JEE Main 2025 — 24 Jan Shift 2 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If 7=5+17(5+α)+172(5+2α)+173(5+3α)+…∞7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2\alpha)+\frac{1}{7^3}(5+3\alpha)+\ldots\infty, then the value of α\alpha is :

  1. A.1
  2. B.6
  3. C.17\frac{1}{7}
  4. D.67\frac{6}{7}

Correct answer: B

Explanation: 5⋅76+α1/7(6/7)2=75\cdot\frac{7}{6}+\alpha\frac{1/7}{(6/7)^2}=7 ⇒ 7α36=76\frac{7\alpha}{36}=\frac{7}{6} ⇒ α = 6.

Question 2

Mathematics · 4 marks · −1

The number of real solution(s) of the equation x2+3x+2=min⁡{∣x−3∣,∣x+2∣}x^2+3x+2=\min\{|x-3|, |x+2|\} is :

  1. A.0
  2. B.3
  3. C.2
  4. D.1

Correct answer: C

Question 3

Mathematics · 4 marks · −1

Let (2, 3) be the largest open interval in which the function f(x)=2log⁡e(x−2)−x2+ax+1f(x)=2\log_e(x-2)-x^2+ax+1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x)=(x−1)3(x+2−a)2g(x)=(x-1)^3(x+2-a)^2 is strictly decreasing. Then 100(a+b−c)100(a+b-c) is equal to :

  1. A.280
  2. B.360
  3. C.420
  4. D.160

Correct answer: B

Explanation: f(ˊ3)=0f\'(3)=0: 2−6+a=02-6+a=0 ⇒ a = 5. g=(x−1)3(x−3)2g=(x-1)^3(x-3)^2, g=ˊ(x−1)2(x−3)(5x−11)g\'=(x-1)^2(x-3)(5x-11): decreasing on (11/5, 3). 100(5+2.2−3)=420100(5+2.2-3)=420.

Question 4

Mathematics · 4 marks · −1

Let a⃗=3i^−j^+2k^\vec{a}=3\hat{i}-\hat{j}+2\hat{k}, b⃗=a⃗×(i^−2k^)\vec{b}=\vec{a}\times(\hat{i}-2\hat{k}) and c⃗=b⃗×k^\vec{c}=\vec{b}\times\hat{k}. Then the projection of c⃗−2j^\vec{c}-2\hat{j} on a⃗\vec{a} is :

  1. A.272\sqrt{7}
  2. B.373\sqrt{7}
  3. C.14\sqrt{14}
  4. D.2142\sqrt{14}

Correct answer: D

Explanation: b⃗=(2,8,1)\vec b=(2,8,1), c⃗=(8,−2,0)\vec c=(8,-2,0); c⃗−2j^=(8,−4,0)\vec c-2\hat j=(8,-4,0); projection 2814=214\frac{28}{\sqrt{14}}=2\sqrt{14}.

Question 5

Mathematics · 4 marks · −1

Let the position vectors of three vertices of a triangle be 4p⃗+q⃗−3r⃗4\vec{p}+\vec{q}-3\vec{r}, −5p⃗+q⃗+2r⃗-5\vec{p}+\vec{q}+2\vec{r} and 2p⃗−q⃗+2r⃗2\vec{p}-\vec{q}+2\vec{r}. If the position vectors of the orthocenter and the circumcenter of the triangle are p⃗+q⃗+r⃗4\frac{\vec{p}+\vec{q}+\vec{r}}{4} and αp⃗+βq⃗+γr⃗\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r} respectively, then α+2β+5γ\alpha+2\beta+5\gamma is equal to :

  1. A.6
  2. B.3
  3. C.4
  4. D.1

Correct answer: B

Explanation: Centroid G=p⃗+q⃗+r⃗3G=\frac{\vec p+\vec q+\vec r}{3}; O=3G−H2=38(p⃗+q⃗+r⃗)O=\frac{3G-H}{2}=\frac38(\vec p+\vec q+\vec r): 38(1+2+5)=3\frac38(1+2+5)=3.

Question 6

Mathematics · 4 marks · −1

Suppose A and B are the coefficients of 30th30^{th} and 12th12^{th} terms respectively in the binomial expansion of (1+x)2n−1(1+x)^{2n-1}. If 2A = 5B, then n is equal to :

  1. A.20
  2. B.19
  3. C.22
  4. D.21

Correct answer: D

Question 7

Mathematics · 4 marks · −1

If the equation of the parabola with vertex V(32,3)V\left(\frac{3}{2}, 3\right) and the directrix x+2y=0x+2y=0 is αx2+βy2−γxy−30x−60y+225=0\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0, then α+β+γ\alpha+\beta+\gamma is equal to :

  1. A.7
  2. B.8
  3. C.9
  4. D.6

Correct answer: C

Explanation: Focus (3, 6); 5[(x−3)2+(y−6)2]=(x+2y)25[(x-3)^2+(y-6)^2]=(x+2y)^2 ⇒ 4x2+y2−4xy−30x−60y+225=04x^2+y^2-4xy-30x-60y+225=0: 4 + 1 + 4 = 9.

