JEE Main 2025 — 4 Apr Shift 1 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
73
Total marks
292
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

1+3+52+7+92+…1+3+5^2+7+9^2+\ldots upto 40 terms is equal to

  1. A.40870
  2. B.43890
  3. C.41880
  4. D.33980

Correct answer: C

Explanation: Odd positions: (4k−3)2(4k-3)^2, k = 1..20, sum 16(2870)−24(210)+180=4106016(2870)-24(210)+180=41060; even positions: 4k−14k-1, sum 820; total 41880.

Question 2

Mathematics · 4 marks · −1

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is

  1. A.35\frac{3}{5}
  2. B.215\frac{2}{15}
  3. C.1115\frac{11}{15}
  4. D.2875\frac{28}{75}

Correct answer: D

Explanation: P(0)=21/45, P(1)=21/45, P(2)=3/45; E = 3/5, E(X²) = 11/15; Var = 11/15 − 9/25 = 28/75.

Question 3

Mathematics · 4 marks · −1

The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is

  1. A.103182\frac{103}{182}
  2. B.129182\frac{129}{182}
  3. C.1926\frac{19}{26}
  4. D.1726\frac{17}{26}

Correct answer: B

Question 4

Mathematics · 4 marks · −1

Consider the equation x2+4x−n=0x^2+4x-n=0, where n∈[20,100]n\in[20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to

  1. A.6
  2. B.5
  3. C.8
  4. D.7

Correct answer: A

Explanation: n+4n+4 must be a perfect square in [24, 104]: 25, 36, 49, 64, 81, 100 → 6.

Question 5

Mathematics · 4 marks · −1

Consider two vectors u⃗=3i^−j^\vec{u}=3\hat{i}-\hat{j} and v⃗=2i^+j^−λk^\vec{v}=2\hat{i}+\hat{j}-\lambda\hat{k}, λ>0\lambda>0. The angle between them is given by cos⁡−1(527)\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right). Let v⃗=v⃗1+v⃗2\vec{v}=\vec{v}_1+\vec{v}_2, where v⃗1\vec{v}_1 is parallel to u⃗\vec{u} and v⃗2\vec{v}_2 is perpendicular to u⃗\vec{u}. Then the value ∣v⃗1∣2+∣v⃗2∣2|\vec{v}_1|^2+|\vec{v}_2|^2 is equal to

  1. A.252\frac{25}{2}
  2. B.232\frac{23}{2}
  3. C.14
  4. D.10

Correct answer: C

Explanation: 5105+λ2=527\frac{5}{\sqrt{10}\sqrt{5+\lambda^2}}=\frac{\sqrt5}{2\sqrt7} ⇒ λ2=9\lambda^2=9; ∣v∣2=14|v|^2=14.

Question 6

Mathematics · 4 marks · −1

Let A={1,6,11,16,…}A=\{1, 6, 11, 16, \ldots\} and B={9,16,23,30,…}B=\{9, 16, 23, 30, \ldots\} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B)n(A\cup B) is

  1. A.3814
  2. B.4027
  3. C.3761
  4. D.4003

Correct answer: C

Question 7

Mathematics · 4 marks · −1

For an integer n≥2n\ge2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3(x+y)^{2n-3} is 16, then the distance of the point P(2n−1,n2−4n)(2n-1, n^2-4n) from the line x+y=8x+y=8 is

  1. A.222\sqrt{2}
  2. B.323\sqrt{2}
  3. C.2\sqrt{2}
  4. D.525\sqrt{2}

Correct answer: B

Explanation: 22n−32n−2=16\frac{2^{2n-3}}{2n-2}=16 ⇒ n = 5: P(9, 5); distance ∣9+5−8∣2=32\frac{|9+5-8|}{\sqrt2}=3\sqrt2.

Question 8

Mathematics · 4 marks · −1

In the expansion of (23+133)n\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n∈Nn\in\mathbf{N}, if the ratio of 15th15^{th} term from the beginning to the 15th15^{th} term from the end is 16\frac{1}{6}, then the value of nC3{}^nC_3 is

  1. A.2300
  2. B.4960
  3. C.4060
  4. D.1040

Correct answer: A

Explanation: Ratio =6(n−28)/3=6−1=6^{(n-28)/3}=6^{-1} ⇒ n = 25; 25C3=2300{}^{25}C_3=2300.

Question 9

Mathematics · 4 marks · −1

The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, −3) and eccentricity is 45\frac{4}{5}, is

  1. A.65\frac{6}{5}
  2. B.503\frac{50}{3}
  3. C.103\frac{10}{3}
  4. D.185\frac{18}{5}

Correct answer: D

Explanation: be=4be=4 (major axis vertical), b=5b=5, a2=25(1−1625)=9a^2=25(1-\frac{16}{25})=9; LR =2a2b=185=\frac{2a^2}{b}=\frac{18}{5}.

Question 10

Mathematics · 4 marks · −1

Let the three sides of a triangle are on the lines 4x−7y+10=04x-7y+10=0, x+y=5x+y=5 and 7x+4y=157x+4y=15. Then the distance of its orthocentre from the orthocentre of the tringle formed by the lines x=0x=0, y=0y=0 and x+y=1x+y=1 is

  1. A.5\sqrt{5}
  2. B.20\sqrt{20}
  3. C.5
  4. D.20

Correct answer: A

Explanation: 4x−7y+10=0⊥7x+4y=154x-7y+10=0\perp7x+4y=15, so the orthocentre is their intersection (1, 2); the second orthocentre is (0, 0): distance 5\sqrt5.

