JEE Main 2025 — 23 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

Let the shortest distance from (a, 0), a > 0, to the parabola y2=4xy^2=4x be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is :

  1. A.x2+y2−10x+9=0x^2+y^2-10x+9=0
  2. B.x2+y2−6x+5=0x^2+y^2-6x+5=0
  3. C.x2+y2−8x+7=0x^2+y^2-8x+7=0
  4. D.x2+y2−4x+3=0x^2+y^2-4x+3=0

Correct answer: B

Explanation: The normal from (a, 0) to y2=4xy^2=4x gives shortest distance² =4(a−1)=16=4(a-1)=16 ⇒ a = 5. The circle through (5, 0) and the focus (1, 0) with centre on the x-axis has centre (3, 0), radius 2: x2+y2−6x+5=0x^2+y^2-6x+5=0.

Question 2

Mathematics · 4 marks · −1

The system of equations x+y+z=6x+y+z=6, x+2y+5z=9x+2y+5z=9, x+5y+λz=μx+5y+\lambda z=\mu, has no solution if

  1. A.λ=15,μ≠17\lambda=15, \mu\neq17
  2. B.λ≠17,μ≠18\lambda\neq17, \mu\neq18
  3. C.λ=17,μ=18\lambda=17, \mu=18
  4. D.λ=17,μ≠18\lambda=17, \mu\neq18

Correct answer: D

Explanation: D=0⇒λ=17D=0\Rightarrow\lambda=17; consistency needs μ=18\mu=18.

Question 3

Mathematics · 4 marks · −1

Let A=[aij]A=[a_{ij}] be a 3×33\times3 matrix such that A[010]=[001]A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}, A[413]=[010]A\begin{bmatrix}4\\1\\3\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix} and A[212]=[100]A\begin{bmatrix}2\\1\\2\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}, then a23a_{23} equals :

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

Correct answer: A

Question 4

Mathematics · 4 marks · −1

The number of complex numbers z, satisfying ∣z∣=1|z|=1 and ∣zzˉ+zˉz∣=1\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1, is :

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

Correct answer: D

Explanation: ∣2cos⁡2θ∣=1|2\cos2\theta|=1: cos⁡2θ=±12\cos2\theta=\pm\frac12 gives 8 values of θ in [0,2π)[0,2\pi).

Question 5

Mathematics · 4 marks · −1

If I=∫0π/2sin⁡3/2xsin⁡3/2x+cos⁡3/2xdxI=\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}dx, then ∫02Ixsin⁡xcos⁡xsin⁡4x+cos⁡4xdx\int_0^{2I}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}dx equals :

  1. A.π28\frac{\pi^2}{8}
  2. B.π24\frac{\pi^2}{4}
  3. C.π212\frac{\pi^2}{12}
  4. D.π216\frac{\pi^2}{16}

Correct answer: D

Explanation: I=π/4I=\pi/4, so upper limit π/2\pi/2; the integral =π4∫0π/2sin⁡xcos⁡xsin⁡4x+cos⁡4xdx=π4⋅π4=π216=\frac\pi4\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\frac\pi4\cdot\frac\pi4=\frac{\pi^2}{16}.

Question 6

Mathematics · 4 marks · −1

If the square of the shortest distance between the lines x−21=y−12=z+3−3\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3} and x+12=y+34=z+5−5\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5} is mn\frac{m}{n}, where m, n are coprime numbers, then m + n is equal to :

  1. A.6
  2. B.14
  3. C.9
  4. D.21

Correct answer: C

Question 7

Mathematics · 4 marks · −1

Let the range of the function f(x)=6+16cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6xf(x)=6+16\cos x\cdot\cos\left(\frac{\pi}{3}-x\right)\cdot\cos\left(\frac{\pi}{3}+x\right)\cdot\sin3x\cdot\cos6x, x∈Rx\in\mathbf{R} be [α,β][\alpha, \beta]. Then the distance of the point (α,β)(\alpha, \beta) from the line 3x+4y+12=03x+4y+12=0 is :

  1. A.11
  2. B.10
  3. C.8
  4. D.9

Correct answer: A

Explanation: f=6+4cos⁡3xsin⁡3xcos⁡6x=6+sin⁡12xf=6+4\cos3x\sin3x\cos6x=6+\sin12x: range [5, 7]; distance of (5,7): 15+28+125=11\frac{15+28+12}{5}=11.

Question 8

Mathematics · 4 marks · −1

The length of the chord of the ellipse x24+y22=1\frac{x^2}{4}+\frac{y^2}{2}=1, whose mid-point is (1,12)\left(1, \frac{1}{2}\right), is :

  1. A.5315\frac{5}{3}\sqrt{15}
  2. B.2315\frac{2}{3}\sqrt{15}
  3. C.1315\frac{1}{3}\sqrt{15}
  4. D.15\sqrt{15}

Correct answer: B

Explanation: Chord with midpoint (1, 1/2) (T = S₁): 2x+2y=32x+2y=3. Substituting in the ellipse and taking the distance between the intersection points gives length 2315\frac23\sqrt{15}.