Question 8

Mathematics · 4 marks · −1

If the system of equations x+2y−3z=2x+2y-3z=2, 2x+λy+5z=52x+\lambda y+5z=5, 14x+3y+μz=3314x+3y+\mu z=33 has infinitely many solutions, then λ+μ\lambda+\mu is equal to :

  1. A.13
  2. B.10
  3. C.12
  4. D.11

Correct answer: C

Question 9

Mathematics · 4 marks · −1

The function f:(−∞,∞)→(−∞,1)f:(-\infty, \infty)\to(-\infty, 1), defined by f(x)=2x−2−x2x+2−xf(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}} is :

  1. A.Both one-one and onto
  2. B.Neither one-one nor onto
  3. C.One-one but not onto
  4. D.Onto but not one-one

Correct answer: C

Explanation: f is strictly increasing with range (−1, 1) ≠ codomain.

Question 10

Mathematics · 4 marks · −1

The equation of the chord, of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1, whose mid-point is (3, 1) is :

  1. A.25x+101y=17625x+101y=176
  2. B.5x+16y=315x+16y=31
  3. C.48x+25y=16948x+25y=169
  4. D.4x+122y=1344x+122y=134

Correct answer: C

Explanation: T=S1T=S_1: 3x25+y16=925+116\frac{3x}{25}+\frac{y}{16}=\frac{9}{25}+\frac{1}{16} ⇒ 48x+25y=16948x+25y=169.

Question 11

Mathematics · 4 marks · −1

Let f:(0,∞)→Rf:(0, \infty)\to\mathbf{R} be a function which is differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2f'(x)=2xf(x)+3, with f(1)=4f(1)=4. Then 2f(2) is equal to :

  1. A.29
  2. B.19
  3. C.39
  4. D.23

Correct answer: C

Explanation: (f/x2)=ˊ3/x4(f/x^2)\'=3/x^4 ⇒ f=−1x+5x2f=-\frac1x+5x^2; 2f(2)=2(20−12)=392f(2)=2(20-\frac12)=39.

Question 12

Mathematics · 4 marks · −1

Let A=[aij]A=[a_{ij}] be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is :

  1. A.58\frac{5}{8}
  2. B.38\frac{3}{8}
  3. C.316\frac{3}{16}
  4. D.18\frac{1}{8}

Correct answer: B

Explanation: 6 of the 16 matrices have ad − bc ≠ 0: 3/8.

Question 13

Mathematics · 4 marks · −1

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to :

  1. A.9100
  2. B.8575
  3. C.8750
  4. D.8925

Correct answer: D

Question 14

Mathematics · 4 marks · −1

Let the points (112,α)\left(\frac{11}{2}, \alpha\right) lie on or inside the triangle with sides x+y=11x+y=11, x+2y=16x+2y=16 and 2x+3y=292x+3y=29. Then the product of the smallest and the largest values of α\alpha is equal to :

  1. A.33
  2. B.55
  3. C.22
  4. D.44

Correct answer: A

Explanation: Vertices (6, 5), (4, 7), (10, 3). On x = 11/2 the triangle lies between x + y = 11 (y = 5.5) and 2x + 3y = 29 (y = 6): product 33.

Question 15

Mathematics · 4 marks · −1

In an arithmetic progression, if S40=1030S_{40}=1030 and S12=57S_{12}=57, then S30−S10S_{30}-S_{10} is equal to :

  1. A.505
  2. B.525
  3. C.510
  4. D.515

Correct answer: D

Explanation: 2a+39d=51.52a+39d=51.5, 2a+11d=9.52a+11d=9.5 ⇒ d = 1.5, a = −24.5; S30−S10=10(2a+39d)=515S_{30}-S_{10}=10(2a+39d)=515.

Question 16

Mathematics · 4 marks · −1

The area of the region enclosed by the curves y=exy=e^x, y=∣ex−1∣y=|e^x-1| and y-axis is :

  1. A.log⁡e2\log_e2
  2. B.1+log⁡e21+\log_e2
  3. C.2log⁡e2−12\log_e2-1
  4. D.1−log⁡e21-\log_e2

Correct answer: D

Explanation: The bounded region lies in x ≤ 0 between y=exy=e^x (above) and y=1−exy=1-e^x (below) from x=−ln⁡2x=-\ln2 (where they meet) to x = 0: ∫−ln⁡20(2ex−1)dx=1−ln⁡2\int_{-\ln2}^{0}(2e^x-1)dx=1-\ln2.

Question 17

Mathematics · 4 marks · −1

Let A={x∈(0,π)−{π2}:log⁡(2/π)∣sin⁡x∣+log⁡(2/π)∣cos⁡x∣=2}A=\left\{x\in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\} and B={x≥0:x(x−4)−3∣x−2∣+6=0}B=\{x\ge0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}. Then n(A∪B)n(A\cup B) is equal to :

  1. A.4
  2. B.8
  3. C.2
  4. D.6

Correct answer: B

Question 18

Mathematics · 4 marks · −1

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+∣x−2∣f(x)=[x]+|x-2|, −2<x<3-2<x<3, is not continuous and not differentiable. Then m + n is equal to :

  1. A.8
  2. B.9
  3. C.6
  4. D.7

Correct answer: A

Explanation: Discontinuous at −1, 0, 1, 2 (m = 4); non-differentiable at those 4 points (x = 2 also a kink) (n = 4): 8.

