be best distinguished by
A. oxidation with alkaline KMnO4 followed by reaction with Fehling’s solution.

B. oxidation with acidic dichromate followed by reaction with Fehling’s solution
C. oxidation by heating copper followed by reaction with Fehling’s solution.
D. oxidation with concentrated H2SO4 followed by reaction with Fehling’s solution.
Answer: (C) oxidation by heating copper followed by reaction with Fehling’s solution.
Solution:
Cu
CH3 −CH2 −CH2 −OH → CH3 −CH2 −CHO
573K Gives red ppt. with
Fehling′s solution
Cu
CH3 −CHOH −CH3 → CH3 −CO −CH3
573K
(Does not give Fehling′s solution test)
Q7 What is the correct IUPAC name of
A. Dioxybutane
B. 2,6−dioxybutane
C. 1,2−dimethoxyethane
D. Dimethylethylether
Answer: (C ) 1,2−dimethoxyethane
Solution:
It is a diether with two methoxy groups at C−1 and C−2 of parent ethane.
-OR (alkoxy) does not have any priority order and is considered as alkyl substituents
while numbering.
The correct IUPAC name is 1,2−dimethoxyethane.
109
Q8 Which of the following statements are true?
A. Only one type of interactions between particles of noble gases are due to weak
dispersion forces
B. Ionisation enthalpy of molecular oxygen is very close to that of xenon
C. Hydrolysis of XeF6 is a redox reaction
D. Xenon fluorides are not reactive.
Answer: (A) Only one type of interactions between particles of noble gases are due to
weak dispersion forces
Solution: Weak dispersion forces are present between particles of noble gases.
Ionisation enthalpy of molecular oxygen is very close to that of the Xenon.
Q9 Glucose molecule reacts with ‘X’ number of molecules of phenylhydrazine to yield

osazone. The value of ‘X’ is
A. Four
B. One
C. Two
D. Three
Answer: (D) Three
Solution: One glucose molecule reacts with the 3 molecules of phenylhydrazine to

give osazone.
Q10 Which stoichiometric defect does not change the density of the crystal?
A. Frenkel defect
B. Schottky defect
C. Interstitial defect
D. F-centres
Answer: (A) Frenkel defect
Solution: In the Frenkel defect, one of the ions is missing from the lattice site and
occupies an interstitial site. So, the density of the crystal does not change.
110
Q11 The reaction conditions leading to the best yields of C2H5Cl are
A. C2H5Cl (Excess) + Cl2 →

B. C2H5Cl + Cl2+ dark, room temperature →
C. C2H5Cl + Cl2 (Excess) →
D. C2H5Cl + Cl2 →
Answer: (A) C2H5Cl (Excess) + Cl2
Solution: In the presence of sunlight, free radical halogenation occurs. Excess of

ethane gives monochlorinated product whereas excess of Cl2 can result into
hexachloroethane as the final product.
Q12 Considering the formation, breaking and strength of hydrogen bonds, predict which of
the following mixture will show a positive deviation from Raoult's law?
A. Methanol and acetone

B. Chloroform and acetone
C. Nitric acid and water
D. Phenol and aniline.
Answer: (A) Methanol and acetone
Solution: Mixture of methanol and acetone exhibits positive deviation because

methanol- methanol and acetone-acetone interactions is more than methanol-acetone.
The more number of hydrogen bonds are broken the less number of new hydrogen
bonds are formed.
Q13 Which of the following has least polar C−X bond?
A. CH3−CH=CH −Cl
B. HO−CH=CH −Cl
C. O2N−CH=CH −Cl
D. HS−CH=CH −Cl
Answer: C) O2N−CH=CH −Cl
Solution: When strong withdrawing group is attached to the compound containing C-

Cl bond, it decreases the polarity of C-Cl bond by pulling the shared pair of electrons
towards itself.
The EWG order is -NO2 > -OH > -SH > -CH3
Thus, the least polar C-X bond is in O2N−CH=CH −Cl
111
Q14 On heating with concentrated NaOH solution in an inert atmosphere of CO, white
phosphorus gives a gas. Which of the following statements is incorrect about the gas?
A. It is highly poisonous and has smell like rotten fish.

B. Its solution in water decomposes in the presence of light.
C. It is more basic than NH3.
D. It is less basic than NH3.
Answer: (C) It is more basic than NH3.
Solution: PH3 is less basic than NH3.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2

(Phosphine)
Q15 What is the correct name of the following compound?
A. D- (+)-fructose
B. D- (-)-fructose
C. L- (+)-fructose
D. L- (-)-fructose
Answer: (C ) L-(+)-fructose.
Solution: Since the -OH group on the lowest chiral carbon adjacent to the last CH2OH
group is on the left handside, it is L configuration. Also, L-fructose is a dextrorotatory
compound, hence (+) sign.
Thus, the given fructose is correctly named as L-(+)-fructose. Hence, C is the correct
option.
Q16 Saccharic acid is obtained by oxidation of glucose. Sum of the number of C, H and O
present in saccharic acid is:
A. 24
B. 18
C. 16
D. 32
112
Answer: (A) 24.
Solution: When Glucose is treated with nitric acid, oxidation of glucose takes place
and it forms Saccharic acid, also called glucaric acid.
The sum of C, H and O present in saccharic acid add up to 24 (6 carbons, 8 oxygens

and 10 hydrogens).
Q17 In Finkelstein Reaction which reactants are used
A. NaI + C2H5OH
B. NaF + acetone
C. NaBr + CH3OH
D. NaI + C2H5Br
Answer: (D) NaI + C2H5Br.
Solution: The Finkelstein Reaction is an organic reaction where an alkyl halide is

converted into another alkyl halide by reacting with NaI.
NaI + C2H5Br → C2H5I + NaBr
This reaction is called Finkelstein Reaction''
Q18 Which of the following elements can be involved in pπ–dπ bonding?
A. Carbon
B. Nitrogen
C. Phosphorus
D. Boron
Answer: C) Phosphorus
Solution: pπ–dπ bonding is present in Phosphorus due to the presence of vacant d-

orbitals and in carbon (C), nitrogen (N) and boron (B) do not have d orbitals.
NaI + C2H5Br → C2H5I + NaBr
113
Q19 Which of the following aqueous solutions should have the highest boiling point?
A. 1.0 M NaOH
B. 1.0 M 𝑁𝑎2 𝑆𝑂4
C. 1.0 M 𝑁𝐻4 𝑁𝑂3
D. 1.0 M 𝐾𝑁𝑂3
Answer: (B) 1.0 M 𝑁𝑎2 𝑆𝑂4
Solution: 𝑁𝑎2 𝑆𝑂4 will release 3 moles of ions/ moles of 𝑁𝑎2 𝑆𝑂4 in the aqueous
solution, and boiling point being a colligative property, the boiling point of this solution
will be the highest as other solutions release only 2 ions each.
Q20 Hydrogen fluoride is liquid unlike other hydrogen halides because
A. fluorine atoms are smaller in size

B. HF is the weakest acid
C. fluorine is highly reactive
D. HF molecules form intermolecular H-bonds.
Answer: (D) HF molecules form intermolecular H-bonds.
Solution: HF molecules are associated due to intermolecular H – bonding and thus

exist as liquid.
Q21 Which of the following alcohol has highest solubility in water?
A. Tertiary butyl alcohol

B. Secondary butyl alcohol
C. Ethylene glycol
D. Glycerol
Answer: (D) Glycerol
Solution:
114
More the number of hydroxyl groups in a compound more is the tendency to form
hydrogen bonds with water and hence, more will be the solubility.
Thus, among the given, glycerol, because of the presence of more hydroxyl groups,
readily dissolves in water. Hence it has the highest solubility in water.
Q22 Main source of helium is

A. Air
B. Radium
C. Monazite
D. Water
Answer: (C) Monazite
Solution: Monazite is the main source of Helium.

Q23 Which of the following statements are incorrect?
A. Among halogens, radius ratio between iodine and fluorine is maximum.

B. Leaving F - F bond, all halogens have weaker X-X bond than X-X’ bond in inter-
halogens.
C. Among inter-halogen compounds maximum. number of atoms ate present in
iodine fluoride.
D. Inter-halogen compounds are more reactive than halogen compounds.
Answer: (B) Leaving F - F bond, all halogens have stronger X-X bond than X-X’ bond
in inter-halogens.
Solution: In case of halogens radius ratio between iodine and fluorine is maximum
radius because iodine has maximum radius while fluorine has minimum radius. Also,
due to highest ratio maximum numbers of atoms are present in iodine fluoride. Inter-
halogen compounds are more reactive than halogen compounds because A-B bond of
dissimilar halogen is weaker than A-A or B-B bond of halogens.
Q24 For a dilute solution, Raoult's law states that
A. The lowering of vapour pressure is equal to the mole fraction of solute.

B. The relative lowering of vapour pressure is equal to the mole fraction of solute.
C. The relative lowering of vapour pressure is proportional to the amount of solute
in solution.
D. The vapour pressure of the solution is equal to the mole fraction of the solute.
Answer: (B) The relative lowering of vapour pressure is equal to the mole fraction of
solute.
Solution: According to Raoult’s law, for a dilute solution, the relative lowering of
vapour pressure is equal to the mole fraction of solute.
𝑃𝐴 ∘ −𝑃𝐴
= 𝑋𝐵
𝑃𝐴 ∘
115
Where,
𝑃𝐴 ∘−𝑃𝐴
= Relative lowering of vapour pressure
𝑃𝐴 ∘
𝑋𝐵 = mole fraction of solute
Q25 Which is the best reaction for preparation of t-butyl ethyl ether?
A.
B.
C.
D.
Answer: (B)
Solution: Williamson Ether Reactions involve an alkoxide that reacts with a primary
haloalkane to produce ether. Alkoxides consist of the conjugate base of an alcohol and
are composed of an R group bonded to an oxygen atom. They are often written as RO-
, where R is the organic substituent.
Only option (b) has primary alkyl halide thus it is used in preparation of t-butyl ethyl
ether.
116
Q26 Value of Henry's constant 𝐾𝐻 is _______.
A. Increases with increase in temperature.

B. Decreases with increase in temperature
C. Remains constant
D. First increases then decreases.
Answer: (A) Increases with increase in temperature.
Solution: Value of Henry's constant increases with increase in temperature.
Q27 Which one of the following compounds is most acidic?
A.
B.
C.
D.
Answer: (C)
117
Solution: Phenols are more acidic than alcohol & water due to
(i) the higher electronegativity of sp2 carbon to which OH group is attached.
(ii) the electron withdrawing nature of the benzene ring.
Conjugate base of phenol is stabilised by conjugation of negative charge in the ring.

Acidic strength is inversely proportional to EDGs and directly proportional to EWGs.
Hence, it is the most acidic since NO2 has strong -I and -R effects. So, being a strong
electron withdrawing group, it makes removal of H+ easier and makes the phenoxide
ion more stable.
Therefore, the correct answer is option (c).
Q28 When 𝑋𝑒𝐹6 is partially hydrolysed, it yields:
A. 𝑋𝑒𝑆𝑂3
B. 𝑋𝑒𝑂3
C. 𝑋𝑒𝑂𝐹4
D. 𝑋𝑒𝐹2
Answer: (C) 𝑋𝑒𝑂𝐹4
Solution: Partial hydrolysis of 𝑋𝑒𝐹6 gives oxyfluorides, 𝑋𝑒𝑂𝐹4 and 𝑋𝑒𝑂2 𝐹2.
𝑋𝑒𝐹6 + 𝐻2 𝑂 → 𝑋𝑒𝑂𝐹4 + 2𝐻𝐹
Xenon oxytetrafluoride
𝑋𝑒𝐹6 + 2𝐻2 𝑂 → 𝑋𝑒𝑂2 𝐹2 + 4𝐻𝐹
Xenon dioxydifluoride
Q29 If the basic formula of an α-amino acid is R-CH(NH2)-COOH, where R is the side
chain, what is the primary point of distinction between any two proteins?
A. Number of amino groups

B. Number of carboxyl groups
C. The side chain R
D. Relative positions of amino, carboxyl groups and R
Answer: (C) The side chain R

Solution: α-amino acids are the constituents of proteins. Different proteins are formed
by polymerisation of different α-amino acids, which are formed due to the difference in
the side chain substituted group R, which may be as simple as hydrogen or as
complex as imidazole.
118
Q30 Which of the following bases is not present in DNA?
A. Adenine
B. Guanine
C. Thymine
D. Uracil
Answer: (D). Uracil
Solution: DNA contains four bases namely adenine (A), guanine (G), cytosine (C)
and thymine (T). RNA also contains four bases, adenine (A), guanine (G), thymine (T)
and uracil (U). Hence (D) is correct.
Q31 Which of the following statements are correct?
A. S – S bond is present in𝐻2 𝑆2 𝑂7.

