JEE Main 2025 — 22 Jan 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

Let E: x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>ba>b and H: x2A2−y2B2=1\frac{x^2}{A^2}-\frac{y^2}{B^2}=1. Let the distance between the foci of E and the foci of H be 232\sqrt{3}. If a−A=2a-A=2, and the ratio of the eccentricities of E and H is 13\frac{1}{3}, then the sum of the lengths of their latus rectums is equal to :

  1. A.10
  2. B.8
  3. C.9
  4. D.7

Correct answer: B

Question 2

Mathematics · 4 marks · −1

The perpendicular distance, of the line x−12=y+2−1=z+32\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2} from the point P(2, −10-10, 1), is :

  1. A.353\sqrt{5}
  2. B.6
  3. C.434\sqrt{3}
  4. D.525\sqrt{2}

Correct answer: A

Question 3

Mathematics · 4 marks · −1

Let a⃗\vec{a} and b⃗\vec{b} be two unit vectors such that the angle between them is π3\frac{\pi}{3}. If λa⃗+2b⃗\lambda\vec{a}+2\vec{b} and 3a⃗−λb⃗3\vec{a}-\lambda\vec{b} are perpendicular to each other, then the number of values of λ\lambda in [−1,3][-1, 3] is :

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

Correct answer: A

Explanation: 3λ−λ22+3−2λ=0⇒λ2−2λ−6=03\lambda-\frac{\lambda^2}{2}+3-2\lambda=0\Rightarrow\lambda^2-2\lambda-6=0, λ=1±7≈3.65,−1.65\lambda=1\pm\sqrt7\approx3.65,-1.65; none lies in [−1,3][-1,3], so 0.

Question 4

Mathematics · 4 marks · −1

Let f(x)=∫0x2t2−8t+15etdtf(x)=\int_0^{x^2}\frac{t^2-8t+15}{e^t}dt, x∈Rx\in\mathbf{R}. Then the numbers of local maximum and local minimum points of f, respectively, are :

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

Correct answer: A

Question 5

Mathematics · 4 marks · −1

If lim⁡x→∞((e1−e)(1e−x1+x))x=α\lim_{x\to\infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha, then the value of log⁡eα1+log⁡eα\frac{\log_e\alpha}{1+\log_e\alpha} equals :

  1. A.e−2e^{-2}
  2. B.e2e^2
  3. C.e−1e^{-1}
  4. D.ee

Correct answer: D

Question 6

Mathematics · 4 marks · −1

Let a line pass through two distinct points P(−2-2, −1-1, 3) and Q, and be parallel to the vector 3i^+2j^+2k^3\hat{i}+2\hat{j}+2\hat{k}. If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of △PQR\triangle PQR is equal to :

  1. A.136
  2. B.148
  3. C.140
  4. D.144

Correct answer: A

Question 7

Mathematics · 4 marks · −1

Suppose that the number of terms in an A.P. is 2k, k∈Nk\in\mathbf{N}. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to :

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

Correct answer: D

Explanation: kd=15kd=15 and (2k−1)d=27(2k-1)d=27 give d=3d=3, k=5k=5.

Question 8

Mathematics · 4 marks · −1

If ∫ex(xsin⁡−1x1−x2+sin⁡−1x(1−x2)3/2+x1−x2)dx=g(x)+C\int e^x\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}\right)dx=g(x)+C, where C is the constant of integration, then g(12)g\left(\frac{1}{2}\right) equals :

  1. A.π6e3\frac{\pi}{6}\sqrt{\frac{e}{3}}
  2. B.π6e2\frac{\pi}{6}\sqrt{\frac{e}{2}}
  3. C.π4e3\frac{\pi}{4}\sqrt{\frac{e}{3}}
  4. D.π4e2\frac{\pi}{4}\sqrt{\frac{e}{2}}

Correct answer: A

Question 9

Mathematics · 4 marks · −1

Let A={1,2,3,4}A=\{1, 2, 3, 4\} and B={1,4,9,16}B=\{1, 4, 9, 16\}. Then the number of many-one functions f:A→Bf:A\to B such that 1∈f(A)1\in f(A) is equal to :

  1. A.163
  2. B.151
  3. C.127
  4. D.139

Correct answer: B

Explanation: Functions with 1 in range: 44−34=1754^4-3^4=175; subtract the 4!=244!=24 bijections: 151.

Question 10

Mathematics · 4 marks · −1

For a 3×33\times3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3×33\times3 matrix such that ∣A∣=12|A|=\frac{1}{2} and trace (A) = 3. If B = adj(adj(2A)), then the value of ∣B∣|B| + trace (B) equals :

  1. A.174
  2. B.56
  3. C.132
  4. D.280

Correct answer: D

Explanation: adj(adj M)=∣M∣M\mathrm{adj}(\mathrm{adj}\,M)=|M|M with ∣2A∣=4|2A|=4: B=8AB=8A, ∣B∣=512⋅12=256|B|=512\cdot\frac12=256, trace =24=24; total 280.