Question 19

Mathematics · 4 marks · −1

For some a, b, let f(x)=∣a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx∣f(x)=\begin{vmatrix}a+\frac{\sin x}{x} & 1 & b\\ a & 1+\frac{\sin x}{x} & b\\ a & 1 & b+\frac{\sin x}{x}\end{vmatrix}, x≠0x\neq0, lim⁡x→0f(x)=λ+μa+νb\lim_{x\to0}f(x)=\lambda+\mu a+\nu b. Then (λ+μ+ν)2(\lambda+\mu+\nu)^2 is equal to :

  1. A.25
  2. B.36
  3. C.9
  4. D.16

Correct answer: D

Explanation: With t=sin⁡xx→1t=\frac{\sin x}{x}\to1: f=t2(t+a+1+b)→2+a+bf=t^2(t+a+1+b)\to2+a+b, so λ = 2, μ = ν = 1: 42=164^2=16.

Question 20

Mathematics · 4 marks · −1

If α>β>γ>0\alpha>\beta>\gamma>0, then the expression cot⁡−1{β+(1+β2)(α−β)}+cot⁡−1{γ+(1+γ2)(β−γ)}+cot⁡−1{α+(1+α2)(γ−α)}\cot^{-1}\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\}+\cot^{-1}\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\}+\cot^{-1}\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\} is equal to :

  1. A.0
  2. B.π2−(α+β+γ)\frac{\pi}{2}-(\alpha+\beta+\gamma)
  3. C.3π3\pi
  4. D.π\pi

Correct answer: D

Question 21

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation 2cos⁡xdydx=sin⁡2x−4ysin⁡x2\cos x\frac{dy}{dx}=\sin2x-4y\sin x, x∈(0,π2)x\in\left(0, \frac{\pi}{2}\right). If y(π3)=0y\left(\frac{\pi}{3}\right)=0, then y′(π4)+y(π4)y'\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right) is equal to ______.

Correct answer: 1

Question 22

Mathematics · 4 marks · −1

If ∫2x2+5x+9x2+x+1dx=xx2+x+1+αx2+x+1+βlog⁡e∣x+12+x2+x+1∣+C\int\frac{2x^2+5x+9}{\sqrt{x^2+x+1}}dx=x\sqrt{x^2+x+1}+\alpha\sqrt{x^2+x+1}+\beta\log_e\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C, where C is the constant of integration, then α+2β\alpha+2\beta is equal to ______.

Correct answer: 16

Question 23

Mathematics · 4 marks · −1

Number of functions f:{1,2,…,100}→{0,1}f:\{1, 2, \ldots, 100\}\to\{0, 1\}, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to ______.

Correct answer: 392

Explanation: Choose the one among 1..98 (98 ways); 99 and 100 free (4 ways): 392.

Question 24

Mathematics · 4 marks · −1

Let H1:x2a2−y2b2=1H_1: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 and H2:−x2A2+y2B2=1H_2: -\frac{x^2}{A^2}+\frac{y^2}{B^2}=1 be two hyperbolas having length of latus rectums 15215\sqrt{2} and 12512\sqrt{5} respectively. Let their ecentricities be e1=52e_1=\sqrt{\frac{5}{2}} and e2e_2 respectively. If the product of the lengths of their transverse axes is 10010100\sqrt{10}, then 25e2225e_2^2 is equal to ______.

Correct answer: 55

Question 25

Mathematics · 4 marks · −1

Let P be the image of the point Q(7, −2, 5) in the line L: x−12=y+13=z4\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} and R(5, p, q) be a point on L. Then the square of the area of △PQR\triangle PQR is ______.

Correct answer: 957

Question 26

Physics · 4 marks · −1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R) : The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  2. B.(A) is true but (R) is false
  3. C.Both (A) and (R) are true and (R) is the correct explanation of (A)
  4. D.(A) is false but (R) is true

Correct answer: C

Question 27

Physics · 4 marks · −1

The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) :

Figure for question 27
  1. A.zero
  2. B.40π40\pi
  3. C.5π5\pi
  4. D.10π10\pi

Correct answer: C

Explanation: Area of the semicircle =π2(100×103 Pa)(100×10−6 m3)=5π=\frac{\pi}{2}(100\times10^3\,Pa)(100\times10^{-6}\,m^3)=5\pi J.

Question 28

Physics · 4 marks · −1

The output of the circuit is low (zero) for : [X and Y feed an OR gate whose output feeds both inputs of a NAND gate] (A) X=0, Y=0 (B) X=0, Y=1 (C) X=1, Y=0 (D) X=1, Y=1. Choose the correct answer from the options given below :

Figure for question 28
  1. A.(A), (B) and (C) only
  2. B.(B), (C) and (D) only
  3. C.(A), (B) and (D) only
  4. D.(A), (C) and (D) only

Correct answer: B

Explanation: Output = NOR(X, Y): low whenever any input is 1.

Question 29

Physics · 4 marks · −1

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−84\times10^{-8}C charge and then removed to a distance such that the force of repulsion between them is 9×10−39\times10^{-3} N. The distance between them is (Take 14πϵ0\frac{1}{4\pi\epsilon_0} as 9×1099\times10^9 in SI units)

  1. A.4 cm
  2. B.2 cm
  3. C.3 cm
  4. D.1 cm

Correct answer: B

Explanation: 9×10−3=9×109(2×10−8)2r29\times10^{-3}=\frac{9\times10^9(2\times10^{-8})^2}{r^2} ⇒ r = 2 cm.

