JEE Main 2025 — 2 Apr 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

Given three indentical bags each containing 10 balls, whose colours are as follows : Bag I: Red 3, Blue 2, Green 5; Bag II: Red 4, Blue 3, Green 3; Bag III: Red 5, Blue 1, Green 4. A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is q, then the value of (1p+1q)\left(\frac{1}{p}+\frac{1}{q}\right) is :

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

Correct answer: A

Explanation: p=312p=\frac{3}{12}, q=412q=\frac{4}{12}; 1p+1q=4+3=7\frac1p+\frac1q=4+3=7.

Question 2

Mathematics · 4 marks · −1

Let (a, b) be the point of intersection of the curve x2=2yx^2=2y and the straight line y−2x−6=0y-2x-6=0 in the second quadrant. Then the integral I=∫ab9x21+5xdxI=\int_a^b\frac{9x^2}{1+5^x}dx is equal to :

  1. A.24
  2. B.18
  3. C.21
  4. D.27

Correct answer: A

Explanation: x2=4x+12⇒x=−2x^2=4x+12\Rightarrow x=-2, point (−2,2)(-2,2); I=∫−229x21+5xdx=12∫−229x2dx=24I=\int_{-2}^{2}\frac{9x^2}{1+5^x}dx=\frac12\int_{-2}^2 9x^2dx=24.

Question 3

Mathematics · 4 marks · −1

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q(α,β,γ\alpha, \beta, \gamma), then α+β+γ\alpha+\beta+\gamma is equal to :

  1. A.463\frac{46}{3}
  2. B.18
  3. C.473\frac{47}{3}
  4. D.13

Correct answer: A

Question 4

Mathematics · 4 marks · −1

If θ∈[−7π6,4π3]\theta\in\left[-\frac{7\pi}{6}, \frac{4\pi}{3}\right], then the number of solutions of 3 cosec2θ−2(3−1)cosec θ−4=0\sqrt{3}\,\mathrm{cosec}^2\theta-2(\sqrt{3}-1)\mathrm{cosec}\,\theta-4=0, is equal to :

  1. A.8
  2. B.7
  3. C.6
  4. D.10

Correct answer: C

Question 5

Mathematics · 4 marks · −1

Let the point P of the focal chord PQ of the parabola y2=16xy^2=16x be (1,−4)(1, -4). If the focus of the parabola divides the chord PQ in the ratio m : n, gcd⁡(m,n)=1\gcd(m, n)=1, then m2+n2m^2+n^2 is equal to :

  1. A.10
  2. B.37
  3. C.17
  4. D.26

Correct answer: C

Explanation: a=4a=4, P=(at2,2at)P=(at^2,2at) with t=−1/2t=-1/2; QQ has t=ˊ2t\'=2: Q=(16,16)Q=(16,16). SP=a+1=5SP=a+1=5, SQ=a+16=20SQ=a+16=20: ratio 1 : 4, m2+n2=17m^2+n^2=17.

Question 6

Mathematics · 4 marks · −1

If the system of equations 2x+λy+3z=52x+\lambda y+3z=5, 3x+2y−z=73x+2y-z=7, 4x+5y+μz=94x+5y+\mu z=9 has infinitely many solutions, then (λ2+μ2)(\lambda^2+\mu^2) is equal to :

  1. A.18
  2. B.22
  3. C.30
  4. D.26

Correct answer: D

Question 7

Mathematics · 4 marks · −1

If the domain of the function f(x)=110+3x−x2+1x+∣x∣f(x)=\frac{1}{\sqrt{10+3x-x^2}}+\frac{1}{\sqrt{x+|x|}} is (a, b), then (1+a)2+b2(1+a)^2+b^2 is equal to :

  1. A.25
  2. B.30
  3. C.26
  4. D.29

Correct answer: C

Explanation: Need −2<x<5-2<x<5 and x>0x>0: domain (0,5)(0,5); 1+25=261+25=26.

Question 8

Mathematics · 4 marks · −1

4∫01(13+x2+1+x2)dx−3log⁡e(3)4\int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right)dx-3\log_e(\sqrt{3}) is equal to :

  1. A.2+2+log⁡e(1+2)2+\sqrt{2}+\log_e(1+\sqrt{2})
  2. B.2−2+log⁡e(1+2)2-\sqrt{2}+\log_e(1+\sqrt{2})
  3. C.2−2−log⁡e(1+2)2-\sqrt{2}-\log_e(1+\sqrt{2})
  4. D.2+2−log⁡e(1+2)2+\sqrt{2}-\log_e(1+\sqrt{2})

Correct answer: C

Question 9

Mathematics · 4 marks · −1

The line L1L_1 is parallel to the vector a⃗=−3i^+2j^+4k^\vec{a}=-3\hat{i}+2\hat{j}+4\hat{k} and passes through the point (7, 6, 2) and the line L2L_2 is parallel to the vector b⃗=2i^+j^+3k^\vec{b}=2\hat{i}+\hat{j}+3\hat{k} and passes through the point (5, 3, 4). The shortest distance between the lines L1L_1 and L2L_2 is :

  1. A.2138\frac{21}{\sqrt{38}}
  2. B.2157\frac{21}{\sqrt{57}}
  3. C.2357\frac{23}{\sqrt{57}}
  4. D.2338\frac{23}{\sqrt{38}}

Correct answer: D

Explanation: a⃗×b⃗=(2,17,−7)\vec a\times\vec b=(2,17,-7), ∣⋅∣=342=338|\cdot|=\sqrt{342}=3\sqrt{38}; (2,3,−2)⋅(2,17,−7)=69(2,3,-2)\cdot(2,17,-7)=69; SD =69/(338)=23/38=69/(3\sqrt{38})=23/\sqrt{38}.