Question 11

Mathematics · 4 marks · −1

Let P(4,43)(4, 4\sqrt{3}) be a point on the parabola y2=4axy^2=4ax and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

  1. A.26338\frac{263\sqrt{3}}{8}
  2. B.17317\sqrt{3}
  3. C.3433\frac{34\sqrt{3}}{3}
  4. D.34338\frac{343\sqrt{3}}{8}

Correct answer: D

Question 12

Mathematics · 4 marks · −1

If A and B are two events such that P(A∩B)=0.1P(A\cap B)=0.1, and P(A∣B)P(A|B) and P(B∣A)P(B|A) are the roots of the equation 12x2−7x+1=012x^2-7x+1=0, then the value of P(Aˉ∪Bˉ)P(Aˉ∩Bˉ)\frac{P(\bar{A}\cup\bar{B})}{P(\bar{A}\cap\bar{B})} is :

  1. A.74\frac{7}{4}
  2. B.53\frac{5}{3}
  3. C.43\frac{4}{3}
  4. D.94\frac{9}{4}

Correct answer: D

Question 13

Mathematics · 4 marks · −1

Let α,β,γ\alpha, \beta, \gamma and δ\delta be the coefficients of x7,x5,x3x^7, x^5, x^3 and x respectively in the expansion of (x+x3−1)5+(x−x3−1)5\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1x>1. If u and v satisfy the equations αu+βv=18\alpha u+\beta v=18, γu+δv=20\gamma u+\delta v=20, then u + v equals :

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

Correct answer: D

Question 14

Mathematics · 4 marks · −1

The sum of all values of θ∈[0,2π]\theta\in[0, 2\pi] satisfying 2sin⁡2θ=cos⁡2θ2\sin^2\theta=\cos2\theta and 2cos⁡2θ=3sin⁡θ2\cos^2\theta=3\sin\theta is

  1. A.π2\frac{\pi}{2}
  2. B.π\pi
  3. C.5π6\frac{5\pi}{6}
  4. D.4π4\pi

Correct answer: B

Explanation: Second equation gives sin⁡θ=12\sin\theta=\frac12; first gives sin⁡2θ=14\sin^2\theta=\frac14. So θ=π6,5π6\theta=\frac\pi6,\frac{5\pi}6, sum π\pi.

Question 15

Mathematics · 4 marks · −1

In a group of 3 girls and 4 boys, there are two boys B1B_1 and B2B_2. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B1B_1 and B2B_2 are not adjacent to each other, is :

  1. A.120
  2. B.72
  3. C.144
  4. D.96

Correct answer: C

Explanation: 2!×3!×(4!−3! 2!)=2×6×12=1442!\times3!\times(4!-3!\,2!)=2\times6\times12=144.

Question 16

Mathematics · 4 marks · −1

The area of the region enclosed by the curves y=x2−4x+4y=x^2-4x+4 and y2=16−8xy^2=16-8x is :

  1. A.83\frac{8}{3}
  2. B.5
  3. C.43\frac{4}{3}
  4. D.8

Correct answer: A

Question 17

Mathematics · 4 marks · −1

Let the curve z(1+i)+zˉ(1−i)=4z(1+i)+\bar{z}(1-i)=4, z∈Cz\in\mathbf{C}, divide the region ∣z−3∣≤1|z-3|\le1 into two parts of areas α\alpha and β\beta. Then ∣α−β∣|\alpha-\beta| equals :

  1. A.1+π41+\frac{\pi}{4}
  2. B.1+π61+\frac{\pi}{6}
  3. C.1+π31+\frac{\pi}{3}
  4. D.1+π21+\frac{\pi}{2}

Correct answer: D

Explanation: Line x−y=2x-y=2 is at distance 1/21/\sqrt2 from centre (3,0); minor segment area π4−12\frac\pi4-\frac12, so ∣α−β∣=π−2(π4−12)=π2+1|\alpha-\beta|=\pi-2(\frac\pi4-\frac12)=\frac\pi2+1.

Question 18

Mathematics · 4 marks · −1

If x=f(y)x=f(y) is the solution of the differential equation (1+y2)+(x−2etan⁡−1y)dydx=0(1+y^2)+\left(x-2e^{\tan^{-1}y}\right)\frac{dy}{dx}=0, y∈(−π2,π2)y\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) with f(0)=1f(0)=1, then f(13)f\left(\frac{1}{\sqrt{3}}\right) is equal to :

  1. A.eπ/12e^{\pi/12}
  2. B.eπ/4e^{\pi/4}
  3. C.eπ/3e^{\pi/3}
  4. D.eπ/6e^{\pi/6}

Correct answer: D

Explanation: dxdy+x1+y2=2etan⁡−1y1+y2\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{\tan^{-1}y}}{1+y^2}: xet=e2t+cxe^{t}=e^{2t}+c with t=tan⁡−1yt=\tan^{-1}y, c=0c=0; x=et=eπ/6x=e^{t}=e^{\pi/6}.