Question 30

Physics · 4 marks · −1

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P1P_1 and P2P_2 are orthogonal to each other. The polarizer P3P_3 covers both the slits with its transmission axis at 45° to those of P1P_1 and P2P_2. An unpolarized light of wavelength λ\lambda and intensity I0I_0 is incident on P1P_1 and P2P_2. The intensity at a point after P3P_3 where the path difference between the light waves from s1s_1 and s2s_2 is λ3\frac{\lambda}{3}, is

Figure for question 30
  1. A.I03\frac{I_0}{3}
  2. B.I04\frac{I_0}{4}
  3. C.I02\frac{I_0}{2}
  4. D.I0I_0

Correct answer: B

Explanation: Each beam: I0/2I_0/2 after P1P_1/P2P_2, I0/4I_0/4 after P3P_3, now parallel. Phase 2π/32\pi/3: I=2⋅I04(1+cos⁡120∘)=I04I=2\cdot\frac{I_0}{4}(1+\cos120^\circ)=\frac{I_0}{4}.

Question 31

Physics · 4 marks · −1

The temperature of a body in air falls from 40°C to 24°C in 4 minutes. The temperature of the air is 16°C. The temperature of the body in the next 4 minutes will be :

  1. A.563\frac{56}{3} °C
  2. B.423\frac{42}{3} °C
  3. C.283\frac{28}{3} °C
  4. D.143\frac{14}{3} °C

Correct answer: A

Explanation: Average form: 164=k(32−16)\frac{16}{4}=k(32-16) ⇒ k=14k=\frac14; 24−T4=14(24+T2−16)\frac{24-T}{4}=\frac14\left(\frac{24+T}{2}-16\right) ⇒ T=563T=\frac{56}{3}.

Question 32

Physics · 4 marks · −1

Which of the following figure represents the relation between Celsius and Fahrenheit temperatures ?

Figure for question 32
  1. A.Straight line with positive slope, C-intercept negative (crosses the F-axis at a positive F)
  2. B.Straight line with negative slope through the origin
  3. C.Straight line with negative slope, positive C-intercept
  4. D.Straight line with positive slope through the origin

Correct answer: A

Explanation: C=59(F−32)C=\frac59(F-32): positive slope, C = 0 at F = 32 (C-intercept −160/9).

Question 33

Physics · 4 marks · −1

In the first configuration (1) as shown in the figure, four identical charges (q0q_0) are kept at the corners A, B, C and D of square of side length 'a'. In the second configuration (2), the same charges are shifted to mid points G, E, H and F, of the square. If K=14πϵ0K=\frac{1}{4\pi\epsilon_0}, the difference between the potential energies of configuration (2) and (1) is given by :

Figure for question 33
  1. A.Kq02a(3−2)\frac{Kq_0^2}{a}(3-\sqrt{2})
  2. B.Kq02a(32−2)\frac{Kq_0^2}{a}(3\sqrt{2}-2)
  3. C.Kq02a(4−22)\frac{Kq_0^2}{a}(4-2\sqrt{2})
  4. D.Kq02a(42−2)\frac{Kq_0^2}{a}(4\sqrt{2}-2)

Correct answer: B

Explanation: U1=Kq2a(4+2)U_1=\frac{Kq^2}{a}(4+\sqrt2); midpoints form a square of side a/2a/\sqrt2: U2=Kq2a(42+2)U_2=\frac{Kq^2}{a}(4\sqrt2+2); difference (32−2)(3\sqrt2-2).

Question 34

Physics · 4 marks · −1

Arrange the following in the ascending order of wavelength (λ\lambda) : (A) Microwaves (λ1\lambda_1) (B) Ultraviolet rays (λ2\lambda_2) (C) Infrared rays (λ3\lambda_3) (D) X-rays (λ4\lambda_4). Choose the most appropriate answer from the options given below :

  1. A.λ4<λ2<λ3<λ1\lambda_4<\lambda_2<\lambda_3<\lambda_1
  2. B.λ4<λ3<λ1<λ2\lambda_4<\lambda_3<\lambda_1<\lambda_2
  3. C.λ3<λ4<λ2<λ1\lambda_3<\lambda_4<\lambda_2<\lambda_1
  4. D.λ4<λ3<λ2<λ1\lambda_4<\lambda_3<\lambda_2<\lambda_1

Correct answer: A

Question 35

Physics · 4 marks · −1

A long straight wire of a circular cross-section with radius 'a' carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by

Figure for question 35
  1. A.B decreases exponentially from B0B_0 at r = 0 (to B0/eB_0/e at r = a)
  2. B.B = 0 inside, jumps at r = a and then decreases as 1/r
  3. C.B increases linearly from 0 to r = a and then decreases as 1/r
  4. D.B constant inside and then decreases as 1/r