Question 9

Mathematics · 4 marks · −1

A rod of length eight units moves such that its ends A and B always lie on the lines x−y+2=0x-y+2=0 and y+2=0y+2=0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2 : 1 is 9(x2+αy2+βxy+γx+28y)−76=09(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0, then α−β−γ\alpha-\beta-\gamma is equal to :

  1. A.22
  2. B.23
  3. C.24
  4. D.21

Correct answer: B

Question 10

Mathematics · 4 marks · −1

If the area of the region {(x,y):−1≤x≤1,0≤y≤a+e∣x∣−e−x,a>0}\{(x, y): -1\le x\le1, 0\le y\le a+e^{|x|}-e^{-x}, a>0\} is e2+8e+1e\frac{e^2+8e+1}{e}, then the value of a is :

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

Correct answer: B

Explanation: Area =2a+∫−11(e∣x∣−e−x)dx=2a+2(e−1)−(e−e−1)=2a+e+1e−2=2a+\int_{-1}^1(e^{|x|}-e^{-x})dx=2a+2(e-1)-(e-e^{-1})=2a+e+\frac1e-2; equate: a=5a=5.

Question 11

Mathematics · 4 marks · −1

lim⁡x→∞(2x2−3x+5)(3x−1)x/2(3x2+5x+4)(3x+2)x\lim_{x\to\infty}\frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}} is equal to :

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

Correct answer: A

Explanation: 23(3x−13x+2)x/2→23e−1/2\frac23\left(\frac{3x-1}{3x+2}\right)^{x/2}\to\frac23e^{-1/2}.

Question 12

Mathematics · 4 marks · −1

Let the point A divide the line segment joining the points P(−1-1, −1-1, 2) and Q(5, 5, 10) internally in the ratio r : 1 (r > 0). If O is the origin and (OQ⃗⋅OA⃗)−15∣OP⃗×OA⃗∣2=10(\vec{OQ}\cdot\vec{OA})-\frac{1}{5}|\vec{OP}\times\vec{OA}|^2=10, then the value of r is :

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

Correct answer: B

Question 13

Mathematics · 4 marks · −1

Let A={(x,y)∈R×R:∣x+y∣≥3}A=\{(x, y)\in\mathbf{R}\times\mathbf{R}: |x+y|\ge3\} and B={(x,y)∈R×R:∣x∣+∣y∣≤3}B=\{(x, y)\in\mathbf{R}\times\mathbf{R}: |x|+|y|\le3\}. If C={(x,y)∈A∩B:x=0 or y=0}C=\{(x, y)\in A\cap B: x=0 \text{ or } y=0\}, then ∑(x,y)∈C∣x+y∣\sum_{(x, y)\in C}|x+y| is :

  1. A.15
  2. B.24
  3. C.12
  4. D.18

Correct answer: C

Explanation: On the axes, A∩BA\cap B gives only (±3,0),(0,±3)(\pm3,0),(0,\pm3): sum 12.

Question 14

Mathematics · 4 marks · −1

If in the expansion of (1+x)p(1−x)q(1+x)^p(1-x)^q, the coefficients of x and x2x^2 are 1 and −2-2, respectively, then p2+q2p^2+q^2 is equal to :

  1. A.13
  2. B.8
  3. C.20
  4. D.18

Correct answer: A

Explanation: p−q=1p-q=1, (p−q)2−(p+q)2=−2⇒p+q=5\frac{(p-q)^2-(p+q)}{2}=-2\Rightarrow p+q=5: p = 3, q = 2; 13.

Question 15

Mathematics · 4 marks · −1

Let X=R×RX=\mathbf{R}\times\mathbf{R}. Define a relation R on X as : (a1,b1)R(a2,b2)⇔b1=b2(a_1, b_1)R(a_2, b_2)\Leftrightarrow b_1=b_2. Statement I : R is an equivalence relation. Statement II : For some (a,b)∈X(a, b)\in X, the set S={(x,y)∈X:(x,y)R(a,b)}S=\{(x, y)\in X: (x, y)R(a, b)\} represents a line parallel to y=xy=x. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: C

Explanation: S is the horizontal line y = b, not parallel to y = x.

Question 16

Mathematics · 4 marks · −1

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 cm3cm^3/min and the thickness of the ice-cream layer decreases at the rate of 14π\frac{1}{4\pi} cm/min. The surface area (in cm2cm^2) of the chocolate ball (without the ice-cream layer) is :

  1. A.196 π\pi
  2. B.128 π\pi
  3. C.225 π\pi
  4. D.256 π\pi

Correct answer: D

Explanation: 4πR2⋅14π=81⇒R=94\pi R^2\cdot\frac{1}{4\pi}=81\Rightarrow R=9 (outer); chocolate radius 8: area 256π256\pi.