Question 11

Mathematics · 4 marks · −1

If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x\to1^+}\frac{(x-1)(6+\lambda\cos(x-1))+\mu\sin(1-x)}{(x-1)^3}=-1, where λ,μ∈R\lambda, \mu\in\mathbf{R}, then λ+μ\lambda+\mu is equal to

  1. A.19
  2. B.17
  3. C.20
  4. D.18

Correct answer: D

Explanation: With t=x−1t=x-1 the numerator is (6+λ−μ)t+(μ6−λ2)t3+…(6+\lambda-\mu)t+\left(\frac{\mu}{6}-\frac{\lambda}{2}\right)t^3+\ldots: μ=6+λ\mu=6+\lambda and μ6−λ2=−1\frac{\mu}{6}-\frac{\lambda}{2}=-1 ⇒ λ = 6, μ = 12; sum 18.

Question 12

Mathematics · 4 marks · −1

Let A and B be two distinct points on the line L: x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}. Both A and B are at a distance 2172\sqrt{17} from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then OA⃗⋅OB⃗\vec{OA}\cdot\vec{OB} is equal to

  1. A.47
  2. B.49
  3. C.21
  4. D.62

Correct answer: A

Question 13

Mathematics · 4 marks · −1

Let the shortest distance between the lines x−33=y−α−1=z−31\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1} and x+3−3=y+72=z−β4\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4} be 3303\sqrt{30}. Then the positive value of 5α+β5\alpha+\beta is

  1. A.40
  2. B.46
  3. C.48
  4. D.42

Correct answer: B

Question 14

Mathematics · 4 marks · −1

Considering the principal values of the inverse trigonometric functions, sin⁡−1(32x+121−x2)\sin^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-x^2}\right), −12<x<12-\frac{1}{2}<x<\frac{1}{\sqrt{2}}, is equal to

  1. A.π4+sin⁡−1x\frac{\pi}{4}+\sin^{-1}x
  2. B.π6+sin⁡−1x\frac{\pi}{6}+\sin^{-1}x
  3. C.5π6−sin⁡−1x\frac{5\pi}{6}-\sin^{-1}x
  4. D.−5π6−sin⁡−1x\frac{-5\pi}{6}-\sin^{-1}x

Correct answer: B

Explanation: With x=sin⁡θx=\sin\theta: sin⁡(θ+π6)\sin(\theta+\frac\pi6) and θ+π6∈(0,5π12)\theta+\frac\pi6\in(0,\frac{5\pi}{12}).

Question 15

Mathematics · 4 marks · −1

If 10sin⁡4θ+15cos⁡4θ=610\sin^4\theta+15\cos^4\theta=6, then the value of 27 cosec6θ+8sec⁡6θ16sec⁡8θ\frac{27\,\mathrm{cosec}^6\theta+8\sec^6\theta}{16\sec^8\theta} is

  1. A.15\frac{1}{5}
  2. B.35\frac{3}{5}
  3. C.25\frac{2}{5}
  4. D.34\frac{3}{4}

Correct answer: C

Explanation: sin⁡2θ=35\sin^2\theta=\frac35, cos⁡2θ=25\cos^2\theta=\frac25: numerator 27⋅12527+8⋅1258=25027\cdot\frac{125}{27}+8\cdot\frac{125}{8}=250; denominator 16⋅62516=62516\cdot\frac{625}{16}=625: 2/5.

Question 16

Mathematics · 4 marks · −1

Consider the sets A={(x,y)∈R×R:x2+y2=25}A=\{(x, y)\in\mathbf{R}\times\mathbf{R}: x^2+y^2=25\}, B={(x,y)∈R×R:x2+9y2=144}B=\{(x, y)\in\mathbf{R}\times\mathbf{R}: x^2+9y^2=144\}, C={(x,y)∈Z×Z:x2+y2≤4}C=\{(x, y)\in\mathbf{Z}\times\mathbf{Z}: x^2+y^2\le4\} and D=A∩BD=A\cap B. The total number of one-one functions from the set D to the set C is:

  1. A.15120
  2. B.18290
  3. C.19320
  4. D.17160

Correct answer: D

Explanation: D has 4 points; C has 13 lattice points: 13⋅12⋅11⋅10=1716013\cdot12\cdot11\cdot10=17160.

Question 17

Mathematics · 4 marks · −1

Let f:[0,∞)→Rf:[0, \infty)\to\mathbf{R} be a differentiable function such that f(x)=1−2x+∫0xex−tf(t) dtf(x)=1-2x+\int_0^xe^{x-t}f(t)\,dt for all x∈[0,∞)x\in[0, \infty). Then the area of the region bounded by y=f(x)y=f(x) and the coordinate axes is

  1. A.2
  2. B.12\frac{1}{2}
  3. C.2\sqrt{2}
  4. D.5\sqrt{5}

Correct answer: B

Explanation: Differentiating gives f=ˊ2f−3+2xf\'=2f-3+2x with f(0)=1f(0)=1, whose solution is f(x)=1−xf(x)=1-x; area of the triangle with the axes =12=\frac12.