Question 10

Mathematics · 4 marks · −1

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 212\frac{21}{2}. Then the number of terms which are integers in the A.P. is :

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

Correct answer: A

Explanation: nd=6nd=6 (n pairs), (2n−1)d=21/2(2n-1)d=21/2 give d=3/2d=3/2, n=4n=4: 8 terms; a=3/2a=3/2, terms 32,3,92,6,…,12\frac32,3,\frac92,6,\ldots,12 → 4 integers.

Question 11

Mathematics · 4 marks · −1

Let a⃗=2i^−3j^+k^\vec{a}=2\hat{i}-3\hat{j}+\hat{k}, b⃗=3i^+2j^+5k^\vec{b}=3\hat{i}+2\hat{j}+5\hat{k} and a vector c⃗\vec{c} be such that (a⃗−c⃗)×b⃗=−18i^−3j^+12k^(\vec{a}-\vec{c})\times\vec{b}=-18\hat{i}-3\hat{j}+12\hat{k} and a⃗⋅c⃗=3\vec{a}\cdot\vec{c}=3. If b⃗×c⃗=d⃗\vec{b}\times\vec{c}=\vec{d}, then ∣a⃗⋅d⃗∣|\vec{a}\cdot\vec{d}| is equal to :

  1. A.18
  2. B.15
  3. C.12
  4. D.9

Correct answer: B

Question 12

Mathematics · 4 marks · −1

Let f:[1,∞)→[2,∞)f:[1, \infty)\to[2, \infty) be a differentiable function. If 10∫1xf(t)dt=5xf(x)−x5−910\int_1^x f(t)dt=5xf(x)-x^5-9 for all x≥1x\ge1, then the value of f(3)f(3) is :

  1. A.18
  2. B.26
  3. C.32
  4. D.22

Correct answer: C

Question 13

Mathematics · 4 marks · −1

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

  1. A.319\frac{3}{\sqrt{19}}
  2. B.316\frac{\sqrt{3}}{16}
  3. C.57\frac{\sqrt{5}}{7}
  4. D.417\frac{4}{\sqrt{17}}

Correct answer: D

Explanation: 2b=2ae4⇒b=ae/42b=\frac{2ae}{4}\Rightarrow b=ae/4; 1−e2=e2/16⇒e=4/171-e^2=e^2/16\Rightarrow e=4/\sqrt{17}.

Question 14

Mathematics · 4 marks · −1

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a+b+aba+b+ab is equal to :

  1. A.103
  2. B.105
  3. C.106
  4. D.100

Correct answer: A

Explanation: a+b=19a+b=19; ∑x2=8(9.25+81)=722\sum x^2=8(9.25+81)=722, known squares 529, so a2+b2=193a^2+b^2=193, ab=84ab=84; 19+84=10319+84=103.

Question 15

Mathematics · 4 marks · −1

If lim⁡x→0cos⁡(2x)+acos⁡(4x)−bx4\lim_{x\to0}\frac{\cos(2x)+a\cos(4x)-b}{x^4} is finite, then (a + b) is equal to :

  1. A.34\frac{3}{4}
  2. B.-1
  3. C.0
  4. D.12\frac{1}{2}

Correct answer: D

Explanation: x2x^2 term: −2−8a=0⇒a=−14-2-8a=0\Rightarrow a=-\frac14; constant: 1+a−b=0⇒b=341+a-b=0\Rightarrow b=\frac34; a+b=12a+b=\frac12.

Question 16

Mathematics · 4 marks · −1

Let A={1,2,3,…,100}A=\{1, 2, 3, \ldots, 100\} and R be a relation on A such that R={(a,b):a=2b+1}R=\{(a, b): a=2b+1\}. Let (a1,a2),(a2,a3),(a3,a4),…,(ak,ak+1)(a_1, a_2), (a_2, a_3), (a_3, a_4), \ldots, (a_k, a_{k+1}) be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k, for which such a sequence exists, is equal to :

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

Correct answer: D

Explanation: Each step halves roughly: the longest chain inside A is 63→31→15→7→3→1 (next would need 127), giving k = 5 pairs.

Question 17

Mathematics · 4 marks · −1

If ∑r=010(10r+1−110r)⋅11Cr+1=α11−11111010\sum_{r=0}^{10}\left(\frac{10^{r+1}-1}{10^r}\right)\cdot{}^{11}C_{r+1}=\frac{\alpha^{11}-11^{11}}{10^{10}}, then α\alpha is equal to :

  1. A.15
  2. B.24
  3. C.11
  4. D.20

Correct answer: D

Explanation: Sum =10∑11Cr+1−∑11Cr+110r=10(211−1)−10[(1.1)11−1]=2011−11111010=10\sum{}^{11}C_{r+1}-\sum\frac{{}^{11}C_{r+1}}{10^r}=10(2^{11}-1)-10\left[(1.1)^{11}-1\right]=\frac{20^{11}-11^{11}}{10^{10}}; α=20\alpha=20.

