JEE Main 2025 — 3 Apr Shift 2 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
74
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The shortest distance between the curves y2=8xy^2=8x and x2+y2+12y+35=0x^2+y^2+12y+35=0 is:

  1. A.23−12\sqrt{3}-1
  2. B.22−12\sqrt{2}-1
  3. C.2\sqrt{2}
  4. D.32−13\sqrt{2}-1

Correct answer: B

Explanation: Circle centre (0,−6), r = 1. Normal to parabola at (2t2,4t)(2t^2,4t) through (0,−6): t=−1t=-1, point (2,−4), distance 222\sqrt2; answer 22−12\sqrt2-1.

Question 2

Mathematics · 4 marks · −1

Line L1L_1 of slope 2 and line L2L_2 of slope 12\frac{1}{2} intersect at the origin O. In the first quadrant, P1,P2,…,P12P_1, P_2, \ldots, P_{12} are 12 points on line L1L_1 and Q1,Q2,…,Q9Q_1, Q_2, \ldots, Q_9 are 9 points on line L2L_2. Then the total number of triangles, that can be formed having vertices at three of the 22 points O, P1,P2,…,P12P_1, P_2, \ldots, P_{12}, Q1,Q2,…,Q9Q_1, Q_2, \ldots, Q_9, is:

  1. A.1026
  2. B.1080
  3. C.1134
  4. D.1188

Correct answer: C

Explanation: (223)−(133)−(103)=1540−286−120=1134\binom{22}{3}-\binom{13}{3}-\binom{10}{3}=1540-286-120=1134.

Question 3

Mathematics · 4 marks · −1

The number of solutions of the equation (4−3)sin⁡x−23cos⁡2x=−41+3(4-\sqrt{3})\sin x-2\sqrt{3}\cos^2x=-\frac{4}{1+\sqrt{3}}, x∈[−2π,5π2]x\in\left[-2\pi, \frac{5\pi}{2}\right] is

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

Correct answer: A

Question 4

Mathematics · 4 marks · −1

If the domain of the function f(x)=log⁡7(1−log⁡4(x2−9x+18))f(x)=\log_7(1-\log_4(x^2-9x+18)) is (α,β)∪(γ,δ)(\alpha, \beta)\cup(\gamma, \delta), then α+β+γ+δ\alpha+\beta+\gamma+\delta is equal to

  1. A.18
  2. B.17
  3. C.16
  4. D.15

Correct answer: A

Explanation: 0<x2−9x+18<40<x^2-9x+18<4: (2,3)∪(6,7)(2,3)\cup(6,7); sum 18.

Question 5

Mathematics · 4 marks · −1

If the probability that the random variable X takes the value x is given by P(X=x)=k(x+1)3−xP(X=x)=k(x+1)3^{-x}, x = 0, 1, 2, 3, ..., where k is a constant, then P(X≥3)P(X\ge3) is equal to

  1. A.827\frac{8}{27}
  2. B.727\frac{7}{27}
  3. C.49\frac{4}{9}
  4. D.19\frac{1}{9}

Correct answer: D

Explanation: ∑(x+1)3−x=94\sum(x+1)3^{-x}=\frac{9}{4} so k=49k=\frac49; P(0)+P(1)+P(2)=49(1+23+13)=89P(0)+P(1)+P(2)=\frac49\left(1+\frac23+\frac13\right)=\frac89, hence P(X≥3)=19P(X\ge3)=\frac19.

Question 6

Mathematics · 4 marks · −1

The integral ∫0π8x dx4cos⁡2x+sin⁡2x\int_0^{\pi}\frac{8x\,dx}{4\cos^2x+\sin^2x} is equal to

  1. A.4π24\pi^2
  2. B.π2\pi^2
  3. C.3π22\frac{3\pi^2}{2}
  4. D.2π22\pi^2

Correct answer: D

Explanation: =4π∫0πdx4cos⁡2x+sin⁡2x=4π⋅2⋅π4=2π2=4\pi\int_0^\pi\frac{dx}{4\cos^2x+\sin^2x}=4\pi\cdot2\cdot\frac{\pi}{4}=2\pi^2.

Question 7

Mathematics · 4 marks · −1

Consider the lines x(3λ+1)+y(7λ+2)=17λ+5x(3\lambda+1)+y(7\lambda+2)=17\lambda+5, λ\lambda being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of d2d^2 is

  1. A.30
  2. B.10
  3. C.20
  4. D.15

Correct answer: C

Explanation: P = (1, 2); L ⟂ OP: x+2y=5x+2y=5; distance of (3,6): 10/510/\sqrt5, d2=20d^2=20.

