JEE Main 2025 — 3 Apr Shift 1 (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 A be a matrix of order 3 x 3 and ∣A∣=5|A|=5. If ∣2 adj(3A adj(2A))∣=2α⋅3β⋅5γ|2\,\mathrm{adj}(3A\,\mathrm{adj}(2A))|=2^\alpha\cdot3^\beta\cdot5^\gamma, α,β,γ∈N\alpha, \beta, \gamma\in\mathbf{N}, then α+β+γ\alpha+\beta+\gamma is equal to

  1. A.26
  2. B.28
  3. C.27
  4. D.25

Correct answer: C

Explanation: ∣2A∣=40|2A|=40, adj(2A)\mathrm{adj}(2A) has det 40240^2; ∣3A adj(2A)∣=27⋅5⋅1600=263353|3A\,\mathrm{adj}(2A)|=27\cdot5\cdot1600=2^6 3^3 5^3; adj squares it: 21236562^{12}3^65^6; times 232^3: 21536562^{15}3^65^6 → 27.

Question 2

Mathematics · 4 marks · −1

The sum of all rational terms in the expansion of (2+3)8(2+\sqrt{3})^8 is

  1. A.18817
  2. B.33845
  3. C.16923
  4. D.3763

Correct answer: A

Explanation: (2+3)8+(2−3)82=18817\frac{(2+\sqrt3)^8+(2-\sqrt3)^8}{2}=18817.

Question 3

Mathematics · 4 marks · −1

The radius of the smallest circle which touches the parabolas y=x2+2y=x^2+2 and x=y2+2x=y^2+2 is

  1. A.728\frac{7\sqrt{2}}{8}
  2. B.724\frac{7\sqrt{2}}{4}
  3. C.7216\frac{7\sqrt{2}}{16}
  4. D.722\frac{7\sqrt{2}}{2}

Correct answer: A

Explanation: The parabolas are mirror images in y=xy=x; nearest points have slope 1: x=12x=\frac12, point (12,94)(\frac12,\frac94), distance to y=xy=x is 742=728\frac{7}{4\sqrt2}=\frac{7\sqrt2}{8}.

Question 4

Mathematics · 4 marks · −1

If y(x)=∣sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111∣y(x)=\begin{vmatrix}\sin x & \cos x & \sin x+\cos x+1\\ 27 & 28 & 27\\ 1 & 1 & 1\end{vmatrix}, x∈Rx\in\mathbf{R}, then d2ydx2+y\frac{d^2y}{dx^2}+y is equal to

  1. A.-1
  2. B.1
  3. C.27
  4. D.28

Correct answer: A

Explanation: y=sin⁡x(1)−cos⁡x(0)+(sin⁡x+cos⁡x+1)(−1)=−cos⁡x−1y=\sin x(1)-\cos x(0)+(\sin x+\cos x+1)(-1)=-\cos x-1; y\'\'+y=-1.

Question 5

Mathematics · 4 marks · −1

Let α\alpha and β\beta be the roots of x2+3x−16=0x^2+\sqrt{3}x-16=0, and γ\gamma and δ\delta be the roots of x2+3x−1=0x^2+3x-1=0. If Pn=αn+βnP_n=\alpha^n+\beta^n and Qn=γn+δnQ_n=\gamma^n+\delta^n, then P25+3P242P23+Q25−Q23Q24\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}} is equal to

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

Correct answer: C

Explanation: P25+3P24=16P23P_{25}+\sqrt3P_{24}=16P_{23} → 8; Q25=−3Q24+Q23Q_{25}=-3Q_{24}+Q_{23} → −3; total 5.