Question 17

Mathematics · 4 marks · −1

The distance of the line x−22=y−63=z−34\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} from the point (1, 4, 0) along the line x1=y−22=z+33\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} is:

  1. A.13\sqrt{13}
  2. B.17\sqrt{17}
  3. C.14\sqrt{14}
  4. D.15\sqrt{15}

Correct answer: C

Question 18

Mathematics · 4 marks · −1

Let ∫x3sin⁡x dx=g(x)+C\int x^3\sin x\,dx=g(x)+C, where C is the constant of integration. If 8\left(g\left(\frac{\pi}{2}\right)+g\'\left(\frac{\pi}{2}\right)\right)=\alpha\pi^3+\beta\pi^2+\gamma, α,β,γ∈Z\alpha, \beta, \gamma\in\mathbf{Z}, then α+β−γ\alpha+\beta-\gamma equals :

  1. A.47
  2. B.62
  3. C.55
  4. D.48

Correct answer: C

Question 19

Mathematics · 4 marks · −1

Let x=x(y)x=x(y) be the solution of the differential equation y=(x−ydxdy)sin⁡(xy)y=\left(x-y\frac{dx}{dy}\right)\sin\left(\frac{x}{y}\right), y>0y>0 and x(1)=π2x(1)=\frac{\pi}{2}. Then cos⁡(x(2))\cos(x(2)) is equal to :

  1. A.1−2(log⁡e2)1-2(\log_e2)
  2. B.1−2(log⁡e2)21-2(\log_e2)^2
  3. C.2(log⁡e2)−12(\log_e2)-1
  4. D.2(log⁡e2)2−12(\log_e2)^2-1

Correct answer: D

Question 20

Mathematics · 4 marks · −1

A board has 16 squares as shown in the figure (a 4 × 4 grid). Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

  1. A.2330\frac{23}{30}
  2. B.45\frac{4}{5}
  3. C.710\frac{7}{10}
  4. D.35\frac{3}{5}

Correct answer: B

Explanation: Adjacent pairs = 24 of (162)=120\binom{16}{2}=120: 1−24120=451-\frac{24}{120}=\frac45.

Question 21

Mathematics · 4 marks · −1

The focus of the parabola y2=4x+16y^2=4x+16 is the centre of the circle C of radius 5. If the values of λ\lambda, for which C passes through the point of intersection of the lines 3x−y=03x-y=0 and x+λy=4x+\lambda y=4, are λ1\lambda_1 and λ2\lambda_2, λ1<λ2\lambda_1<\lambda_2, then 12λ1+29λ212\lambda_1+29\lambda_2 is equal to ______.

Correct answer: 15

Question 22

Mathematics · 4 marks · −1

The variance of the numbers 8, 21, 34, 47, ..., 320 is ______.

Correct answer: 8788

Explanation: 25 terms with d = 13: variance =d2n2−112=169×52=8788=d^2\frac{n^2-1}{12}=169\times52=8788.

Question 23

Mathematics · 4 marks · −1

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is ______.

Correct answer: 17280

Explanation: Together: 5!⋅5!=144005!\cdot5!=14400; no two: 4!⋅5!=28804!\cdot5!=2880; total 17280.

Question 24

Mathematics · 4 marks · −1

The roots of the quadratic equation 3x2−px+q=03x^2-px+q=0 are 10th10^{th} and 11th11^{th} terms of an arithmetic progression with common difference 32\frac{3}{2}. If the sum of the first 11 terms of this arithmetic progression is 88, then q−2pq-2p is equal to ______.

Correct answer: 474

Explanation: a6=8a_6=8, so a10=14a_{10}=14, a11=15.5a_{11}=15.5; p=3(29.5)=88.5p=3(29.5)=88.5, q=3(217)=651q=3(217)=651; q−2p=474q-2p=474.

Question 25

Mathematics · 4 marks · −1

Let α,β\alpha, \beta be the roots of the equation x2−ax−b=0x^2-ax-b=0 with Im(α)<Im(β)\mathrm{Im}(\alpha)<\mathrm{Im}(\beta). Let Pn=αn−βnP_n=\alpha^n-\beta^n. If P3=−57iP_3=-5\sqrt{7}i, P4=−37iP_4=-3\sqrt{7}i, P5=117iP_5=11\sqrt{7}i and P6=457iP_6=45\sqrt{7}i, then ∣α4+β4∣|\alpha^4+\beta^4| is equal to ______.

Correct answer: 31

Explanation: Pn+2=aPn+1+bPnP_{n+2}=aP_{n+1}+bP_n: 11=−3a−5b11=-3a-5b, 45=11a−3b45=11a-3b → a = 3, b = −4. Then α+β=3\alpha+\beta=3, αβ=4\alpha\beta=4: α2+β2=1\alpha^2+\beta^2=1, α4+β4=1−32=−31\alpha^4+\beta^4=1-32=-31 → 31.

Question 26

Physics · 4 marks · −1

The refractive index of the material of a glass prism is 3\sqrt{3}. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism ?

  1. A.48°
  2. B.50°
  3. C.60°
  4. D.58°

Correct answer: C

Explanation: 3=sin⁡Asin⁡(A/2)=2cos⁡(A/2)⇒A=60∘\sqrt3=\frac{\sin A}{\sin(A/2)}=2\cos(A/2)\Rightarrow A=60^\circ.

Question 27

Physics · 4 marks · −1

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2V, what is the wavelength of the em-wave ? (Given hc = 1242 eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

  1. A.300 nm
  2. B.200 nm
  3. C.400 nm
  4. D.600 nm

Correct answer: A

Explanation: E=4.14E=4.14 eV, λ=1242/4.14=300\lambda=1242/4.14=300 nm.

Question 28

Physics · 4 marks · −1

A galvanometer having a coil of resistance 30 Ω\Omega need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be 30X Ω\frac{30}{X}\ \Omega, where X is

  1. A.149
  2. B.596
  3. C.298
  4. D.447

Correct answer: A

Explanation: S=0.02×302.98=30149S=\frac{0.02\times30}{2.98}=\frac{30}{149}.