B. In peroxosulphuric acid (𝐻2 𝑆𝑂5 )sulphur is in +6 oxidation state.
C. Iron powder along with 𝐴𝑙2 𝑂3 and 𝐾2 𝑂is used as a catalyst in the preparation of
𝑁𝐻3 by Haber’s process.
D. Change in enthalpy is positive for the preparation of 𝑆𝑂3 by catalytic oxidation
of 𝑆𝑂2 .
Answer: (B) In peroxosulphuric acid (𝐻2 𝑆𝑂5 )sulphur is in +6 oxidation state.
Solution: In 𝐻2 𝑆𝑂5, there is a peroxo-linkage:
(O in peroxide linkage has oxidation state – 1)

Q32 Silver crystallises in f.c.c. Lattice. It’s edge length of the unit cells is 4.07× 10−8 cm
and density is 10.5 g𝑐𝑚−3. Calculate the atomic mass of silver.
A. 144 g/mol
B. 125 g/mol
C. 106.6 g/mol
D. 213 g/mol
Answer: (C) 106.6 g/mol
𝑍×𝑀
Solution: 𝑑 =
𝑎3 ×𝑁𝐴
𝑑×𝑁𝐴 ×𝑎3
𝑀= 𝑍
10.5×6.022×1023 ×(4.07×10−8 )𝑔𝑐𝑚−3
= 4
= 106.6 𝑔𝑚𝑜𝑙 −1
119
Q33 The solubility of haloalkanes/haloarenes is highest in:
A. Water
B. Non-polar solvents
C. Polar Solvents
D. Acetic acid
Answer: (B) Non-polar solvents
Solution: Energy required to overcome attraction between R-X molecules and energy
required to break H-bond between water molecules is greater than energy released
when new bonds are set up between R-X and water. Hence, haloalkanes are slightly
soluble in water.
Energy required to overcome attraction between R-X molecules and energy required
to overcome interactions between organic solvents is nearly the same as energy
released when new bonds are set up between R-X and organic solvents. Hence,
haloalkanes are highly soluble in organic solvents.
Q34 What are the products obtained when ammonia is reacted with excess chlorine?
A. 𝑁2 and 𝑁𝐶𝑙3
B. 𝑁2 and HCl
C. 𝑁2 and 𝑁𝐻4 𝐶𝑙
D. 𝑁𝐶𝑙3 and HCl
Answer: (D) 𝑁𝐶𝑙3 and HCl
Solution: 𝑁𝐻3 + 3𝐶𝑙2 (𝑒𝑥𝑐𝑒𝑠𝑠) → 𝑁𝐶𝑙3 + 3𝐻𝐶𝑙
Q35 A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes
place when a small amount of ‘A’ is added to the solution. The solution is _________.
A. Saturated
B. Supersaturated
C. Unsaturated
D. Concentrated
Answer: (B) Supersaturated
Solution: When a small amount of solute is added to its solution, it does not dissolve
and get precipitated then this type of solution is called a supersaturated solution.
120
Q36 Which of the following can not be the product of this reaction?
A. 1
B. 2
C. 3
D. 4
Answer: (D) .4
Solution:
In the mechanism of dehydration, after the removal of water, elimination of a hydrogen

atom only occurs from an α- carbon atom. Thus, it is quite evident that (4) can not be
the product of this reaction.
Q37 If 𝐴𝑙 3+ ions replace 𝑁𝑎+ ions at the edge centres of NaCl lattice, then the number of
vacancies in one mole of NaCl will be
A. 3.01 × 1023
B. 6.02 × 1023
C. 9.03 × 1023
D. 12.04 × 1023
Answer: (A) 3.01 × 1023
Solution: There are 12 edge centres.

1
Contribution of 𝑁𝑎+ ions at edge centres = 12 × = 3
4
Besides, there is one𝑁𝑎+ ion at body centre.
Thus, there are four 𝑁𝑎+ ions per unit cell.
3
are replaced by 𝐴𝑙 3+ ions.
4
Total 𝑁𝑎+ ions in 1 mole of NaCl = 6.023 × 1023
121
3
𝑁𝑎+ ion replaced by 𝐴𝑙 3+ = × 6.023 × 1023
4
1 𝐴𝑙 3+ replaces 3𝑁𝑎+ ions, thereby creating 2 vacancies.
2 3
Therefore, no. of vacancies created = × ( × 6.023 × 1023 )
3 4
= 3.01 × 1023
Q38 Ratio of the total volume of bcc to simple cubic structure is
A. 3√3: 8
B. 8: 3√3
C. 24√3: 1
D. 1: 24√3
Answer: (B) 8 ∶ 3√3
Solution: Volume of unit cell = 𝑎3

√3 4𝑟
For bcc, r = 𝑎 or a =
4 √3
4𝑟 3 64𝑟3
Thus, volume of bcc unit cell = ( ) =( )
√3 3√3
𝑎
For simple cubic, r = or a = 2r
2
Thus, volume of simple cubic unit cell = (2𝑟)3 =8𝑟 3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑏𝑐𝑐 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 64𝑟 3 /3√3 64 8

= =8×3√3=3√3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑖𝑚𝑝𝑙𝑒 𝑐𝑢𝑏𝑖𝑐 8𝑟 3
Q39 Which of the following are peroxoacids of sulphur?
A. 𝐻2 𝑆𝑂5 and 𝐻2 𝑆2 𝑂8
B. 𝐻2 𝑆𝑂5 and 𝐻2 𝑆2 𝑂7
C. 𝐻2 𝑆2 𝑂7and 𝐻2 𝑆2 𝑂8
D. 𝐻2 𝑆2 𝑂6and 𝐻2 𝑆2 𝑂7
Answer: (A) 𝐻2 𝑆𝑂5 and 𝐻2 𝑆2 𝑂8
Solution: Peroxymonosulphuric acid and Peroxydisulphuric acid are peroxoacids of

sulphur.
122
Q40 Fibrous and globular shapes come under which structure of proteins?
A. Primary
B. Secondary
C. Tertiary
D. Quaternary
Answer: (C) Tertiary
Solution: Secondary structures of proteins arise due to the regular folding of the
backbone of the polypeptide chain. Further folding of 20 structures of protein gives
molecular shape to the 30 structure proteins. Fibrous and globular shapes come under
tertiary structure of proteins. Globular proteins have spherical structure. Hence, the
correct answer is option (C).
Q41 The following tripeptide can be represented as:
A. Tyr - Val - Thr

B. Phe - Ala - Ser
C. Phe - Leu - Cys
D. Lys - Ala - Ser
Answer: (B) Phe - Ala - Ser
Solution:
The sequence of amino acids are read from left side containing NH2 group of an
amino acid and its right part contains COOH group which either form an amide bond
(for non-terminal amino acid) or remain at right side of the chain (for terminal amino
acid in the right side).
The given tripeptide can be represented as
Phe - Ala - Ser
Phe = phenylalanine
Ala = alanine
Ser = serine
Hence, (B) is correct.
123
Q42 Which one of the following oxides of nitrogen is blue solid?
A. NO
B. 𝑁2 𝑂3
C. 𝑁2 𝑂
D. 𝑁2 𝑂5
Answer: (B) 𝑁2 𝑂3
Solution: 𝑁2 𝑂3 exists as blue solid at a very low temperature.
Q43 Assertion: Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene

Reason: Alkyl halides are less reactive than acyl halides.
A. if both assertion and reason are correct and the reason explains the Assertion.
B. if both Assertion and reason are correct but reason is not the correct
explanation of the Assertion.
C. if the assertion is correct but the reason is incorrect.
D. if both the Assertion and reason are incorrect.
Answer: (C) if the assertion is correct but the reason is incorrect
Solution: Alkyl halides give polyalkylation products
Q44 Assertion: The bond angle in alcohol is slightly less than the tetrahedral angle.
Reason: In alcohols, the oxygen of the -OH group is attached to sp3 hybridized carbon
A. if both assertion and reason are correct and the reason explains the
Assertion.
C. if the assertion is correct but the reason is incorrect.
Answer: (A) if both assertion and reason are correct and the reason explains the
Assertion
Solution: The C-O-H bond angle in alcohol is slightly less than the tetrahedral angle
because of the repulsion between the two lone pairs of electrons on oxygen atom as
these pair pushes the C-O bond closer to each other
124
Q45 Assertion (A): F2 is a strong oxidizing agent than chlorine
Reason (R): Electron gain enthalpy of fluorine is less negative than chlorine
B. if both (A) and (R) are correct but (R) is not the correct explanation of (A).
D. A is false and R is True
Answer: (B) if both (A) and (R) are correct but (R) is not the correct explanation of (A)
Solution: Fluorine is the best oxidizing agent because it has more reduction potential
(more ability to lose the electrons) which is attributed to its high electronegativity.
Q46 Assertion: SN2 reaction of an optically active aryl halide with an aqueous solution of
KOH always gives an alcohol with opposite sign of rotation.
Reason: SN2 reactions always proceed with inversion configuration
A. if both assertion and reason are correct and the reason explains the Assertion.
C. if the assertion is incorrect but the reason is correct
Answer: (C) if the assertion is incorrect but the reason is correct
Solution: (C) Assertion is false, because aryl halides do not undergo nucleophilic
substitution under ordinary conditions. This is due to resonance, because of the which
the carbon-chloride bond acquires partial double bond character, hence it become
shorter and thus cannot be replaced by nucleophilic, however reason is true.
125
Q47 Assertion(A): 0.1 M Solution of KCl has greater osmotic pressure than 0.1 M solution
of glucose at the same temperature.
Reason(R): in solution KCl dissociates to produce more number of particles
Answer: (A) if both (A) and (R) are correct and the (R) explains the (A).
Solution: KCl is an ionic compound, hence dissociates into ions but glucose is a
covalent compound which does not dissociate into ions.
Q48 Assertion (A): Vapour pressure increases with increase in temperature.

Reason (R): With increase in temperature, more molecules of the liquid can go into
the vapour phase.
D. (A) is incorrect but the (R) is correct.
Answer: (A)
Solution: Vapour pressure increases with increased temperature because more

molecules of the liquid can go into the vapour phase with increase in temperature.
Q49 Assertion (A): N2 is less reactive than P4

Reason (R): nitrogen gas has more electron gain enthalpy than phosphorus
A. if both (A) and (R) are correct and the (R) explains the (A)
D. (A) is incorrect but the (R) is correct.
Answer: (C) (A) is correct but the (R) is incorrect
Solution: Due to the high bond dissociation energy of the triple bond between the two
N atoms, nitrogen (N) is less reactive than P4 and its electron gain enthalpy is less
than phosphorus.
126
Q50 Match the orbital overlap figures shown in List-I with the description given in List-II and
select the correct combination using the code provided in the options.
A. A-R; B-P; C-S; D-Q

B. A-R; B-P; C-Q; D-S
C. A-P; B-R; C-Q; D-S
D. A-P; B-R; C-S; D-Q
Answer: (B) A-R; B-P; C-Q; D-S
Solution:
(Waves are in phase)
127
(Waves are out of phase)
(Waves are in phase)
(Waves are out of phase)
Q51 Which of the following analogies is correct:

A. F2 > Cl2 > Br2 > I2: Bond dissociation energy
B. F2 > Cl2 > Br2 > I2: Electronegativity
C. F2 < Cl2 < Br2 < I2: Oxidising power
D. F2 < Cl2 < Br2 > I2: Electron gain enthalpy
Answer: (B) F2 > Cl2 > Br2 > I2: Electronegativity

Solution:
In case of diatomic molecules (X2) of halogens, the bond dissociation energy

decreases in the order : Cl2 > Br2 > F2 > I2. This is due to relatively large electron-
electron repulsion among the lone pairs is F2 than in case of Cl2. The oxidising power,
electronegativity and reactivity decrease in the order : F2 > Cl2 > Br2 > I2. Electron
gain enthalpy of halogens follows the given order : Cl2 > F2 > Br2 > I2. The low value of
electron gain enthalpy of fluorine is probably due to the small size of fluorine atoms.
128
Q52 Which of the following analogies is correct:
A. XeF4- Planar
B. XeOF2 - Planar
C. XeO2F2 - Planar
D. XeO4 - Planar
Answer: (A) XeF4- Planar

Solution:
In XeF4 the ‘Xe’ atom is sp3d2 hybridised, which contains two lone pair orbitals and
four bond pair orbitals. Therefore, the shape of the XeF4 molecule is square planar,
with one lone pair orbital over and the other below the plane.
Q53 I. Read the passage given below and answer the following question: All real
structures are three dimensional structures. They can be obtained by stacking two
dimensional layers one above the other while placing the second square closed
packed layer above the first, we follow the same rule that was followed when one row
was placed adjacent to the other. The second layer is placed over the first layer such
that the spheres of the upper layer are exactly above there of the first layer. In this
arrangement spheres of both the layers are perfectly aligned horizontally as well as
vertically. A metallic element crystallises into a lattice having a ABC ABC pattern and
packing and packing of spheres leaves out voids in the lattice.
What types of structures are formed by this arrangement?