Question 19

Mathematics · 4 marks · −1

If the system of linear equations : x+y+2z=6x+y+2z=6, 2x+3y+az=a+12x+3y+az=a+1, −x−3y+bz=2b-x-3y+bz=2b, where a,b∈Ra, b\in\mathbf{R}, has infinitely many solutions, then 7a+3b7a+3b is equal to :

  1. A.16
  2. B.22
  3. C.9
  4. D.12

Correct answer: A

Question 20

Mathematics · 4 marks · −1

Let αθ\alpha_\theta and βθ\beta_\theta be the distinct roots of 2x2+(cos⁡θ)x−1=02x^2+(\cos\theta)x-1=0, θ∈(0,2π)\theta\in(0, 2\pi). If m and M are the minimum and the maximum values of αθ4+βθ4\alpha_\theta^4+\beta_\theta^4, then 16(M + m) equals :

  1. A.27
  2. B.25
  3. C.24
  4. D.17

Correct answer: B

Question 21

Mathematics · 4 marks · −1

Let A(6, 8), B(10cos⁡α10\cos\alpha, −10sin⁡α-10\sin\alpha) and C(−10sin⁡α-10\sin\alpha, 10cos⁡α10\cos\alpha), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a−3h+6k+100sin⁡2α)(5a-3h+6k+100\sin2\alpha) is equal to ______.

Correct answer: 145

Question 22

Mathematics · 4 marks · −1

Let y=f(x)y=f(x) be the solution of the differential equation dydx+xyx2−1=x6+4x1−x2\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}, −1<x<1-1<x<1 such that f(0)=0f(0)=0. If 6∫−1/21/2f(x)dx=2π−α6\int_{-1/2}^{1/2}f(x)dx=2\pi-\alpha then α2\alpha^2 is equal to ______.

Correct answer: 27

Question 23

Mathematics · 4 marks · −1

If ∑r=130r2(30Cr)230Cr−1=α×229\sum_{r=1}^{30}\frac{r^2\left({}^{30}C_r\right)^2}{{}^{30}C_{r-1}}=\alpha\times2^{29}, then α\alpha is equal to ______.

Correct answer: 465

Question 24

Mathematics · 4 marks · −1

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)2(QR)^2 is equal to ______.

Correct answer: 28

Question 25

Mathematics · 4 marks · −1

Let A={1,2,3}A=\{1, 2, 3\}. The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is ______.

Correct answer: 3

Question 26

Physics · 4 marks · −1

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I0I_0. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

  1. A.I0I_0
  2. B.I02\frac{I_0}{2}
  3. C.2I02I_0
  4. D.I02\frac{I_0}{\sqrt{2}}

Correct answer: B

Explanation: At resonance I=E/RI=E/R, so doubling R halves the current.

Question 27

Physics · 4 marks · −1

A tube of length L is shown in the figure. The radius of cross section at the point (1) is 2 cm and at the point (2) is 1 cm, respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be

Figure for question 27
  1. A.4 m/s
  2. B.6 m/s
  3. C.8 m/s
  4. D.2 m/s

Correct answer: C

Explanation: A1v1=A2v2⇒v2=2×4=8A_1v_1=A_2v_2\Rightarrow v_2=2\times4=8 m/s.

Question 28

Physics · 4 marks · −1

Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and C stand for unit of mass, length, time and charge,

  1. A.[F]=[CM−2L−2T−2][F]=[CM^{-2}L^{-2}T^{-2}]
  2. B.[F]=[C2M−1L−2T2][F]=[C^2M^{-1}L^{-2}T^2]
  3. C.[F]=[CM−1L−2T2][F]=[CM^{-1}L^{-2}T^2]
  4. D.[F]=[C2M−2L2T2][F]=[C^2M^{-2}L^2T^2]

Correct answer: B

Explanation: C=Q2/UC=Q^2/U: [C2M−1L−2T2][C^2M^{-1}L^{-2}T^2].

Question 29

Physics · 4 marks · −1

A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)

  1. A.Mg2\frac{Mg}{2}
  2. B.Mg
  3. C.2 Mg
  4. D.32\frac{3}{2} Mg

Correct answer: A

Explanation: Fv=Mg−buoyancy=Mg−Mg/2=Mg/2F_v=Mg-\text{buoyancy}=Mg-Mg/2=Mg/2.