Question 29

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 binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170. Reason (R) : Nuclear force is long range. 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 true but (R) is false
  3. C.(A) is false but (R) is true
  4. D.Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct answer: B

Question 30

Physics · 4 marks · −1

A concave mirror of focal length f in air is dipped in a liquid of refractive index μ\mu. Its focal length in the liquid will be :

  1. A.μf\mu f
  2. B.f
  3. C.f(μ−1)\frac{f}{(\mu-1)}
  4. D.fμ\frac{f}{\mu}

Correct answer: B

Explanation: Mirror focal length depends only on R.

Question 31

Physics · 4 marks · −1

A massless spring gets elongated by amount x1x_1 under a tension of 5 N. Its elongation is x2x_2 under the tension of 7 N. For the elongation of (5x1−2x2)(5x_1-2x_2), the tension in the spring will be,

  1. A.39 N
  2. B.15 N
  3. C.11 N
  4. D.20 N

Correct answer: C

Explanation: T=5(5)−2(7)=11T=5(5)-2(7)=11 N.

Question 32

Physics · 4 marks · −1

If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected.

  1. A.81 days
  2. B.27 days
  3. C.3 days
  4. D.1 day

Correct answer: D

Explanation: T∝r3/2T\propto r^{3/2}: 27/27=127/27=1 day.

Question 33

Physics · 4 marks · −1

Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P1P_1. The reading of the pressure gauge falls to P2P_2 when the valve is opened. The speed of water flowing in the pipe is proportional to

  1. A.P1−P2P_1-P_2
  2. B.P1−P2\sqrt{P_1-P_2}
  3. C.(P1−P2)4(P_1-P_2)^4
  4. D.(P1−P2)2(P_1-P_2)^2

Correct answer: B

Question 34

Physics · 4 marks · −1

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 Vm−1E_y=9.3\ Vm^{-1}. Then, the magnetic field vector of the wave at that point is

  1. A.Bz=9.3×10−8B_z=9.3\times10^{-8} T
  2. B.Bz=1.55×10−8B_z=1.55\times10^{-8} T
  3. C.Bz=3.1×10−8B_z=3.1\times10^{-8} T
  4. D.Bz=6.2×10−8B_z=6.2\times10^{-8} T

Correct answer: C

Explanation: B=E/c=3.1×10−8B=E/c=3.1\times10^{-8} T.

Question 35

Physics · 4 marks · −1

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t2−8t\theta(t)=5t^2-8t, where θ(t)\theta(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t = 2 s ?

  1. A.108 MR2MR^2
  2. B.72 MR2MR^2
  3. C.8 MR2MR^2
  4. D.60 MR2MR^2

Correct answer: D

Explanation: α=10\alpha=10, ω(2)=12\omega(2)=12; P=MR22⋅10⋅12=60MR2P=\frac{MR^2}{2}\cdot10\cdot12=60MR^2.

Question 36

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Permeability of free space (B) Magnetic field (C) Magnetic moment (D) Torsional constant. List - II: (I) [ML2T−2][ML^2T^{-2}] (II) [MT−2A−1][MT^{-2}A^{-1}] (III) [MLT−2A−2][MLT^{-2}A^{-2}] (IV) [L2A][L^2A]. Choose the correct answer from the options given below :

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

Correct answer: C

Question 37

Physics · 4 marks · −1

The energy of a system is given as E(t)=α3e−βtE(t)=\alpha^3e^{-\beta t}, where t is the time and β=0.3 s−1\beta=0.3\ s^{-1}. The errors in the measurement of α\alpha and t are 1.2% and 1.6%, respectively. At t = 5 s, maximum percentage error in the energy is :

  1. A.6%
  2. B.11.6%
  3. C.4%
  4. D.8.4%

Correct answer: A

Explanation: 3(1.2)+βt(1.6)=3.6+1.5(1.6)=6%3(1.2)+\beta t(1.6)=3.6+1.5(1.6)=6\%.

Question 38

Physics · 4 marks · −1

Two point charges −4 μ-4\ \muc and 4 μ4\ \muc, constituting an electric dipole, are placed at (−9-9, 0, 0) cm and (9, 0, 0) cm in a uniform electric field of strength 10410^4 NC−1C^{-1}. The work done on the dipole in rotating it from the equilibrium through 180° is :

  1. A.14.4 mJ
  2. B.18.4 mJ
  3. C.12.4 mJ
  4. D.16.4 mJ

Correct answer: A

Explanation: W=2pE=2(4×10−6)(0.18)(104)=14.4W=2pE=2(4\times10^{-6})(0.18)(10^4)=14.4 mJ.

Question 39

Physics · 4 marks · −1

Water of mass m gram is slowly heated to increase the temperature from T1T_1 to T2T_2. The change in entropy of the water, given specific heat of water is 1 Jkg−1K−1kg^{-1}K^{-1}, is :

  1. A.zero
  2. B.m(T2−T1)m(T_2-T_1)
  3. C.mln⁡(T1T2)m\ln\left(\frac{T_1}{T_2}\right)
  4. D.mln⁡(T2T1)m\ln\left(\frac{T_2}{T_1}\right)

Correct answer: D

Question 40

Physics · 4 marks · −1

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4 then what is the value of x ? (Assume that the field strength varies according to the slit width.)