Question 6

Mathematics · 4 marks · −1

Line L1L_1 passes through the point (1, 2, 3) and is parallel to z-axis. Line L2L_2 passes through the point (λ\lambda, 5, 6) and is parallel to y-axis. Let for λ=λ1,λ2\lambda=\lambda_1, \lambda_2, λ2<λ1\lambda_2<\lambda_1, the shortest distance between the two lines be 3. Then the square of the distance of the point (λ1\lambda_1, λ2\lambda_2, 7) from the line L1L_1 is

  1. A.32
  2. B.37
  3. C.25
  4. D.40

Correct answer: C

Explanation: SD =∣λ−1∣=3=|\lambda-1|=3: λ1=4,λ2=−2\lambda_1=4,\lambda_2=-2; distance² of (4,−2,7) from L1L_1: 32+42=253^2+4^2=25.

Question 7

Mathematics · 4 marks · −1

Let z∈Cz\in\mathbf{C} be such that z2+3iz−2+i=2+3i\frac{z^2+3i}{z-2+i}=2+3i. Then the sum of all possible values of z2z^2 is

  1. A.-19 + 2i
  2. B.19 + 2i
  3. C.19 − 2i
  4. D.-19 − 2i

Correct answer: D

Explanation: z2−(2+3i)z+(7+7i)=0z^2-(2+3i)z+(7+7i)=0: s=2+3is=2+3i, p=7+7ip=7+7i; s2−2p=−19−2is^2-2p=-19-2i.

Question 8

Mathematics · 4 marks · −1

Let a line passing through the point (4, 1, 0) intersect the line L1:x−12=y−23=z−34L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} at the point A(α,β,γ\alpha, \beta, \gamma) and the line L2:x−6=y=−z+4L_2: x-6=y=-z+4 at the point B(a, b, c). Then ∣101αβγabc∣\begin{vmatrix}1 & 0 & 1\\ \alpha & \beta & \gamma\\ a & b & c\end{vmatrix} is equal to

  1. A.16
  2. B.8
  3. C.6
  4. D.12

Correct answer: B

Question 9

Mathematics · 4 marks · −1

The sum 1 + 3 + 11 + 25 + 45 + 71 + ... upto 20 terms, is equal to

  1. A.8124
  2. B.7240
  3. C.6982
  4. D.7130

Correct answer: B

Explanation: Successive differences 2, 8, 14, … form an AP (d = 6), so Tn=3n2−7n+5T_n=3n^2-7n+5 and ∑n=120Tn=3(2870)−7(210)+100=7240\sum_{n=1}^{20}T_n=3(2870)-7(210)+100=7240.

Question 10

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a G.P. of increasing positive numbers. If a3a5=729a_3a_5=729 and a2+a4=1114a_2+a_4=\frac{111}{4}, then 24(a1+a2+a3)24(a_1+a_2+a_3) is equal to

  1. A.128
  2. B.131
  3. C.130
  4. D.129

Correct answer: D

Explanation: a4=27a_4=27, a2=34a_2=\frac34, r=6r=6, a1=18a_1=\frac18: 24(18+34+92)=12924(\frac18+\frac34+\frac92)=129.

Question 11

Mathematics · 4 marks · −1

Let A = {−3, −2, −1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if 0≤x2+2y≤40\le x^2+2y\le4. Let ll be the number of elements in R and m be the minimum number of elements required to be added in R to make it a reflexive relation. Then l+ml+m is equal to

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

Correct answer: B

Question 12

Mathematics · 4 marks · −1

Let the domain of the function f(x)=log⁡2log⁡4log⁡6(3+4x−x2)f(x)=\log_2\log_4\log_6(3+4x-x^2) be (a, b). If ∫0b−a[x2]dx=p−q−r\int_0^{b-a}[x^2]dx=p-\sqrt{q}-\sqrt{r}, p,q,r∈Np, q, r\in\mathbf{N}, gcd⁡(p,q,r)=1\gcd(p, q, r)=1, where [⋅][\cdot] is the greatest integer function, then p + q + r is equal to

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

Correct answer: B

Question 13

Mathematics · 4 marks · −1

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines L1:2x+y+6=0L_1: 2x+y+6=0 and L2:4x+2y−p=0L_2: 4x+2y-p=0, p > 0, at the points A and B, respectively. If AB=92AB=\frac{9}{\sqrt{2}} and the foot of the perpendicular from the point A on the line L2L_2 is M, then AMBM\frac{AM}{BM} is equal to