Question 30

Physics · 4 marks · −1

The torque due to the force (2i^+j^+2k^)(2\hat{i}+\hat{j}+2\hat{k}) about the origin, acting on a particle whose position vector is (i^+j^+k^)(\hat{i}+\hat{j}+\hat{k}), would be

  1. A.i^−j^+k^\hat{i}-\hat{j}+\hat{k}
  2. B.j^+k^\hat{j}+\hat{k}
  3. C.i^+k^\hat{i}+\hat{k}
  4. D.i^−k^\hat{i}-\hat{k}

Correct answer: D

Explanation: r⃗×F⃗=(1⋅2−1⋅1)i^−(1⋅2−1⋅2)j^+(1⋅1−1⋅2)k^=i^−k^\vec r\times\vec F=(1\cdot2-1\cdot1)\hat i-(1\cdot2-1\cdot2)\hat j+(1\cdot1-1\cdot2)\hat k=\hat i-\hat k.

Question 31

Physics · 4 marks · −1

A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550 nm. [Assume that the light is incident nearly perpendicular to the glass surface.]

  1. A.137.5 nm
  2. B.275 nm
  3. C.68.7 nm
  4. D.94.8 nm

Correct answer: A

Explanation: Only the top reflection has a phase change, so reflection is minimum (transmission maximum) when 2μt=λ2\mu t=\lambda: t=550/4=137.5t=550/4=137.5 nm.

Question 32

Physics · 4 marks · −1

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

  1. A.hπeB\sqrt{\frac{h}{\pi eB}}
  2. B.h2πeB\sqrt{\frac{h}{2\pi eB}}
  3. C.4hπeB\sqrt{\frac{4h}{\pi eB}}
  4. D.2hπeB\sqrt{\frac{2h}{\pi eB}}

Correct answer: A

Explanation: mvr=nh2πmvr=\frac{nh}{2\pi} and r=mveBr=\frac{mv}{eB} give r2=nh2πeBr^2=\frac{nh}{2\pi eB}; n = 2: r=hπeBr=\sqrt{\frac{h}{\pi eB}}.

Question 33

Physics · 4 marks · −1

A ball of mass 100 g is projected with velocity 20 m/s at 60° with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

  1. A.5 J
  2. B.zero
  3. C.20 J
  4. D.15 J

Correct answer: D

Explanation: ΔK=12(0.1)(20sin⁡60∘)2=15\Delta K=\frac12(0.1)(20\sin60^\circ)^2=15 J.

Question 34

Physics · 4 marks · −1

A force F⃗=2i^+bj^+k^\vec{F}=2\hat{i}+b\hat{j}+\hat{k} is applied on a particle and it undergoes a displacement i^−2j^−k^\hat{i}-2\hat{j}-\hat{k}. What will be the value of b, if work done on the particle is zero.

  1. A.12\frac{1}{2}
  2. B.13\frac{1}{3}
  3. C.2
  4. D.0

Correct answer: A

Explanation: 2−2b−1=0⇒b=122-2b-1=0\Rightarrow b=\frac12.

Question 35

Physics · 4 marks · −1

For a diatomic gas, if γ1=(CpCv)\gamma_1=\left(\frac{C_p}{C_v}\right) for rigid molecules and γ2=(CpCv)\gamma_2=\left(\frac{C_p}{C_v}\right) for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct ? (CpC_p and CvC_v are specific heats of the gas at constant pressure and volume)

  1. A.γ2<γ1\gamma_2<\gamma_1
  2. B.γ2>γ1\gamma_2>\gamma_1
  3. C.2γ2=γ12\gamma_2=\gamma_1
  4. D.γ2=γ1\gamma_2=\gamma_1

Correct answer: A

Explanation: More degrees of freedom lower γ\gamma: 7/5 vs 9/7.

Question 36

Physics · 4 marks · −1

A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnate field, the plot of magnitude of induced emf (ε\varepsilon) with time (t) is given by

Figure for question 36
  1. A.Graph: ε\varepsilon increases linearly from zero with t
  2. B.Graph: ε\varepsilon decreases (decaying curve) with t
  3. C.Graph: ε\varepsilon constant (horizontal line) with t
  4. D.Graph: ε\varepsilon increases non-linearly (upward curving) with t

Correct answer: C

Explanation: ε=Blv\varepsilon=Blv is constant while the loop is leaving the field.

Question 37

Physics · 4 marks · −1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. Reason (R) : The mass of the pendulum remains unchanged at Earth and the other planet. 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.Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  4. D.(A) is true but (R) is false

Correct answer: C

Explanation: g∝M/R2g\propto M/R^2 is unchanged, so T is same (A true); mass also unchanged (R true) but T does not depend on bob mass.

Question 38

Physics · 4 marks · −1

A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is

Figure for question 38
  1. A.4D
  2. B.2D
  3. C.D
  4. D.8D

Correct answer: B

Explanation: Each piece is plano-convex: power halves to 2D (cutting along the axis does not change power).