Correct answer: C

Question 36

Physics · 4 marks · −1

The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation

  1. A.E2=pc2+m2c2E^2=pc^2+m^2c^2
  2. B.E2=p2c2+m2c2E^2=p^2c^2+m^2c^2
  3. C.E2=p2c2+m2c4E^2=p^2c^2+m^2c^4
  4. D.E2=pc2+m2c4E^2=pc^2+m^2c^4

Correct answer: C

Question 37

Physics · 4 marks · −1

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t1t_1 and t2t_2, respectively, then

  1. A.t1=2t2t_1=2t_2
  2. B.t1<t2t_1<t_2
  3. C.t1>t2t_1>t_2
  4. D.t1=t2t_1=t_2

Correct answer: B

Question 38

Physics · 4 marks · −1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statements, choose the correct answer from the options given below :

  1. A.(A) is false but (R) is true
  2. B.Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. C.Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  4. D.(A) is true but (R) is false

Correct answer: A

Question 39

Physics · 4 marks · −1

In photoelectric effect, the stopping potential (V0V_0) v/s frequency (ν\nu) curve is plotted. (h is the Planck's constant and ϕ0\phi_0 is work function of metal) (A) V0V_0 v/s ν\nu is linear. (B) The slope of V0V_0 v/s ν\nu curve =ϕ0h=\frac{\phi_0}{h} (C) h constant is related to the slope of V0V_0 v/s ν\nu line. (D) The value of electric charge of electron is not required to determine h using the V0V_0 v/s ν\nu curve. (E) The work function can be estimated without knowing the value of h. Choose the correct answer from the options given below :

  1. A.(A), (B) and (C) only
  2. B.(D) and (E) only
  3. C.(C) and (D) only
  4. D.(A), (C) and (E) only

Correct answer: D

Explanation: V0=heν−ϕ0eV_0=\frac{h}{e}\nu-\frac{\phi_0}{e}: linear, slope h/e; ϕ0=e×\phi_0=e\times(−intercept) needs no h.

Question 40

Physics · 4 marks · −1

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

  1. A.43\frac{4}{3}
  2. B.34\frac{3}{4}
  3. C.25\frac{2}{5}
  4. D.52\frac{5}{2}

Correct answer: D

Explanation: 1/21/5=52\frac{1/2}{1/5}=\frac52.

Question 41

Physics · 4 marks · −1

Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m, will be :

  1. A.0.63 mm
  2. B.0.33 mm
  3. C.0.46 mm
  4. D.0.23 mm

Correct answer: D

Explanation: β=690×10−9×0.721.44×1.5×10−3=0.23\beta=\frac{690\times10^{-9}\times0.72}{1.44\times1.5\times10^{-3}}=0.23 mm.

Question 42

Physics · 4 marks · −1

A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm × 2 cm and the area of the landscape photographed is 400 km2km^2. The focal length of the lens in the drone camera is :

  1. A.2.5 cm
  2. B.0.9 cm
  3. C.1.8 cm
  4. D.2.8 cm

Correct answer: C

Explanation: Linear magnification 2 cm20 km=10−6\frac{2\text{ cm}}{20\text{ km}}=10^{-6}; f≈18 km×10−6=1.8f\approx18\text{ km}\times10^{-6}=1.8 cm.

Question 43

Physics · 4 marks · −1

The position vector of a moving body at any instant of time is given as r⃗=(5t2i^−5tj^)\vec{r}=(5t^2\hat{i}-5t\hat{j})m. The magnitude and direction of velocity at t = 2s is,

  1. A.5155\sqrt{15} m/s, making an angle of tan⁡−14\tan^{-1}4 with −ve Y axis
  2. B.5155\sqrt{15} m/s, making an angle of tan⁡−14\tan^{-1}4 with +ve X axis
  3. C.5175\sqrt{17} m/s, making an angle of tan⁡−14\tan^{-1}4 with +ve X axis
  4. D.5175\sqrt{17} m/s, making an angle of tan⁡−14\tan^{-1}4 with −ve Y axis

Correct answer: D

Explanation: v⃗=(20,−5)\vec v=(20,-5): 5175\sqrt{17}, tan⁡−14\tan^{-1}4 from the −y axis.

Question 44

Physics · 4 marks · −1

A particle oscillates along the x-axis according to the law, x(t)=x0sin⁡2(t2)x(t)=x_0\sin^2\left(\frac{t}{2}\right) where x0x_0 = 1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph

Figure for question 44
  1. A.K is maximum at x = 0 and x = 1 and zero at x = 1/2 (U-shaped parabola)
  2. B.K is zero at x = 0 and x = 1 and maximum at x = 1/2 (inverted parabola)
  3. C.K flat near x = 0 and 1, dropping to a cusp at zero at x = 1/2
  4. D.K zero at x = 0 and 1 with a cusp maximum at x = 1/2

Correct answer: B

Explanation: x=12(1−cos⁡t)x=\frac12(1-\cos t): SHM about x = 1/2 with amplitude 1/2, so K∝14−(x−12)2K\propto\frac14-(x-\frac12)^2 — an inverted parabola.

Question 45

Physics · 4 marks · −1

N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity ω\omega as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, IA−IBI_A-I_B, for the given Amperian loops is

Figure for question 45
  1. A.N2πqω\frac{N}{2\pi}q\omega
  2. B.2πNqω\frac{2\pi}{N}q\omega
  3. C.Nπqω\frac{N}{\pi}q\omega
  4. D.N22πqω\frac{N^2}{2\pi}q\omega

Correct answer: A

Explanation: Loop A: current through a section of the ring =Nqω2π=\frac{Nq\omega}{2\pi}; loop B: net 0.