A. ccp
B. hcp
C. ccp / hcp
Answer: (C) ccp / hcp
Solution: A metallic element crystallises into a lattice having a ABC ABC pattern and
packing and packing of spheres leaves out voids in the lattice. This type of
arrangement is ccp.
129
Q54 Name the non-stoichiometric point defect responsible for colour in alkali metal halides.
A. Frenkel defect
B. Interstitial defect
C. Schottky defect
D. F-centres
Answer: (D) F-centres
Solution: The non-stoichiometric point defect responsible for colour in alkali metal
halides is F-center.
Q55 What is the total volume of the atom in a fcc centred cubic unit cell of metal?
A. 16/3 𝜋𝑟 3
B. 𝜋𝑟 3
C. 24/3 𝜋𝑟 3
D. 12/3 𝜋𝑟 3
Answer: (A) 16/3 𝜋𝑟 3
Solution:
Number of atoms per unit cell in fcc = 4
Therefore, Total volume of atoms present in the fcc unit cell = 4 x (4/3) 𝜋𝑟 3 = (16/3)
𝜋𝑟 3
130
BIOLOGY
Time: 90 Minutes Max. Marks 35

1. The Question Paper contains three sections.
SECTION - A
Section – A consists of 24 questions. Attempt any 20 questions from this section. The first attempted 20
questions would be evaluated.
1 The plant known as ‘terror of Bengal’ reproduces vegetatively by: 0.70

A. offset
B. sucker
C. rhizome
D. bulbil
2 The image given below depicts the transport of ovum, fertilisation and passage of the 0.70
growing embryo through the fallopian tube of the human female reproductive system.
Read the statements given below and select the correct option regarding I-V given in the
131
image.
A. I - secondary oocyte. It is released from the ovary due to the rupture of corpus
luteum
B. II - zygote. It is formed in the infundibulum of the fallopian tube
C. III - morula. Morula has 8-16 blastomeres
D. V - blastocyst. It gets implanted into the myometrium of the uterus
3 Which of the following statement(s) is/are correct with respect to ART? 0.70
i. ZIFT stands for zygote intra fallopian transfer.
ii. In ZIFT, zygote is transferred into the uterus after in vitro fertilisation.
iii. GIFT stands for gamete inter fallopian transfer.
A. Only i
B. Both i and iii
C. Only iii
D. Only ii
4 Microsporangium in angiosperms is surrounded by four wall layers. 0.70

Select the option that has the correct sequence of layers from outside to inside
A. Epidermis, wall layers, tapetum, endothecium
B. Epidermis, tapetum, middle layers, endothecium
C. Epidermis, endothecium, middle layers, tapetum
D. Tapetum, endothecium, middle layers, epidermis
5 Endosperm persists in the mature seeds of 0.70

(i) pea
(ii) groundnut
(iii) maize
(iv) castor
A. (i) and (ii)
B. (iii) and (iv)
C. (i), (ii) and (iv)
D. Only (iii)
6 Which of the following is the correct developmental sequence during microsporogenesis? 0.70
A. Pollen grain → microspore tetrad → pollen mother cell → male gametes
B. Male gametes → pollen mother cell → pollen grain → microspore tetrad
C. Microspore tetrad → pollen mother cell → male gametes → pollen grain
D. Pollen mother cell → microspore tetrad → pollen grain → male gametes
132
7 The image given below is the longitudinal and cross section anatomy of an apple. 0.70
Read the following statements and select the incorrect statements

1. The fruit of apple is an example of pseudocarp
2. The part labelled as A in the image is the fleshy edible part of the apple developed from
the ovary
3. The part labelled as B in the image is fertilised ovule
4. The part labelled as C in the image is the mesocarp, the innermost layer of the fruit wall
(pericarp)
A. 1 and 2
B. 2 and 3
C. 2 and 4
D. 2, 3 and 4
8 Which of the following hormones is not correctly matched with its site of secretion and its 0.70
function?
A. GnRH - hypothalamus - stimulates anterior pituitary
B. FSH - anterior pituitary - stimulates the developing follicles
C. LH - posterior pituitary - ovulation
D. Progesterone - corpus luteum - maintenance of endometrium
133
9 Name the hormones involved at each stage in the following flow chart. 0.70
A. 1 - GnRH, 2 - LH, 3 - FSH

B. 1 - GnRH, 2 - LH, 3 - Testosterone
C. 1 - LH, 2 - FSH, 3 - GnRH
D. 1 - FSH, 2 - LH, 3 - GnRH
10 Which of the following cells are haploid? 0.70

I) primary spermatocyte
II) secondary spermatocyte
III) primary oocyte
IV) secondary oocyte
A. II and IV
B. I and III
C. I and II
D. III and IV
11 Which one of the following genotypes can form only 2 different types of gametes? 0.70
A. Aa Bb cc Dd
B. Aa Bb cc DD Ee
C. AA Bb CC dd EE
D. Aa Bb Cc DD Ee
12 In garden peas, the dominant seed colour is ______(i)_________ and the recessive fruit 0.70
colour is ________(ii)________.
A. (i)- yellow, (ii)-green
B. (i)- green, (ii)-yellow
C. (i)- yellow, (ii)-yellow
D. (i)- green, (ii)-green
134
13 How many of the following statements are correct? 0.70
I. Three scientists independently rediscovered Mendel’s work.
II. Coloured bodies called lysosomes can be visualised inside the nucleus by staining
cells.
III. By 1890, the chromosome movement during meiosis had been studied.
IV. Chromosomes and genes occur in pairs.
V. Two alleles of a gene are located on a homologous site on homologous
chromosomes.
A. 4
B. 2
C. 5
D. 3
14 If a man with blood group A marries a woman with blood group O, the possible blood 0.70
groups of their children would be:
A. A only
B. A and O
C. O only
D. A, AB and O
15 In a cross between red-flowered (RR) and white- flowered (rr) plants of snapdragon, the 0.70
F1 plants were all pink-flowered. What is the genotype of the F1 and the reason for this
phenotype?
A. Rr; Codominance
B. rr; Multiple alleles
C. rr; Codominance
D. Rr; Incomplete dominance
16 Choose the correct statement regarding sickle-cell anaemia. 0.70

A. Valine is replaced by glutamic acid at the 6th position of beta chain in people
suffering from sickle-cell anaemia
B. Sickle-cell anaemia is an autosomal dominant trait
C. Valine is replaced by glutamic acid at the 6th position of alpha chain in
people suffering from sickle-cell anaemia
D. Glutamic acid is replaced by valine at 6th position of the beta chain in people
135
17 A normal sperm in humans possess 0.70
A. X chromosome and Y chromosome
B. X or Y chromosome
C. Two X chromosomes
D. Two Y chromosomes
18 In the biochemical characterisation of the transforming principle, it was found that the 0.70
process of transformation is not affected by which of the following enzymes?
I. DNase
II. RNase
III. Protease
IV. Lipase
A. I, II and IV
B. I, III and IV
C. II, III and IV
D. I and IV
19 DNA differs from RNA in: 0.70

A. nature of sugar alone
B. nature of purines alone
C. nature of sugar and pyrimidines
D. nature of sugar and purines
20 The sequence of the coding strands in the transcription units are as follows: 0.70
(a) 5'-TGAACTGTAGCATGC-3'
(b) 5'-TGATACTGGTACATTC-3'
Find out the correct sequences of the mRNAs transcribed from these transcription units.
A. (a) 5'-UGAACUGUAGCAUGC-3'
(b) 5'-UGAUACUGGUACAUUC-3'
B. (a) 3'-CGUACGAUGACAAGU-5'
C. (a) 3'-CGUACGAUGUCAAGU-5'
(b) 3’-TGATACUGGTACAUTC-5'
D. (a) 5'-UGACUGUAGCUUGCC-3'
136
21 How many of the following statements are correct regarding functions of RNA? 0.70
I. It is a carrier of genetic information from DNA to ribosomes for synthesising
polypeptides
II. It carries amino acids to the ribosomes
III. It is a constituent component of ribosomes
IV. It contains the genetic makeup of most of the cells
A. 2
B. 3
C. 4
D. 1
22 How many of the following amino acids are coded by only one codon? 0.70
Tryptophan, arginine, methionine, valine, leucine.
A. 1
B. 2
C. 3
D. 4
23 Select the correct statements regarding transcription in prokaryotes and eukaryotes. 0.70
1. Capping and tailing takes place in prokaryotes.
2. Splicing takes place only in eukaryotes.
3. Transcription takes place in the cytoplasm in prokaryotes.
4. Prokaryotes have introns.
A. 1 and 4
B. 2 and 4
C. 2 and 3
D. 3 and 4
24 The image given below represents the regulation of lac operon in absence of the inducer. 0.70
______(i)______binds to the operator region (o) and prevents _____(ii)________from

transcribing the operon.
137
A. (i)- Inducer, (ii)- RNA polymerase
B. (i)- Repressor, (ii)- RNA polymerase
C. (i)- Repressor, (ii)- DNA polymerase
D. (i)- Promoter, (ii)- RNA polymerase
SECTION - B
Section - B consists of 24 questions (Sl. No.25 to 48). Attempt any 20 questions from this section. The
first attempted 20 questions would be evaluated.
Question No. 25 to 28 consists of two statements – Assertion (A) and Reason (R). Answer these
questions selecting the appropriate option given below:
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true and R is not the correct explanation of A
C. A is true but R is false
D. A is False but R is true
25 Assertion: Pollen grains are well preserved as fossils. 0.70

Reason: The outer layer of pollen grains called the exine is made up of
sporopollenin.
26 Assertion: The decision that the plant is going to flower is taken much 0.70
before the actual flowering takes place.
Reason: Various hormonal and structural changes take place only after
initiation of flowering.
27 Assertion: In eukaryotes, both introns and exons are transcribed to form 0.70
hnRNA.
Reason: The exons are removed to make the final transcript by splicing.
138
28 Assertion: Lac z is one of the structural genes in Lac operon of E.coli. 0.70
Reason: Lac z is essential for the uptake of lactose from the medium.
29 Choose the incorrect statement. 0.70

A. The end of juvenile/vegetative phase marks the beginning of the reproductive
phase
B. Many mammals are reproductively active throughout their reproductive phase
and hence are called continuous breeders
C. In monkeys and apes, the cyclical changes that occur during the reproductive
phase are called oestrus cycle
D. Adventitious buds arise from the notches present at margins of leaves of
Bryophyllum
30 If the meiocyte of an organism has 24 chromosomes, then how many chromosomes will 0.70
be present in its gamete cell?
A. 24
B. 12
C. 48
D. 23
31 The chromosome number in a root cell of a plant X is 30. What would be the chromosome 0.70
number in the cell of (i) endothecium, (ii) primary sporogenous cells of microsporangium,
(iii) microspore?
A. i-30, ii-30, iii-30
B. i-30, ii-30, iii-15
C. i-30, ii-15, iii-15
D. i-15, ii-15, iii-15
32 A bisexual plant produces 150 megaspore mother cells (MMC) and 250 pollen mother 0.70
cells (PMC) in a single season. Assuming the same number of MMCs and PMCs are
formed by the plant each season, how many mature embryo sacs and pollen grains would
be formed over 10 seasons by the plant?
A. 1500 embryo sacs, 10000 pollen grains
B. 6000 embryo sacs, 2500 pollen grains
C. 1500 embryo sacs, 2500 pollen grains
D. 6000 embryo sacs, 10000 pollen grains
139
33 Choose the correct statements from the following. 0.70
I) Both wind and water pollinated flowers are very colourful and produce nectar.
II) Wind-pollinated flowers have a large feathery stigma to easily trap air-borne pollen
grains.
III) Birds are the most dominant biotic pollinating agents.
IV) Pollination by water is quite rare in flowering plants and is limited to about 30 genera,
mostly monocotyledons.
A. I and III are correct
B. II and III are correct
C. II and IV are correct
D. I and IV are correct
34 Which of the following is/are situated inside the scrotum? 0.70

A. Penis
B. Seminal vesicle
C. Testes
D. All of the above
35 Select the incorrectly matched pairs from the following options. 0.70
A. Ovary- Gonads
B. Mons pubis- Internal genital organ
C. Mammary gland- Pectoralis muscle
D. Cervical canal along with vagina - Birth canal
36 What happens when many sperms reach closer to the ovum? 0.70
A. Zona pellucida binds with all the sperms and induces acrosome reaction.
B. Only two sperms are able to penetrate the zona pellucida.
C. Acrosomal secretion from one sperm helps it to enter inside the ovum by
penetrating the zona pellucida.
D. All sperms except one loses their tail in order to penetrate the zona pellucida.
37 Which of the following hormones are produced by a woman only during pregnancy? 0.70
A. Estrogen and progesterone
B. Progesterone and thyroxine
C. Relaxin and estrogen
D. hCG and hPL
140
38 All the statements given below indicate improved reproductive health of the society, 0.70
except:
A. better awareness about sex related aspects.
B. better detection and cure of STDs
C. increased number of couples with small families
D. increased maternal mortality rate (MMR) and infant mortality rate (IMR)
39 Given below are four methods of contraception (Column I). Select the correct match with 0.70
options given in Column II.
Method Mode of Action
a. Natural method i. Cover the penis
b. Male condoms ii. Absence of ovulation
c. Female condoms iii. Do not rely on use of any pills or devices
d. Lactational amenorrhea iv. Cover cervix
A. a- i, b- iii, c- ii, d- iv
B. a- iii, b- i, c- ii, d- iv
C. a- iii, b- i, c- iv, d- ii
D. a-iii, b- ii, c- i, d- iv
40 The image given below depicts: 0.70
A. vasectomy, a minor surgery performed in males.