Question 18

Mathematics · 4 marks · −1

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

Figure for question 18
  1. A.840
  2. B.5880
  3. C.5760
  4. D.960

Correct answer: C

Explanation: Choose 5 boxes with every row non-empty: (85)−(55)−(55)−(65)=56−1−1−6=48\binom85-\binom55-\binom55-\binom65=56-1-1-6=48 (rows of sizes 3,3,2); arrangements 48×5!=576048\times5!=5760.

Question 19

Mathematics · 4 marks · −1

Let the area of the triangle formed by a straight line L : x+by+c=0x+by+c=0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of b2+c2b^2+c^2 is :

  1. A.90
  2. B.97
  3. C.93
  4. D.83

Correct answer: B

Explanation: Normal at 45° ⇒ b = 1; intercepts |c| each, area c2/2=48c^2/2=48, c2=96c^2=96; b2+c2=97b^2+c^2=97.

Question 20

Mathematics · 4 marks · −1

Let A be a 3×33\times3 real matrix such that A2(A−2I)−4(A−I)=OA^2(A-2I)-4(A-I)=O, where I and O are the identity and null matrices, respectively. If A5=αA2+βA+γIA^5=\alpha A^2+\beta A+\gamma I, where α\alpha, β\beta, and γ\gamma are real constants, then α+β+γ\alpha+\beta+\gamma is equal to :

  1. A.20
  2. B.76
  3. C.12
  4. D.4

Correct answer: C

Question 21

Mathematics · 4 marks · −1

If the set of all a∈R−{1}a\in\mathbf{R}-\{1\}, for which the roots of the equation (1−a)x2+2(a−3)x+9=0(1-a)x^2+2(a-3)x+9=0 are positive is (−∞,−α]∪[β,γ)(-\infty, -\alpha]\cup[\beta, \gamma), then 2α+β+γ2\alpha+\beta+\gamma is equal to ______.

Correct answer: 7

Question 22

Mathematics · 4 marks · −1

Let A(4, −2-2), B(1, 1) and C(9, −3-3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ______.

Correct answer: 3

Explanation: Area(ABC) =12∣(−3)(−1)−(3)(5)∣=6=\frac12|(-3)(-1)-(3)(5)|=6; maximum inscribed parallelogram = half of it = 3.

Question 23

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation dydx+2ysec⁡2x=2sec⁡2x+3tan⁡x⋅sec⁡2x\frac{dy}{dx}+2y\sec^2x=2\sec^2x+3\tan x\cdot\sec^2x such that y(0)=54y(0)=\frac{5}{4}. Then 12(y(π4)−e−2)12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right) is equal to ______.

Correct answer: 21

Question 24

Mathematics · 4 marks · −1

If the sum of the first 10 terms of the series 4⋅11+4⋅14+4⋅21+4⋅24+4⋅31+4⋅34+…\frac{4\cdot1}{1+4\cdot1^4}+\frac{4\cdot2}{1+4\cdot2^4}+\frac{4\cdot3}{1+4\cdot3^4}+\ldots is mn\frac{m}{n}, where gcd⁡(m,n)=1\gcd(m, n)=1, then m + n is equal to ______.

Correct answer: 441

Explanation: 4r1+4r4=12r2−2r+1−12r2+2r+1\frac{4r}{1+4r^4}=\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}; sum =1−1221=220221=1-\frac{1}{221}=\frac{220}{221}; 441.

Question 25

Mathematics · 4 marks · −1

If y=cos⁡(π3+cos⁡−1x2)y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right), then (x−y)2+3y2(x-y)^2+3y^2 is equal to ______.

Correct answer: 3

Explanation: y=x4−321−x24y=\frac{x}{4}-\frac{\sqrt3}{2}\sqrt{1-\frac{x^2}{4}}; then (x−y)2+3y2=x2−2xy+4y2=3(x-y)^2+3y^2=x^2-2xy+4y^2=3.

Question 26

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R) : In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: C

Question 27

Physics · 4 marks · −1

A solenoid having area A and length 'll' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

  1. A.B2AlB^2Al
  2. B.B2Al2μ0\frac{B^2Al}{2\mu_0}
  3. C.B2Alμ0\frac{B^2Al}{\mu_0}
  4. D.B2Al4μ0\frac{B^2Al}{4\mu_0}

Correct answer: D

Explanation: U=B22μrμ0Al=B2Al4μ0U=\frac{B^2}{2\mu_r\mu_0}Al=\frac{B^2Al}{4\mu_0}.

Question 28

Physics · 4 marks · −1

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x = 0 at t = 0 sec and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?

  1. A.y=2sin⁡(0.83x−3.5t)y=2\sin(0.83x-3.5t) cm
  2. B.y=2cos⁡(3.35x−0.83t)y=2\cos(3.35x-0.83t) cm
  3. C.y=2cos⁡(0.13x−0.5t)y=2\cos(0.13x-0.5t) cm
  4. D.y=2cos⁡(0.83x−3.35t)y=2\cos(0.83x-3.35t) cm

Correct answer: D

Explanation: k=2π/7.5=0.84k=2\pi/7.5=0.84 cm−1^{-1}, v=4v=4 cm/s, ω=kv=3.35\omega=kv=3.35 s−1^{-1}; crest at origin at t = 0 → cosine.