Question 46

Physics · 4 marks · −1

Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ______ g.

Correct answer: 9

Explanation: g ∝ 1/R²: 9g.

Question 47

Physics · 4 marks · −1

The increase in pressure required to decrease the volume of a water sample by 0.2% is P×105 Nm−2P\times10^5\ Nm^{-2}. Bulk modulus of water is 2.15×109 Nm−22.15\times10^9\ Nm^{-2}. The value of P is ______.

Correct answer: 43

Explanation: ΔP=2.15×109×0.002=43×105\Delta P=2.15\times10^9\times0.002=43\times10^5.

Question 48

Physics · 4 marks · −1

The ratio of the power of a light source S1S_1 to that the light source S2S_2 is 2. S1S_1 is emitting 2×10152\times10^{15} photons per second at 600 nm. If the wavelength of the source S2S_2 is 300 nm, then the number of photons per second emitted by S2S_2 is ______ ×1014\times10^{14}.

Correct answer: 5

Explanation: n∝Pλn\propto P\lambda: n2=2×1015×12×12=5×1014n_2=2\times10^{15}\times\frac12\times\frac12=5\times10^{14}.

Question 49

Physics · 4 marks · −1

A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes (3π)\left(\frac{3}{\pi}\right) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ______ ML.

Figure for question 49

Correct answer: 36

Explanation: T=Mω2LT=M\omega^2L with ω=2π⋅3π=6\omega=2\pi\cdot\frac3\pi=6: 36 ML.

Question 50

Physics · 4 marks · −1

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is ______. [Take mass of electron me=9×10−31m_e=9\times10^{-31} kg, charge of electron ∣qe∣=1.6×10−19|q_e|=1.6\times10^{-19} C, μ0=4π×10−7 NA2\mu_0=4\pi\times10^{-7}\ \frac{N}{A^2}, 1 ns = 10−910^{-9} s]

Correct answer: 250

Explanation: B=2πmeT=μ0nIB=\frac{2\pi m}{eT}=\mu_0nI ⇒ n=2π(9×10−31)1.6×10−19(75×10−9)(4π×10−7)(1.5)=250n=\frac{2\pi(9\times10^{-31})}{1.6\times10^{-19}(75\times10^{-9})(4\pi\times10^{-7})(1.5)}=250.

Question 51

Chemistry · 4 marks · −1

When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be :

  1. A.Green → Pale Blue → Blue → Violet
  2. B.Violet → Blue → Pale Blue → Green
  3. C.Pale Blue → Blue → Green → Violet
  4. D.Pale Blue → Blue → Violet → Green

Correct answer: A

Question 52

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : [graph of t1/2t_{1/2} vs [R]0[R]_0 is a horizontal line] is valid for first order reaction. Statement (II) : [graph of log⁡[R][R]0\log\frac{[R]}{[R]_0} vs time is a straight line through the origin with positive slope = k2.303\frac{k}{2.303}] is valid for first order reaction. In the light of the above statements, choose the correct answer from the options given below :

Figure for question 52
  1. A.Both Statement I and Statement II are false
  2. B.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are true
  4. D.Statement I is true but Statement II is false

Correct answer: D

Explanation: log⁡[R][R]0=−kt2.303\log\frac{[R]}{[R]_0}=-\frac{kt}{2.303} has negative slope, so II is false.

Question 53

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) Adenine (B) Cytosine (C) Thymine (D) Uracil. List - II: (I) pyrimidine-2,4-dione (uracil skeleton, no methyl) (II) 5-methylpyrimidine-2,4-dione (III) 6-aminopurine (IV) 4-aminopyrimidin-2-one. Choose the correct answer from the options given below :

Figure for question 53
  1. A.(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. B.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  3. C.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. D.(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct answer: A

Question 54

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : The first ionization energy of Pb is greater than that of Sn. Statement (II) : The first ionization energy of Ge is greater than that of Si. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is true but Statement II is false
  2. B.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are false
  4. D.Both Statement I and Statement II are true

Correct answer: A

Explanation: IE₁: Pb 715 > Sn 708 kJ/mol; Ge 762 < Si 786.

Question 55

Chemistry · 4 marks · −1

For the reaction, H2(g)+I2(g)⇌2HI(g)H_2(g)+I_2(g)\rightleftharpoons2HI(g). Attainment of equilibrium is predicted correctly by :

Figure for question 55
  1. A.H2H_2 and I2I_2 curves fall from high values and level off (H2H_2 above I2I_2); HI rises from 0 and levels off below both
  2. B.All three (H2H_2, I2I_2, HI) rise from 0 and level off
  3. C.H2H_2 rises and levels off at top; I2I_2 and HI fall and level off
  4. D.I2I_2 rises and levels off at top; H2H_2 falls; HI rises and levels off below

Correct answer: A

Question 56

Chemistry · 4 marks · −1

In the given structure, number of sp and sp2sp^2 hybridized carbon atoms present respectively are : CH3−CO−CH=CH−CH2−C≡C−CH=CH−C≡NCH_3-CO-CH=CH-CH_2-C\equiv C-CH=CH-C\equiv N

Figure for question 56
  1. A.4 and 6
  2. B.4 and 5
  3. C.3 and 6
  4. D.3 and 5

Correct answer: D

Explanation: sp: two alkyne C + nitrile C = 3; sp²: C=O carbon + 4 alkene C = 5.