Question 18

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a continuous function satisfying f(0)=1f(0)=1 and f(2x)−f(x)=xf(2x)-f(x)=x for all x∈Rx\in\mathbf{R}. If lim⁡n→∞{f(x)−f(x2n)}=G(x)\lim_{n\to\infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=G(x), then ∑r=110G(r2)\sum_{r=1}^{10}G(r^2) is equal to

  1. A.420
  2. B.215
  3. C.385
  4. D.540

Correct answer: C

Explanation: Telescoping: G(x)=xG(x)=x; ∑r2=385\sum r^2=385.

Question 19

Mathematics · 4 marks · −1

The value of ∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−xdx\int_{-1}^{1}\frac{\left(1+\sqrt{|x|-x}\right)e^x+\left(\sqrt{|x|-x}\right)e^{-x}}{e^x+e^{-x}}dx is equal to

  1. A.1−2231-\frac{2\sqrt{2}}{3}
  2. B.1+2231+\frac{2\sqrt{2}}{3}
  3. C.2+2232+\frac{2\sqrt{2}}{3}
  4. D.3−2233-\frac{2\sqrt{2}}{3}

Correct answer: B

Explanation: =∫−11exex+e−xdx+∫−11∣x∣−x dx=1+∫−10−2x dx=1+223=\int_{-1}^1\frac{e^x}{e^x+e^{-x}}dx+\int_{-1}^{1}\sqrt{|x|-x}\,dx=1+\int_{-1}^0\sqrt{-2x}\,dx=1+\frac{2\sqrt2}{3}.

Question 20

Mathematics · 4 marks · −1

Let f,g:(1,∞)→Rf, g:(1, \infty)\to\mathbf{R} be defined as f(x)=2x+35x+2f(x)=\frac{2x+3}{5x+2} and g(x)=2−3x1−xg(x)=\frac{2-3x}{1-x}. If the range of the function f∘g:[2,4]→Rf\circ g:[2, 4]\to\mathbf{R} is [α,β][\alpha, \beta], then 1β−α\frac{1}{\beta-\alpha} is equal to

  1. A.68
  2. B.56
  3. C.29
  4. D.2

Correct answer: B

Question 21

Mathematics · 4 marks · −1

Let A={z∈C:∣z−2−i∣=3}A=\{z\in\mathbf{C}: |z-2-i|=3\}, B={z∈C:Re(z−iz)=2}B=\{z\in\mathbf{C}: \mathrm{Re}(z-iz)=2\} and S=A∩BS=A\cap B. Then ∑z∈S∣z∣2\sum_{z\in S}|z|^2 is equal to ______.

Correct answer: 22

Question 22

Mathematics · 4 marks · −1

Let m and n be the number of points at which the function f(x)=max⁡{x,x3,x5,…,x21}f(x)=\max\{x, x^3, x^5, \ldots, x^{21}\}, x∈Rx\in\mathbf{R}, is not differentiable and not continuous, respectively. Then m + n is equal to ______.

Correct answer: 3

Explanation: f = x on (−∞,−1], x21x^{21} on [−1,0], x on [0,1], x21x^{21} on [1,∞): kinks at −1, 0, 1 → m = 3, n = 0.

Question 23

Mathematics · 4 marks · −1

If the area of the region {(x,y):∣x−5∣≤y≤4x}\{(x, y): |x-5|\le y\le4\sqrt{x}\} is A, then 3A is equal to ______.

Correct answer: 368

Question 24

Mathematics · 4 marks · −1

Let C be the circle x2+(y−1)2=2x^2+(y-1)^2=2, E1E_1 and E2E_2 be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line x+y=3x+y=3 touch the curves C, E1E_1 and E2E_2 at P(x1x_1, y1y_1), Q(x2x_2, y2y_2) and R(x3x_3, y3y_3) respectively. Given that P is the mid point of the line segment QR and PQ=223PQ=\frac{2\sqrt{2}}{3}, the value of 9(x1y1+x2y2+x3y3)9(x_1y_1+x_2y_2+x_3y_3) is equal to ______.

Correct answer: 46

Question 25

Mathematics · 4 marks · −1

Let A=[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]A=\begin{bmatrix}\cos\theta & 0 & -\sin\theta\\ 0 & 1 & 0\\ \sin\theta & 0 & \cos\theta\end{bmatrix}. If for some θ∈(0,π)\theta\in(0, \pi), A2=ATA^2=A^T, then the sum of the diagonal elements of the matrix (A+I)3+(A−I)3−6A(A+I)^3+(A-I)^3-6A is equal to ______.

Correct answer: 6

Question 26

Physics · 4 marks · −1

In an experiment with a closed organ pipe, it is filled with water by (15)\left(\frac{1}{5}\right)th of its volume. The frequency of the fundamental note will change by

  1. A.25%
  2. B.−20%
  3. C.−25%
  4. D.20%

Correct answer: A

Explanation: Air column becomes 4L/5: f∝1/Lf\propto1/L increases by 25%.