B. tubectomy, a minor surgery performed in males due to which gametes are
not produced.
C. tubectomy, a minor surgery performed in females due to which ovulation is
inhibited.
D. tubectomy, a minor surgery performed in females in which the fallopian tubes
are tied and cut.
141
41 An embryologist was involved in a test tube baby programme. He successfully induced 0.70
zygote formation under simulated laboratory conditions and kept it safely on Friday. On
Tuesday, he came back to the lab and found the fertilised specimen to be in a healthy
condition.
Which of the following procedures should be used to successfully transfer the specimen
into a female?
A. ZIFT
B. GIFT
C. IUT
D. ICSI
42 1 : 1 phenotypic ratio is obtained in the offspring when __________ for the character 0.70
considered in the cross.
A. both the parents carry dissimilar alleles
B. one parent carries both the dominant alleles and the other parent carries both
the recessive alleles
C. one parent carries both the dominant alleles and the other parent carries
dissimilar alleles
D. one parent carries both the recessive alleles and the other parent carries
dissimilar alleles
43 XX-XO type of sex determination is found in ____________. 0.70

A. hens
B. Drosophila
C. parrots
D. grasshoppers
44 Which of the following statements is incorrect with respect to Morgan’s dihybrid cross? 0.70
A. Morgan carried out several dihybrid crosses in Drosophila to study genes that
were sex-linked.
B. He observed that the two genes segregate independently of each other and
the F2 ratio is 9:3:3:1.
C. He coined the term linkage to describe the physical association of genes on a
chromosome.
D. Morgan conducted his experiment on Drosophila melanogaster.
45 Monosomic females with rudimentary ovaries and lacking secondary sexual characters 0.70
suffer from _____________.
A. Turner’s syndrome
B. Klinefelter’s syndrome
C. Down’s syndrome
D. Edwards syndrome
142
46 Which of the following statements about the Hershey- Chase experiment are incorrect? 0.70
I: Bacteria infected with bacteriophage radiolabeled with 35P were radioactive
II: Bacteria infected with bacteriophage radiolabeled with 35S were radioactive
III: Bacteria infected with bacteriophage radiolabeled with 32P were radioactive
IV: Bacteria infected with bacteriophage radiolabeled with 35P and 32S were radioactive
V: Bacteria infected with bacteriophage radiolabeled with 35S were not radioactive
A. Statements I, II, III are incorrect.

B. Statements I, II, V are incorrect.
C. Statements I, II, IV are incorrect
D. Statements II, III, IV are incorrect
47 The total number of base pairs in the human genome and the total estimated cost of the 0.70
Human genome project are ________ and ________ respectively.
A. 6 х 10⁸ ; US $ 9 billion
B. 3 х 10⁹ ; US $ 3 billion
C. 3 х 10⁸ ; US $ 9 billion
D. 3 х 10⁹ ; US $ 9 billion
48 If a nucleoside monophosphate is used instead of nucleotide triphosphate in 0.70

polymerisation reaction during DNA replication, select the most appropriate consequence:
A. The nucleoside monophosphate is not incorporated into the existing DNA
strand
B. The nucleoside monophosphate is incorporated into the existing DNA strand
C. The nucleoside is incorporated into the existing DNA strand after the removal
of phosphate group
D. The nucleoside monophosphate gets an additional phosphate and nucleoside
bisphosphate gets added to the existing DNA
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SECTION - C
Section-C consists of one case followed by 6 questions linked to this case (Q.No.49 to 54). Besides this, 6
more questions are given. Attempt any 10 questions in this section. The first attempted 10 questions
would be evaluated.
Case To answer the questions, study the pedigrees given below.
a.
b.
49 Given that the traits are rare, choose the correct option about the inheritance pattern in the 0.70
pedigree ‘a’ is
A. X-linked dominant inheritance
B. Y-linked inheritance
C. Autosomal dominant inheritance
D. X-linked recessive inheritance
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50 The inheritance pattern exhibited in the pedigree ‘b’ is 0.70
51 If the couple in the F1 generation of pedigree ‘b’ had a fourth child. What would be the 0.70
probability of the child being an affected male?
A. 0
B. 0.25
C. 0.5
D. 1
52 If the couple in the F1 generation of pedigree ‘b’ had a fourth child. What would be the 0.70
probability of the child being an affected female?
A. 0
B. 0.25
C. 0.5
D. 1
53 The pedigree ‘b’ can represent a case of: 0.70

A. Colour blindness
B. Turner’s syndrome
C. Klinefelter’s syndrome
D. Thalassemia
54 Given below is the karyotype obtained during the amniocentesis of a pregnant female. 0.70
What can you infer from this? Choose the correct option.
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A. Nondisjunction during meiosis in the husband
B. Nondisjunction during meiosis in the lady
C. Perfectly normal and healthy karyotype and embryo
D. Absence of crossing over during meiosis in the lady
55 Due to global climate change, dumping of wastes and excessive pollution, a lake 0.70
experiences abnormal changes in temperature and pH. This lake has several species
including sexually reproducing frogs, water fleas that multiply by parthenogenesis, hydra
that multiply by budding, and sponges that multiply by fragmentation. Which of these
species will most likely survive the changing conditions of the lake.
A. Sexually reproducing frogs
B. Parthenogenetic water fleas
C. Hydra that buds
D. Fragmenting sponges
56 Study the graph below showing the levels of various hormones during a 28 day menstrual 0.70
cycle.
Peaks of which hormone - A, B, C or D indicate ovulation?

A. D
B. A
C. C
D. B
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57 The figure represents a Drosophila linkage map for genes A-E. 0.70
The numbers between the gene loci are the relative map units between each gene. Based
on the linkage map, predict and identify the two genes that are most likely to segregate
together.
A. A and B
B. B and C
C. C and D
D. D and E
58 Given below is the double helical structure of DNA. If there are 6000 base pairs in this 0.70
DNA, its total length would be:
A. 1040 mm
B. 1020 nm
C. 4020 mm
D. 2040 nm
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59 Study the process happening in the given image. 0.70
The product of this process has 15% Guanine. What will be the percentage of cytosine?
A. 0%
B. 15%
C. 30%
60 Observe this experiment. 0.70
If Meselson and Stahl's experiment is continued for 4 generations in E. coli, then the ratio of
15N/15N : 15N/14N : 14N/14N
A. 0:1:3
B. 0:1:15
C. 0:1:7
D. 0:1:31
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ANSWER KEYS
Q1 A. Q21 B. Q41 C.
Q2 C. Q22 B. Q42 D.
Q3 A. Q23 C. Q43 D.
Q4 C. Q24 B. Q44 B.
Q5 B. Q25 A. Q45 A.
Q6 D. Q26 C. Q46 C.
Q7 C. Q27 C. Q47 D.
Q8 C. Q28 C. Q48 A.
Q9 B. Q29 C. Q49 B.
Q10 A. Q30 B. Q50 D.
Q11 C. Q31 B. Q51 B.
Q12 C. Q32 A. Q52 A.
Q13 D. Q33 C. Q53 A.
Q14 B. Q34 C. Q54 B.
Q15 D. Q35 B. Q55 A.
Q16 D. Q36 C. Q56 C.
Q17 B. Q37 D. Q57 C.
Q18 C. Q38 D. Q58 D.
Q19 C. Q39 C. Q59 D.
Q20 A. Q40 D. Q60 C.
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SOLUTIONS
SECTION - A
Q1 The plant known as ‘terror of Bengal’ reproduces vegetatively by

A. offset
B. sucker
C. rhizome
D. bulbil
Answer: (A) offset
Solution:
Offset, sucker, rhizome and bulbil are vegetative propagules which are capable of giving rise to
new offspring.
The aquatic plant, water hyacinth, is known as the ‘terror of Bengal’. It is one of the most invasive
weeds and it drains oxygen from the water bodies which leads to the death of fishes. It
reproduces vegetatively by means offset.
Q2 The image given below depicts the transport of ovum, fertilisation and passage of the growing
embryo through the fallopian tube of the human female reproductive system.
Read the statements given below and select the correct option regarding I-V given in the image
A. I - secondary oocyte. It is released from the ovary due to the rupture of corpus luteum
B. II - zygote. It is formed in the infundibulum of the fallopian tube
C. III - morula. Morula has 8-16 blastomeres
D. V - blastocyst. It gets implanted into the myometrium of the uterus.
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Answer: (C) III - morula. Morula has 8-16 blastomeres
Solution:
LH surge leads to the rupture of the Graafian follicle and the secondary oocyte (I) is released from
the ovary. After the release of the secondary oocyte, the ruptured Graafian follicle transforms into
corpus luteum.
The secondary oocyte (ovum) gets fertilised in the ampullary region of the fallopian tube if the
sperms also reach the fallopian tube simultaneously and thus the zygote is formed (II).
The zygote further divides by mitosis and forms 2, 4, 8, 16 daughter cells called blastomeres. The
embryo with 8-16 blastomeres is called morula (III). Morula divides and transforms into a
blastocyst (IV) as it moves further into the uterus. The blastocyst gets implanted in the
endometrium of the uterus (V).
Q3 Which of the following statement(s) is/are correct with respect to ART?

i. ZIFT stands for zygote intra fallopian transfer.
ii. In ZIFT, zygote is transferred into the uterus after in vitro fertilisation.
iii. GIFT stands for gamete inter fallopian transfer
A. Only i
B. Both i and iii
C. Only iii
D. Only ii
Answer: (A) Only i
Solution:
There are several assisted reproductive technologies (ART) which help infertile couples to have
children.
In IVF (in vitro fertilisation) followed by ET (embryo transfer) method, also called test tube baby
programme, ova from the female and sperms from male are collected and induced to form zygote
in the laboratory. This is called in vitro fertilisation.
The zygote is then transferred into the fallopian tube. This is termed as zygote intra fallopian
transfer (ZIFT).
Gamete intra fallopian transfer (GIFT) is a technique in which the female gamete (ova) is
transferred from a donor into the fallopian tube of the female who cannot produce ova but can
provide a suitable environment for fertilisation and development.
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Q4 Microsporangium in angiosperms is surrounded by four wall layers.
Select the option that has the correct sequence of layers from outside to inside
A. Epidermis, wall layers, tapetum, endothecium

B. Epidermis, tapetum, middle layers, endothecium
C. Epidermis, endothecium, middle layers, tapetum
D. Tapetum, endothecium, middle layers, epidermis
Answer: (C) Epidermis, endothecium, middle layers, tapetum
Solution:
Microsporangia are surrounded by four wall layers. The outermost layer is the epidermis.
Epidermis is followed by endothecium and middle layers. The innermost layer is the tapetum.
Epidermis, endothecium and middle layers provide protection and also help in the dehiscence of
anthers. Tapetum provides nourishment to the developing pollen grains.
Enlarged view of one microsporangium
Q5 Endosperm persists in the mature seeds of

(i) pea
(ii) groundnut
(iii) maize
(iv) castor
A. (i) and (ii)
B. (iii) and (iv)
C. (i), (ii) and (iv)
D. Only (iii)
Answer: (B) (iii) and (iv)
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Solution:
Endosperm is the triploid nutritive tissue that nourishes the developing embryo in flowering plants.
In some plants the endosperm is completely consumed by the developing embryo and hence the
mature seeds lack endosperm. Such seeds are called non-albuminous or ex-albuminous seeds.
For example: pea, groundnut and beans.
In some plants the endosperm persists even in the mature seeds. Such seeds are called
albuminous seeds.
For example: castor, maize and coconut.
Q6 Which of the following is the correct developmental sequence during microsporogenesis?

A. Pollen grain → microspore tetrad → pollen mother cell → male gametes
B. Male gametes → pollen mother cell → pollen grain → microspore tetrad
C. Microspore tetrad → pollen mother cell → male gametes → pollen grain
D. Pollen mother cell → microspore tetrad → pollen grain → male gametes
Answer: (D) Pollen mother cell → microspore tetrad → pollen grain → male gametes
Solution:
The correct developmental sequence during microsporogenesis is pollen mother cell →
microspore tetrad → pollen grain → male gametes
The pollen mother cell (PMC) undergoes meiosis and forms a haploid microspore tetrad. The
microspores develop into pollen grains.
Each pollen grain divides mitotically and forms a vegetative and a generative cell.
The generative cell undergoes mitosis and forms two male gametes.
Q7 The image given below is the longitudinal and cross section anatomy of an apple
Read the following statements and select the incorrect statements

1. The fruit of apple is an example of pseudocarp
2. The part labelled as A in the image is the fleshy edible part of the apple developed from the
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ovary
3. The part labelled as B in the image is a fertilised ovule
4. The part labelled as C in the image is the mesocarp, the innermost layer of the fruit wall
(pericarp).
A. 1 and 2
B. 2 and 3
C. 2 and 4
D. 2, 3 and 4
Answer: (C) 2 and 4
Solution:
Apple is a false fruit (pseudocarp) as thalamus also contributes to the formation of fruit along with
the ovary. The edible fleshy part in the apple is thalamus and it encloses the true fruit that
develops from the ovary.
True fruit has a fruit wall (pericarp) differentiated into three layers. The outer epicarp, middle
mesocarp and an inner endocarp. Endocarp surrounds the seeds.
In the given image, A- thalamus, B- seeds (fertilised ovules), C- Endocarp
Q8 Which of the following hormones is not correctly matched with its site of secretion and its
function?
A. GnRH - hypothalamus - stimulates anterior pituitary

B. FSH - anterior pituitary - stimulates the developing follicles.
C. LH - posterior pituitary - ovulation
D. Progesterone - corpus luteum - maintenance of endometrium
Answer: (C) LH - posterior pituitary - ovulation
Solution:
Gonadotropin releasing hormone (GnRH) is secreted by the hypothalamus, and it stimulates the
anterior pituitary to secrete the gonadotropins, FSH and LH.
Follicle stimulating hormone (FSH) is secreted by the anterior pituitary and it stimulates the
growth and development of follicles inside the ovary in females.
Luteinizing hormone (LH) is secreted by the anterior pituitary, not posterior pituitary, and it
induces ovulation in females. Ovulation is the release of the secondary oocyte from the Graafian
follicle.
Progesterone is secreted by the corpus luteum and it helps in the maintenance of the
endometrium. Endometrium is the innermost layer of the wall of the uterus. It is the site of
implantation.
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Q9 Name the hormones involved at each stage in the following flow chart.
A. 1 - GnRH, 2 - LH, 3 - FSH

B. 1 - GnRH, 2 - LH, 3 - Testosterone
C. 1 - LH, 2 - FSH, 3 - GnRH
D. 1 - FSH, 2 - LH, 3 - GnRH
Answer: (B) 1 - GnRH, 2 - LH, 3 - Testosterone
Solution:
Spermatogenesis is the process of production of the sperms and it is under hormonal control.
Hypothalamus secretes gonadotropin releasing hormone or GnRH.