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

Correct answer: A

Question 14

Mathematics · 4 marks · −1

A line passing through the point P(5\sqrt{5}, 5\sqrt{5}) intersects the ellipse x236+y225=1\frac{x^2}{36}+\frac{y^2}{25}=1 at A and B such that (PA)·(PB) is maximum. Then 5(PA2+PB2)5(PA^2+PB^2) is equal to :

  1. A.218
  2. B.377
  3. C.290
  4. D.338

Correct answer: D

Question 15

Mathematics · 4 marks · −1

If the domain of the function f(x)=log⁡e(2x−35+4x)+sin⁡−1(4+3x2−x)f(x)=\log_e\left(\frac{2x-3}{5+4x}\right)+\sin^{-1}\left(\frac{4+3x}{2-x}\right) is [α,β)[\alpha, \beta), then α2+4β\alpha^2+4\beta is equal to

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

Correct answer: B

Explanation: sin⁡−1\sin^{-1} needs x∈[−3,−12]x\in[-3,-\frac12]; log needs x<−54x<-\frac54 or x>32x>\frac32: domain [−3,−54)[-3,-\frac54); 9−5=49-5=4.

Question 16

Mathematics · 4 marks · −1

Let g be a differentiable function such that ∫0xg(t)dt=x−∫0xtg(t)dt\int_0^xg(t)dt=x-\int_0^xtg(t)dt, x≥0x\ge0 and let y=y(x)y=y(x) satisfy the differential equation dydx−ytan⁡x=2(x+1)sec⁡x g(x)\frac{dy}{dx}-y\tan x=2(x+1)\sec x\,g(x), x∈[0,π2)x\in\left[0, \frac{\pi}{2}\right). If y(0)=0y(0)=0, then y(π3)y\left(\frac{\pi}{3}\right) is equal to

  1. A.4π3\frac{4\pi}{3}
  2. B.2π33\frac{2\pi}{3\sqrt{3}}
  3. C.2π3\frac{2\pi}{3}
  4. D.4π33\frac{4\pi}{3\sqrt{3}}

Correct answer: A

Explanation: g=11+xg=\frac{1}{1+x}; ddx(ycos⁡x)=2\frac{d}{dx}(y\cos x)=2 so y=2xsec⁡xy=2x\sec x; y(π/3)=4π3y(\pi/3)=\frac{4\pi}{3}.

Question 17

Mathematics · 4 marks · −1

The number of solutions of the equation 2x+3tan⁡x=π2x+3\tan x=\pi, x∈[−2π,2π]−{±π2,±3π2}x\in[-2\pi, 2\pi]-\left\{\pm\frac{\pi}{2}, \pm\frac{3\pi}{2}\right\} is:

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

Correct answer: C

Question 18

Mathematics · 4 marks · −1

Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/2−2(x+c)1/3−2,x>0f(x)=\begin{cases}(1+ax)^{1/x}, & x<0\\ 1+b, & x=0\\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2}, & x>0\end{cases} be continuous at x = 0. Then eabce^abc is equal to:

  1. A.36
  2. B.48
  3. C.64
  4. D.72

Correct answer: B

Explanation: c = 8 (so denominator → 0), right limit 1/41/12=3\frac{1/4}{1/12}=3; ea=3e^a=3, b = 2; eabc=3⋅2⋅8=48e^abc=3\cdot2\cdot8=48.

Question 19

Mathematics · 4 marks · −1

Let f(x)=∫x33−x2 dxf(x)=\int x^3\sqrt{3-x^2}\,dx. If 5f(2)=−45f(\sqrt{2})=-4, then f(1)f(1) is equal to

  1. A.−625-\frac{6\sqrt{2}}{5}
  2. B.−825-\frac{8\sqrt{2}}{5}
  3. C.−425-\frac{4\sqrt{2}}{5}
  4. D.−225-\frac{2\sqrt{2}}{5}