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

Correct answer: C

Explanation: 1+xx−1=32⇒x=5\frac{1+x}{x-1}=\frac32\Rightarrow x=5.

Question 41

Physics · 4 marks · −1

The equation of a transverse wave travelling along a string is y(x,t)=4.0sin⁡[20×10−3x+600t]y(x, t)=4.0\sin[20\times10^{-3}x+600t] mm, where x is in mm and t is in second. The velocity of the wave is :

  1. A.+30 m/s
  2. B.+60 m/s
  3. C.−60 m/s
  4. D.−30 m/s

Correct answer: D

Explanation: v=ω/k=600/0.02=30000v=\omega/k=600/0.02=30000 mm/s = 30 m/s along −x.

Question 42

Physics · 4 marks · −1

Using the given P-V diagram, the work done by an ideal gas along the path ABCD is :

Figure for question 42
  1. A.3 P0V0P_0V_0
  2. B.−4 P0V0P_0V_0
  3. C.−3 P0V0P_0V_0
  4. D.4 P0V0P_0V_0

Correct answer: C

Explanation: P0V0−2P0(2V0)=−3P0V0P_0V_0-2P_0(2V_0)=-3P_0V_0.

Question 43

Physics · 4 marks · −1

Two charges 7 μ\muc and −4-4 μ\muc are placed at (−7-7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, ϵ0=8.85×10−12 C2N−1m−2\epsilon_0=8.85\times10^{-12}\ C^2N^{-1}m^{-2}, the electrostatic potential energy of the charge configuration is :

  1. A.−1.2 J
  2. B.−2.0 J
  3. C.−1.5 J
  4. D.−1.8 J

Correct answer: D

Explanation: U=9×109(7)(−4)×10−120.14=−1.8U=\frac{9\times10^9(7)(-4)\times10^{-12}}{0.14}=-1.8 J.

Question 44

Physics · 4 marks · −1

What is the current through the battery in the circuit shown below ?

Figure for question 44
  1. A.1.0 A
  2. B.0.5 A
  3. C.1.5 A
  4. D.0.25 A

Correct answer: B

Explanation: Both branches conduct: 20 Ω ∥ 20 Ω = 10 Ω, I = 5/10 = 0.5 A.

Question 45

Physics · 4 marks · −1

A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?

  1. A.(KE)2\frac{(KE)}{2}
  2. B.(KE)4\frac{(KE)}{4}
  3. C.(KE)16\frac{(KE)}{16}
  4. D.(KE)8\frac{(KE)}{8}

Correct answer: B

Explanation: cos⁡260∘=14\cos^260^\circ=\frac14.

Question 46

Physics · 4 marks · −1

At steady state the charge on the capacitor, as shown in the circuit below, is ______ μ\muC.

Figure for question 46

Correct answer: 16

Explanation: I = 0.2 A, V10=2V_{10}=2 V, Q = 16 μC.

Question 47

Physics · 4 marks · −1

An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 J/m2m^2 and density 10310^3 kg/m3m^3. The difference between pressure inside the bubble and atmospheric pressure is ______ N/m2m^2. (Take g = 10 m/s2m/s^2)

Correct answer: 2190

Explanation: ρgh+2Tr=2000+190=2190\rho gh+\frac{2T}{r}=2000+190=2190.

Question 48

Physics · 4 marks · −1

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μ\muF. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be ______ Vs−1Vs^{-1}.

Correct answer: 100

Explanation: dV/dt=I/C=100dV/dt=I/C=100.

Question 49

Physics · 4 marks · −1

In a series LCR circuit, a resistor of 300 Ω\Omega, a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is ______ ×104\times10^4 radians s−1s^{-1}.

Correct answer: 2

Explanation: ω=1/0.1×25×10−9=2×104\omega=1/\sqrt{0.1\times25\times10^{-9}}=2\times10^4.

Question 50

Physics · 4 marks · −1

A satellite of mass M2\frac{M}{2} is revolving around earth in a circular orbit at a height of R3\frac{R}{3} from earth surface. The angular momentum of the satellite is MGMRxM\sqrt{\frac{GMR}{x}}. The value of x is ______, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant)

Correct answer: 3

Explanation: L=M2GM⋅4R3=MGMR3L=\frac M2\sqrt{GM\cdot\frac{4R}{3}}=M\sqrt{\frac{GMR}{3}}, x = 3.

Question 51

Chemistry · 4 marks · −1

Consider the reaction X2Y(g)⇌X2(g)+12Y2(g)X_2Y(g)\rightleftharpoons X_2(g)+\frac{1}{2}Y_2(g). The equation representing correct relationship between the degree of dissociation (x) of X2Y(g)X_2Y(g) with its equilibrium constant Kp is ______. Assume x to be very very small.

  1. A.x=Kpp3x=\sqrt[3]{\frac{Kp}{p}}
  2. B.x=2Kp2p3x=\sqrt[3]{\frac{2Kp^2}{p}}
  3. C.x=2Kpp3x=\sqrt[3]{\frac{2Kp}{p}}
  4. D.x=Kp2p3x=\sqrt[3]{\frac{Kp}{2p}}

Correct answer: B

Explanation: Kp≈x(x2)1/2p1/2K_p\approx x\left(\frac{x}{2}\right)^{1/2}p^{1/2} ⇒ x3=2Kp2px^3=\frac{2K_p^2}{p}.