Question 39

Physics · 4 marks · −1

For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are V0V_0 and E0E_0, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by

Figure for question 39
  1. A.zero and E08\frac{E_0}{8}
  2. B.V02\frac{V_0}{2} and E016\frac{E_0}{16}
  3. C.zero and E016\frac{E_0}{16}
  4. D.V0V_0 and E04\frac{E_0}{4}

Correct answer: C

Explanation: Equatorial point: V = 0; EB=kp(2r)3=116⋅2kpr3=E016E_B=\frac{kp}{(2r)^3}=\frac{1}{16}\cdot\frac{2kp}{r^3}=\frac{E_0}{16}.

Question 40

Physics · 4 marks · −1

Given are statements for certain thermodynamic variables, (A) Internal energy, volume (V) and mass (M) are extensive variables. (B) Pressure (P), temperature (T) and density (ρ\rho) are intensive variables. (C) Volume (V), temperature (T) and density (ρ\rho) are intensive variables. (D) Mass (M), temperature (T) and internal energy are extensive variables. Choose the correct answer from the options given below :

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

Correct answer: B

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) : In Young's double slit experiment, the fringes produced by red light are closer as compared to those produced by blue light. Reason (R) : The fringe width is directly proportional to the wavelength of light. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: D

Question 42

Physics · 4 marks · −1

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is : (Take acceleration due to gravity as 10 m/s2m/s^2)

Figure for question 42
  1. A.2+23\frac{2+\sqrt{2}}{3}
  2. B.3+32\frac{3+\sqrt{3}}{2}
  3. C.2+33\frac{2+\sqrt{3}}{3}
  4. D.3−22\frac{3-\sqrt{2}}{2}

Correct answer: B

Explanation: hB=R(1−cos⁡30∘)h_B=R(1-\cos30^\circ), hC=R(1+sin⁡30∘)h_C=R(1+\sin30^\circ) above A. KB∝100−40(1−32)=60+203K_B\propto100-40(1-\frac{\sqrt3}{2})=60+20\sqrt3, KC∝100−40(1.5)=40K_C\propto100-40(1.5)=40; ratio 3+32\frac{3+\sqrt3}{2}.

Question 43

Physics · 4 marks · −1

A light source of wavelength λ\lambda illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength λ2\frac{\lambda}{2}, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)

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

Correct answer: C

Explanation: Photon energy 3 eV → 6 eV; Kmax=6−1=5K_{max}=6-1=5 eV.

Question 44

Physics · 4 marks · −1

The maximum percentage error in the measurment of density of a wire is [Given, mass of wire = (0.60±0.003)(0.60\pm0.003)g, radius of wire = (0.50±0.01)(0.50\pm0.01)cm, length of wire = (10.00±0.05)(10.00\pm0.05)cm]

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

Correct answer: D

Explanation: Δρρ=0.5%+2(2%)+0.5%=5%\frac{\Delta\rho}{\rho}=0.5\%+2(2\%)+0.5\%=5\%.

Question 45

Physics · 4 marks · −1

A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is FαM\sqrt{\frac{F}{\alpha M}} in SI unit. The value of α\alpha is ______.

Correct answer: 1

Explanation: F=∫ω2x dm=2Mω2L2=Mω2F=\int\omega^2x\,dm=2M\omega^2\frac{L}{2}=M\omega^2 (L = 1 m), so α=1\alpha=1.

Question 46

Physics · 4 marks · −1

The net current flowing in the given circuit is ______ A.

Figure for question 46

Correct answer: 1

Explanation: In steady state the capacitor blocks everything beyond it. Current = 2/4 (through 4 Ω) + 2/(2+3‖6) = 0.5 + 0.5 = 1 A.

Question 47

Physics · 4 marks · −1

Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is x×10−5x\times10^{-5} T. The value of x is ______. Take permeability of free space as μ0=4π×10−7\mu_0=4\pi\times10^{-7} SI units.

Figure for question 47

Correct answer: 1

Explanation: BY=μ042π(0.04)=2×10−5B_Y=\frac{\mu_0 4}{2\pi(0.04)}=2\times10^{-5}, BX=μ052π(0.1)=10−5B_X=\frac{\mu_0 5}{2\pi(0.1)}=10^{-5}, opposite: net 10−510^{-5} T, x = 1.

Question 48

Physics · 4 marks · −1

A parallel plate capacitor of area A = 16 cm2cm^2 and separation between the plates 10 cm, is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2A_0=3.2\ cm^2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through A0A_0 is ______ mA.

Correct answer: 1200

Explanation: Id=6×3.216=1.2I_d=6\times\frac{3.2}{16}=1.2 A = 1200 mA.