GnRH stimulates the anterior pituitary to produce two gonadotropins, luteinizing hormone (LH)
and follicle stimulating hormone (FSH).
LH or luteinizing hormone is also called interstitial cell-stimulating hormone (ICSH) in males. It

stimulates Leydig cells (Interstitial cells) of the testes to produce testosterone and other
androgens. Testosterone supports the process of spermatogenesis.
FSH or follicle stimulating hormone acts on Sertoli cells. Sertoli cells produce androgen binding
protein (ABP) which binds to testosterone and maintains high levels of androgens essential for
spermatogenesis.
Q10 Which of the following cells are haploid?

I) primary spermatocyte
II) secondary spermatocyte
III) primary oocyte
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IV) secondary oocyte
A. II and IV
B. I and III
C. I and II
D. III and IV
Answer: (A) II and IV
Solution:
Haploid cells have a single set of chromosomes whereas diploid cells have two sets of
chromosomes.
In males, primary spermatocytes are formed from spermatogonia as a result of mitosis and
differentiation. Spermatogonia are diploid cells. Thus, the primary spermatocyte is also diploid.
The primary spermatocyte undergoes meiosis to produce secondary spermatocytes. Since the
primary spermatocyte is diploid, meiosis of this cell would result in the production of haploid
secondary spermatocytes.
Similarly in females, the diploid oogonia undergoes mitosis and differentiation to produce the
diploid primary oocyte. Further the primary oocyte will undergo meiosis to produce haploid
secondary oocyte.
Hence, the secondary spermatocyte and secondary oocyte are haploid cells.
Q11 Which one of the following genotypes can form only 2 different types of gametes?
A. Aa Bb cc Dd
B. Aa Bb cc DD Ee
C. AA Bb CC dd EE
D. Aa Bb Cc DD Ee
Answer: (C) AA Bb CC dd EE
Solution:
Formula to calculate different types of gametes:- 2n
Where n is the number of heterozygotes.
AA Bb CC dd EE genotype has only one heterozygote (Bb), hence it will form 2 types of gametes,
ABCdE and AbCdE.
AA Bb CC dd EE produces only 2 types of gametes as shown below:
AA Bb CC dd EE
↓
ABCdE & AbCdE
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Option A- Aa Bb cc Dd forms 8 types of gametes
Option B- Aa Bb cc DD Ee forms 8 types of gametes
Option D- Aa Bb Cc DD Ee forms 16 types of gametes.
Q12 In garden peas, the dominant seed colour is ______(i)_________ and the recessive fruit colour is
________(ii)________.
A. (i)- yellow, (ii)-green
B. (i)- green, (ii)-yellow
C. (i)- yellow, (ii)-yellow
D. (i)- green, (ii)-green
Answer: (C) (i)- yellow, (ii)-yellow
Solution:
The colour of the seed and the fruit (pod) were two of the seven characters that Mendel selected
in pea plants to perform hybridisation experiments.
Each of the characters had two traits - one dominant and one recessive. The dominant trait will
express itself in both homozygous and heterozygous conditions while the recessive trait will be
able to express itself only in a homozygous condition.
For the seed colour, yellow is the dominant trait and green is a recessive trait while for the fruit
(pod) colour, green is dominant and yellow is recessive.
Q13 How many of the following statements are correct?

I. Three scientists independently rediscovered Mendel’s work.
II. Coloured bodies called lysosomes can be visualised inside the nucleus by staining cells.
III. By 1890, the chromosome movement during meiosis had been studied.
IV. Chromosomes and genes occur in pairs.
V. Two alleles of a gene are located on a homologous site on homologous chromosomes.
A. 4
B. 2
C. 5
D. 3
Answer: (D) 3
Solution:
- In 1900, three scientists (Hugo DeVries, Carl Correns and Erich von Tschermak) independently
rediscovered Mendel’s work.
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- Chromosomes are cell structures (bodies) found inside the nucleus that can be visualized when
stained. Lysosomes are cell organelles found in the cytoplasm outside the nucleus . They are
commonly called as the ‘suicide bags’ of the cells.
- By 1902, the chromosomal movement during meiosis had been worked and studied due to
advancements in microscopy.
- Chromosomes and genes occur in pairs. Each cell in our bodies has 23 pairs of chromosomes,
one half from the father and the other half from the mother. The gametes carry only one set of
chromosomes. The X and the Y chromosomes help in the determination of the sex. Similarly,
even genes occur in pairs. Each parent will possess a pair of genes for a particular trait and only
one of them gets passed onto us during gamete formation and fertilisation.
- Two alleles of a gene are located on a homologous site on homologous chromosomes. This is
because the two alleles for a particular character are present on the same loci on the homologous
pair of chromosomes.
Hence, three statements are correct (I, IV and V).
Q14 If a man with blood group A marries a woman with blood group O, the possible blood groups of
their children would be:
A. A only
B. A and O
C. O only
D. A, AB and O
Answer: (B) A and O
Solution:
ABO blood groups are controlled by three alleles of the gene I (IA,IB, i). Out of the three alleles, IA
and IB alleles are dominant over the i allele. Therefore, i is the recessive allele. The blood group A
can be due to either genotype - IAIA or IAi and the blood group O is due to genotype ii.
In the above mentioned case, the genotypes have not been specified, and hence we must see
both possible genotypes for the man with blood group A.
If he is homozygous (IAIA), the couple will have children with A blood group only, but if he is
heterozygous (IAi), the couple could have either A or O blood group children.
Q15 In a cross between red-flowered (RR) and white- flowered (rr) plants of snapdragon, the F1 plants
were all pink-flowered. What is the genotype of the F1 and the reason for this phenotype?
A. Rr; Codominance
B. rr; Multiple alleles
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C. rr; Codominance
D. Rr; Incomplete dominance
Answer: (D) Rr; Incomplete dominance
Solution:
The allele ‘R’ imparts red colour to the petals of the flower of snapdragon and the allele ‘r’ imparts
a white colour. However, the allele ‘R’ is not completely dominant over ‘r’ and due to this, a cross
between a true-breeding ‘RR’ (red) plant and ‘rr’ (white) plant results in ‘Rr’ (pink) flowered plants
and not a red flowered plant as expected according to Mendel’s law of dominance. This
phenomenon is known as incomplete dominance. In this phenomenon, the phenotype of F1 is an
intermediate between the phenotypes of the parents.
In the case of codominance, the F1 generation resembles both parents. Eg. AB blood group.
Multiple alleles exist in a population when there are many variations of a single gene. However, in
an individual, only two alleles are present. Eg. ABO blood group
Q16 Choose the correct statement regarding sickle-cell anaemia

A. Valine is replaced by glutamic acid at the 6th position of the beta chain in people suffering
from sickle-cell anaemia
B. Sickle-cell anaemia is an autosomal dominant trait
C. Valine is replaced by glutamic acid at the 6th position of the alpha chain in people
D. Glutamic acid is replaced by valine at 6th position of the beta chain in people suffering
from sickle-cell anaemia
Answer: (D) Glutamic acid is replaced by valine at 6th position of the beta chain in people
Solution:
Sickle-cell anaemia is an autosomal recessive disorder characterized by sickle shaped RBCs.
This genetic disorder results as the glutamic acid gets replaced by valine at the 6th position of the
beta chain of haemoglobin. A haemoglobin molecule is made up of four polypeptide chains which
include two alpha chains and two beta chains.
Q17 A normal sperm in humans possess

A. X chromosome and Y chromosome
B. X or Y chromosome
C. Two X chromosomes
D. Two Y chromosomes
Answer: (B) X or Y chromosome
Solution:
A human sperm cell is a haploid male gamete produced by germ cells through the process of
meiosis. Human sperm consists of 22 autosomes along with an allosome i.e., X or Y.
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Q18 In the biochemical characterisation of the transforming principle, it was found that the process of
transformation is not affected by which of the following enzymes?
I. DNase
II. RNase
III. Protease
IV. Lipase
A. I, II and IV
B. I, III and IV
C. II, III and IV
D. I and IV
Answer: (C) II, III and IV
Solution:
Transformation is the genetic alteration of a cell that happens as a result of direct uptake and
incorporation of genetic material from the surroundings through the cell membrane.
Griffith discovered the transforming principle but was unaware of its biochemical nature as to
whether it was a protein, DNA or RNA. Avery, MacLeod and McCarty worked to find out the
biochemical characterisation of the transforming principle.They used three experimental tubes to
which RNases, proteases and DNases were added along with heat-killed S-strain bacteria and
live R-strain bacteria.
Among the three sets of experimental tubes, those to which RNAse and protease were added,
still retained the ability to transform R-strain to the live S-strain. Therefore, neither RNA nor
protein could be the transforming principle. It was only in the tubes to which DNase was added,
that transformation could not happen. It was thus concluded that DNA is the genetic material.
Q19 DNA differs from RNA in:

A. nature of sugar alone
B. nature of purines alone
C. nature of sugar and pyrimidines
D. nature of sugar and purines
Answer: (C) nature of sugar and pyrimidines
Solution:
DNA stands for deoxyribonucleic acid and RNA stands for ribonucleic acid. They are
polynucleotide chains.
Each nucleotide has a 5-carbon sugar, nitrogenous base and a phosphate group.
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The nucleotides of DNA differ from the nucleotides of RNA in the nature of sugar and pyrimidines.
The nucleotides of DNA have deoxyribose sugar, while nucleotides of RNA have ribose sugar.
Nitrogenous bases are of two types - Purines and pyrimidines.
Among purine bases, adenine and guanine are present both in DNA and RNA.
Among pyrimidine bases, DNA has thymine and cytosine, while RNA has uracil and cytosine.
Q20 The sequence of the coding strands in the transcription units are as follows:
(a) 5'-TGAACTGTAGCATGC-3'
Find out the correct sequences of the mRNAs transcribed from these transcription units.
A. (a) 5'-UGAACUGUAGCAUGC-3'
B. (a) 3'-CGUACGAUGACAAGU-5'
C. (a) 3'-CGUACGAUGUCAAGU-5'
(b) 3’-TGATACUGGTACAUTC-5'
D. (a) 5'-UGACUGUAGCUUGCC-3'
Answer: (A) (a) 5'-UGAACUGUAGCAUGC-3'

Solution:
The genetic information from one strand of the DNA is copied into RNA during transcription.
Hence, synthesis of RNA occurs during transcription.
In the transcription unit, the strand that has the polarity 3'→5' acts as a template for the synthesis
of mRNA, and it is also referred to as a template strand. The other strand that has the polarity
5'→3' is called a coding strand.
If the coding strand in a transcription unit is having the following sequence:
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5'-TGAACTGTAGCATGC-3', then, it is known that the sequence of mRNA transcribed is the
same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5'-UGAACUGUAGCAUGC-3'.
Similarly, if the coding strand in a transcription unit is having the following sequence:
5'-TGATACTGGTACATTC-3', then the sequence of mRNA will be
5'-UGAUACUGGUACAUUC-3'
Q21 How many of the following statements are correct regarding functions of RNA?
I. It is a carrier of genetic information from DNA to ribosomes for synthesising polypeptides
II. It carries amino acids to the ribosomes
III. It is a constituent component of ribosomes
IV. It contains the genetic makeup of most of the cells
A. 2
B. 3
C. 4
D. 1
Answer: (B) 3
Solution:
I, II and III are correct statements regarding the functions of RNA.
RNA stands for ribonucleic acid. It is a polynucleotide chain made of many 5-C compound ribose
sugars, phosphates and nitrogenous bases (purines and pyrimidines).
There are different types of RNA such as mRNA, rRNA, tRNA etc.
The messenger RNA (mRNA) gets transcribed from the DNA and carries the information in the
form of codons to ribosomes. These codons code for the amino acids, which join by peptide
bonds and make polypeptide chains.
The transfer RNA (tRNA), also known as adapter molecule or soluble RNA, carries amino acids
from the cytoplasm to the ribosomes and mRNA for protein synthesis.
The ribosomal RNA (rRNA) is an important component of ribosomes.
The genetic makeup of most of the organisms is DNA. RNA forms the genetic makeup of some
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viruses like HIV.
Q22 How many of the following amino acids are coded by only one codon?
Tryptophan, arginine, methionine, valine, leucine.
A. 1
B. 2
C. 3
D. 4
Answer: (B) 2
Solution:
The genetic code is a triplet code. Three consecutive nucleotides in mRNA constitute a codon
which codes for a particular amino acid.
Mostly the amino acids are coded by more than one codon.
For example, leucine is an amino acid which is coded by six different codons (UUA, UUG, CUU,
CUC, CUA, CUG).
Arginine is also coded by six different codons (CGU, CGC, CGA, CGG, AGA, AGG).
Valine is coded by four different codons (GUC, GUU, GUG, GUA).
Amino acids like tryptophan and methionine are coded by only one codon.
UGG codes for tryptophan.
AUG codes for methionine.
Q23 Select the correct statements regarding transcription in prokaryotes and eukaryotes.
1. Capping and tailing takes place in prokaryotes.