Question 27

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 photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below

  1. A.A is true but R is false
  2. B.Both A and R are true but R is NOT the correct explanation of A
  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: D

Question 28

Physics · 4 marks · −1

Two infinite identical charged sheets and a charged spherical body of charge density 'ρ\rho' are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is:

Figure for question 28
  1. A.E⃗A>E⃗B\vec{E}_A>\vec{E}_B; E⃗C=E⃗D\vec{E}_C=\vec{E}_D
  2. B.E⃗C≠E⃗D\vec{E}_C\neq\vec{E}_D; E⃗A>E⃗B\vec{E}_A>\vec{E}_B
  3. C.∣E⃗A∣=∣E⃗B∣|\vec{E}_A|=|\vec{E}_B|; E⃗C>E⃗D\vec{E}_C>\vec{E}_D
  4. D.E⃗A=E⃗B\vec{E}_A=\vec{E}_B; E⃗C=E⃗D\vec{E}_C=\vec{E}_D

Correct answer: B

Question 29

Physics · 4 marks · −1

The Boolean expression Y=ABˉC+AˉCˉY=A\bar{B}C+\bar{A}\bar{C} can be realised with which of the following gate configurations. A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate, B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate C. 3-input OR gate, 3 NOT gates and one 2-input AND gate. Choose the correct answer from the options given below:

  1. A.A, C Only
  2. B.A, B Only
  3. C.A, B, C Only
  4. D.B, C Only

Correct answer: B

Explanation: A: direct SOP. B: AˉCˉ=A+C‾\bar A\bar C=\overline{A+C} (NOR) and ABˉCA\bar BC with one NOT; then OR.

Question 30

Physics · 4 marks · −1

An alternating current is represented by the equation, i=1002sin⁡(100πt)i=100\sqrt{2}\sin(100\pi t) ampere. The RMS value of current and the frequency of the given alternating current are

  1. A.1002\frac{100}{\sqrt{2}}A, 100 Hz
  2. B.1002100\sqrt{2}A, 100 Hz
  3. C.100 A, 50 Hz
  4. D.50250\sqrt{2}A, 50 Hz

Correct answer: C

Question 31

Physics · 4 marks · −1

When an object is placed 40 cm away from a spherical mirror an image of magnification 12\frac{1}{2} is produced. To obtain an image with magnification of 13\frac{1}{3}, the object is to be moved :

  1. A.80 cm away from the mirror.
  2. B.20 cm towards the mirror.
  3. C.20 cm away from the mirror.
  4. D.40 cm away from the mirror.

Correct answer: D

Explanation: Convex mirror: m=ff−um=\frac{f}{f-u}: 12=ff+40\frac12=\frac{f}{f+40} ⇒ f = 40; 13\frac13 needs |u| = 80: move 40 cm away.

Question 32

Physics · 4 marks · −1

Current passing through a wire as function of time is given as I(t)=0.02t+0.01I(t)=0.02t+0.01 A. The charge that will flow through the wire from t = 1 s to t = 2 s is

  1. A.0.06 C
  2. B.0.04 C
  3. C.0.02 C
  4. D.0.07 C

Correct answer: B

Explanation: ∫12(0.02t+0.01)dt=0.03+0.01=0.04\int_1^2(0.02t+0.01)dt=0.03+0.01=0.04 C.

Question 33

Physics · 4 marks · −1

A body of mass m is suspended by two strings making angles θ1\theta_1 and θ2\theta_2 with the horizontal ceiling with tensions T1T_1 and T2T_2 simultaneously. T1T_1 and T2T_2 are related by T1=3T2T_1=\sqrt{3}T_2, the angles θ1\theta_1 and θ2\theta_2 are

  1. A.θ1=30∘\theta_1=30^\circ θ2=60∘\theta_2=60^\circ with T2=4mg5T_2=\frac{4mg}{5}
  2. B.θ1=60∘\theta_1=60^\circ θ2=30∘\theta_2=30^\circ with T2=mg2T_2=\frac{mg}{2}
  3. C.θ1=45∘\theta_1=45^\circ θ2=45∘\theta_2=45^\circ with T2=3mg4T_2=\frac{3mg}{4}
  4. D.θ1=30∘\theta_1=30^\circ θ2=60∘\theta_2=60^\circ with T2=3mg4T_2=\frac{3mg}{4}

Correct answer: B

Explanation: T1cos⁡θ1=T2cos⁡θ2T_1\cos\theta_1=T_2\cos\theta_2 with T1=3T2T_1=\sqrt3T_2 fits θ₁ = 60°, θ₂ = 30°; vertical balance gives T2=mg/2T_2=mg/2.

Question 34

Physics · 4 marks · −1

The mean free path and the average speed of oxygen molecules at 300K and 1 atm are 3×10−73\times10^{-7} m and 600 m/s, respectively. Find the frequency of its collisions.

  1. A.2×10102\times10^{10} /s
  2. B.9×1059\times10^5 /s
  3. C.2×1092\times10^9 /s
  4. D.5×1085\times10^8 /s

Correct answer: C

Explanation: 600/3×10−7=2×109600/3\times10^{-7}=2\times10^9 /s.

Question 35

Physics · 4 marks · −1

A small mirror of mass m is suspended by a massless thread of length ll. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c = speed of light in vacuum and g = acceleration due to gravity)

  1. A.θ=Emcgl\theta=\frac{E}{mc\sqrt{gl}}
  2. B.θ=E2mcgl\theta=\frac{E}{2mc\sqrt{gl}}
  3. C.θ=3E4mcgl\theta=\frac{3E}{4mc\sqrt{gl}}
  4. D.θ=2Emcgl\theta=\frac{2E}{mc\sqrt{gl}}

Correct answer: D

Explanation: Momentum given 2E/c2E/c (reflection): v=2Emcv=\frac{2E}{mc}; v=θglv=\theta\sqrt{gl} ⇒ θ=2Emcgl\theta=\frac{2E}{mc\sqrt{gl}}.