Question 8

Mathematics · 4 marks · −1

Let A = {−2, −1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if y = max{x, 1}. Let ll be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+nl+m+n is equal to

  1. A.12
  2. B.14
  3. C.13
  4. D.11

Correct answer: A

Explanation: R = {(−2,1),(−1,1),(0,1),(1,1),(2,2),(3,3)}: l = 6; reflexive needs 3 more, symmetric needs 3: 12.

Question 9

Mathematics · 4 marks · −1

Each of the angles β\beta and γ\gamma that a given line makes with the positive y- and z-axes, respectively, is half of the angle that this line makes with the positive x-axes. Then the sum of all possible values of the angle β\beta is

  1. A.π2\frac{\pi}{2}
  2. B.3π2\frac{3\pi}{2}
  3. C.3π4\frac{3\pi}{4}
  4. D.π\pi

Correct answer: C

Explanation: cos⁡22β+2cos⁡2β=1\cos^22\beta+2\cos^2\beta=1 ⇒ cos⁡2β(cos⁡2β+1)=0\cos2\beta(\cos2\beta+1)=0: β=π4\beta=\frac\pi4 or π2\frac\pi2; sum 3π4\frac{3\pi}{4}.

Question 10

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation dydx+3(tan⁡2x)y+3y=sec⁡2x\frac{dy}{dx}+3(\tan^2x)y+3y=\sec^2x, y(0)=13+e3y(0)=\frac{1}{3}+e^3. Then y(π4)y\left(\frac{\pi}{4}\right) is equal to

  1. A.23+e3\frac{2}{3}+e^3
  2. B.23\frac{2}{3}
  3. C.43\frac{4}{3}
  4. D.43+e3\frac{4}{3}+e^3

Correct answer: C

Explanation: y+ˊ3sec⁡2x y=sec⁡2xy\'+3\sec^2x\,y=\sec^2x: y=13+Ce−3tan⁡xy=\frac13+Ce^{-3\tan x}, C = e3e^3; y(π/4)=13+1=43y(\pi/4)=\frac13+1=\frac43.

Question 11

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a function defined by f(x)=∣∣x+2∣−2∣x∣∣f(x)=||x+2|-2|x||. If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is

  1. A.3
  2. B.4
  3. C.5
  4. D.2

Correct answer: A

Question 12

Mathematics · 4 marks · −1

Let the equation x(x+2)(12−k)=2x(x+2)(12-k)=2 have equal roots. Then the distance of the point (k,k2)\left(k, \frac{k}{2}\right) from the line 3x+4y+5=03x+4y+5=0 is

  1. A.15
  2. B.12
  3. C.535\sqrt{3}
  4. D.15515\sqrt{5}

Correct answer: A

Explanation: (12−k)x2+2(12−k)x−2=0(12-k)x^2+2(12-k)x-2=0; D=4(12−k)2+8(12−k)=0⇒k=14D=4(12-k)^2+8(12-k)=0\Rightarrow k=14; distance of (14, 7): 42+28+55=15\frac{42+28+5}{5}=15.

Question 13

Mathematics · 4 marks · −1

If the four distinct points (4, 6), (−1-1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then 10k+r210k+r^2 is equal to

  1. A.33
  2. B.35
  3. C.32
  4. D.34

Correct answer: B

Question 14

Mathematics · 4 marks · −1

Let C be the circle of minimum area enclosing the ellipse E: x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with eccentricity 12\frac{1}{2} and foci (±2\pm2, 0). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 2a is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is :

  1. A.8(2+3)8(2+\sqrt{3})
  2. B.6(3+2)6(3+\sqrt{2})
  3. C.8(3+2)8(3+\sqrt{2})
  4. D.6(2+3)6(2+\sqrt{3})

Correct answer: A

Explanation: a = 4, b=23b=2\sqrt3, C: radius 4. QR on y=−23y=-2\sqrt3, length 8; max height 4+234+2\sqrt3; area 4(4+23)=8(2+3)4(4+2\sqrt3)=8(2+\sqrt3).

Question 15

Mathematics · 4 marks · −1

If z1,z2,z3∈Cz_1, z_2, z_3\in\mathbf{C} are the vertices of an equilateral triangle, whose centroid is z0z_0, then ∑k=13(zk−z0)2\sum_{k=1}^{3}(z_k-z_0)^2 is equal to

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

Correct answer: C

Explanation: zk−z0=rωkeiθz_k-z_0=r\omega^k e^{i\theta}; ∑ω2k=0\sum\omega^{2k}=0.