Question 49

Physics · 4 marks · −1

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms−12\times10^5\ ms^{-1}. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is x×104x\times10^4 N/C. The value of x is ______. Take the mass of the proton = 1.6×10−271.6\times10^{-27} kg.

Correct answer: 2

Explanation: B=mvqr=1.6×10−27⋅2×1051.6×10−19⋅0.02=0.1B=\frac{mv}{qr}=\frac{1.6\times10^{-27}\cdot2\times10^5}{1.6\times10^{-19}\cdot0.02}=0.1 T; E=vB=2×104E=vB=2\times10^4, x = 2.

Question 50

Chemistry · 4 marks · −1

The maximum covalency of a non-metallic group 15 element 'E' with weakest E–E bond is :

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

Correct answer: B

Explanation: N–N single bond is the weakest (lone-pair repulsion); nitrogen has maximum covalency 4.

Question 51

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : An element in the extreme left of the periodic table forms acidic oxides. Statement (II) : Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table. 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 true
  3. C.Both Statement I and Statement II are false
  4. D.Statement I is false but Statement II is true

Correct answer: D

Question 52

Chemistry · 4 marks · −1

Identify the number of structure/s from the following which can be correlated to D-glyceraldehyde. (A) Fischer chain: CHO / C–OH (right) / C–OH (right) / CH2CH_2OH. (B) CHO / C–OH (right) / HO–C (left) / C–OH (right) / CH2CH_2OH. (C) CHO / C–OH (right) / HO–C (left) / HO–C (left) / CH2CH_2OH. (D) drawn upside down: CH2CH_2OH (top) / HO–C (left) / C–OH (right) / HO–C (left) / CHO (bottom).

Figure for question 52
  1. A.four
  2. B.two
  3. C.three
  4. D.one

Correct answer: C

Explanation: D-series if the OH on the chiral carbon farthest from CHO is on the right (after rotating D by 180° it reads CHO/OH-right/HO-left/OH-right): A, B and D are D; C is L.

Question 53

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Nitrogen, sulphur, halogen and phosphorus present in an organic compound are detected by Lassaigne's Test. Statement (II) : The elements present in the compound are converted from covalent form into ionic form by fusing the compound with Magnesium in Lassaigne's test. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: A

Explanation: Fusion is done with sodium, not magnesium.

Question 54

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : A spectral line will be observed for a 2px→2py2p_x\to2p_y transition. Statement (II) : 2px2p_x and 2py2p_y are degenerate orbitals. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: B

Question 55

Chemistry · 4 marks · −1

Consider the given figure and choose the correct option :

Figure for question 55
  1. A.Activation energy of backward reaction is E1E_1 and product is more stable than reactant.
  2. B.Activation energy of both forward and backward reaction is E1+E2E_1+E_2 and reactant is more stable than product.
  3. C.Activation energy of forward reaction is E1+E2E_1+E_2 and product is more stable than reactant.
  4. D.Activation energy of forward reaction is E1+E2E_1+E_2 and product is less stable than reactant.

Correct answer: D

Question 56

Chemistry · 4 marks · −1

The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3(PO43−)4(Zr^{4+})_3(PO_4^{3-})_4 is given by relation :

  1. A.(Ksp9612)1/3\left(\frac{K_{sp}}{9612}\right)^{1/3}
  2. B.(Ksp8435)1/7\left(\frac{K_{sp}}{8435}\right)^{1/7}
  3. C.(Ksp5348)1/6\left(\frac{K_{sp}}{5348}\right)^{1/6}
  4. D.(Ksp6912)1/7\left(\frac{K_{sp}}{6912}\right)^{1/7}

Correct answer: D

Explanation: Ksp=(3s)3(4s)4=6912s7K_{sp}=(3s)^3(4s)^4=6912s^7.

Question 57

Chemistry · 4 marks · −1

The most stable carbocation from the following is :

  1. A.p-Methoxybenzyl cation (pp-MeO–C6H4C_6H_4–C+H2\overset{+}{C}H_2)
  2. B.m-Methoxybenzyl cation (mm-MeO–C6H4C_6H_4–C+H2\overset{+}{C}H_2)
  3. C.Benzyl cation (C6H5C_6H_5–C+H2\overset{+}{C}H_2)
  4. D.p-Methylbenzyl cation (pp-H3CH_3C–C6H4C_6H_4–C+H2\overset{+}{C}H_2)

Correct answer: A

Question 58

Chemistry · 4 marks · −1

Arrange the following compounds in increasing order of their dipole moment : HBr, H2SH_2S, NF3NF_3 and CHCl3CHCl_3

  1. A.HBr<H2S<NF3<CHCl3HBr<H_2S<NF_3<CHCl_3
  2. B.NF3<HBr<H2S<CHCl3NF_3<HBr<H_2S<CHCl_3
  3. C.CHCl3<NF3<HBr<H2SCHCl_3<NF_3<HBr<H_2S
  4. D.H2S<HBr<NF3<CHCl3H_2S<HBr<NF_3<CHCl_3

Correct answer: B

Explanation: NF3NF_3 0.23 D < HBr 0.80 D < H2SH_2S 0.95 D < CHCl3CHCl_3 1.04 D.