Question 36

Physics · 4 marks · −1

Two small spherical balls of mass 10g each with charges −2μC-2\mu C and 2μC2\mu C, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of 100μC/m2100\mu C/m^2 such that length of rod makes an angle of 30° with electric field generated by charge sheet. Net torque acting on the rod is: (Take ϵ0\epsilon_0 : 8.85×10−12 C2/Nm28.85\times10^{-12}\ C^2/Nm^2)

  1. A.11.2 Nm
  2. B.1.12 Nm
  3. C.112 Nm
  4. D.2.24 Nm

Correct answer: B

Explanation: E=σ2ϵ0=5.65×106E=\frac{\sigma}{2\epsilon_0}=5.65\times10^6; τ=pEsin⁡30∘=(4×10−7)(5.65×106)(0.5)=1.13\tau=pE\sin30^\circ=(4\times10^{-7})(5.65\times10^6)(0.5)=1.13 N m.

Question 37

Physics · 4 marks · −1

Consider the sound wave travelling in ideal gases of He, CH4CH_4, and CO2CO_2. All the gases have the same ratio Pρ\frac{P}{\rho}, where P is the pressure and ρ\rho is the density. The ratio of the speed of sound through the gases vHe:vCH4:vCO2v_{He}:v_{CH_4}:v_{CO_2} is given by

  1. A.43:53:75\sqrt{\frac{4}{3}}:\sqrt{\frac{5}{3}}:\sqrt{\frac{7}{5}}
  2. B.53:43:43\sqrt{\frac{5}{3}}:\sqrt{\frac{4}{3}}:\sqrt{\frac{4}{3}}
  3. C.53:43:75\sqrt{\frac{5}{3}}:\sqrt{\frac{4}{3}}:\sqrt{\frac{7}{5}}
  4. D.75:53:43\sqrt{\frac{7}{5}}:\sqrt{\frac{5}{3}}:\sqrt{\frac{4}{3}}

Correct answer: B

Explanation: v=γP/ρv=\sqrt{\gamma P/\rho}: γ = 5/3 (He), 4/3 (CH₄, non-linear polyatomic), 7/5 (CO₂ linear, rigid).

Question 38

Physics · 4 marks · −1

Two simple pendulums having lengths l1l_1 and l2l_2 with negligible string mass undergo angular displacements θ1\theta_1 and θ2\theta_2, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

  1. A.θ1l22=θ2l12\theta_1l_2^2=\theta_2l_1^2
  2. B.θ1l2=θ2l1\theta_1l_2=\theta_2l_1
  3. C.θ1l1=θ2l2\theta_1l_1=\theta_2l_2
  4. D.θ1l12=θ2l22\theta_1l_1^2=\theta_2l_2^2

Correct answer: B

Explanation: α=glθ\alpha=\frac{g}{l}\theta.

Question 39

Physics · 4 marks · −1

Two liquids A and B have θA\theta_A and θB\theta_B as contact angles in a capillary tube. If K=cos⁡θA/cos⁡θBK=\cos\theta_A/\cos\theta_B, then identify the correct statement:

  1. A.K is zero, then liquid A has convex meniscus and liquid B has concave meniscus.
  2. B.K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.
  3. C.K is negative, then liquid A and liquid B have convex meniscus.
  4. D.K is negative, then liquid A and liquid B have concave meniscus.

Correct answer: B

Question 40

Physics · 4 marks · −1

Considering the Bohr model of hydrogen like atoms, the ratio of the radius of 5th5^{th} orbit of the electron in Li2+Li^{2+} and He+He^+ is

  1. A.49\frac{4}{9}
  2. B.94\frac{9}{4}
  3. C.23\frac{2}{3}
  4. D.32\frac{3}{2}

Correct answer: C

Explanation: r∝1/Zr\propto1/Z: 2/3.

Question 41

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: The kinetic energy needed to project a body of mass m from earth surface to infinity is 12\frac{1}{2}mgR, where R is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface. 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.Both A and R are true and R is the correct explanation of A
  3. C.A is true but R is false
  4. D.A is false but R is true

Correct answer: D

Explanation: Escape KE = mgR (not ½mgR); PE at infinity (its maximum) is zero.

Question 42

Physics · 4 marks · −1

In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is :

  1. A.25%
  2. B.0%
  3. C.100%
  4. D.50%

Correct answer: D

Explanation: β ∝ 1/d: halves, a 50% change.

Question 43

Physics · 4 marks · −1

If L⃗\vec{L} and P⃗\vec{P} represent the angular momentum and linear momentum respectively of a particle of mass 'm' having position vector as r⃗=a(i^cos⁡ωt+j^sin⁡ωt)\vec{r}=a(\hat{i}\cos\omega t+\hat{j}\sin\omega t). The direction of force is

  1. A.Opposite to the direction of r⃗\vec{r}
  2. B.Opposite to the direction of L⃗\vec{L}
  3. C.Opposite to the direction of P⃗\vec{P}
  4. D.Opposite to the direction of L⃗×P⃗\vec{L}\times\vec{P}

Correct answer: A

Question 44

Physics · 4 marks · −1

In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of MPLQTRASM^PL^QT^RA^S, where value of 'Q' and 'R' are

  1. A.(−2, 2)
  2. B.(3, −5)
  3. C.(1, −1)
  4. D.(−2, 1)

Correct answer: C

Explanation: ϕE/ϕB\phi_E/\phi_B has the dimensions of velocity: Q = 1, R = −1.