Question 52

Chemistry · 4 marks · −1

Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is :

  1. A.82
  2. B.6
  3. C.14
  4. D.50

Correct answer: D

Explanation: Sn (Z = 50) has the lowest melting point (505 K) in group 14.

Question 53

Chemistry · 4 marks · −1

Identify the coordination complexes in which the central metal ion has d4d^4 configuration. (A) [FeO4]2−[FeO_4]^{2-} (B) [Mn(CN)6]3−[Mn(CN)_6]^{3-} (C) [Fe(CN)6]3−[Fe(CN)_6]^{3-} (D) Cr2(O−CO−Me)4(H2O)2Cr_2(O-CO-Me)_4(H_2O)_2 (E) [NiF6]2−[NiF_6]^{2-}. Choose the correct answer from the options given below :

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

Correct answer: A

Explanation: Mn3+Mn^{3+} d4d^4; Cr2+Cr^{2+} in chromium(II) acetate d4d^4; FeO42−FeO_4^{2-} is d2d^2, Fe3+Fe^{3+} d5d^5, Ni4+Ni^{4+} d6d^6.

Question 54

Chemistry · 4 marks · −1

The α\alpha - Helix and β\beta - Pleated sheet structures of protein are associated with its :

  1. A.tertiary structure
  2. B.secondary structure
  3. C.quaternary structure
  4. D.primary structure

Correct answer: B

Question 55

Chemistry · 4 marks · −1

Consider a binary solution of two volatile liquid components 1 and 2. x1x_1 and y1y_1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1x1\frac{1}{x_1} vs 1y1\frac{1}{y_1} are given respectively as :

  1. A.P10P20,P20−P10P20\frac{P_1^0}{P_2^0}, \frac{P_2^0-P_1^0}{P_2^0}
  2. B.P20P10,P10−P20P20\frac{P_2^0}{P_1^0}, \frac{P_1^0-P_2^0}{P_2^0}
  3. C.P10P20,P10−P20P20\frac{P_1^0}{P_2^0}, \frac{P_1^0-P_2^0}{P_2^0}
  4. D.P20P10,P20−P10P20\frac{P_2^0}{P_1^0}, \frac{P_2^0-P_1^0}{P_2^0}

Correct answer: A

Explanation: 1x1=P10P20⋅1y1+P20−P10P20\frac{1}{x_1}=\frac{P_1^0}{P_2^0}\cdot\frac{1}{y_1}+\frac{P_2^0-P_1^0}{P_2^0}.

Question 56

Chemistry · 4 marks · −1

Standard electrode potentials for a few half cells are mentioned below : ECu2+/Cu∘=0.34E^\circ_{Cu^{2+}/Cu}=0.34 V, EZn2+/Zn∘=−0.76E^\circ_{Zn^{2+}/Zn}=-0.76 V, EAg+/Ag∘=0.80E^\circ_{Ag^+/Ag}=0.80 V, EMg2+/Mg∘=−2.37E^\circ_{Mg^{2+}/Mg}=-2.37 V. Which one of the following cells gives the most negative value of ΔG∘\Delta G^\circ ?

  1. A.Zn∣Zn2+(1M)∣∣Mg2+(1M)∣MgZn|Zn^{2+}(1M)||Mg^{2+}(1M)|Mg
  2. B.Cu∣Cu2+(1M)∣∣Ag+(1M)∣AgCu|Cu^{2+}(1M)||Ag^+(1M)|Ag
  3. C.Zn∣Zn2+(1M)∣∣Ag+(1M)∣AgZn|Zn^{2+}(1M)||Ag^+(1M)|Ag
  4. D.Ag∣Ag+(1M)∣∣Mg2+(1M)∣MgAg|Ag^+(1M)||Mg^{2+}(1M)|Mg

Correct answer: C

Explanation: Zn/Ag: E∘=1.56E^\circ=1.56 V, n = 2 → most negative ΔG°.

Question 57

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : The boiling points of alcohols and phenols increase with increase in the number of C-atoms. Statement (II) : The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. 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: A

Question 58

Chemistry · 4 marks · −1

Consider the following reactions K2Cr2O7→−H2OKOH[A]→−H2OH2SO4[B]+K2SO4K_2Cr_2O_7\xrightarrow[-H_2O]{KOH}[A]\xrightarrow[-H_2O]{H_2SO_4}[B]+K_2SO_4. The products [A] and [B], respectively are :

  1. A.K2CrO4K_2CrO_4 and Cr2O3Cr_2O_3
  2. B.K2CrO4K_2CrO_4 and K2Cr2O7K_2Cr_2O_7
  3. C.K2CrO4K_2CrO_4 and CrO
  4. D.K2Cr(OH)6K_2Cr(OH)_6 and Cr2O3Cr_2O_3

Correct answer: B

Question 59

Chemistry · 4 marks · −1

The ascending order of relative rate of solvolysis of following compounds is : (A) 1-bromo-1,4,4-trimethylcyclohexane (tertiary bromide) (B) 1-bromo-4-methyl-1,2,3,4-tetrahydronaphthalene (secondary benzylic bromide) (C) bromodiphenylmethane (Ph2CHBrPh_2CHBr) (D) 2-bromo-3-methylpentane (secondary alkyl bromide)