Question 16

Mathematics · 4 marks · −1

The area of the region {(x,y):∣x−y∣≤y≤4x}\{(x, y): |x-y|\le y\le4\sqrt{x}\} is

  1. A.5123\frac{512}{3}
  2. B.10243\frac{1024}{3}
  3. C.20483\frac{2048}{3}
  4. D.512

Correct answer: B

Question 17

Mathematics · 4 marks · −1

The distance of the point (7, 10, 11) from the line x−41=y−40=z−23\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3} along the line x−92=y−133=z−176\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6} is

  1. A.12
  2. B.14
  3. C.18
  4. D.16

Correct answer: B

Explanation: Point (7+2t,10+3t,11+6t)(7+2t,10+3t,11+6t) on first line needs y=4y=4: t = −2 → (3,4,−1), which satisfies z−2=3(x−4)z-2=3(x-4); distance 2×7=142\times7=14.

Question 18

Mathematics · 4 marks · −1

Let f be a function such that f(x)+3f(24x)=4xf(x)+3f\left(\frac{24}{x}\right)=4x, x≠0x\neq0. Then f(3)+f(8)f(3)+f(8) is equal to

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

Correct answer: D

Explanation: f(3)+3f(8)=12f(3)+3f(8)=12, f(8)+3f(3)=32f(8)+3f(3)=32: sum = 44/4 = 11.

Question 19

Mathematics · 4 marks · −1

Let the Mean and Variance of five observations x1=1x_1=1, x2=3x_2=3, x3=ax_3=a, x4=7x_4=7 and x5=bx_5=b, a > b, be 5 and 10 respectively. Then the Variance of the observations n+xnn+x_n, n = 1, 2, ..., 5 is

  1. A.17.4
  2. B.16.4
  3. C.17
  4. D.16

Correct answer: D

Explanation: a+b=14a+b=14, a2+b2=5(35)−59=116a^2+b^2=5(35)-59=116 ⇒ a = 10, b = 4. New data 2, 5, 13, 11, 9: mean 8, variance 36+9+25+9+15=16\frac{36+9+25+9+1}{5}=16.

Question 20

Mathematics · 4 marks · −1

The sum 1+1+32!+1+3+53!+1+3+5+74!+…1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots upto ∞\infty terms, is equal to

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

Correct answer: D

Explanation: ∑n2n!=2e\sum\frac{n^2}{n!}=2e.

Question 21

Mathematics · 4 marks · −1

If lim⁡x→0(tan⁡xx)1x2=p\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}=p, then 96log⁡ep96\log_ep is equal to ______.

Correct answer: 32

Explanation: ln⁡p=lim⁡1x2⋅x23=13\ln p=\lim\frac{1}{x^2}\cdot\frac{x^2}{3}=\frac13; answer 32.

Question 22

Mathematics · 4 marks · −1

If the equation of the hyperbola with foci (4, 2) and (8, 2) is 3x2−y2−αx+βy+γ=03x^2-y^2-\alpha x+\beta y+\gamma=0, then α+β+γ\alpha+\beta+\gamma is equal to ______.

Correct answer: 141

Explanation: Centre (6,2), c = 2, b2=3a2b^2=3a^2 ⇒ a=1a=1, b2=3b^2=3: 3(x−6)2−(y−2)2=33(x-6)^2-(y-2)^2=3 → 3x2−y2−36x+4y+101=03x^2-y^2-36x+4y+101=0; 36 + 4 + 101 = 141.

Question 23

Mathematics · 4 marks · −1

Let a⃗=i^+2j^+k^\vec{a}=\hat{i}+2\hat{j}+\hat{k}, b⃗=3i^−3j^+3k^\vec{b}=3\hat{i}-3\hat{j}+3\hat{k}, c⃗=2i^−j^+2k^\vec{c}=2\hat{i}-\hat{j}+2\hat{k} and d⃗\vec{d} be a vector such that b⃗×d⃗=c⃗×d⃗\vec{b}\times\vec{d}=\vec{c}\times\vec{d} and a⃗⋅d⃗=4\vec{a}\cdot\vec{d}=4. Then ∣(a⃗×d⃗)∣2|(\vec{a}\times\vec{d})|^2 is equal to ______.

Correct answer: 128

Explanation: d⃗=λ(b⃗−c⃗)=λ(1,−2,1)\vec d=\lambda(\vec b-\vec c)=\lambda(1,-2,1); a⃗⋅d⃗=−2λ=4\vec a\cdot\vec d=-2\lambda=4 ⇒ d⃗=(−2,4,−2)\vec d=(-2,4,-2); ∣a⃗×d⃗∣2=∣a∣2∣d∣2−(a⋅d)2=6⋅24−16=128|\vec a\times\vec d|^2=|a|^2|d|^2-(a\cdot d)^2=6\cdot24-16=128.