Correct answer: A

Question 20

Mathematics · 4 marks · −1

If ∑r=19(r+32r)⋅9Cr=α(32)9−β\sum_{r=1}^{9}\left(\frac{r+3}{2^r}\right)\cdot{}^9C_r=\alpha\left(\frac{3}{2}\right)^9-\beta, α,β∈N\alpha, \beta\in\mathbf{N}, then (α+β)2(\alpha+\beta)^2 is equal to

  1. A.18
  2. B.27
  3. C.81
  4. D.9

Correct answer: C

Explanation: ∑r 9Cr2−r=92(32)8=3(32)9\sum r\,{}^9C_r2^{-r}=\frac92(\frac32)^8=3(\frac32)^9, 3∑9Cr2−r=3((32)9−1)3\sum{}^9C_r2^{-r}=3((\frac32)^9-1): total 6(32)9−36(\frac32)^9-3; (6+3)2=81(6+3)^2=81.

Question 21

Mathematics · 4 marks · −1

The area of the region bounded by the curve y=max⁡{∣x∣,x∣x−2∣}y=\max\{|x|, x|x-2|\}, the x-axis and the lines x=−2x=-2 and x=4x=4 is equal to ______.

Correct answer: 12

Question 22

Mathematics · 4 marks · −1

Let the product of the focal distances of the point P(4, 232\sqrt{3}) on the hyperbola H: x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2+q2p^2+q^2 is equal to ______.

Correct answer: 120

Question 23

Mathematics · 4 marks · −1

Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}, b⃗=3i^+2j^−k^\vec{b}=3\hat{i}+2\hat{j}-\hat{k}, c⃗=λj^+μk^\vec{c}=\lambda\hat{j}+\mu\hat{k} and d^\hat{d} be a unit vector such that a⃗×d^=b⃗×d^\vec{a}\times\hat{d}=\vec{b}\times\hat{d} and c⃗⋅d^=1\vec{c}\cdot\hat{d}=1. If c⃗\vec{c} is perpendicular to a⃗\vec{a}, then ∣3λd^+μc⃗∣2|3\lambda\hat{d}+\mu\vec{c}|^2 is equal to ______.

Correct answer: 5

Question 24

Mathematics · 4 marks · −1

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by WnW_n. Let the probability P(Wn)P(W_n) of choosing the word WnW_n satisfy P(Wn)=2P(Wn−1)P(W_n)=2P(W_{n-1}), n > 1. If P(CDBEA)=2α2β−1P(CDBEA)=\frac{2^\alpha}{2^\beta-1}, α,β∈N\alpha, \beta\in\mathbf{N}, then α+β\alpha+\beta is equal to : ______

Correct answer: 183

Explanation: Rank of CDBEA = 24+24+6+6+2+1+1 = 64; P(Wn)=2n−12120−1P(W_n)=\frac{2^{n-1}}{2^{120}-1}, so α=63\alpha=63, β=120\beta=120: 183.

Question 25

Mathematics · 4 marks · −1

If the number of seven-digit numbers, such that the sum of their digits is even, is m⋅n⋅10nm\cdot n\cdot10^n; m,n∈{1,2,3,…,9}m, n\in\{1, 2, 3, \ldots, 9\}, then m + n is equal to ______.

Correct answer: 14

Explanation: Exactly half of the 9×1069\times10^6 numbers: 45×105=9⋅5⋅10545\times10^5=9\cdot5\cdot10^5, so n = 5, m = 9: 14.

Question 26

Physics · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. 01n+92235U→54140Xe+3894Sr+201n{}^1_0n+{}^{235}_{92}U\to{}^{140}_{54}Xe+{}^{94}_{38}Sr+2{}^1_0n B. 2H2+O2→2H2O2H_2+O_2\to2H_2O C. 12H+12H→23He+01n{}^2_1H+{}^2_1H\to{}^3_2He+{}^1_0n D. 11H+13H→12H+12H{}^1_1H+{}^3_1H\to{}^2_1H+{}^2_1H. LIST-II: I. Chemical reaction II. Fusion with +ve Q value III. Fission IV. Fusion with −ve Q value. Choose the correct answer from the options given below:

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

Correct answer: C

Question 27

Physics · 4 marks · −1

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

  1. A.Yellow
  2. B.Green
  3. C.Red
  4. D.Blue

Correct answer: D

Explanation: λ<1240/3≈413\lambda<1240/3\approx413 nm: only blue/violet end qualifies.