Question 29

Physics · 4 marks · −1

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions T1T_1 and T2T_2, respectively, are (in N) : (Take acceleration due to gravity 10 m/s2m/s^2)

Figure for question 29
  1. A.535\sqrt{3}, 535\sqrt{3}
  2. B.5, 5
  3. C.5, 535\sqrt{3}
  4. D.535\sqrt{3}, 5

Correct answer: D

Explanation: T1cos⁡60=T2cos⁡30T_1\cos60=T_2\cos30 and T1sin⁡60+T2sin⁡30=10T_1\sin60+T_2\sin30=10 give T1=53T_1=5\sqrt3, T2=5T_2=5.

Question 30

Physics · 4 marks · −1

An electron with mass 'm' with an initial velocity (t=0) v⃗=v0i^\vec{v}=v_0\hat{i} (v0>0v_0>0) enters a magnetic field B⃗=B0j^\vec{B}=B_0\hat{j}. If the initial de-Broglie wavelength at t = 0 is λ0\lambda_0 then its value after time 't' would be :

  1. A.λ01+e2B02t2m2\lambda_0\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}
  2. B.λ0\lambda_0
  3. C.λ01+e2B02t2m2\frac{\lambda_0}{\sqrt{1+\frac{e^2B_0^2t^2}{m^2}}}
  4. D.λ01−e2B02t2m2\frac{\lambda_0}{\sqrt{1-\frac{e^2B_0^2t^2}{m^2}}}

Correct answer: B

Explanation: Magnetic force does no work; speed and hence wavelength are unchanged.

Question 31

Physics · 4 marks · −1

Energy released when two deuterons (1H2{}_1H^2) fuse to form a helium nucleus (2He4{}_2He^4) is : (Given : Binding energy per nucleon of 1H2{}_1H^2 = 1.1 MeV and binding energy per nucleon of 2He4{}_2He^4 = 7.0 MeV)

  1. A.23.6 MeV
  2. B.26.8 MeV
  3. C.8.1 MeV
  4. D.5.9 MeV

Correct answer: A

Explanation: 4×7.0−2×(2×1.1)=28−4.4=23.64\times7.0-2\times(2\times1.1)=28-4.4=23.6 MeV.

Question 32

Physics · 4 marks · −1

Consider a circular loop that is uniformly charged and has a radius a2a\sqrt{2}. Find the position along the positive z-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in xy plane at the origin :

  1. A.0
  2. B.a/2a/\sqrt{2}
  3. C.a/2a/2
  4. D.a

Correct answer: D

Explanation: EE is maximum at z=R/2=az=R/\sqrt2=a.

Question 33

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Heat capacity of body (B) Specific heat capacity of body (C) Latent heat (D) Thermal conductivity. List - II: (I) J kg−1J\,kg^{-1} (II) J K−1J\,K^{-1} (III) J kg−1K−1J\,kg^{-1}K^{-1} (IV) J m−1K−1s−1J\,m^{-1}K^{-1}s^{-1}. Choose the correct answer from the options given below :

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

Correct answer: D

Question 34

Physics · 4 marks · −1

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :

Figure for question 34
  1. A.34\frac{3}{4}V
  2. B.1 V
  3. C.25\frac{2}{5}V
  4. D.14\frac{1}{4}V

Correct answer: C

Explanation: ΔV=V10×4=2V5\Delta V=\frac{V}{10}\times4=\frac{2V}{5}.

Question 35

Physics · 4 marks · −1

In the digital circuit shown in the figure, for the given inputs the P and Q values are :

Figure for question 35
  1. A.P = 0, Q = 0
  2. B.P = 0, Q = 1
  3. C.P = 1, Q = 1
  4. D.P = 1, Q = 0

Correct answer: A

Question 36

Physics · 4 marks · −1

Two water drops each of radius 'r' coalesce to form a bigger drop. If 'T' is the surface tension, the surface energy released in this process is :

  1. A.4πr2T[2−22/3]4\pi r^2T\left[2-2^{2/3}\right]
  2. B.4πr2T[2−21/3]4\pi r^2T\left[2-2^{1/3}\right]
  3. C.4πr2T[2−1]4\pi r^2T\left[\sqrt{2}-1\right]
  4. D.4πr2T[1+2]4\pi r^2T\left[1+\sqrt{2}\right]

Correct answer: A

Explanation: R=21/3rR=2^{1/3}r: released =4πT(2r2−22/3r2)=4\pi T(2r^2-2^{2/3}r^2).

Question 37

Physics · 4 marks · −1

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

  1. A.32Mr2\frac{3}{2}Mr^2
  2. B.12Mr2\frac{1}{2}Mr^2
  3. C.38Mr2\frac{3}{8}Mr^2
  4. D.2Mr22Mr^2

Correct answer: C

Explanation: 32MR2\frac32MR^2 with R=r/2R=r/2: 38Mr2\frac38Mr^2.

Question 38

Physics · 4 marks · −1

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of Li++Li^{++} ion in its ground state is given by 1Xa0\frac{1}{X}a_0, where X = ______. (Where a0a_0 is the first Bohr's radius.)

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

Correct answer: C

Explanation: r=a0/Zr=a_0/Z with Z = 3.

Question 39

Physics · 4 marks · −1

A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (R1≠R2R_1\neq R_2), without any change in lens power then possible combination of R1R_1 and R2R_2 is :

  1. A.13\frac{1}{3} cm and 13\frac{1}{3} cm
  2. B.16\frac{1}{6} cm and 19\frac{1}{9} cm
  3. C.15\frac{1}{5} cm and 17\frac{1}{7} cm
  4. D.13\frac{1}{3} cm and 17\frac{1}{7} cm

Correct answer: C

Explanation: Need 1R1+1R2=12\frac1{R_1}+\frac1{R_2}=12: 5 + 7 = 12.