Question 24

Mathematics · 4 marks · −1

Let (1+x+x2)10=a0+a1x+a2x2+…+a20x20(1+x+x^2)^{10}=a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}. If (a1+a3+a5+…+a19)−11a2=121k(a_1+a_3+a_5+\ldots+a_{19})-11a_2=121k, then k is equal to ______.

Correct answer: 239

Question 25

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 : If Oxygen ion (O−2O^{-2}) and Hydrogen ion (H+H^+) enter normal to the magnetic field with equal momentum, then the path of O−2O^{-2} ion has a smaller curvature than that of H+H^+. Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: C

Explanation: r=p/(qB)r=p/(qB): with equal momentum the doubly charged O2−O^{2-} has half the radius of H+H^+; a proton and an electron with equal momentum have equal radii, so R is false.

Question 26

Physics · 4 marks · −1

A particle moves along the x–axis and has its displacement x varying with time t according to the equation: x=c0(t2−2)+c(t−2)2x=c_0(t^2-2)+c(t-2)^2 where c0c_0 and c are constants of appropriate dimensions. Then, which of the following statements is correct?

  1. A.the acceleration of the particle is 2c02c_0
  2. B.the acceleration of the particle is 2c
  3. C.the acceleration of the particle is 2(c+c0)2(c+c_0)
  4. D.the initial velocity of the particle is 4c

Correct answer: C

Question 27

Physics · 4 marks · −1

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

  1. A.(3+22):(3−22)(3+2\sqrt{2}):(3-2\sqrt{2})
  2. B.3 : 1
  3. C.9 : 1
  4. D.(22+1):(22−1)(2\sqrt{2}+1):(2\sqrt{2}-1)

Correct answer: A

Explanation: I∝I\propto width: amplitudes 1:21:\sqrt2; (2+1)2(2−1)2=3+223−22\frac{(\sqrt2+1)^2}{(\sqrt2-1)^2}=\frac{3+2\sqrt2}{3-2\sqrt2}.

Question 28

Physics · 4 marks · −1

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is: (in pF)

  1. A.600
  2. B.200
  3. C.100
  4. D.400

Correct answer: B

Explanation: 6000=(100+C)20⇒C=2006000=(100+C)20\Rightarrow C=200 pF.

Question 29

Physics · 4 marks · −1

A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45×10−22.45\times10^{-2} m/s while falling under gravity through an oil of density 925 kg m−3m^{-3}. Take density of steel as 7825 kg m−3m^{-3} and g as 9.8 m/s2m/s^2. The viscosity of the oil in SI unit is

  1. A.1.99
  2. B.1.68
  3. C.2.18
  4. D.2.38

Correct answer: A

Explanation: η=2r2(ρ−σ)g9v=2(1.8×10−3)2(6900)(9.8)9(2.45×10−2)≈1.99\eta=\frac{2r^2(\rho-\sigma)g}{9v}=\frac{2(1.8\times10^{-3})^2(6900)(9.8)}{9(2.45\times10^{-2})}\approx1.99.

Question 30

Physics · 4 marks · −1

A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is 91.6%, then the loss of power in units of cal/s is

  1. A.6.2
  2. B.2
  3. C.4
  4. D.8.4

Correct answer: B

Explanation: Loss = 8.4 W = 8.4/4.2 = 2 cal/s.

Question 31

Physics · 4 marks · −1

Consider two blocks A and B of masses m1m_1 = 10 kg and m2m_2 = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constant k = 3000 N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

Figure for question 31
  1. A.0.1 m
  2. B.0.3 m
  3. C.0.4 m
  4. D.0.2 m

Correct answer: A

Explanation: 12μv2=12kx2\frac12\mu v^2=\frac12kx^2 with μ=10/3\mu=10/3: x2=(10/3)(9)3000=0.01x^2=\frac{(10/3)(9)}{3000}=0.01, x = 0.1 m.

Question 32

Physics · 4 marks · −1

A monochromatic light of frequency 5×10145\times10^{14} Hz travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

  1. A.500 nm
  2. B.600 nm
  3. C.300 nm
  4. D.400 nm

Correct answer: C

Explanation: λ0=600\lambda_0=600 nm; in medium 300 nm.

Question 33

Physics · 4 marks · −1

The truth table corresponding to the circuit given below is : [A and B feed an OR gate; the OR output and A feed an AND gate whose output is C]

Figure for question 33
  1. A.A B C: 0 0 0; 1 0 0; 0 1 0; 1 1 1
  2. B.A B C: 0 0 0; 0 1 0; 1 0 1; 1 1 1
  3. C.A B C: 0 0 1; 0 1 0; 1 0 0; 1 1 0
  4. D.A B C: 0 0 1; 1 0 0; 0 1 0; 1 1 0

Correct answer: B

Explanation: C=A⋅(A+B)=AC=A\cdot(A+B)=A.