Question 57

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Experimentally determined oxygen-oxygen bond lengths in the O3O_3 are found to be same and the bond length is greater than that of a O=O (double bond) but less than that of a single (O−O) bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O=O) but more than that of a single bond (O−O). In the light of the above statements, choose the correct answer from the options given below :

  1. A.Both Statement I and Statement II are true
  2. B.Statement I is true but Statement II is false
  3. C.Both Statement I and Statement II are false
  4. D.Statement I is false but Statement II is true

Correct answer: B

Question 58

Chemistry · 4 marks · −1

For hydrogen atom, the orbital/s with lowest energy is/are : (A) 4s (B) 3px3p_x (C) 3dx2−y23d_{x^2-y^2} (D) 3dz23d_{z^2} (E) 4pz4p_z. Choose the correct answer from the options given below :

  1. A.(A) only
  2. B.(B) only
  3. C.(A) and (E) only
  4. D.(B), (C) and (D) only

Correct answer: D

Question 59

Chemistry · 4 marks · −1

S(g)+32O2(g)→SO3(g)+2xS(g)+\frac{3}{2}O_2(g)\to SO_3(g)+2x kcal; SO2(g)+12O2(g)→SO3(g)+ySO_2(g)+\frac{1}{2}O_2(g)\to SO_3(g)+y kcal. The heat of formation of SO2(g)SO_2(g) is given by :

  1. A.y − 2x kcal
  2. B.2xy\frac{2x}{y} kcal
  3. C.x + y kcal
  4. D.2x + y kcal

Correct answer: A

Explanation: ΔHf(SO2)=−2x−(−y)=y−2x\Delta H_f(SO_2)=-2x-(-y)=y-2x.

Question 60

Chemistry · 4 marks · −1

The conditions and consequence that favours the t2g3eg1t_{2g}^3e_g^1 configuration in a metal complex are :

  1. A.strong field ligand, high spin complex
  2. B.strong field ligand, low spin complex
  3. C.weak field ligand, low spin complex
  4. D.weak field ligand, high spin complex

Correct answer: D

Question 61

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) RCN→(i) SnCl2,HCl; (ii) H3O+RCHORCN\xrightarrow{\text{(i) } SnCl_2, HCl;\ \text{(ii) } H_3O^+}RCHO (B) Benzoyl chloride →H2,Pd−BaSO4\xrightarrow{H_2, Pd-BaSO_4} benzaldehyde (C) Toluene →(i) CrO2Cl2,CS2; (ii) H3O+\xrightarrow{\text{(i) } CrO_2Cl_2, CS_2;\ \text{(ii) } H_3O^+} benzaldehyde (D) Benzene →(i) CO, HCl; (ii) anhydrous AlCl3/CuCl\xrightarrow{\text{(i) CO, HCl; (ii) anhydrous } AlCl_3/CuCl} benzaldehyde. List - II: (I) Etard reaction (II) Gatterman-Koch reaction (III) Rosenmund reduction (IV) Stephen reaction. Choose the correct answer from the options given below :

  1. A.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  2. B.(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  3. C.(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  4. D.(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Correct answer: D

Question 62

Chemistry · 4 marks · −1

The structure of the major product formed in the following reaction is : 3-bromo-5-iodobenzyl chloride →AgCN\xrightarrow{AgCN} major product

Figure for question 62
  1. A.1-bromo-3-(chloromethyl)-5-cyanobenzene (CN replaces ring I)
  2. B.3-bromo-5-iodobenzyl cyanide (ArCH2CNArCH_2CN)
  3. C.3-bromo-5-iodobenzyl isocyanide (ArCH2NCArCH_2NC)
  4. D.3-iodo-5-cyanobenzyl chloride (CN replaces ring Br)

Correct answer: C

Explanation: AgCN (covalent) attacks through N at the benzylic carbon: isocyanide; aryl halides are unreactive.

Question 63

Chemistry · 4 marks · −1

Based on the data given below : ECr2O72−/Cr3+∘=1.33E^\circ_{Cr_2O_7^{2-}/Cr^{3+}}=1.33 V, ECl2/Cl−∘=1.36E^\circ_{Cl_2/Cl^-}=1.36 V, EMnO4−/Mn2+∘=1.51E^\circ_{MnO_4^-/Mn^{2+}}=1.51 V, ECr3+/Cr∘=−0.74E^\circ_{Cr^{3+}/Cr}=-0.74 V, the strongest reducing agent is :

  1. A.MnO4−MnO_4^-
  2. B.Mn2+Mn^{2+}
  3. C.Cl−Cl^-
  4. D.Cr

Correct answer: D

Question 64

Chemistry · 4 marks · −1

The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ/mol, respectively. By using the above values predict the group in which the above element is present :

  1. A.Group 13
  2. B.Group 4
  3. C.Group 2
  4. D.Group 14

Correct answer: A

Explanation: Big jump after the 3rd IE: 3 valence electrons (boron).