Question 40

Physics · 4 marks · −1

Given a charge q, current I and permeability of vacuum μ0\mu_0. Which of the following quantity has the dimension of momentum ?

  1. A.q2μ0Iq^2\mu_0I
  2. B.qI/μ0qI/\mu_0
  3. C.qμ0/Iq\mu_0/I
  4. D.qμ0Iq\mu_0I

Correct answer: D

Explanation: μ0I\mu_0 I has dimensions of (magnetic field × length), so qμ0Iq\mu_0 I has dimensions of qBL=(qvB)(L/v)qBL=(qvB)(L/v) = force × time = momentum.

Question 41

Physics · 4 marks · −1

If μ0\mu_0 and ϵ0\epsilon_0 are the permeability and permittivity of free space, respectively, then the dimension of (1μ0ϵ0)\left(\frac{1}{\mu_0\epsilon_0}\right) is :

  1. A.L/T2L/T^2
  2. B.T2/LT^2/L
  3. C.L2/T2L^2/T^2
  4. D.T2/L2T^2/L^2

Correct answer: C

Explanation: 1/μ0ϵ0=c21/\mu_0\epsilon_0=c^2.

Question 42

Physics · 4 marks · −1

Identify the characteristics of an adiabatic process in a monoatomic gas. (A) Internal energy is constant. (B) Work done in the process is equal to the change in internal energy. (C) The product of temperature and volume is a constant. (D) The product of pressure and volume is a constant. (E) The work done to change the temperature from T1T_1 to T2T_2 is proportional to (T2−T1)(T_2-T_1). Choose the correct answer from the options given below :

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

Correct answer: C

Question 43

Physics · 4 marks · −1

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

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

Correct answer: D

Explanation: m=ff−um=\frac{f}{f-u}: convex 66+18=14\frac{6}{6+18}=\frac14, concave −6−6+18=−12\frac{-6}{-6+18}=-\frac12; ratio 1/2.

Question 44

Physics · 4 marks · −1

A sportsman runs around a circular track of radius r such that he traverses the path ABAB. The distance travelled and displacement, respectively, are

Figure for question 44
  1. A.3πr3\pi r, πr\pi r
  2. B.2r2r, 3πr3\pi r
  3. C.3πr3\pi r, 2r2r
  4. D.πr\pi r, 3r3r

Correct answer: C

Explanation: Three half-circles: 3πr3\pi r; displacement AB = 2r.

Question 45

Physics · 4 marks · −1

In a moving coil galvanometer, two moving coils M1M_1 and M2M_2 have the following particulars : R1=5 ΩR_1=5\,\Omega, N1=15N_1=15, A1=3.6×10−3 m2A_1=3.6\times10^{-3}\ m^2, B1=0.25B_1=0.25 T; R2=7 ΩR_2=7\,\Omega, N2=21N_2=21, A2=1.8×10−3 m2A_2=1.8\times10^{-3}\ m^2, B2=0.50B_2=0.50 T. Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M1M_1 and M2M_2 ?

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

Correct answer: C

Explanation: Vs=NBAkRV_s=\frac{NBA}{kR}: both give NBA/R=2.7×10−3NBA/R=2.7\times10^{-3} → 1 : 1.

Question 46

Physics · 4 marks · −1

The internal energy of air in 4 m × 4 m × 3 m sized room at 1 atmospheric pressure will be ______ ×106\times10^6 J. (Consider air as diatomic molecule)

Correct answer: 12

Explanation: U=52PV=2.5×1.013×105×48≈12×106U=\frac52PV=2.5\times1.013\times10^5\times48\approx12\times10^6 J.

Question 47

Physics · 4 marks · −1

The length of a light string is 1.4 m when the tension on it is 5 N. If the tension increases to 7 N, the length of the string is 1.56 m. The original length of the string is ______ m.

Correct answer: 1

Explanation: L0+5k=1.4L_0+5k=1.4, L0+7k=1.56L_0+7k=1.56: k=0.08k=0.08, L0=1.0L_0=1.0 m.

Question 48

Physics · 4 marks · −1

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad/s2rad/s^2. Moment of intertia of the wheel is ______ kg m2m^2. (Acceleration due to gravity = 10 m/s2m/s^2)

Figure for question 48

Correct answer: 1

Explanation: I=τ/α=10×0.2/2=1I=\tau/\alpha=10\times0.2/2=1.

Question 49

Physics · 4 marks · −1

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to 60°. The refractive index of the prism material is 2\sqrt{2}. The angle of incidence (in degrees) is ______.

Correct answer: 45

Explanation: r=30∘r=30^\circ, sin⁡i=2sin⁡30∘\sin i=\sqrt2\sin30^\circ → i=45∘i=45^\circ.