2. Splicing takes place only in eukaryotes.
3. Transcription takes place in the cytoplasm in prokaryotes.
4. Prokaryotes have introns.
A. 1 and 4
B. 2 and 4
C. 2 and 3
D. 3 and 4
Answer: (C) 2 and 3
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Solution:
Transcription is the process of formation of RNA from one strand of DNA.
Transcription occurs inside the nucleus in a eukaryotic cell, whereas the prokaryotic cell lacks a
well-defined nucleus and the genetic material is found in the cytoplasm. Thus, transcription in
prokaryotes takes place in the cytoplasm.
Transcription is different in eukaryotes as compared to prokaryotes as they have a more complex

cellular organisation. The primary transcript (hnRNA) formed after transcription in eukaryotes
contains coding (exons) as well as non-coding (introns) sequences. The primary transcript of
eukaryotes undergoes post-transcriptional changes namely,
• Capping - addition of methyl guanosine triphosphate to the 5’ end.
• Tailing - addition of adenylate residues at the 3’ end.
• Splicing - removal of introns and linking of the exons.
The prokaryotic DNA has only exons, introns are absent. Hence they do not undergo the
processes of capping, splicing and tailing.
Q24 The image given below represents the regulation of lac operon in absence of the inducer.
______(i)______binds to the operator region (o) and prevents _____(ii)________from

transcribing the operon.
A. (i)- Inducer, (ii)- RNA polymerase
B. (i)- Repressor, (ii)- RNA polymerase
C. (i)- Repressor, (ii)- DNA polymerase
D. (i)- Promoter, (ii)- RNA polymerase
Answer: (B) (i)- Repressor, (ii)- RNA polymerase

Solution:
The lac operon contains three structural genes: lacZ, lacY and lacA. All three structural genes are
essential for the entry and breakdown of lactose into its smaller monomeric units such as the
glucose and galactose.
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The repressor protein is coded by the regulatory gene (i gene). The main function of the repressor
protein is to prevent the process of transcription of the structural genes when glucose is present.
Therefore, the lac repressor protein binds to the operator region of the operon and thereby
prevents the RNA polymerase from transcribing the operon.
However, in the presence of inducers like lactose, the repressor protein gets inactivated due to
the interaction with the inducer thus allowing the transcription of the structural genes (z, y and a).
SECTION - B
Question No. 25 to 28 consists of two statements – Assertion (A) and Reason (R). Answer
these questions selecting the appropriate option given below:
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true and R is not the correct explanation of A
C. A is true but R is false
D. A is False but R is true
Q25 Assertion: Pollen grains are well preserved as fossils.

Reason: The outer layer of pollen grains called the exine is made up of
sporopollenin.
Answer: (A) Both A and R are true and R is the correct explanation of A
Solution:
Pollen grains are well preserved as fossils because their outer wall exine is made up of a highly
resistant organic material called sporopollenin.
Sporopollenin is capable of withstanding high temperatures and strong acids and alkali. It cannot
be degraded by any enzyme known so far.
Therefore, both assertion and reason are true and reason is the correct explanation of assertion.
Q26 Assertion: The decision that the plant is going to flower is taken much before the actual
flowering takes place.
Reason: Various hormonal and structural changes take place only after initiation of flowering.
Answer: (C) A is true but R is false

Solution:
The decision that a plant is going to flower takes place much before the actual flowering takes
place. Several hormonal and structural changes are initiated prior to the initiation of flowering.
Floral primordia grow into floral buds that are borne on inflorescence and flowering occurs.
Thus only assertion is correct and the reason is incorrect.
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Q27 Assertion: In eukaryotes, both introns and exons are transcribed to form hnRNA.
Reason: The exons are removed to make the final transcript by splicing.
Solution:
The genetic information from one strand of the DNA is copied into RNA during transcription.
Hence, synthesis of RNA occurs during transcription.
In eukaryotes, the monocistronic structural genes contain both exons and introns. Exons are the
coding sequences (code for amino acids) and introns are the non-coding sequences.
During transcription, both the exons and introns are transcribed to form the pre-mRNA or
heterogeneous nuclear RNA (hnRNA). Thus, the assertion holds true.
The hnRNA undergoes post-transcriptional modifications to form the functional mRNA.

The post-transcriptional modifications include a process called splicing in which there is removal
of the non-coding introns and subsequent joining of the coding exons. Therefore, the mature or
functional mRNA contains only exons. Hence the reason is false.
Q28 Assertion: Lac z is one of the structural genes in Lac operon of E.coli.
Reason: Lac z is essential for the uptake of lactose from the medium.
Solution:
The lac operon contains three structural genes: lac z, lac y and lac a. All three structural genes
are essential for the metabolism of lactose.
The lac z gene codes for the enzyme beta-galactosidase (β-gal) which catalyses the hydrolysis of
the disaccharide, lactose into its monomeric units, galactose and glucose. The Lac y gene codes
for the enzyme permease, which increases the permeability of the bacterial cell to β-galactosides
such as lactose. Hence it is the Laz y gene that is essential for the uptake of lactose and not Lac
z.The Lac a gene encodes a transacetylase.
Hence the assertion is true and the reason is false.
Q29 Choose the incorrect statement.

A.The end of juvenile/vegetative phase marks the beginning of the reproductive phase.
B. Many mammals are reproductively active throughout their reproductive phase and hence are
called continuous breeders.
C. In monkeys and apes, the cyclical changes that occur during the reproductive phase are called
oestrus cycle.
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D. Adventitious buds arise from the notches present at margins of leaves of Bryophyllum.
Answer: (C) In monkeys and apes, the cyclical changes that occur during the reproductive phase
are called oestrus cycle.
Solution:
Monkeys and apes are the primates and the cyclical changes that occur during the reproductive
phase in primate females is called the menstrual cycle. Oestrus cycle is the cyclical changes that
occur in the activities of ovaries, accessory ducts as well as hormones during the reproductive
phase in non-primate females.
Juvenile/vegetative phase is the period of growth and development before the organism attains
reproductive maturity. The end of juvenile/vegetative phase marks the beginning of the
reproductive phase.
Adventitious buds are the vegetative propagules of Bryophyllum. They arise from the notches
present at margins of leaves. Vegetative propagules are units of vegetative propagation capable
of giving rise to a new plant.
Continuous breeders are those animals that can reproduce throughout the year after attaining
reproductive maturity. They include humans and apes (chimpanzees, gorillas).
Q30 If the meiocyte of an organism has 24 chromosomes, then how many chromosomes will be
present in its gamete cell?
A. 24
B. 12
C. 48
D. 23
Answer: (B) 12
Solution:
Meiocyte is the specialised gamete mother cell of the diploid organisms. Meiocyte is diploid and
has 2n number of chromosomes. It undergoes meiosis to produce gametes that will have 'n'
number of chromosomes where ‘n’ represents a single set of chromosomes.
Therefore, if the meiocyte of an organism has 24 chromosomes, then the gamete will have 12
chromosomes.
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Q31 The chromosome number in a root cell of a plant X is 30. What would be the chromosome
number in the cell of (i) endothecium, (ii) sporogenous cells of microsporangium, (iii)
microspore?
A. i-30, ii-30, iii-30
B. i-30, ii-30, iii-15
C. i-30, ii-15, iii-15
D. i-15, ii-15, iii-15
Answer: (B) i-30, ii-30, iii-15
Solution:
Root cells are diploid (2n) in nature. Microspores are haploid plant spores that develop into male
gametophytes in angiosperms. Hence, they will have 15 (n) chromosomes.
The sporogenous cells of the microsporangium are diploid. Hence, these cells will have 30
chromosomes.
Endothecium is a protective layer present in the wall of the microsporangium made up of diploid
cells. The endothecial cells will thus have 30 chromosomes.
Q32 A bisexual plant produces 150 megaspore mother cells (MMC) and 250 pollen mother cells
(PMC) in a single season. Assuming the same number of MMCs and PMCs are formed by the
plant each season, how many mature embryo sacs and pollen grains would be formed over 10
seasons by the plant?
A. 1500 embryo sacs, 10000 pollen grains
B. 6000 embryo sacs, 2500 pollen grains
C. 1500 embryo sacs, 2500 pollen grains
D. 6000 embryo sacs, 10000 pollen grains
Answer: (A) 1500 embryo sacs, 10000 pollen grains
Solution:
Each diploid megaspore mother cell - MMC, gives rise to four haploid megaspores. However,
three of them degenerate and only one of them remains functional which develops into an
embryo sac. As a result, one diploid MMC results in the formation of only one mature embryo sac.
Since the given plant produces 150 MMCs in a season, it will give rise to 150 x 10 i.e. 1500
mature embryo sacs over 10 seasons.
One diploid pollen mother cell (PMC) gives rise to four haploid microspores. All of them are
functional and hence each of them gives rise to four haploid pollen grains in turn. Since the given
plant produces 250 PMCs in a season it will give rise to 250 x 4 x 10 i.e. 10000 pollen grains
over 10 seasons.
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Q33 Choose the correct statements from the following.
I) Both wind and water pollinated flowers are very colourful and produce nectar.
II) Wind-pollinated flowers have a large feathery stigma to easily trap air-borne pollen grains.
III) Birds are the most dominant biotic pollinating agents.
IV) Pollination by water is quite rare in flowering plants and is limited to about 30 genera, mostly
monocotyledons.
A. I and III are correct
B. II and III are correct
C. II and IV are correct
D. I and IV are correct
Answer: (C) II and IV are correct
Solution:
Wind pollinated flowers have a large feathery and well exposed stigma so that it can easily trap
the pollen grains present in the air.
Water pollination is very rare and seen in only about 30 genera.
Both wind and water pollinated flowers are not colourful and also do not produce nectar. Bees are
the most dominant biotic pollinating agents.
Q34 Which of the following is/are situated inside the scrotum?