Question 34

Physics · 4 marks · −1

Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by 1°C. Its initial temperature must be:

  1. A.2500 K
  2. B.250°C
  3. C.250 K
  4. D.25°C

Correct answer: C

Explanation: ΔP/P=ΔT/T\Delta P/P=\Delta T/T: T=1/0.004=250T=1/0.004=250 K.

Question 35

Physics · 4 marks · −1

Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

  1. A.3:1
  2. B.9:1
  3. C.8:1
  4. D.4:1

Correct answer: D

Explanation: (1+3)2(3−1)2=4\frac{(1+3)^2}{(3-1)^2}=4.

Question 36

Physics · 4 marks · −1

Two cylindrical vessels of equal cross sectional area of 2m22m^2 contain water upto heights 10m and 6m, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is (Density of water is 10310^3 kg/m3m^3 and g = 10 m/s2m/s^2)

  1. A.1×1051\times10^5 J
  2. B.4×1044\times10^4 J
  3. C.6×1046\times10^4 J
  4. D.8×1048\times10^4 J

Correct answer: D

Explanation: Final level 8 m; PE change =ρgA(100+36)−2(64)2=104⋅2⋅4=8×104=\rho gA\frac{(100+36)-2(64)}{2}=10^4\cdot2\cdot4=8\times10^4 J.

Question 37

Physics · 4 marks · −1

A magnetic dipole experiences a torque of 80380\sqrt{3} N m when placed in uniform magnetic field in such a way that dipole moment makes angle of 60° with magnetic field. The potential energy of the dipole is :

  1. A.−403-40\sqrt{3} J
  2. B.−80 J
  3. C.80 J
  4. D.−60 J

Correct answer: B

Explanation: mB=160mB=160; U=−mBcos⁡60∘=−80U=-mB\cos60^\circ=-80 J.

Question 38

Physics · 4 marks · −1

In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

  1. A.360 m/s
  2. B.370 m/s
  3. C.340 m/s
  4. D.335 m/s

Correct answer: A

Explanation: v4(1−11.2)=15⇒v=360\frac{v}{4}\left(1-\frac{1}{1.2}\right)=15\Rightarrow v=360 m/s.

Question 39

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 Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R : The formulation of Bohr model does not include repulsive force between electrons. 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 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 false but R is true

Correct answer: B

Question 40

Physics · 4 marks · −1

A particle is projected with velocity u so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as nu225g\frac{nu^2}{25g}, where value of n is: (Given, \'g\' is the acceleration due to gravity.)

  1. A.18
  2. B.6
  3. C.24
  4. D.12

Correct answer: C

Explanation: R=3H⇒tan⁡θ=4/3R=3H\Rightarrow\tan\theta=4/3; R=u2(2⋅45⋅35)g=24u225gR=\frac{u^2(2\cdot\frac45\cdot\frac35)}{g}=\frac{24u^2}{25g}.

Question 41

Physics · 4 marks · −1

An electric bulb rated as 100W-220V is connected to an ac source of rms voltage 220V. The peak value of current through the bulb is :

  1. A.0.45 A
  2. B.0.64 A
  3. C.0.32 A
  4. D.2.2 A

Correct answer: B

Explanation: Irms=0.4545I_{rms}=0.4545, peak =0.64=0.64 A.

Question 42

Physics · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. Boltzmann constant B. Coefficient of viscosity C. Planck's constant D. Thermal conductivity. LIST-II: I. ML2T−1ML^2T^{-1} II. MLT−3K−1MLT^{-3}K^{-1} III. ML2T−2K−1ML^2T^{-2}K^{-1} IV. ML−1T−1ML^{-1}T^{-1}. Choose the correct answer from the options given below:

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

Correct answer: D

Question 43

Physics · 4 marks · −1

An ideal gas exists in a state with pressure P0P_0, volume V0V_0. It is isothermally expanded to 4 times of its initial volume (V0V_0), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

  1. A.P0V0(ln⁡2−0.25)P_0V_0(\ln2-0.25)
  2. B.P0V0(ln⁡2−0.75)P_0V_0(\ln2-0.75)
  3. C.P0V0(2ln⁡2−0.75)P_0V_0(2\ln2-0.75)
  4. D.P0V0(2ln⁡2−0.25)P_0V_0(2\ln2-0.25)

Correct answer: C

Explanation: Net Q = net W =P0V0ln⁡4−P04(3V0)=P0V0(2ln⁡2−0.75)=P_0V_0\ln4-\frac{P_0}{4}(3V_0)=P_0V_0(2\ln2-0.75).