Question 28

Physics · 4 marks · −1

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

Figure for question 28
  1. A.3.5 m/s2m/s^2
  2. B.0.35 m/s2m/s^2
  3. C.2.5 m/s2m/s^2
  4. D.0.25 m/s2m/s^2

Correct answer: A

Explanation: a=2Fm(1+2/5)=9828=3.5 m/s2a=\frac{2F}{m(1+2/5)}=\frac{98}{28}=3.5\ m/s^2.

Question 29

Physics · 4 marks · −1

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2m away from it, is

  1. A.3×10−83\times10^{-8} Pascals
  2. B.0
  3. C.6×10−86\times10^{-8} Pascals
  4. D.1.5×10−81.5\times10^{-8} Pascals

Correct answer: C

Explanation: I=4504π⋅4≈9I=\frac{450}{4\pi\cdot4}\approx9 W/m2^2; P=2I/c=6×10−8P=2I/c=6\times10^{-8} Pa.

Question 30

Physics · 4 marks · −1

A person measures mass of 3 different particles as 435.42 g, 226.3 g and 0.125 g. According to the rules for arithmetic operations with significant figures, the addition of the masses of 3 particles will be.

  1. A.661.845 g
  2. B.662 g
  3. C.661.84 g
  4. D.661.8 g

Correct answer: D

Question 31

Physics · 4 marks · −1

The angle of projection of a particle is measured from the vertical axis as ϕ\phi and the maximum height reached by the particle is hmh_m. Here hmh_m as function of ϕ\phi can be presented as

Figure for question 31
  1. A.Graph: hmh_m increases (concave up) from 0 at ϕ=0\phi=0 to maximum at ϕ=90∘\phi=90^\circ
  2. B.Graph: hmh_m decreases (concave up) from a maximum at ϕ=0\phi=0 towards 0 at ϕ=90∘\phi=90^\circ
  3. C.Graph: hmh_m rises from 0 to a maximum and falls back to 0 at ϕ=90∘\phi=90^\circ (arch)
  4. D.Graph: hmh_m decreases from a maximum at ϕ=0\phi=0 to 0 at ϕ=90∘\phi=90^\circ, concave down (cosine-squared shape)

Correct answer: D

Explanation: hm=u2cos⁡2ϕ2gh_m=\frac{u^2\cos^2\phi}{2g}: maximum at 0, zero at 90°, flat-topped (concave down initially).

Question 32

Physics · 4 marks · −1

A piston of mass M is hung from a massless spring whose restoring force law goes as F=−kx3F=-kx^3, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L0L_0 to L1L_1, the total energy delivered by the filament is :(Assume spring to be in its natural length before heating)

Figure for question 32
  1. A.nRTln⁡(L1L0)+Mg(L1−L0)+k4(L14−L04)nRT\ln\left(\frac{L_1}{L_0}\right)+Mg(L_1-L_0)+\frac{k}{4}(L_1^4-L_0^4)
  2. B.nRTln⁡(L12L02)+Mg2(L1−L0)+k4(L14−L04)nRT\ln\left(\frac{L_1^2}{L_0^2}\right)+\frac{Mg}{2}(L_1-L_0)+\frac{k}{4}(L_1^4-L_0^4)
  3. C.nRTln⁡(L1L0)+Mg(L1−L0)+3k4(L14−L04)nRT\ln\left(\frac{L_1}{L_0}\right)+Mg(L_1-L_0)+\frac{3k}{4}(L_1^4-L_0^4)
  4. D.3nRTln⁡(L1L0)+2Mg(L1−L0)+k3(L13−L03)3nRT\ln\left(\frac{L_1}{L_0}\right)+2Mg(L_1-L_0)+\frac{k}{3}(L_1^3-L_0^3)

Correct answer: A

Question 33

Physics · 4 marks · −1

During the melting of a slab of ice at 273 K at atmospheric pressure:

  1. A.Internal energy of the ice-water system decreases.
  2. B.Positive work is done on the ice-water system by the atmosphere.
  3. C.Positive work is done by the ice-water system on the atmosphere.
  4. D.Internal energy of ice-water system remains unchanged.