Figure for question 59
  1. A.(D) < (A) < (B) < (C)
  2. B.(C) < (D) < (B) < (A)
  3. C.(C) < (B) < (A) < (D)
  4. D.(D) < (B) < (A) < (C)

Correct answer: A

Question 60

Chemistry · 4 marks · −1

Given below are two statements about X-ray spectra of elements : Statement (I) : A plot of ν\sqrt{\nu} (ν\nu = frequency of X-rays emitted) vs atomic mass is a straight line. Statement (II) : A plot of ν\nu (ν\nu = frequency of X-rays emitted) vs atomic number is a straight line. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: B

Explanation: Moseley: ν\sqrt\nu vs atomic number is linear; neither statement is correct.

Question 61

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : For a given shell, the total number of allowed orbitals is given by n2n^2. Statement (II) : For any subshell, the spatial orientation of the orbitals is given by −l-l to +l+l values including zero. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: C

Question 62

Chemistry · 4 marks · −1

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

  1. A.0.6
  2. B.0.2
  3. C.0.8
  4. D.0.4

Correct answer: A

Explanation: Δp=p0xB\Delta p=p^0x_B: p0=50p^0=50; xB=0.4x_B=0.4, solvent 0.6.

Question 63

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Isomers of C10H14C_{10}H_{14}, all hexahydro/tetrahydro-indene skeletons with a C=C at the ring fusion C3a=C7a): (A) extra C=C in the six-membered ring conjugated with the fusion bond (C6=C7), methyl on C2 of the saturated five-membered ring; (B) extra C=C in the five-membered ring (C1=C2) with the methyl on C2, six-membered ring otherwise saturated; (C) extra C=C in the six-membered ring (C4=C5) with a methyl on the adjacent sp3^3 carbon C6, five-membered ring saturated; (D) extra C=C in the five-membered ring (C2=C3) with the methyl on C3 (next to the fusion carbon), six-membered ring otherwise saturated. List - II (Ozonolysis product): (I) OHC−CH(CH3)−CH2−CO−(CH2)3−CO−CHOOHC-CH(CH_3)-CH_2-CO-(CH_2)_3-CO-CHO (II) CH3−CO−CO−(CH2)4−CO−CH2−CHOCH_3-CO-CO-(CH_2)_4-CO-CH_2-CHO (III) OHC−CH2CH2−CO−CH2−CH(CH3)−CH2−CO−CHOOHC-CH_2CH_2-CO-CH_2-CH(CH_3)-CH_2-CO-CHO (IV) OHC−CO−(CH2)4−CO−CH2−CO−CH3OHC-CO-(CH_2)_4-CO-CH_2-CO-CH_3. Choose the correct answer from the options given below :

Figure for question 63
  1. A.(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  2. B.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. C.(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  4. D.(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct answer: B

Explanation: (D): the fusion C=C and the C(CH3)=CH bond cleave to CH3-CO-CO-(CH2)4-CO-CH2-CHO (II); (B) likewise gives OHC-CO-(CH2)4-CO-CH2-COCH3 (IV).

Question 64

Chemistry · 4 marks · −1

Identify the products [A] and [B], respectively in the following reaction : Chlorobenzene →(i) NaOH, 623 K, 300 atm; (ii) H+\xrightarrow{\text{(i) NaOH, 623 K, 300 atm; (ii) } H^+} [A] →Na2Cr2O7/H2SO4\xrightarrow{Na_2Cr_2O_7/H_2SO_4} [B]

Figure for question 64
  1. A.[A] benzene, [B] cyclohexa-2,4-dienone
  2. B.[A] benzene, [B] phenol
  3. C.[A] phenol, [B] p-benzoquinone
  4. D.[A] phenol, [B] sodium phenoxide

Correct answer: C

Question 65

Chemistry · 4 marks · −1

pH of water is 7 at 25°C. If water is heated to 80°C., it's pH will :

  1. A.Remains the same
  2. B.H+H^+ concentration increases, OH−OH^- concentration decreases
  3. C.Increase
  4. D.Decrease

Correct answer: D

Question 66

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) Bronze (B) Brass (C) UK silver coin (D) Stainless steel. List - II: (I) Cu, Ni (II) Fe, Cr, Ni, C (III) Cu, Zn (IV) Cu, Sn. Choose the correct answer from the options given below :

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

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Given below are two statements : Consider the following reaction R−CO−R+H2O⇌KR−C(OH)2−RR-CO-R+H_2O\overset{K}{\rightleftharpoons}R-C(OH)_2-R. Statement (I) : In the case of formaldehyde (HCHO), K is about 2280, due to small substituents, hydration is faster. Statement (II) : In the case of trichloro acetaldehyde (Cl3C−CHOCl_3C-CHO), K is about 2000 due to −I effect of −Cl. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: D

Question 68

Chemistry · 4 marks · −1

Identify A, B and C in the given below reaction sequence: [A] →HNO3\xrightarrow{HNO_3} Pb(NO3)2Pb(NO_3)_2 →H2SO4\xrightarrow{H_2SO_4} [B] →(1) Ammonium acetate, (2) Acetic acid, (3) K2CrO4\xrightarrow{\text{(1) Ammonium acetate, (2) Acetic acid, (3) } K_2CrO_4} [C] Yellow ppt