Question 44

Physics · 4 marks · −1

A block of mass 1 kg, moving along x with speed viv_i = 10 m/s enters a rough region ranging from x = 0.1 m to x = 1.9 m. The retarding force acting on the block in this range is Fr=−kxF_r=-kx N, with k = 10 N/m. Then the final speed of the block as it crosses rough region is.

  1. A.4 m/s
  2. B.6 m/s
  3. C.10 m/s
  4. D.8 m/s

Correct answer: D

Explanation: W=−5(1.92−0.12)=−18W=-5(1.9^2-0.1^2)=-18 J; 12v2=50−18\frac12v^2=50-18, v = 8 m/s.

Question 45

Physics · 4 marks · −1

An electron in the hydrogen atom initially in the fourth excited state makes a transition to nthn^{th} energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ______.

Correct answer: 2

Explanation: E5=−0.544E_5=-0.544 eV; −0.544−2.86=−3.4-0.544-2.86=-3.4 eV ⇒ n = 2.

Question 46

Physics · 4 marks · −1

Two cells of emfs 1 V and 2 V and internal resistances 2 Ω\Omega and 1 Ω\Omega, respectively, are connected in series with an external resistance of 6 Ω\Omega. The total current in the circuit is I1I_1. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is I2I_2. The value of (I1I2)\left(\frac{I_1}{I_2}\right) is x3\frac{x}{3}. The value of x is ______.

Correct answer: 4

Explanation: I1=3/9=13I_1=3/9=\frac13; parallel: E=1/2+2/13/2=53E=\frac{1/2+2/1}{3/2}=\frac53, r=23r=\frac23, I2=5/320/3=14I_2=\frac{5/3}{20/3}=\frac14; ratio 43\frac43, x = 4.

Question 47

Physics · 4 marks · −1

Light from a point source in air falls on a spherical glass surface (refractive index, μ\mu = 1.5 and radius of curvature = 50 cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ______ m.

Correct answer: 4

Explanation: 1.5200−1u=0.550⇒u=−400\frac{1.5}{200}-\frac{1}{u}=\frac{0.5}{50}\Rightarrow u=-400 cm = 4 m.

Question 48

Physics · 4 marks · −1

The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is n times the volume of the bubble B, then, the value of n is ______.

Correct answer: 8

Explanation: rA=2rBr_A=2r_B, volume ratio 8.

Question 49

Physics · 4 marks · −1

A physical quantity C is related to four other quantities p, q, r and s as follows C=pq2r3sC=\frac{pq^2}{r^3\sqrt{s}}. The percentage errors in the measurement of p, q, r and s are 1%, 2%, 3% and 2%, respectively. The percentage error in the measurement of C will be ______ %.

Correct answer: 15

Explanation: 1 + 4 + 9 + 1 = 15%.

Question 50

Chemistry · 4 marks · −1

The correct orders among the following are A. Atomic radius : B < Al < Ga < In < Tl B. Electronegativity : Al < Ga < In < Tl < B C. Density : Tl < In < Ga < Al < B D. 1st Ionisation Energy : In < Al < Ga < Tl < B. Choose the correct answer from the options given below:

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

Correct answer: C

Question 51

Chemistry · 4 marks · −1

The major product (P) in the following reaction is : Ph−CO−CHO→ΔKOHPh-CO-CHO\xrightarrow[\Delta]{KOH} P (Major Product)

  1. A.Ph−CH(OH)−COO−K+Ph-CH(OH)-COO^-K^+
  2. B.Ph−CO−COO−K+Ph-CO-COO^-K^+
  3. C.Ph−CH(OH)−CH2OHPh-CH(OH)-CH_2OH
  4. D.Ph−CO−CH2OHPh-CO-CH_2OH

Correct answer: A

Explanation: Intramolecular Cannizzaro of phenylglyoxal gives the mandelate salt.

Question 52

Chemistry · 4 marks · −1

Identify the diamagnetic octahedral complex ions from below ; A. [Mn(CN)6]3−[Mn(CN)_6]^{3-} B. [Co(NH3)6]3+[Co(NH_3)_6]^{3+} C. [Fe(CN)6]4−[Fe(CN)_6]^{4-} D. [Co(H2O)3F3][Co(H_2O)_3F_3]. Choose the correct answer from the options given below:

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

Correct answer: C

Question 53

Chemistry · 4 marks · −1

In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K = 15 mm Hg)

  1. A.15.71 %
  2. B.7.85 %
  3. C.20.95 %
  4. D.17.46 %

Correct answer: A

Explanation: N2N_2 mass 0.0628 g → 15.71%.