A. Penis
B. Seminal vesicle
C. Testes
D. All of the above
Answer: (C) Testes
Solution:
Male reproductive system includes a pair of testes, accessory ducts, glands and external
genitalia.
Scrotum is an extra-abdominal pouch within which the testes are present. Testes produce
sperms. Testes are located outside the abdominal cavity in order to maintain a temperature of 2-
2.5 degrees Celsius less than the human body temperature which is optimum for the production
of sperms, i.e spermatogenesis.
Penis is the male copulatory organ. It is located outside the scrotum.
A pair of seminal vesicles are the accessory glands located in the pelvis region lying between the
rectum and the base of the bladder.
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Q35 Select the incorrectly matched pairs from the following options.
A. Ovary- Gonads
B. Mons pubis- Internal genital organ
C. Mammary gland- Pectoralis muscle
D. Cervical canal along with vagina - Birth canal
Answer: (B) Mons pubis- Internal genital organ
Solution:
Mons pubis is a pad of fatty tissue which is covered by the skin and the pubic hair. It is an
external genital organ in females.
Gonads are reproductive organs that produce gametes. The male gonads are the testes and the
female gonads are the ovaries.
Mammary glands (breasts) lie over the pectoralis muscle in the thoracic region.
The cervical canal along with the vagina forms the birth canal through which the baby passes
during parturition.
Q36 What happens when many sperms reach closer to the ovum?
A. Zona pellucida binds with all the sperms and induces acrosome reaction.
B. Only two sperms are able to penetrate the zona pellucida.
C. Acrosomal secretion from one sperm helps it to enter inside the ovum by penetrating the
zona pellucida.
D. All sperms except one loses their tail in order to penetrate the zona pellucida.
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Answer: (C) Acrosomal secretion from one sperm helps it to enter inside the ovum by penetrating
the zona pellucida
Solution:
Millions of sperms are released during ejaculation inside the female reproductive tract. They swim
and reach the fallopian tube.If an ovum is present in the fallopian tube then one of the sperms
enters inside the ovum due to acrosomal reaction.
Acrosome reaction allows the sperm to penetrate the zona pellucida of the ovum by releasing
acrosomal enzymes. The zona pellucida is a thick extracellular covering that surrounds all
mammalian eggs.
After the sperm enters inside the ovum, the oocyte releases cortical granules in the perivitelline
space which hardens the zona pellucida so that other sperms cannot enter the ovum. This is
called a cortical reaction.
Only one sperm enters the ovum and loses its tail.
Hence not all sperms entering the female reproductive tract can enter inside the ovum.
Q37 Which of the following hormones are produced by a woman only during her pregnancy?
A. Estrogen and progesterone
B. Progesterone and thyroxine
C. Relaxin and estrogen
D. hCG and hPL
Answer: (D) hCG and hPL
Solution:
hCG (human chorionic gonadotropin) and hPL (human placental lactogen) are the hormones
produced by women only during pregnancy. hCG helps in maintaining the pregnancy and hPL
helps in providing energy to the foetus by breaking down the fats in the mother.
Hormones like progesterone, estrogen and relaxin are produced in non-pregnant women as
well, but the levels of these hormones increase several fold during pregnancy.
Estrogen and progesterone are hormones that are continuously secreted from puberty until
menopause.
Thyroxine is secreted by the thyroid gland throughout life and helps various physiological
functions.
Q38 All the statements given below indicate improved reproductive health of the society, except:
A. better awareness about sex related aspects.
B. better detection and cure of STDs
C. increased number of couples with small families
D. increased maternal mortality rate (MMR) and infant mortality rate (IMR)
Answer: (D) increased maternal mortality rate (MMR) and infant mortality rate (IMR)
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Solution:
Currently various action plans and programmes are in operation to attain total reproductive health
as a social goal.
These programmes cover wider reproduction related areas and create awareness among people
about various aspects of reproduction for building up a reproductively healthy society.
A reproductively healthy society has better awareness about sex related matters, increased
number of medically assisted deliveries, better post-natal care leading to decreased maternal
mortality rate (MMR) and infant mortality rate(IMR), increased number of couples with small
families and better detection and cure of STDs.
Hence, increased maternal mortality rate (MMR) and infant mortality rate (IMR) does not indicate
improved reproductive health of the society.
Q39 Given below are four methods of contraception (Column I). Select the correct match with options
given in Column II.
Method Mode of Action
a. Natural method i. i. Cover the penis
b. Male condoms ii. Absence of ovulation
c. Female condoms iii. Do not rely on use of any pills or devices
d. Lactational amenorrhea iv. Cover cervix
A. a- i, b- iii, c- ii, d- iv
B. a- iii, b- i, c- ii, d- iv
C. a- iii, b- i, c- iv, d- ii
D. a-iii, b- ii, c- i, d- iv
Answer: (C) a- iii, b- i, c- iv, d- ii
Solution:
Natural/traditional methods of contraception do not involve use of any device, pills, etc.
Male condoms are barriers made up of rubber/thin latex. They cover the penis to prevent the
introduction of sperms into the female genital tract. Similarly, female condoms are also made of
rubber/thin latex but they are inserted into the female reproductive tract to cover the cervix during
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coitus.
After parturition, during the intense lactation period, the mother does not ovulate due to the
absence of the menstrual cycle. So the chances of pregnancy during this period (around 6
months after delivery) is very less.
Q40 The image given below depicts:
A. vasectomy, a minor surgery performed in males.

B. tubectomy, a minor surgery performed in males due to which gametes are not produced.
C. tubectomy, a minor surgery performed in females due to which ovulation is inhibited.
D. tubectomy, a minor surgery performed in females in which the fallopian tubes are tied and
cut
Answer: (D) tubectomy, a minor surgery performed in females in which the fallopian tubes are
tied and cut
Solution:
The image given in the question represents the human female reproductive system in which the
fallopian tubes are tied and cut. It is a surgical method of contraception performed in females and
is known as tubectomy or tubal ligation.
As the fallopian tubes are blocked after surgery, the egg released from the ovary is prevented
from meeting the sperm. Thus tubectomy prevents conception.
Tubectomy does not interfere with the production of ova and also does not inhibit ovulation.
Vasectomy is a surgical method of contraception and involves tying and cutting of vas deferens in
males. Thus vasectomy blocks the transport of sperm but does not stop the production of sperm.
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Q41 An embryologist was involved in a test tube baby programme. On Friday he successfully induced
zygote formation under simulated laboratory conditions and kept it safely. On Tuesday, he came
back to the lab and found the fertilised specimen to be in a healthy condition.
Which of the following procedures should be used to successfully transfer the specimen into a
female?
A. ZIFT
B. GIFT
C. IUT
D. ICSI
Answer: (C) IUT
Solution:
During the in vitro fertilisation technique (fertilisation outside the body of the female) followed by
embryo transfer (ET), also called test tube baby programme, ova from the female and sperm from
male are collected and induced to form zygote in the laboratory.
The zygote or early embryo (with upto 8 blastomeres) are transferred into the fallopian tube and
this is called ZIFT(zygote intra fallopian transfer).
The transfer of embryos with more than 8 blastomeres into the uterus is called IUT (intrauterine
transfer).
According to the information provided in the question, fertilisation took place on a friday. The
embryo develops into 16 celled stage in 3-4 days after fertilisation and is termed morula. Hence,
after 4 days, the embryo would be at 16 celled stage, making it suitable for IUT.
Gamete intra fallopian transfer (GIFT) is a technique in which the female gamete is transferred
from a donor into the fallopian tube of the female who cannot produce ova but can provide a
suitable environment for fertilisation and development.
In intra cytoplasmic sperm injection (ICSI), a tiny needle is used to inject sperm directly into the
egg and an embryo is formed in the laboratory.
Q42 1:1 phenotypic ratio is obtained in the offspring when __________ for the character considered
in the cross.
A. both the parents carry dissimilar alleles

B. one parent carries both the dominant alleles and the other parent carries both the
recessive alleles
C. one parent carries both the dominant alleles and the other parent carries dissimilar alleles
D. one parent carries both the recessive alleles and the other parent carries dissimilar
alleles
Answer: (D) one parent carries both the recessive alleles and the other parent carries dissimilar
alleles
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Solution:
1 : 1 phenotypic ratio is obtained in the offspring when one parent carries both the recessive
alleles and the other parent carries dissimilar alleles(heterozygous) for the character considered
in the cross.
This can be understood by considering the monohybrid cross in which the parents differ in height.
Tall and dwarf are two contrasting traits for this character in the pea plant.
The tall trait is dominant over the dwarf trait.
‘T’ represents the allele responsible for tallness while ‘t’ represents the allele responsible for
dwarfness.
When the pea plant with dissimilar alleles (Tt) is crossed with the pea plant carrying recessive
alleles (tt), tall and dwarf offspring are obtained in the ratio of 1 : 1.
Hence option d is correct.
Option a: When both the parents carry dissimilar alleles (Tt), then the phenotypic ratio of 3 : 1 is
obtained.
Option b: When one parent carries dominant alleles (TT) and the other parent carries recessive
alleles (tt), then all the offspring produced from this cross show dominant phenotype (tall).
Option c: When one parent carries dominant alleles (TT) and the other parent carries dissimilar
alleles (Tt), then all the offspring produced from this cross show dominant phenotype (tall).
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Q43 XX-XO type of sex determination is found in ____________.
A. hens
B. Drosophila
C. parrots
D. grasshoppers
Answer: (D) grasshoppers
Solution:
Grasshoppers have an XX-XO sex-determination system. Males only have one X chromosome
(XO), while females have two (XX).
In hens and parrots, ZW-ZZ system is found. The ovum determines the sex of the offspring.
Males are the homogametic sex (ZZ), while females are the heterogametic sex (ZW).
Sex determination in Drosophila is achieved by a balance in the male and female determinants on
the X chromosomes and autosomes. Drosophila has either one or two X chromosomes and two
sets of autosomes (A).
If there is one X chromosome in a diploid cell (OX:AA), the fly is male.
If there are two X chromosomes in a diploid cell (XX:AA), the fly is female.
Thus, XO Drosophila are sterile males. The Y chromosome carries genes for the formation of the
male gametes in adult flies, it is not involved in determining sex.
Q44 Which of the following statements is incorrect with respect to Morgan’s dihybrid cross?
A. Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-
linked.
B. He observed that the two genes segregate independently of each other and the F2 ratio is
9:3:3:1.
C. He coined the term linkage to describe the physical association of genes on a
chromosome.
D. Morgan conducted his experiment on Drosophila melanogaster.
Answer: (B) He observed that the two genes segregate independently of each other and the F2
ratio is 9:3:3:1.
Solution:
T.H. Morgan conducted his experiment on Drosophila melanogaster (commonly known as fruit fly).
He carried out several dihybrid crosses using Drosophila to study the pattern of inheritance of
genes. The crosses were similar to the dihybrid crosses carried out by Mendel in peas, however
the outcomes were quite different.
He observed that the two genes did not segregate independently of each other while he crossed
brown bodied, red eyed males with yellow bodied, white eyed females. After selfing F1, he
observed that these genes were not independently assorted and the F2 phenotypic ratio deviated
very significantly from the 9:3:3:1 ratio (expected when the two genes get assorted
independently). It was observed that in the F2 generation, there were more offspring of parental
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type (98.7%) and less percentage of offspring of recombinant type (1.3%).
His observations further provided a concrete evidence of the fact that certain genes were sex-
linked. Morgan also coined the term linkage to describe this physical association of genes on a
chromosome.
Q45 Monosomic females with rudimentary ovaries and lacking secondary sexual characters suffer
from _____________.
A. Turner’s syndrome
B. Klinefelter’s syndrome
C. Down’s syndrome
D. Edwards’s syndrome
Answer: (A) Turner’s syndrome
Solution:
Edwards syndromeTurner’s syndrome is a chromosomal disorder in human beings characterised
by the absence of one of the X chromosomes in the females. Normal females have two X
chromosomes. The affected females are sterile with rudimentary ovaries and lack secondary
sexual characters.
The absence of one chromosome is called monosomy (2n-1) and females exhibiting monosomy
are called monosomic females.
Klinefelter’s syndrome is a chromosomal disorder characterised by the presence of an extra X

chromosome in the human males and hence the affected individuals have three sex
chromosomes (XXY).
This condition is termed as trisomy of sex chromosomes.
The affected individuals display overall masculine development, however, the feminine characters
like development of breast, i.e., gynecomastia are also seen. The affected individuals are sterile.
Down’s syndrome is a chromosomal disorder which is characterised by the trisomy of the

$21^{st}$ chromosome. The affected individuals have a short stature with rounded head,
furrowed tongue, wide palm with distinct palm crease and retarded physical, mental and
psychomotor development.
Edwards syndrome, a chromosomal disorder which is characterised by the trisomic condition of

chromosome 18. The affected individuals have small and abnormally shaped head and jaws,
arched spine, and birth defects in many organs.
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Q46 Which of the following statements about the Hershey- Chase experiment are incorrect?
I: Bacteria infected with bacteriophage radiolabeled with 35P were radioactive
II: Bacteria infected with bacteriophage radiolabeled with 35S were radioactive
III: Bacteria infected with bacteriophage radiolabeled with 32P were radioactive
IV: Bacteria infected with bacteriophage radiolabeled with 35P and 32S were radioactive
V: Bacteria infected with bacteriophage radiolabeled with 35S were not radioactive
A. Statements I, II, III are incorrect.

B. Statements I, II, V are incorrect.
C. Statements I, II, IV are incorrect
D. Statements II, III, IV are incorrect
Answer: (C) Statements I, II, IV are incorrect
Solution:
Alfred Hershey and M. Chase (1952) gave unequivocal proof that DNA is the genetic material.
In their experiment, radioactive phosphorus (32P) and radioactive sulphur (35S) were used to
culture bacteriophages. The DNA of bacteriophage was radiolabeled with 32P and protein with
35
S.
These radiolabeled bacteriophages were then made to infect E. coli. It was seen that the bacterial
cells contained radioactive viral DNA (32P), but not radioactive viral protein(35S). This means that
the bacteriophages transferred their DNA to the E. coli bacteria.
It can be concluded that the presence of radioactive 32P in the bacteriophage enables the bacteria
to become radioactive. As radioactive 32P incorporated in the DNA of bacteriophage is passed
down to the bacteria in the culture medium.
Thus, in the given question only statements III and V are correct, the rest are incorrect.
Q47 The total number of base pairs in the human genome and the total estimated cost of the Human
genome project are ________ and ________ respectively.
A. 6 х 10⁸ ; US $ 9 billion
B. 3 х 10⁹ ; US $ 3 billion
C. 3 х 10⁸ ; US $ 9 billion
D. 3 х 10⁹ ; US $ 9 billion
Answer: (D) 3 х 10⁹ ; US $ 9 billion
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Solution:
Human genome project aims at finding out the sequence of the entire human genome. Genome
refers to the haploid set of chromosomes in a cell.
In total, the entire human genome is said to have approximately 3 х 10⁹ base pairs. If the cost of
sequencing is estimated to be about $ 3 for each base pair, the estimated cost of the entire
project turns out to be: (3 х 10⁹) х $ 3 = $ 9 billion.
Q48 If a nucleoside monophosphate is used instead of nucleotide triphosphate in polymerisation