Question 50

Physics · 4 marks · −1

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth\'s surface. Kinetic energy of the satellite in this orbit is ______ ×1010\times10^{10} J. (Mass of earth = 6×10246\times10^{24} kg, Radius of earth = 6.4×1066.4\times10^6 m, Gravitational constant = 6.67×10−11 Nm2kg−26.67\times10^{-11}\ Nm^2kg^{-2})

Correct answer: 3

Explanation: K=GMm2r=6.67×10−11⋅6×1024⋅10002×6.67×106=3×1010K=\frac{GMm}{2r}=\frac{6.67\times10^{-11}\cdot6\times10^{24}\cdot1000}{2\times6.67\times10^6}=3\times10^{10} J.

Question 51

Chemistry · 4 marks · −1

Which of the following statements are true ? (A) The subsidiary quantum number ll describes the shape of the orbital occupied by the electron. (B) [a two-lobed dumbbell drawn along the x-axis] is the boundary surface diagram of the 2px2p_x orbital. (C) The + and − signs in the wave function of the 2px2p_x orbital refer to charge. (D) The wave function of 2px2p_x orbital is zero everywhere in the xy plane. Choose the correct answer from the options given below :

Figure for question 51
  1. A.(A), (B) and (C) only
  2. B.(A) and (B) only
  3. C.(C) and (D) only
  4. D.(B) and (D) only

Correct answer: B

Explanation: The nodal plane of 2px2p_x is the yz plane, not xy; signs are phases, not charges.

Question 52

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Purification technique): (A) Distillation (simple) (B) Fractional distillation (C) Distillation under reduced pressure (D) Steam distillation. List - II (Mixture of organic compounds): (I) Diesel + Petrol (II) Aniline + Water (III) Chloroform + Aniline (IV) Glycerol + Spent-lye. Choose the correct answer from the options given below :

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

Correct answer: B

Question 53

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Reaction): (A) 2C6H5X+2Na→Dry etherC6H5−C6H5+2NaX2C_6H_5X+2Na\xrightarrow{\text{Dry ether}}C_6H_5-C_6H_5+2NaX (B) ArN2+X−→Cu/HClArCl+N2↑+CuXArN_2^+X^-\xrightarrow{Cu/HCl}ArCl+N_2\uparrow+CuX (C) C2H5Br+NaI→Dry acetoneC2H5I+NaBrC_2H_5Br+NaI\xrightarrow{\text{Dry acetone}}C_2H_5I+NaBr (D) CH3C(OH)(CH3)CH3→HCl/ZnCl2CH3C(Cl)(CH3)CH3CH_3C(OH)(CH_3)CH_3\xrightarrow{HCl/ZnCl_2}CH_3C(Cl)(CH_3)CH_3. List - II (Name of reaction): (I) Lucas reaction (II) Finkelstein reaction (III) Fittig reaction (IV) Gatterman reaction. Choose the correct answer from the options given below :

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

Correct answer: C

Question 54

Chemistry · 4 marks · −1

The nature of oxide (TeO2TeO_2) and hydride (TeH2TeH_2) formed by Te, respectively are :

  1. A.Reducing and basic
  2. B.Reducing and acidic
  3. C.Oxidising and basic
  4. D.Oxidising and acidic

Correct answer: D

Question 55

Chemistry · 4 marks · −1

Which among the following molecules is (a) involved in sp3dsp^3d hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom ?

  1. A.SF4SF_4
  2. B.PF5PF_5
  3. C.XeF2XeF_2
  4. D.XeF4XeF_4

Correct answer: A

Question 56

Chemistry · 4 marks · −1

Formation of Na4[Fe(CN)5NOS]Na_4[Fe(CN)_5NOS], a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate extract of salt indicates the presence of :

  1. A.Sulphate ion
  2. B.Sodium ion
  3. C.Sulphite ion
  4. D.Sulphide ion

Correct answer: D

Question 57

Chemistry · 4 marks · −1

Arrange the following in order of magnitude of work done by the system/on the system at constant temperature. (a) ∣wreversible∣|w_{reversible}| for expansion in infinite stages. (b) ∣wirreversible∣|w_{irreversible}| for expansion in single stage. (c) ∣wreversible∣|w_{reversible}| for compression in infinite stages. (d) ∣wirreversible∣|w_{irreversible}| for compression in single stage. Choose the correct answer from the options given below :

  1. A.a > c > b > d
  2. B.a > b > c > d
  3. C.d > c = a > b
  4. D.c = a > d > b

Correct answer: C

Question 58

Chemistry · 4 marks · −1

Reactant A converts to product D through the given mechanism (with the net evolution of heat) : A →\to B slow; ΔH\Delta H = +ve. B →\to C fast; ΔH\Delta H = −ve. C →\to D fast; ΔH\Delta H = −ve. Which of the following represents the above reaction mechanism ?

Figure for question 58
  1. A.Profile 1: the first hump (A→B) is the highest barrier, B lies above A, then two lower humps lead down through C to D, and D lies well below A
  2. B.Profile 2: first hump high, second hump lower, third hump the tallest; D lies slightly above A
  3. C.Profile 3: first hump the tallest, then two lower humps; D lies slightly above A
  4. D.Profile 4: humps rise successively so the last (C→D) barrier is the highest; D lies far below A

Correct answer: A

Explanation: Slow first step = highest barrier; endothermic first step puts B above A; net exothermic puts D below A.

Question 59

Chemistry · 4 marks · −1

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1°C to 25°C. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature ? [Consider the solubility of NaCl remains unchanged over the temperature range]

Figure for question 59
  1. A.M decreases linearly with t
  2. B.M dips slightly then increases with t
  3. C.M increases linearly with t
  4. D.M first rises to a maximum (around 4 °C) and then decreases with t

Correct answer: D

Explanation: Molarity follows the density of water, which is maximum at 4 °C: M rises slightly up to ~4 °C and then falls.