Question 45

Physics · 4 marks · −1

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 12\frac{1}{\sqrt{2}} Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is ______ mV.

Figure for question 45

Correct answer: 10

Explanation: Effective length = vertical separation AE =2×10sin⁡45∘=102=2\times10\sin45^\circ=10\sqrt2 cm; ε=12(0.12)(0.1)=0.01\varepsilon=\frac{1}{\sqrt2}(0.1\sqrt2)(0.1)=0.01 V = 10 mV.

Question 46

Physics · 4 marks · −1

Four capacitors each of capacitance 16μF\mu F are connected as shown in the figure. The capacitance between points A and B is : ______ (in μF\mu F).

Figure for question 46

Correct answer: 64

Explanation: All four capacitors are in parallel between A and B: 4 × 16 = 64 μF.

Question 47

Physics · 4 marks · −1

Two slabs with square cross section of different materials (1, 2) with equal sides (ll) and thickness d1d_1 and d2d_2 such that d2=2d1d_2=2d_1 and l>d2l>d_2. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is θ2=2θ1\theta_2=2\theta_1. If the shear moduli of material 1 is 4×109 N/m24\times10^9\ N/m^2, then shear moduli of material 2 is x×109 N/m2x\times10^9\ N/m^2, where value of x is ______.

Correct answer: 1

Explanation: Shear area l⋅dl\cdot d: θ=Fldη\theta=\frac{F}{ld\eta}; θ2=2θ1\theta_2=2\theta_1 with d2=2d1d_2=2d_1 ⇒ η2=η1/4=1\eta_2=\eta_1/4=1.

Question 48

Physics · 4 marks · −1

Distance between object and its image (magnified by −13-\frac{1}{3}) is 30 cm. The focal length of the mirror used is (x4)\left(\frac{x}{4}\right) cm, where magnitude of value of x is ______

Correct answer: 45

Explanation: v = u/3 (same side), |u| − |v| = 30 ⇒ u = −45, v = −15; f = −45/4, |x| = 45.

Question 49

Chemistry · 4 marks · −1

The major product (A) formed in the following reaction sequence is Nitrobenzene →(i) Sn, HCl; (ii) Ac2O,Pyridine; (iii) Br2,AcOH; (iv) NaOH (aq)\xrightarrow{\text{(i) Sn, HCl; (ii) } Ac_2O, \text{Pyridine; (iii) } Br_2, \text{AcOH; (iv) NaOH (aq)}} A

Figure for question 49
  1. A.2,4-Dibromoaniline
  2. B.3-Bromoaniline
  3. C.4-Bromoaniline
  4. D.2,4,6-Tribromoaniline

Correct answer: C

Question 50

Chemistry · 4 marks · −1

Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (Bohr\'s radius is represented by a0a_0)

  1. A.The electron can be found at a distance 2a02a_0 from the nucleus
  2. B.The probability density of finding the electron is maximum at the nucleus
  3. C.The total energy of the electron is maximum when it is at a distance a0a_0 from the nucleus
  4. D.The 1s orbital is spherically symmetrical

Correct answer: C

Explanation: The total energy of the 1s electron is fixed (−13.6 eV); it does not vary with distance.

Question 51

Chemistry · 4 marks · −1

Identify the pair of reactants that upon reaction, with elimination of HCl will give rise to the dipeptide Gly-Ala.

  1. A.NH2−CH2−COOHNH_2-CH_2-COOH and NH2−CH(CH3)−COClNH_2-CH(CH_3)-COCl
  2. B.NH2−CH2−COClNH_2-CH_2-COCl and N+H3−CH(CH3)−COCl\overset{+}{N}H_3-CH(CH_3)-COCl
  3. C.NH2−CH2−COOHNH_2-CH_2-COOH and NH2−CH(CH3)−COOHNH_2-CH(CH_3)-COOH
  4. D.NH2−CH2−COClNH_2-CH_2-COCl and NH2−CH(CH3)−COOHNH_2-CH(CH_3)-COOH

Correct answer: D

Question 52

Chemistry · 4 marks · −1

XY is the membrane/partition between two chambers 1 and 2 containing sugar solutions of concentration c1c_1 and c2c_2 (c1>c2c_1>c_2) mol L−1L^{-1}. For the reverse osmosis to take place identify the correct condition. (Here p1p_1 and p2p_2 are pressures applied on chamber 1 and 2). A. Membrane/Partition : Cellophane, p1>πp_1>\pi B. Membrane/Partition : Porous, p2>πp_2>\pi C. Membrane/Partition : Parchment paper, p1>πp_1>\pi D. Membrane/Partition : Cellophane, p2>πp_2>\pi. Choose the correct answer from the option given below:

Figure for question 52
  1. A.C Only
  2. B.A and C Only
  3. C.B and D Only
  4. D.A and D Only

Correct answer: B

Question 53

Chemistry · 4 marks · −1

Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. (B < Al), (Al < Ga), (Ga < In) and (In < Tl). Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius (M3+M^{3+}) than the other one. The atomic number of the element (X) is

  1. A.81
  2. B.49
  3. C.13
  4. D.31

Correct answer: D

Explanation: Ga (135 pm) < Al (143 pm), so (Al < Ga) is wrong; Ga3+Ga^{3+} (62 pm) > Al3+Al^{3+} (53.5 pm): Z = 31.