Question 59

Chemistry · 4 marks · −1

Identify the homoleptic complex(es) that is/are low spin. (A) [Fe(CN)5NO]2−[Fe(CN)_5NO]^{2-} (B) [CoF6]3−[CoF_6]^{3-} (C) [Fe(CN)6]4−[Fe(CN)_6]^{4-} (D) [Co(NH3)6]3+[Co(NH_3)_6]^{3+} (E) [Cr(H2O)6]2+[Cr(H_2O)_6]^{2+}. Choose the correct answer from the options given below :

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

Correct answer: A

Question 60

Chemistry · 4 marks · −1

The correct order of the following complexes in terms of their crystal field stabilization energies is :

  1. A.[Co(NH3)4]2+<[Co(NH3)6]2+<[Co(NH3)6]3+<[Co(en)3]3+[Co(NH_3)_4]^{2+}<[Co(NH_3)_6]^{2+}<[Co(NH_3)_6]^{3+}<[Co(en)_3]^{3+}
  2. B.[Co(NH3)4]2+<[Co(NH3)6]2+<[Co(en)3]3+<[Co(NH3)6]3+[Co(NH_3)_4]^{2+}<[Co(NH_3)_6]^{2+}<[Co(en)_3]^{3+}<[Co(NH_3)_6]^{3+}
  3. C.[Co(NH3)6]2+<[Co(NH3)6]3+<[Co(NH3)4]2+<[Co(en)3]3+[Co(NH_3)_6]^{2+}<[Co(NH_3)_6]^{3+}<[Co(NH_3)_4]^{2+}<[Co(en)_3]^{3+}
  4. D.[Co(en)3]3+<[Co(NH3)6]3+<[Co(NH3)6]2+<[Co(NH3)4]2+[Co(en)_3]^{3+}<[Co(NH_3)_6]^{3+}<[Co(NH_3)_6]^{2+}<[Co(NH_3)_4]^{2+}

Correct answer: A

Question 61

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Partial Derivatives): (A) (∂G∂T)P\left(\frac{\partial G}{\partial T}\right)_P (B) (∂H∂T)P\left(\frac{\partial H}{\partial T}\right)_P (C) (∂G∂P)T\left(\frac{\partial G}{\partial P}\right)_T (D) (∂U∂T)V\left(\frac{\partial U}{\partial T}\right)_V. List - II (Thermodynamic Quantity): (I) CpC_p (II) −S-S (III) CvC_v (IV) V. Choose the correct answer from the options given below :

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

Correct answer: B

Question 62

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode. Statement (II) : The rate of corrosion is more in alkaline medium than in acidic medium. 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.Both Statement I and Statement II are true
  4. D.Statement I is false but Statement II is true

Correct answer: B

Question 63

Chemistry · 4 marks · −1

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is :

  1. A.2 m
  2. B.3 m
  3. C.1.79 m
  4. D.2.79 m

Correct answer: D

Explanation: 1 L: mass 1250 g, NaCl 175.5 g, water 1074.5 g; m=3/1.0745=2.79m=3/1.0745=2.79.

Question 64

Chemistry · 4 marks · −1

Match the Compounds (List - I) with the appropriate Catalyst/Reagents (List - II) for their reduction into corresponding amines. List - I (Compounds): (A) R−CO−NH2R-CO-NH_2 (B) Nitrobenzene (C) R−C≡NR-C\equiv N (D) N-alkylphthalimide. List - II (Catalyst/Reagents): (I) NaOH (aqueous) (II) H2H_2/Ni (III) LiAlH4LiAlH_4, H2OH_2O (IV) Sn, HCl. Choose the correct answer from the options given below :

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

Correct answer: C

Question 65

Chemistry · 4 marks · −1

RBr →(i) Mg, dry ether; (ii) H2O\xrightarrow{\text{(i) Mg, dry ether; (ii) } H_2O} 2-Methylbutane. The maximum number of RBr producing 2-methylbutane by above sequence of reactions is ______. (Consider the structural isomers only)

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

Correct answer: B

Explanation: Monobromo derivatives of 2-methylbutane: 1-bromo-2-methyl-, 2-bromo-2-methyl-, 2-bromo-3-methyl- and 1-bromo-3-methylbutane = 4.

Question 66

Chemistry · 4 marks · −1

The species which does not undergo disproportionation reaction is :

  1. A.ClO2−ClO_2^-
  2. B.ClO−ClO^-
  3. C.ClO3−ClO_3^-
  4. D.ClO4−ClO_4^-

Correct answer: D

Explanation: In ClO4−ClO_4^- chlorine is in its highest (+7) state, so it cannot disproportionate.