Correct answer: B

Explanation: Ice contracts on melting, so the atmosphere does positive work on the system.

Question 34

Physics · 4 marks · −1

Choose the correct logic circuit for the given truth table having inputs A and B. Truth table: A=0, B=0, Y=0; A=0, B=1, Y=0; A=1, B=0, Y=1; A=1, B=1, Y=1.

Figure for question 34
  1. A.Circuit 1: A and B feed an OR gate; the OR output and A (direct wire) feed an AND gate giving Y
  2. B.Circuit 2: A and B feed an OR gate; the OR output and NOT(A) feed an AND gate giving Y
  3. C.Circuit 3: A and B feed an OR gate; NOT(OR output) and A feed an AND gate giving Y
  4. D.Circuit 4: A and B feed an OR gate; the OR output and A feed an AND gate giving Y (A taken from above)

Correct answer: D

Explanation: Y = A: A⋅(A+B)=AA\cdot(A+B)=A, realised by feeding A and (A + B) into AND with no inverter.

Question 35

Physics · 4 marks · −1

Which of the following curves possibly represent one-dimensional motion of a particle? A. Phase vs time: a straight line with positive slope crossing the time axis. B. Velocity vs displacement: a circle centred at the origin. C. Velocity vs time: a circle centred at the origin. D. Total distance vs time: a non-decreasing curve starting at the origin. Choose the correct answer from the options given below:

Figure for question 35
  1. A.A and B only
  2. B.A, B and C only
  3. C.A, C and D only
  4. D.A, B and D only

Correct answer: D

Explanation: A (SHM phase), B (SHM v–x ellipse/circle) and D are possible; C would need two velocities at one instant.

Question 36

Physics · 4 marks · −1

Consider a completely full cylindrical water tank of height 1.6 m and of cross-sectional area 0.5 m2m^2. It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is: (g = 10 m/s2m/s^2)

  1. A.2 m/s
  2. B.3 m/s
  3. C.5 m/s
  4. D.4 m/s

Correct answer: D

Explanation: Extra pressure 1000 Pa ≡ 0.1 m of water; head 0.7 + 0.1 = 0.8 m; v=16=4v=\sqrt{16}=4 m/s.

Question 37

Physics · 4 marks · −1

Two blocks of masses m and M, (M > m), are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released, then (μ\mu = coefficient of friction between the two blocks) A. The time period of small oscillation of the two blocks is T=2π(m+M)kT=2\pi\sqrt{\frac{(m+M)}{k}} B. The acceleration of the blocks is a=−kxM+ma=-\frac{kx}{M+m} (x = displacement of the blocks from the mean position) C. The magnitude of the frictional force on the upper block is mμ∣x∣M+m\frac{m\mu|x|}{M+m} D. The maximum amplitude of the upper block, if it does not slip, is μ(M+m)gk\frac{\mu(M+m)g}{k} E. Maximum frictional force can be μ(M+m)g\mu(M+m)g. Choose the correct answer from the options given below:

Figure for question 37
  1. A.A, B, D Only
  2. B.C, D, E Only
  3. C.A, B, C Only
  4. D.B, C, D Only

Correct answer: A

Question 38

Physics · 4 marks · −1

The electrostatic potential on the surface of uniformly charged spherical shell of radius R = 10 cm is 120 V. The potential at the centre of shell, at a distance r = 5 cm from centre, and at a distance r = 15 cm from the centre of the shell respectively, are:

  1. A.0V, 0V, 80V
  2. B.40V, 40V, 80V
  3. C.0V, 120V, 40V
  4. D.120V, 120V, 80V

Correct answer: D

Question 39

Physics · 4 marks · −1

A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

  1. A.S4,3gS2\frac{S}{4}, \frac{3gS}{2}
  2. B.S2,3gS2\frac{S}{2}, \frac{3gS}{2}
  3. C.S2,3gS2\frac{S}{2}, \sqrt{\frac{3gS}{2}}
  4. D.S4,3gS2\frac{S}{4}, \sqrt{\frac{3gS}{2}}

Correct answer: D

Explanation: mgh=14mgS⇒h=S/4mgh=\frac14mgS\Rightarrow h=S/4; v2=2g⋅3S4v^2=2g\cdot\frac{3S}{4}.

Question 40

Physics · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. Gravitational constant B. Gravitational potential energy C. Gravitational potential D. Acceleration due to gravity. LIST-II: I. [LT−2][LT^{-2}] II. [L2T−2][L^2T^{-2}] III. [ML2T−2][ML^2T^{-2}] IV. [M−1L3T−2][M^{-1}L^3T^{-2}]. Choose the correct answer from the options given below:

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

Correct answer: D

Question 41

Physics · 4 marks · −1

Consider following statements for refraction of light through prism, when angle of deviation is minimum. A. The refracted ray inside prism becomes parallel to the base. B. Larger angle prisms provide smaller angle of minimum deviation. C. Angle of incidence and angle of emergence becomes equal. D. There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting. E. Angle of refraction becomes double of prism angle. Choose the correct answer from the options given below:

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

Correct answer: B

Question 42

Physics · 4 marks · −1

The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is

  1. A.1.2
  2. B.1.8
  3. C.1.4
  4. D.1.5

Correct answer: D

Explanation: 112=(μ−1)(110+115)⇒μ=1.5\frac{1}{12}=(\mu-1)\left(\frac1{10}+\frac1{15}\right)\Rightarrow\mu=1.5.

Question 43

Physics · 4 marks · −1

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm3cm^3 and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 cm3cm^3 is: (Take γ\gamma = 1.5 ; γ\gamma is the ratio of specific heats at constant pressure and at constant volume)

  1. A.327 K
  2. B.600 K
  3. C.300 K
  4. D.522 K

Correct answer: C

Explanation: T2=300×40.5=600T_2=300\times4^{0.5}=600 K; change 300 K.

Question 44

Physics · 4 marks · −1

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constants ε1\varepsilon_1 and ε2\varepsilon_2, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are C1C_1 and C2C_2 respectively, then C1C2\frac{C_1}{C_2} is :

Figure for question 44
  1. A.ε0(ε1+ε2)2\frac{\varepsilon_0(\varepsilon_1+\varepsilon_2)}{2}
  2. B.ε1ε2ε1+ε2\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}
  3. C.ε1ε22(ε1+ε2)2\frac{\varepsilon_1\varepsilon_2^2}{(\varepsilon_1+\varepsilon_2)^2}
  4. D.4ε1ε2(ε1+ε2)2\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}

Correct answer: D

Explanation: C1=2ϵ0Adε1ε2ε1+ε2C_1=\frac{2\epsilon_0A}{d}\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}, C2=ϵ0A2d(ε1+ε2)C_2=\frac{\epsilon_0A}{2d}(\varepsilon_1+\varepsilon_2); ratio 4ε1ε2(ε1+ε2)2\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}.

Question 45

Physics · 4 marks · −1

A loop ABCDA, carrying current I = 12 A, is placed in a plane, consists of two semi-circular segments of radius R1=6πR_1=6\pi m and R2=4πR_2=4\pi m. The magnitude of the resultant magnetic field at center O is k×10−7k\times10^{-7} T. The value of k is ______. (Given μ0=4π×10−7\mu_0=4\pi\times10^{-7} TmA−1A^{-1})

Figure for question 45

Correct answer: 1

Explanation: B=μ0I4(1R2−1R1)=π×10−7×12(14π−16π)=1×10−7B=\frac{\mu_0I}{4}\left(\frac{1}{R_2}-\frac{1}{R_1}\right)=\pi\times10^{-7}\times12\left(\frac{1}{4\pi}-\frac{1}{6\pi}\right)=1\times10^{-7} T.