Question 54

Chemistry · 4 marks · −1

Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z). Identify the structure of compound (Z) from the following option.

  1. A.Hydroquinone (1,4-dihydroxybenzene)
  2. B.Catechol (1,2-dihydroxybenzene)
  3. C.Phenol
  4. D.Benzene

Correct answer: D

Explanation: X = benzenesulphonic acid, Y = phenol, Zn dust reduces phenol to benzene.

Question 55

Chemistry · 4 marks · −1

Given below are two statements. Statement I: The dipole moment of C4H3−C3H=C2H−C1H=O\overset{4}{C}H_3-\overset{3}{C}H=\overset{2}{C}H-\overset{1}{C}H=O is greater than C4H3−C3H2−C2H2−C1H=O\overset{4}{C}H_3-\overset{3}{C}H_2-\overset{2}{C}H_2-\overset{1}{C}H=O. Statement II: C1−C2C_1-C_2 bond length of CH3−CH=CH−CH=OCH_3-CH=CH-CH=O is greater than C1−C2C_1-C_2 bond length of CH3−CH2−CH2−CH=OCH_3-CH_2-CH_2-CH=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 false but Statement II is true
  3. C.Both Statement I and Statement II are false
  4. D.Statement I is true but Statement II is false

Correct answer: D

Question 56

Chemistry · 4 marks · −1

Pair of transition metal ions having the same number of unpaired electrons is

  1. A.Ti2+Ti^{2+}, Co2+Co^{2+}
  2. B.Ti3+Ti^{3+}, Mn2+Mn^{2+}
  3. C.V2+V^{2+}, Co2+Co^{2+}
  4. D.Fe3+Fe^{3+}, Cr2+Cr^{2+}

Correct answer: C

Explanation: V2+V^{2+} (d3d^3) and Co2+Co^{2+} (d7d^7): 3 each.

Question 57

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of pπ\pi - pπ\pi bond with oxygen. Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it. In the light of given statements, choose the correct answer from the options given below.

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

Correct answer: D

Question 58

Chemistry · 4 marks · −1

Number of stereoisomers possible for the complexes, [CrCl3(py)3][CrCl_3(py)_3] and [CrCl2(ox)2]3−[CrCl_2(ox)_2]^{3-} are respectively (py = pyridine, ox = oxalate)

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

Correct answer: C

Explanation: MA3B3MA_3B_3: fac + mer = 2; M(AA)2B2M(AA)_2B_2: trans + pair of cis enantiomers = 3.

Question 59

Chemistry · 4 marks · −1

Which of the following molecule(s) show/s paramagnetic behavior? A. O2O_2 B. N2N_2 C. F2F_2 D. S2S_2 E. Cl2Cl_2. Choose the correct answer from the options given below:

  1. A.A & E Only
  2. B.A & C only
  3. C.B Only
  4. D.A & D Only

Correct answer: D

Question 60

Chemistry · 4 marks · −1

One mole of an ideal gas expands isothermally and reversibly from 10 dm3dm^3 to 20 dm3dm^3 at 300 K. ΔU, q and work done in the process respectively are Given: R = 8.3 J K−1K^{-1} mol−1mol^{-1}, ln 10 = 2.3, log 2 = 0.30, log 3 = 0.48

  1. A.0, 1.718 kJ, −1.718 kJ
  2. B.0, −17.18 kJ, 1.718 J
  3. C.0, 21.84 kJ, 21.84 kJ
  4. D.0, 21.84 kJ, −1.726 J

Correct answer: A

Explanation: q=nRTln⁡2=8.3×300×2.3×0.3=1718q=nRT\ln2=8.3\times300\times2.3\times0.3=1718 J; w = −q.

Question 61

Chemistry · 4 marks · −1

Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH and ΔS were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at:

  1. A.T = Te
  2. B.Te = 5T
  3. C.Te > T
  4. D.T > Te

Correct answer: D

Question 62

Chemistry · 4 marks · −1

For A2+B2⇌2ABA_2+B_2\rightleftharpoons2AB, EaE_a for forward and backward reaction are 180 and 200 kJ mol−1mol^{-1} respectively. If catalyst lowers EaE_a for both reaction by 100 kJ mol−1mol^{-1}. Which of the following statement is correct?

  1. A.The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.
  2. B.The enthalpy change for the reaction is +20 kJ mol−1mol^{-1}.
  3. C.Catalyst does not alter the Gibbs energy change of a reaction.
  4. D.Catalyst can cause non-spontaneous reactions to occur.

Correct answer: C

Question 63

Chemistry · 4 marks · −1

Predict the major product of the following reaction sequence:- Methylcyclohexane →(1) Br2/hν; (2) Alcoholic KOH, Δ; (3) H–Br/R–O–O–R, hν\xrightarrow{\text{(1) } Br_2/h\nu;\ \text{(2) Alcoholic KOH, }\Delta;\ \text{(3) H–Br/R–O–O–R, }h\nu} ?

  1. A.Bromocycloheptane
  2. B.1-Bromo-2-methylcyclohexane
  3. C.(Bromomethyl)cyclohexane
  4. D.1-Bromo-1-methylcyclohexane

Correct answer: B

Explanation: Tertiary bromide → 1-methylcyclohexene → anti-Markovnikov HBr addition gives 1-bromo-2-methylcyclohexane.