Question 67

Chemistry · 4 marks · −1

When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is :

Figure for question 67
  1. A.1-Bromo-2-(sec-butyl)cyclohexane (Br on the ring carbon adjacent to the substituent)
  2. B.2-Bromo-3-cyclohexylbutane (Br on the side-chain CH next to the branching CH)
  3. C.1-Bromo-1-(sec-butyl)cyclohexane (Br on the ring carbon bearing the side chain)
  4. D.2-Bromo-2-cyclohexylbutane (Br on the side-chain tertiary carbon)

Correct answer: D

Question 68

Chemistry · 4 marks · −1

The alkane from below having two secondary hydrogens is :

  1. A.2,2,4,5-Tetramethylheptane
  2. B.4-Ethyl-3,4-dimethyloctane
  3. C.2,2,3,3-Tetramethylpentane
  4. D.2,2,4,4-Tetramethylhexane

Correct answer: C

Explanation: 2,2,3,3-Tetramethylpentane has only one CH2CH_2 (C-4): two secondary H.

Question 69

Chemistry · 4 marks · −1

Toluene (excess) →(i) CrO2Cl2,CS2; (ii) H3O+; (iii) NaHSO3\xrightarrow{\text{(i) } CrO_2Cl_2, CS_2;\ \text{(ii) } H_3O^+;\ \text{(iii) } NaHSO_3} Filter →\to Residue (A). Residue (A) + HCl(dil) →\to Compound (B). Structure of residue (A) and compound (B) formed respectively is :

  1. A.A = benzaldehyde (C6H5CHOC_6H_5CHO); B = benzoic acid (C6H5COOHC_6H_5COOH)
  2. B.A = benzaldehyde (C6H5CHOC_6H_5CHO); B = sodium benzoate (C6H5COONaC_6H_5COONa)
  3. C.A = bisulphite adduct C6H5CH(OH)SO3NaC_6H_5CH(OH)SO_3Na; B = benzaldehyde (C6H5CHOC_6H_5CHO)
  4. D.A = chromium complex C6H5CH(OCrOHCl2)2C_6H_5CH(OCrOHCl_2)_2; B = benzaldehyde (C6H5CHOC_6H_5CHO)

Correct answer: C

Explanation: Etard reaction gives benzaldehyde, isolated as its crystalline bisulphite adduct (residue); dilute acid regenerates benzaldehyde.

Question 70

Chemistry · 4 marks · −1

The complex of Ni2+Ni^{2+} ion and dimethyl glyoxime contains ______ number of Hydrogen (H) atoms.

Correct answer: 14

Explanation: [Ni(dmgH)2]=NiC8H14N4O4[Ni(dmgH)_2]=NiC_8H_{14}N_4O_4: 14 H.

Question 71

Chemistry · 4 marks · −1

Niobium (Nb) and ruthenium (Ru) have "x" and "y" number of electrons in their respective 4d orbitals. The value of x + y is ______.

Correct answer: 11

Explanation: Nb: [Kr]4d45s1[Kr]4d^45s^1, Ru: [Kr]4d75s1[Kr]4d^75s^1; 4 + 7 = 11.

Question 72

Chemistry · 4 marks · −1

The compound with molecular formula C6H6C_6H_6, which gives only one monobromo derivative and takes up four moles of hydrogen per mole for complete hydrogenation has ______ π\pi electrons.

Correct answer: 8

Question 73

Chemistry · 4 marks · −1

20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is ______ ×10−2\times10^{-2} M. (Nearest integer)

Correct answer: 57

Explanation: (40+200)/420=0.571(40+200)/420=0.571 M → 57.

Question 74

Chemistry · 4 marks · −1

Consider the following cases of standard enthalpy of reaction (ΔHr∘\Delta H_r^\circ in kJ mol−1mol^{-1}): C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)C_2H_6(g)+\frac{7}{2}O_2(g)\to2CO_2(g)+3H_2O(l) ΔH1∘=−1550\Delta H_1^\circ=-1550; C(graphite) + O2(g)→CO2(g)O_2(g)\to CO_2(g) ΔH2∘=−393.5\Delta H_2^\circ=-393.5; H2(g)+12O2(g)→H2O(l)H_2(g)+\frac{1}{2}O_2(g)\to H_2O(l) ΔH3∘=−286\Delta H_3^\circ=-286. The magnitude of ΔHf∘\Delta H_f^\circ C2H6(g)C_2H_6(g) is ______ kJ mol−1mol^{-1} (Nearest integer).

Correct answer: 95

Explanation: ΔHf=2(−393.5)+3(−286)+1550=−95\Delta H_f=2(-393.5)+3(-286)+1550=-95.

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