Question 54

Chemistry · 4 marks · −1

40 mL of a mixture of CH3COOHCH_3COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?

Figure for question 54
  1. A.The concentration of HCl in the original mixture is 0.005 M
  2. B.CH3COOHCH_3COOH is neutralised first followed by neutralisation of HCl
  3. C.Point 'C' indicates the complete neutralisation of HCl
  4. D.The concentration of CH3COOHCH_3COOH in the original mixture is 0.005 M

Correct answer: A

Explanation: HCl neutralised at B (2 mL): 0.2/40=0.0050.2/40=0.005 M; acetic acid needs 3 mL more: 0.0075 M.

Question 55

Chemistry · 4 marks · −1

The standard cell potential (Ecell⊖E^\ominus_{cell}) of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V. The standard half cell reduction potential for O2O_2 (EO2/H2O∘E^\circ_{O_2/H_2O}) is 1.229 V. Choose the correct statement:

  1. A.Reactants are fed at one go to each electrode.
  2. B.Oxygen is formed at the anode.
  3. C.Reduction of methanol takes place at the cathode.
  4. D.The standard half cell reduction potential for the reduction of CO2CO_2 (ECO2/CH3OH∘E^\circ_{CO_2/CH_3OH}) is 19 mV

Correct answer: D

Explanation: 1.229−1.21=0.0191.229-1.21=0.019 V.

Question 56

Chemistry · 4 marks · −1

Given below are two statements: Statement I : CrO3CrO_3 is a stronger oxidizing agent than MoO3MoO_3 Statement II : Cr(VI) is more stable than Mo(VI). 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.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are false

Correct answer: C

Question 57

Chemistry · 4 marks · −1

Consider the following statements related to temperature dependence of rate constants. Identify the correct statements. A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea >> RT, A and Ea become interdependent. Choose the correct answer from the options given below:

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

Correct answer: D

Question 58

Chemistry · 4 marks · −1

Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is Given: Molar mass of Mg is 24 g mol−1mol^{-1}.

  1. A.236 mg
  2. B.235.7 g
  3. C.2.444 g
  4. D.0.24 mg

Correct answer: A

Explanation: 22022400×24=0.2357\frac{220}{22400}\times24=0.2357 g ≈ 236 mg.

Question 59

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Family): A. Pnictogen (group 15) B. Chalcogen C. Halogen D. Noble gas. LIST-II (Symbol of Element): I. Ts II. Og III. Lv IV. Mc. Choose the correct answer from the options given below:

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

Correct answer: D

Question 60

Chemistry · 4 marks · −1

For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are :

  1. A.h2π\frac{h}{2\pi} and 2h2π\sqrt{2}\frac{h}{2\pi}
  2. B.0 and 6h2π\sqrt{6}\frac{h}{2\pi}
  3. C.0 and 2h2π\sqrt{2}\frac{h}{2\pi}
  4. D.2h2π\sqrt{2}\frac{h}{2\pi} and 0

Correct answer: C

Explanation: l(l+1)ℏ\sqrt{l(l+1)}\hbar: 0 and 2ℏ\sqrt2\hbar.

Question 61

Chemistry · 4 marks · −1

Compounds that should not be used as primary standards in titrimetric analysis are : A. Na2Cr2O7Na_2Cr_2O_7 B. Oxalic acid C. NaOH D. FeSO4⋅6H2OFeSO_4\cdot6H_2O E. Sodium tetraborate. Choose the most appropriate answer from the options given below:

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

Correct answer: D

Explanation: Na2Cr2O7Na_2Cr_2O_7 is hygroscopic (K salt is used), NaOH absorbs moisture/CO2_2, FeSO4FeSO_4 is oxidised by air.

Question 62

Chemistry · 4 marks · −1

10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is :

  1. A.Neutral solution
  2. B.20 M HCl solution
  3. C.0.2 M NaCl solution
  4. D.10 M HCl solution

Correct answer: B

Explanation: The first mixture is neutral; the flask holds 2 mol HCl in 0.1 L = 20 M.

Question 63

Chemistry · 4 marks · −1

Fat soluble vitamins are: A. Vitamin B1B_1 B. Vitamin C C. Vitamin E D. Vitamin B12B_{12} E. Vitamin K. Choose the correct answer from the options given below:

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

Correct answer: A

Question 64

Chemistry · 4 marks · −1

What is the correct IUPAC name of the compound: benzoic acid with OH at C-2, Br at C-3 and NO2NO_2 at C-5 (drawn with CO2HCO_2H top-right, O2NO_2N top-left, OH right, Br bottom)?