Question 64

Chemistry · 4 marks · −1

On charging the lead storage battery, the oxidation state of lead changes from x1x_1 to y1y_1 at the anode and from x2x_2 to y2y_2 at the cathode. The values of x1,y1,x2,y2x_1, y_1, x_2, y_2 are respectively :

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

Correct answer: C

Explanation: Electrodes named as in discharge: at the negative plate (anode) PbSO4PbSO_4 (+2) → Pb (0); at the positive plate (cathode) PbSO4PbSO_4 (+2) → PbO2PbO_2 (+4).

Question 65

Chemistry · 4 marks · −1

An organic compound (X) with molecular formula C3H6OC_3H_6O is not readily oxidised. On reduction it gives C3H8OC_3H_8O(Y) which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2, 3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:

  1. A.CH3COCH3CH_3COCH_3, CH3CH(OH)CH3CH_3CH(OH)CH_3, CH3CH(Br)CH3CH_3CH(Br)CH_3
  2. B.CH3CH2CHOCH_3CH_2CHO, CH3CH2CH2OHCH_3CH_2CH_2OH, CH3CH2CH2BrCH_3CH_2CH_2Br
  3. C.CH3CH2CHOCH_3CH_2CHO, CH3CH=CH2CH_3CH=CH_2, CH3CH(Br)CH3CH_3CH(Br)CH_3
  4. D.CH3COCH3CH_3COCH_3, CH3CH2CH2OHCH_3CH_2CH_2OH, CH3CH(Br)CH3CH_3CH(Br)CH_3

Correct answer: A

Question 66

Chemistry · 4 marks · −1

Aldol condensation is a popular and classical method to prepare α,β\alpha,\beta-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?

Figure for question 66
  1. A.2-Methylene-1-tetralone (exocyclic C=CH2 alpha to the ketone of 1-tetralone)
  2. B.Cyclohex-2-en-1-one
  3. C.Bicyclic enone: octahydronaphthalenone with the C=C at the ring fusion (Δ1,9-octalin-2-one type)
  4. D.Inden-1-one (benzo-fused cyclopentenone)

Correct answer: A

Question 67

Chemistry · 4 marks · −1

Which one of the following complexes will have Δo=0\Delta_o=0 and μ\mu = 5.96 B.M?

  1. A.[FeF6]4−[FeF_6]^{4-}
  2. B.[Fe(CN)6]4−[Fe(CN)_6]^{4-}
  3. C.[Mn(SCN)6]4−[Mn(SCN)_6]^{4-}
  4. D.[Co(NH3)6]3+[Co(NH_3)_6]^{3+}

Correct answer: C

Explanation: High-spin d5d^5 (Mn2+Mn^{2+} with weak-field SCN⁻): CFSE 0, 5 unpaired electrons.

Question 68

Chemistry · 4 marks · −1

Rate law for a reaction between A and B is given by r=k[A]n[B]mr=k[A]^n[B]^m. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction (r2r1)\left(\frac{r_2}{r_1}\right) is

  1. A.12m+n\frac{1}{2^{m+n}}
  2. B.(n − m)
  3. C.2(n−m)2^{(n-m)}
  4. D.(m + n)

Correct answer: C

Question 69

Chemistry · 4 marks · −1

The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is ______ 5'- G - G-C-A-A-A-T-C-G-G-C-T-A-3'

Correct answer: 33

Explanation: G/C = 7 pairs × 3 = 21; A/T = 6 pairs × 2 = 12; total 33.

Question 70

Chemistry · 4 marks · −1

The pH of a 0.01 M weak acid HX (Ka=4×10−10K_a=4\times10^{-10}) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as x×10−4x\times10^{-4}M. The value of x is ______ (nearest integer)

Correct answer: 25

Explanation: [H+]2=Kac[H^+]^2=K_ac: 10−12=4×10−10c10^{-12}=4\times10^{-10}c ⇒ c=2.5×10−3c=2.5\times10^{-3} = 25 × 10⁻⁴.

Question 71

Chemistry · 4 marks · −1

Fortification of food with iron is done using FeSO4⋅7H2OFeSO_4\cdot7H_2O. The mass in grams of the FeSO4⋅7H2OFeSO_4\cdot7H_2O required to achieve 12 ppm of iron in 150 kg of wheat is ______ (Nearest integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16 g mol−1mol^{-1}]

Correct answer: 9

Explanation: Fe = 1.8 g; × 278/56 = 8.94 ≈ 9 g.

Question 72

Chemistry · 4 marks · −1

In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is ______% (nearest integer) (Aqueous tension at 300 K = 15 mm Hg)

Correct answer: 20

Explanation: n=(885/760)(0.15)0.0821×300=7.09×10−3n=\frac{(885/760)(0.15)}{0.0821\times300}=7.09\times10^{-3} mol; 0.1985 g → 20%.

Question 73

Chemistry · 4 marks · −1

KMnO4KMnO_4 acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is ______

Correct answer: 10

Explanation: X = 7 − 2 = 5; basic ferric acetate contains Fe3+Fe^{3+} (d5d^5): 5 + 5 = 10.

View all papers — JEE Main

Practice this paper with a timer

Attempt it online, get scored instantly, and see which subjects are costing you marks.

Start practising free