Question 60

Chemistry · 4 marks · −1

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : (Given molar mass in g mol−1mol^{-1} H : 1, C : 12, N : 14, O : 16, S : 32)

  1. A.330
  2. B.343
  3. C.33
  4. D.66

Correct answer: C

Explanation: Azo dye HO3S−C6H4−N=N−C10H6−NH2HO_3S-C_6H_4-N=N-C_{10}H_6-NH_2 (C16H13N3O3SC_{16}H_{13}N_3O_3S, 327 g/mol) → 32.7 ≈ 33 g.

Question 61

Chemistry · 4 marks · −1

The type of hybridization and the magnetic property of [MnCl6]3−[MnCl_6]^{3-} are,

  1. A.d2sp3d^2sp^3, paramagnetic with two unpaired electrons.
  2. B.d2sp3d^2sp^3, paramagnetic with four unpaired electrons.
  3. C.sp3d2sp^3d^2, paramagnetic with four unpaired electrons.
  4. D.sp3d2sp^3d^2, paramagnetic with two unpaired electrons.

Correct answer: C

Explanation: Mn3+Mn^{3+} d4d^4 with weak-field Cl−Cl^-: high spin, outer orbital sp3d2sp^3d^2, 4 unpaired.

Question 62

Chemistry · 4 marks · −1

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(g)⇌B(g)+C(g)A(g)\rightleftharpoons B(g)+C(g). If p being the total pressure, KpK_p is the pressure equilibrium constant and α\alpha is the degree of dissociation, then which of the following is true at equilibrium ?

  1. A.If KpK_p value is extremely high compared to p, α\alpha becomes much less than unity
  2. B.When p increases α\alpha decreases
  3. C.When p increases α\alpha increases
  4. D.If p value is extremely high compared to KpK_p, α≈1\alpha\approx1

Correct answer: B

Explanation: Kp=α2p1−α2K_p=\frac{\alpha^2p}{1-\alpha^2}: α decreases as p increases.

Question 63

Chemistry · 4 marks · −1

A tetrapeptide, "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is :

  1. A.32
  2. B.24
  3. C.16
  4. D.8

Correct answer: B

Explanation: 4! = 24.

Question 64

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Neopentane forms only one monosubstituted derivative. Statement (II) : Melting point of neopentane is higher than n-pentane. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: B

Question 65

Chemistry · 4 marks · −1

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process ?

Figure for question 65
  1. A.Plot 1: −ΔGRθ/T-\Delta G_R^\theta/T highest (≈ +300 falling), −ΔHRθ/T-\Delta H_R^\theta/T (≈ +100 falling below 0), ΔHRθ\Delta H_R^\theta flat ≈ −100, ΔSRθ\Delta S_R^\theta flat ≈ −200
  2. B.Plot 2: −ΔHRθ-\Delta H_R^\theta ≈ +200 falling, −ΔGRθ/T-\Delta G_R^\theta/T ≈ +100 falling below 0, ΔHRθ/T\Delta H_R^\theta/T ≈ −100, ΔSRθ/T\Delta S_R^\theta/T ≈ −200
  3. C.Plot 3: ΔSRθ\Delta S_R^\theta ≈ +200 falling to 100, ΔHRθ\Delta H_R^\theta ≈ +100, −ΔGRθ/T-\Delta G_R^\theta/T rising from −100 to 0, −ΔHRθ/T-\Delta H_R^\theta/T rising from −300
  4. D.Plot 4: −ΔHRθ/T-\Delta H_R^\theta/T highest (≈ +300 falling), −ΔGRθ/T-\Delta G_R^\theta/T (≈ +100 falling below 0), ΔHRθ\Delta H_R^\theta flat ≈ −100, ΔSRθ\Delta S_R^\theta flat ≈ −200

Correct answer: D

Explanation: For N2+3H2→2NH3N_2+3H_2\to2NH_3: ΔH≈−92\Delta H\approx-92 and ΔS≈−198\Delta S\approx-198 (flat); −ΔH/T-\Delta H/T is large positive and decreasing; −ΔG/T=−ΔH/T+ΔS-\Delta G/T=-\Delta H/T+\Delta S lies about 200 below it and crosses zero.

Question 66

Chemistry · 4 marks · −1

Electronic configuration of four elements A, B, C and D are given below : (A) 1s22s22p31s^22s^22p^3 (B) 1s22s22p41s^22s^22p^4 (C) 1s22s22p51s^22s^22p^5 (D) 1s22s22p21s^22s^22p^2. Which of the following is the correct order of increasing electronegativity (Pauling's scale) ?

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

Correct answer: B

Explanation: C < N < O < F.