reaction during DNA replication, select the most appropriate consequence:
A. The nucleoside monophosphate is not incorporated into the existing DNA strand
B. The nucleoside monophosphate is incorporated into the existing DNA strand
C. The nucleoside is incorporated into the existing DNA strand after the removal of
phosphate group
D. The nucleoside monophosphate gets an additional phosphate and nucleoside
bisphosphate gets added to the existing DNA
Answer: (A) The nucleoside monophosphate is not incorporated into the existing DNA strand
Solution:
During the synthesis of complementary daughter strands, DNA polymerase uses
deoxyribonucleotide triphosphates (dNTPs) as substrates. dNTPs as substrates perform dual
role:-
- They get added to the 3’ OH after losing a molecule of pyrophosphate and help in strand
elongation.
- The two terminal phosphate molecules which get released in the form of PPi provide energy
during the process of polymerization.
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If dNMPs or deoxyribonucleoside monophosphates are utilized during polymerisation reactions, it
can’t be utilized as a substrate by DNA polymerase III. This is because each and every enzyme
has a specific substrate which fits into its active site. Since dNMPs and dNTPs differ in their size
and molecular structures, dNMPs won’t get recognized as a substrate by DNA polymerase
enzymes. Additionally, dNMPs won’t liberate pyrophosphate, hence, won’t yield energy needed
for the polymerization to occur.
SECTION - C
Case To answer the questions, study the pedigrees given below.
a.
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b.
Q49 Given that the traits are rare, choose the correct option about the inheritance pattern in the
pedigree ‘a’ is
Answer: (B) Y-linked inheritance
Solution:
In the given pedigree, only the father is affected in the F1 generation. The F2 generation has 3
normal females and 2 affected males. One of the normal females marry a normal male and have
2 normal daughters and 2 normal sons. One of the affected males marry a normal female and
have 2 normal daughters and one affected son. All the sons in this pedigree are affected and
daughters are not. Hence, this pedigree exhibits a Y-linked inheritance.
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Q50 The inheritance pattern exhibited in the pedigree ‘b’ is
Answer: (D) X-linked recessive inheritance
Solution:
In the given pedigree the F1 generation is normal. In the F2 generation, one son is normal, which
means it is not a Y-linked inheritance. It cannot be X-linked dominant, because if it was then one
of the parents must have been affected too, Since it is not, we can rule it out.
Since the parents are unaffected and only one son is affected, the inheritance can be safely
assumed to be recessive. Trait is affecting only males in this pedigree, so, this must be X-linked
recessive and the female of F1 must be a carrier.
Q51 If the couple in the F1 generation of pedigree ‘b’ had a fourth child. What would be the probability
of the child being an affected male?
A. 0
B. 0.25
C. 0.5
D. 1
Answer: (B) 0.25
Solution:
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The genotype of the fourth male child is XY. As both the parents are unaffected and the trait is
affecting only males in this pedigree, this must be an X-linked recessive disorder. The female
parent must be a carrier with the genotype XXc.
The plotting the details in a punnett square,
X Y
Xc XXc XcY
X XX XY
The probability of an affected male progeny (XcY) - 0.25
Q52 If the couple in the F1 generation of pedigree ‘b’ had a fourth child. What would be the probability
of the child being an affected female?
A. 0
B. 0.25
C. 0.5
D. 1
Answer: (A) 0
Solution:
The genotype of the female child will be XX. Since both the parents are unaffected, it is not a
dominant trait, moreover the trait is affecting only males in this pedigree. So, this must be X-
linked recessive. Thus the female parent is a carrier with the genotype XXc
The plotting the details in a punnett square,
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X Y
Xc XXc XcY
X XX XY
The probability of an affected female progeny (XcXc) - 0
Q53 The pedigree ‘b’ can represent a case of:

A. Colour blindness
B. Turner’s syndrome
C. Klinefelter’s syndrome
D. Thalassemia
Answer: (A) Colour blindness
Solution:
Both the parents in the given pedigree chart are unaffected, making this a recessive disorder.
The trait is affecting only the males, thus the trait must be present on the X chromosome of the
female parent. Hence, pedigree ‘b’ represents an X-linked recessive inheritance.
Inheritance of Mendelian disorders like haemophilia, sickle cell anemia, phenylketonuria, colour
blindness, thalassemia etc. can be analysed by pedigree chart. Colour blindness is an example
for such a disorder.
Turner’s syndrome and Klinefelter’s syndrome are due to abnormal chromosome numbers while
thalassemia is an autosomal recessive disorder.
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Q54 Given below is the karyotype obtained during the amniocentesis of a pregnant female.
What can you infer from this? Choose the correct option.
A. Nondisjunction during meiosis in the husband

B. Nondisjunction during meiosis in the lady
C. Perfectly normal and healthy karyotype and embryo
D. Absence of crossing over during meiosis in the lady
Answer: (B) Nondisjunction during meiosis in the lady
Solution:
The karyotype shows 46+XXY chromosomal constitution. This indicates Klinefelter’s syndrome.
Nondisjunction of X chromosomes during meiosis results in the formation of a female gamete
with XX chromosomes. When this female gamete fuses with a sperm with a Y chromosome,
resultant zygote has XXY chromosome constitution and the baby will be a male and suffer from
Klinefelter syndrome.
Q55 Due to global climate change, dumping of wastes and excessive pollution, a lake experiences
abnormal changes in temperature and pH. This lake has several species including sexually
reproducing frogs, water fleas that multiply by parthenogenesis, hydra that multiply by budding,
and sponges that multiply by fragmentation. Which of these species will most likely survive the
changing conditions of the lake?
A. Sexually reproducing frogs
B. Parthenogenetic water fleas
C. Hydra that buds
D. Fragmenting sponges
Answer: (A) Sexually reproducing frogs
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Solution:
Organisms that reproduce asexually produce offspring that are genetically identical to their
parents. Such organisms lack the potential evolutionary advantage due to variations, which helps
in adapting and surviving in a changing environment.
In the case of sexually reproducing organisms, they have a mix of maternal and paternal genetic
material, which creates variation among the offspring. The evolutionary advantage of variation
due to a mix of genes may help them survive in a lake where temperature and pH are changing
due to global climatic changes.
Generally, in lower organisms, asexual reproduction is preferred in favourable conditions while

sexual reproduction is preferred in unfavourable conditions.
Q56 Study the graph below showing the levels of various hormones during a 28 day menstrual cycle.
Peaks of which hormone - A, B, C or D indicate ovulation?

A. D
B. A
C. C
D. B
Answer: (C) peak C
Solution:
Ovulation is the process of rupture and release of the ootid from the ovary. This takes place on
the 14th day of a 28 day menstrual cycle. The luteinizing hormone places an important role in the
rupture of the Graafian follicle to release the developing ootid.
186
A - Progesterone
B - Estrogen
C - Luteinizing hormone
D - Follicle stimulating Hormone
For ovulation to occur, there needs to be an LH surge. So, the correct option is C.
Q57 The figure represents a Drosophila linkage map for genes A-E.
The numbers between the gene loci are the relative map units between each gene. Based on
the linkage map, predict and identify the two genes that are most likely to segregate together.
A. A and B
B. B and C
C. C and D
D. D and E
Answer: (C) C and D
Solution:
Closer the genes, tightly linked they are and higher the probability of them being segregated
together. In the given map, C and D are the closest at a distance of 3 mu (map unit) and will
segregate together.
Q58 Given below is the double helical structure of DNA. If there are 6000 base pairs in this DNA, its
total length would be:
187
A. 1040 mm
B. 1020 nm
C. 4020 mm
D. 2040 nm
Answer: (D) 2040 nm
Solution:
DNA is made up of nucleotides (deoxynucleotides, deoxyribose sugar and phosphate bonds).
The number of nucleotides present determines the length of the DNA molecule.
Distance between two consecutive base pairs is 0.34 nm
For 6000 base pairs, the total length of the DNA will be
6000✕0.34 = 2040 nm
Q59 Study the process happening in the given image.
The product of this process has 15% Guanine. What will be the percentage of Cytosine?
A. 0%
B. 15%
C. 30%
Answer: (D) Cannot be determined
Solution:
188
The process shown in the image is transcription. It is the process by which a single stranded RNA
is formed from a DNA template. Since the given question is about single-stranded RNA, it is
impossible to determine the percentage of nitrogenous bases. Chargaff’s rule, which is used to
calculate the number of bases, is applicable only to dsDNA. This is because the DNA forms a
pair, unlike the RNA where the transcript synthesized after transcription is a single stranded
molecule.
Hence the data is insufficient to answer this question.
Q60 Observe this experiment.
If Meselson and Stahl's experiment is continued for 4 generations in E. coli, then the ratio
189
of 15N/15N : 15N/14N : 14N/14N
A. 0:1:3
B. 0:1:15
C. 0:1:7
D. 0:1:31
Answer: (C) 0:1:7
Solution:
Matthew Meselson and Franklin Stahl conducted experiments to confirm the semi conservative
nature of DNA replication. They grew Escherichia coli for several generations in a medium
containing 15N (heavy) compounds of nitrogen. These were then transferred to a medium
containing 14N (light) compound.
They observed that the first round of DNA replication produced a hybrid variety of DNA having
one heavy (15N) and one light (14N) chain.
So in the first generation after 20 minutes the ratio was 0:1:0
The second round of replication produced one hybrid and one light type of DNA. so in the second
generation after 40 minutes the ratio was 0:1:1
The third round of replication showed one intermediate and 3 light DNA chains.
So in the third generation after 60 minutes the ratio was 0:1:3
As the bacteria continues to divide, it was observed that DNA replication incorporated the 14N,
light chains, using the nitrogen from the give medium and thus the fourth generation after 80
minutes had one intermediate and 7 light chained DNA molecules i.e. 0:1:7
190

BYJU'S All India Mock Board Exams: CBSE Grade XII - Term I Mock Papers Set - 1

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BYJU'S All India Mock Board Exams: CBSE Grade XII - Term I Mock Papers Set - 1

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BYJU’S All India Mock Board Exams

CBSE Grade XII - Term I

Time Allowed: 90 minutes Maximum Marks: 40

Q1 Let 𝐴 = {1,2,3, ⋯ ,40} and 𝑅 be an equivalence relation on 𝐴 × 𝐴 defined by (1)

Q7 The domain of the function 𝑓(𝑥) = sin−1(−𝑥 2 ) is (1)

Q10 ln(1+3𝑥)−ln(1−2𝑥) (1)

Q18 The derivative of ln 𝑥 with respect to cot 𝑥 is (1)

Q33 (𝑖+𝑗)+(𝑖.𝑗) (1)

Q41 If 𝑍 = 𝑥 + 2𝑦, subject to the constraints: 2𝑥 + 𝑦 ≥ 3, 𝑥 + 2𝑦 ≥ 6, 𝑥, 𝑦 ≥ 0, then the (1)

Which of the following statement(s) is/are correct?

A. Both (I) and (II)

Based on the given information, answer the following questions.

Answer: (𝐴) Reflexive relation, Transitive relation

A. Reflexive and transitive only

Answer: (𝐴) Reflexive and transitive only

As the given relation 𝑅 satisfies the above condition.

Answer: (𝐶) [0, 1)

Q6 The range of the function 𝑓(𝑥) = 4 tan−1 𝑥 + 3 sin−1 𝑥 + sec −1 𝑥 is

Q7 The domain of the function 𝑓(𝑥) = sin−1(−𝑥 2 ) is

Answer: (C) [−1,1]

ln(1 + 3𝑥) − ln(1 − 2𝑥)

A. continuous but non-differentiable at 𝑥 = 1

Answer: (A) continuous but non-differentiable at 𝑥 = 1

A. continuous and differentiable

Answer: (B) continuous but non-differentiable

Answer: (B) cosec 𝑥

Differentiation w.r.t. 𝑥, we get

Q17 𝑒 𝑡 +𝑒 {−𝑡} et −e{−t} dx

Answer: (A) Increasing

Answer: (A) 𝑥 ∈ (−∞, 2] ∪ [4, ∞)

Q23 If 𝑓(𝑥) = cos(cos 𝑥) is an increasing function, then a possible interval of 𝑥 is

Answer: (𝐵) [0, 𝜋/2]

At point (2,1), 𝑚𝑛𝑜𝑟𝑚𝑎𝑙 = −1

Answer: (D) 2𝑎𝑥 cos 𝑥

Answer: (B) [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]

⇒ 𝐴𝐵𝐶 = [𝑎𝑥 2 + ℎ𝑥𝑦 + ℎ𝑥𝑦 + 𝑏𝑦 2 ]

Answer: (B) [25 2

Answer: (C) 5 square units

A. All the values of 𝐾 ∈ ℝ

Answer: (𝐴) All the values of 𝐾 ∈ ℝ

Answer: (𝐵) Infinite solutions

In the given equation, we have

Q39 If the system of equations 𝑥 − 𝑘𝑦 − 𝑧 = 0, 𝑘𝑥 − 𝑦 − 𝑧 = 0 and 𝑥 + 𝑦 − 𝑧 = 0 has

Answer: (𝐷) The LPP has no feasible solution

A. Only one point

Answer: (𝐶) More than two points

From the above table minimum value of 𝑍 = 6

Answer: (𝐴) 740

Now, we get common area as marked in figure

Which of the following statement(s) is/are correct?

A. Both (I) and (II)

Answer: (𝐶) (II) only

Answer: (𝐶) (12,20)

So, 𝑍𝑚𝑎𝑥 occurs at C(12,20).

Based on the given information, answer the following questions.

Differentiating both sides w.r.t. 𝑥, we get

Q11. When a body is charged by induction, then the body (0.77)

Q22. Which of the following is a wrong statement? (0.77)