  1. A.3-Bromo-4-hydroxy-1-nitrobenzoic acid
  2. B.3-Bromo-2-hydroxy-5-nitrobenzoic acid
  3. C.2-Hydroxy-3-bromo-5-nitrobenzoic acid
  4. D.5-Nitro-3-bromo-2-hydroxybenzoic acid

Correct answer: B

Question 65

Chemistry · 4 marks · −1

The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :

  1. A.Nitrobenzene: (i) Br2Br_2, acetic acid; (ii) Sn, HCl
  2. B.Aniline: Br2Br_2, water
  3. C.Benzene: (i) Br2Br_2, AlBr3AlBr_3; (ii) NH3NH_3
  4. D.4-Nitroaniline: (i) Br2Br_2 (excess), acetic acid; (ii) NaNO2NaNO_2, HCl, CuBr; (iii) Sn, HCl

Correct answer: D

Explanation: 4-Nitroaniline brominates at both positions ortho to NH2NH_2; Sandmeyer replaces NH2NH_2 by Br (giving 3,4,5-tribromonitrobenzene); reduction gives 3,4,5-tribromoaniline.

Question 66

Chemistry · 4 marks · −1

Given below are two statements: Statement I : When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II : At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. 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 false
  2. B.Statement I is true but Statement II is false
  3. C.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are true

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Given below are two statements: Statement I : Hyperconjugation is not a permanent effect. Statement II : In general, greater the number of alkyl groups attached to a positively charged C-atom, greater is the hyperconjugation interaction and stabilization of the cation. 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.Both Statement I and Statement II are false
  3. C.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are true

Correct answer: C

Question 68

Chemistry · 4 marks · −1

In the following series of reactions identify the major products A & B respectively: Bromobenzene →SO3/H2SO4\xrightarrow{SO_3/H_2SO_4} A (Major product) →Br2,Fe\xrightarrow{Br_2, Fe} B (Major product)

Figure for question 68
  1. A.A = 2-bromobenzenesulfonic acid; B = 2,5-dibromobenzenesulfonic acid
  2. B.A = 2-bromobenzenesulfonic acid; B = 2,6-dibromobenzenesulfonic acid
  3. C.A = 4-bromobenzenesulfonic acid; B = 3,4-dibromobenzenesulfonic acid (Br ortho to the first Br)
  4. D.A = 4-bromobenzenesulfonic acid; B = 2,4-dibromobenzenesulfonic acid (Br ortho to SO3H)

Correct answer: C

Explanation: Br is o/p-directing: sulfonation para. In B, Br directs ortho to itself and SO3HSO_3H directs meta to itself — both point to C-3 (relative to SO3HSO_3H at C-1): 3,4-dibromobenzenesulfonic acid.

Question 69

Chemistry · 4 marks · −1

Given below are two statements: Statement I : Wet cotton clothes made of cellulose based carbohydrate takes comparatively longer time to get dried than wet nylon polymer based clothes. Statement II : Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes. In the light of above 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 false but Statement II is true
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are true

Correct answer: C

Question 70

Chemistry · 4 marks · −1

A sample of n−octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K−1K^{-1}. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ______ kJ mol−1mol^{-1} (nearest integer).

Correct answer: 2500

Explanation: 25 kJ for 0.01 mol → 2500 kJ/mol.

Question 71

Chemistry · 4 marks · −1

X g of nitrobenzene on nitration gave 4.2 g of m−dinitrobenzene. X = ______ g. (nearest integer) [Given : molar mass (in g mol−1mol^{-1}) C : 12, H : 1, O : 16, N : 14]

Correct answer: 3

Explanation: 4.2/168 = 0.025 mol × 123 = 3.075 ≈ 3 g.

Question 72

Chemistry · 4 marks · −1

A perfect gas (0.1 mol) having CˉV\bar{C}_V = 1.50R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (−) ______ J (nearest integer) [Given: R = 0.082 L atm K−1mol−1K^{-1}mol^{-1}]

Figure for question 72

Correct answer: 304

Explanation: Only 2→3 does work: w=−3 atm×1 L=−3w=-3\text{ atm}\times1\text{ L}=-3 L atm =−304=-304 J.

Question 73

Chemistry · 4 marks · −1

The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula C9H12C_9H_{12} is ______.

Correct answer: 8

Explanation: Propyl, isopropyl, 3 ethyltoluenes, 3 trimethylbenzenes = 8.

Question 74

Chemistry · 4 marks · −1

Among, Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ______ BM (in nearest integer).

Correct answer: 4

Explanation: Co (425 kJ/mol) is highest among these; Co2+Co^{2+} d7d^7: 15=3.87\sqrt{15}=3.87 ≈ 4.

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