Question 67

Chemistry · 4 marks · −1

Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product ? (A) R−C≡N→(i) H+/H2O, mild conditionR-C\equiv N\xrightarrow{\text{(i) } H^+/H_2O,\ \text{mild condition}} (B) R−MgX→(i) CO2; (ii) H3O+R-MgX\xrightarrow{\text{(i) } CO_2;\ \text{(ii) } H_3O^+} (C) R−C≡N→(i) SnCl2/HCl; (ii) H3O+R-C\equiv N\xrightarrow{\text{(i) } SnCl_2/HCl;\ \text{(ii) } H_3O^+} (D) R⋅CH2⋅OH→PCCR\cdot CH_2\cdot OH\xrightarrow{PCC} (E) C6H5COCl→(i) H2/Pd−BaSO4; (ii) Br2 waterC_6H_5COCl\xrightarrow{\text{(i) } H_2/Pd-BaSO_4;\ \text{(ii) } Br_2\ \text{water}}. Choose the correct answer from the options given below :

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

Correct answer: B

Explanation: Mild hydrolysis of nitrile stops at amide; Stephen reduction and PCC give aldehydes; Grignard + CO2 and Rosenmund followed by Br2-water give acids.

Question 68

Chemistry · 4 marks · −1

In 3,3-dimethylhex-1-en-4-yne, there are ______ sp3sp^3, ______ sp2sp^2 and ______ sp hybridised carbon atoms respectively.

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

Correct answer: B

Explanation: CH2=CH−C(CH3)2−C≡C−CH3CH_2=CH-C(CH_3)_2-C\equiv C-CH_3: sp3^3 = C3, two CH3_3, C6 = 4; sp2^2 = 2; sp = 2.

Question 69

Chemistry · 4 marks · −1

The d-orbital electronic configuration of the complex among [Co(en)3]3+[Co(en)_3]^{3+}, [CoF6]3−[CoF_6]^{3-}, [Mn(H2O)6]2+[Mn(H_2O)_6]^{2+} and [Zn(H2O)6]2+[Zn(H_2O)_6]^{2+} that has the highest CFSE is :

  1. A.t2g3eg2t_{2g}^3e_g^2
  2. B.t2g6eg0t_{2g}^6e_g^0
  3. C.t2g6eg4t_{2g}^6e_g^4
  4. D.t2g4eg2t_{2g}^4e_g^2

Correct answer: B

Question 70

Chemistry · 4 marks · −1

In Dumas' method for estimation of nitrogen, 0.5 gram 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 (Aqueous tension at 300 K = 15 mm Hg) is ______ %.

  1. A.1.257
  2. B.18.67
  3. C.20.87
  4. D.12.57

Correct answer: D

Explanation: n=(700/760)×0.0600.0821×300=2.24×10−3n=\frac{(700/760)\times0.060}{0.0821\times300}=2.24\times10^{-3} mol; mass 0.0628 g → 12.57%.

Question 71

Chemistry · 4 marks · −1

0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 mΩ\Omega m. The molar conductivity of the solution will be ______ ×102\times10^2 mS dm2dm^2 mol−1mol^{-1}. (Nearest integer)

Correct answer: 23

Explanation: c = 0.05 M; κ=1/0.87=1.149\kappa=1/0.87=1.149 S/m; Λm=κ/c=1.149/50\Lambda_m=\kappa/c=1.149/50 S m2^2mol−1^{-1} = 2.30 S dm2^2mol−1^{-1} = 2300 mS dm2^2mol−1^{-1} → 23.

Question 72

Chemistry · 4 marks · −1

The spin-only magnetic moment value of Mn+M^{n+} ion formed among Ni, Zn, Mn and Cu that has the least enthalpy of atomisation is ______. (in nearest integer) Here n is equal to the number of diamagnetic complexes among K2[NiCl4]K_2[NiCl_4], [Zn(H2O)6]Cl2[Zn(H_2O)_6]Cl_2, K3[Mn(CN)6]K_3[Mn(CN)_6] and [Cu(PPh3)3I][Cu(PPh_3)_3I]

Correct answer: 0

Explanation: Least atomisation enthalpy: Zn; diamagnetic complexes: Zn(II) and Cu(I) → n = 2; Zn2+Zn^{2+} is d10d^{10}, moment 0.

Question 73

Chemistry · 4 marks · −1

For the reaction A →\to B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g L−1L^{-1} (if the initial concentration of A was 50 g L−1L^{-1}) is ______. (Nearest integer) Given : log 2 = 0.3010

Figure for question 73

Correct answer: 43

Explanation: First order, t1/2=10t_{1/2}=10 s: t=100.301log⁡20=10×1.3010.301≈43t=\frac{10}{0.301}\log20=\frac{10\times1.301}{0.301}\approx43 s.

Question 74

Chemistry · 4 marks · −1

When 1 g each of compounds AB and AB2AB_2 are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is ______ ×10−1\times10^{-1} (Nearest integer) (Given : Molal boiling point elevation constant is 0.5 K kg mol−1mol^{-1})

Correct answer: 25

Explanation: MAB=0.5×10002.7×15=12.35M_{AB}=\frac{0.5\times1000}{2.7\times15}=12.35, MAB2=0.5×10001.5×15=22.22M_{AB_2}=\frac{0.5\times1000}{1.5\times15}=22.22; B = 9.87, A = 2.47 → 25.

Question 75

Chemistry · 4 marks · −1

2-Bromopentane →alcoholic KOH\xrightarrow{\text{alcoholic KOH}} P (Major Product) →Br2\xrightarrow{Br_2} Q. Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is 80% whereas Q is 100%. Mass of product Q obtained is ______ g. (Given molar mass in g mol−1mol^{-1} H : 1, C : 12, O : 16, Br : 80)

Correct answer: 184

Explanation: 1 mol substrate → 0.8 mol pent-2-ene → 0.8 mol 2,3-dibromopentane (230 g/mol) = 184 g.

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