  1. A.PbCl2PbCl_2, Pb(SO4)2Pb(SO_4)_2, PbCrO4PbCrO_4
  2. B.PbCl2PbCl_2, PbSO4PbSO_4, PbCrO4PbCrO_4
  3. C.PbS, PbSO4PbSO_4, Pb(CH3COO)2Pb(CH_3COO)_2
  4. D.PbS, PbSO4PbSO_4, PbCrO4PbCrO_4

Correct answer: D

Question 69

Chemistry · 4 marks · −1

The effect of temperature on spontaneity of reactions are represented as : (A) ΔH +, ΔS −, any T, Non spontaneous; (B) ΔH +, ΔS +, low T, spontaneous; (C) ΔH −, ΔS −, low T, Non spontaneous; (D) ΔH −, ΔS +, any T, spontaneous. The incorrect combinations are :

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

Correct answer: A

Question 70

Chemistry · 4 marks · −1

Which of the following graphs most appropriately represents a zero order reaction ?

Figure for question 70
  1. A.Rate vs time: straight line decreasing
  2. B.Rate vs time: decaying curve
  3. C.Reactant concentration vs time: straight line decreasing
  4. D.Reactant concentration vs time: horizontal line

Correct answer: C

Question 71

Chemistry · 4 marks · −1

Consider the following sequence of reactions. 4-Ethoxyaniline →(i) NaNO2,HCl,0−5∘C\xrightarrow{\text{(i) } NaNO_2, HCl, 0-5^\circ C} (A) →(i) phenol, NaOH; (ii) HCl dil.\xrightarrow{\text{(i) phenol, NaOH; (ii) HCl dil.}} (B) (Molecular formula C14H14N2O2C_{14}H_{14}N_2O_2) →(i) NaOH; (ii) H3CCH2Br\xrightarrow{\text{(i) NaOH; (ii) } H_3CCH_2Br} (C) (Molecular formula C16H18N2O2C_{16}H_{18}N_2O_2). Total number of sp3sp^3 hybridised carbon atoms in the major product C formed is ______.

Correct answer: 4

Explanation: C = 4,4′-diethoxyazobenzene: two OCH2CH3OCH_2CH_3 groups → 4 sp3^3 carbons.

Question 72

Chemistry · 4 marks · −1

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ______. (Nearest integer) Given : Molar mass of Al is 27.0 g mol−1mol^{-1}, Molar mass of O is 16.0 g mol−1mol^{-1}

Correct answer: 153

Explanation: Al 3 mol (limiting; needs 2.25 mol O2O_2 of 4) → 1.5 mol Al2O3Al_2O_3 × 102 = 153 g.

Question 73

Chemistry · 4 marks · −1

0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 g H2OH_2O. Molar mass of (X) is ______ g mol−1mol^{-1}.

Correct answer: 100

Explanation: H = 0.1 g in 0.01 mol → 10 g H per mol = 10% ⇒ M = 100.

Question 74

Chemistry · 4 marks · −1

The bond dissociation enthalpy of X2X_2 ΔHbond\Delta H_{bond} calculated from the given data is ______ kJ mol−1mol^{-1}. (Nearest integer) M+X−(s)→M+(g)+X−(g)M^+X^-(s)\to M^+(g)+X^-(g) ΔHlattice=800\Delta H_{lattice}=800 kJ mol−1mol^{-1}; M(s)→M(g)M(s)\to M(g) ΔHsub=100\Delta H_{sub}=100 kJ mol−1mol^{-1}; M(g)→M+(g)+e−(g)M(g)\to M^+(g)+e^-(g) ΔHi=500\Delta H_i=500 kJ mol−1mol^{-1}; X(g)+e−(g)→X−(g)X(g)+e^-(g)\to X^-(g) ΔHeg=−300\Delta H_{eg}=-300 kJ mol−1mol^{-1}; M(s)+12X2(g)→M+X−(s)M(s)+\frac{1}{2}X_2(g)\to M^+X^-(s) ΔHf=−400\Delta H_f=-400 kJ mol−1mol^{-1}. [Given : M+X−M^+X^- is a pure ionic compound and X forms a diatomic molecule X2X_2 in gaseous state]

Correct answer: 200

Explanation: −400=100+500+12D−300−800⇒D=200-400=100+500+\frac12D-300-800\Rightarrow D=200.

Question 75

Chemistry · 4 marks · −1

A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with KMnO4∣H+KMnO_4|H^+ gives CH3−CO−CH3CH_3-CO-CH_3, CH3−CO−OHCH_3-CO-OH and CH3−CO−CH2CH2−CO−OHCH_3-CO-CH_2CH_2-CO-OH. The total number of σ\sigma bonds present in the compound 'X' is ______.

Correct answer: 27

Explanation: X = (CH3)2C=C(CH3)−CH2−CH2−CH=CH−CH3(CH_3)_2C=C(CH_3)-CH_2-CH_2-CH=CH-CH_3 (C10H18C_{10}H_{18}, a diene taking up 2 mol H2H_2): 9 C–C σ + 18 C–H σ = 27.

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