Question 65

Chemistry · 4 marks · −1

For reaction Aniline →\to 2-Bromoaniline (Major). The correct order of set of reagents for the above conversion is :

  1. A.Ac2OAc_2O, Br2Br_2, H2O(Δ)H_2O(\Delta), NaOH
  2. B.H2SO4H_2SO_4, Ac2OAc_2O, Br2Br_2, H2O(Δ)H_2O(\Delta), NaOH
  3. C.Ac2OAc_2O, H2SO4H_2SO_4, Br2Br_2, NaOH
  4. D.Br2∣FeBr3Br_2|FeBr_3, H2O(Δ)H_2O(\Delta), NaOH

Correct answer: B

Explanation: Sulphonate to block the para position (sulphanilic acid), acetylate the amino group to moderate it, brominate ortho, then hot aqueous hydrolysis removes the SO3HSO_3H and acetyl groups; NaOH liberates 2-bromoaniline.

Question 66

Chemistry · 4 marks · −1

Find the compound 'A' from the following reaction sequences. A →aqua-regia\xrightarrow{\text{aqua-regia}} B →(1) KNO2∣NH4OH; (2) AcOH\xrightarrow{\text{(1) } KNO_2|NH_4OH;\ \text{(2) AcOH}} yellow ppt

  1. A.NiS
  2. B.ZnS
  3. C.MnS
  4. D.CoS

Correct answer: D

Explanation: Yellow K3[Co(NO2)6]K_3[Co(NO_2)_6] confirms cobalt.

Question 67

Chemistry · 4 marks · −1

The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O. If the molar mass of the compound is 132 g mol−1mol^{-1}, the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16]

  1. A.C4H8O2C_4H_8O_2
  2. B.C6H12O6C_6H_{12}O_6
  3. C.C4H9O3C_4H_9O_3
  4. D.C6H12O3C_6H_{12}O_3

Correct answer: D

Explanation: Empirical C2H4OC_2H_4O (44) × 3 = 132.

Question 68

Chemistry · 4 marks · −1

Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature ?

  1. A.30 mL CH3COOHCH_3COOH and 30 mL NaOH
  2. B.45 mL CH3COOHCH_3COOH and 25 mL NaOH
  3. C.30 mL HCl and 30 mL NaOH
  4. D.50 mL HCl and 20 mL NaOH

Correct answer: C

Explanation: Strong acid–strong base, 30 mmol neutralised in 60 mL gives the largest heat per unit volume.

Question 69

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Transition metal ion): (A) Ti3+Ti^{3+} (B) V2+V^{2+} (C) Ni2+Ni^{2+} (D) Sc3+Sc^{3+}. List - II (Spin only magnetic moment (B.M.)): (I) 3.87 (II) 0.00 (III) 1.73 (IV) 2.84. Choose the correct answer from the options given below :

  1. A.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. B.(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. C.(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  4. D.(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Correct answer: A

Question 70

Chemistry · 4 marks · −1

Identify correct statement/s : (A) −OCH3-OCH_3 and −NHCOCH3-NHCOCH_3 are activating group. (B) −CN and −OH are meta directing group. (C) −CN and −SO3H-SO_3H are meta directing group. (D) Activating groups act as ortho - and para directing groups. (E) Halides are activating groups. Choose the correct answer from the options given below :

  1. A.(A) only
  2. B.(A), (C) and (D) only
  3. C.(A), (B) and (E) only
  4. D.(A) and (C) only

Correct answer: B

Question 71

Chemistry · 4 marks · −1

Consider a complex reaction taking place in three steps with rate constants k1k_1, k2k_2 and k3k_3 respectively. The overall rate constant k is given by the expression k=k1k3k2k=\sqrt{\frac{k_1k_3}{k_2}}. If the activation energies of the three steps are 60, 30 and 10 kJ mol−1mol^{-1} respectively, then the overall energy of activation in kJ mol−1mol^{-1} is ______. (Nearest integer)

Correct answer: 20

Explanation: E=12(60+10−30)=20E=\frac12(60+10-30)=20.

Question 72

Chemistry · 4 marks · −1

The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ______.

Correct answer: 4

Explanation: One stereocentre (C-2) and a stereogenic C=C: 4.

Question 73

Chemistry · 4 marks · −1

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ______ ×10−1\times10^{-1}% (Nearest integer). (Given : Molar mass of Ag is 108 and Br is 80 g mol−1mol^{-1})

Correct answer: 255

Explanation: Br = 0.15 × 80/188 = 0.0638 g → 25.5% = 255 × 10⁻¹.

Question 74

Chemistry · 4 marks · −1

The hydrocarbon (X) with molar mass 80 g mol−1mol^{-1} and 90% carbon has ______ degree of unsaturation.

Correct answer: 3

Explanation: C = 72 g (6 C), H = 8: C6H8C_6H_8, DU = 3.

Question 75

Chemistry · 4 marks · −1

The observed and normal molar masses of compound MX2MX_2 are 65.6 and 164 respectively. The percent degree of ionisation of MX2MX_2 is ______%. (Nearest integer)

Correct answer: 75

Explanation: i=164/65.6=2.5=1+2αi=164/65.6=2.5=1+2\alpha ⇒ α = 75%.

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