Question 46

Physics · 4 marks · −1

A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to a uniform magnetic field of strength 0.15 T. The magnetic force on the wire is ______ mN.

Correct answer: 48

Explanation: F=8×0.04×0.15=0.048F=8\times0.04\times0.15=0.048 N = 48 mN.

Question 47

Physics · 4 marks · −1

Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ______ seconds.

Correct answer: 8

Explanation: T∝a3/GmT\propto\sqrt{a^3/Gm}: factor 8/2=2\sqrt{8/2}=2 → 8 s.

Question 48

Physics · 4 marks · −1

In the figure shown below, a resistance of 150.4 Ω\Omega is connected in series to an ammeter A of resistance 240 Ω\Omega. A shunt resistance of 10 Ω\Omega is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

Figure for question 48

Correct answer: 5

Explanation: R=150.4+9.6=160 ΩR=150.4+9.6=160\ \Omega, I = 0.125 A; ammeter share 0.125×10250=50.125\times\frac{10}{250}=5 mA.

Question 49

Physics · 4 marks · −1

Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of x is ______.

Correct answer: 24

Explanation: (2+3)2−(3−2)2=24(2+3)^2-(3-2)^2=24.

Question 50

Chemistry · 4 marks · −1

Which of the following properties will change when system containing solution 1 will become solution 2? Solution 1: 10 mol of solute x + 10 L of water. Solution 2: 1 L of solution 1 + 1 mol of solute x + 1 L of water.

Figure for question 50
  1. A.Molar heat capacity
  2. B.Concentration
  3. C.Gibbs free energy
  4. D.Density

Correct answer: C

Explanation: Both solutions have the same composition (1 mol per L of water), so intensive properties stay the same; the extensive Gibbs energy changes with the amount.

Question 51

Chemistry · 4 marks · −1

Correct order of limiting molar conductivity for cations in water at 298 K is :

  1. A.H+>Na+>Ca2+>Mg2+>K+H^+>Na^+>Ca^{2+}>Mg^{2+}>K^+
  2. B.H+>Ca2+>Mg2+>K+>Na+H^+>Ca^{2+}>Mg^{2+}>K^+>Na^+
  3. C.Mg2+>H+>Ca2+>K+>Na+Mg^{2+}>H^+>Ca^{2+}>K^+>Na^+
  4. D.H+>Na+>K+>Ca2+>Mg2+H^+>Na^+>K^+>Ca^{2+}>Mg^{2+}

Correct answer: B

Explanation: λ∘\lambda^\circ (S cm2^2 mol−1^{-1}): H+H^+ 349.6, Ca2+Ca^{2+} 119, Mg2+Mg^{2+} 106, K+K^+ 73.5, Na+Na^+ 50.1.

Question 52

Chemistry · 4 marks · −1

The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are : A. Cr2+Cr^{2+} B. Fe2+Fe^{2+} C. Fe3+Fe^{3+} D. Co2+Co^{2+} E. Mn3+Mn^{3+}. Choose the correct answer from the options given below:

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

Correct answer: C

Explanation: 4 unpaired electrons: Cr2+Cr^{2+} (d4d^4), Fe2+Fe^{2+} (d6d^6), Mn3+Mn^{3+} (d4d^4).

Question 53

Chemistry · 4 marks · −1

In a reaction A + B →\to C, initial concentrations of A and B are related as [A]0=8[B]0[A]_0=8[B]_0. The half lives of A and B are 10 min and 40 min, respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same ?

  1. A.60 min
  2. B.40 min
  3. C.20 min
  4. D.80 min

Correct answer: B

Explanation: 8(1/2)t/10=(1/2)t/40⇒t10−t40=3⇒t=408(1/2)^{t/10}=(1/2)^{t/40}\Rightarrow\frac{t}{10}-\frac{t}{40}=3\Rightarrow t=40 min.

Question 54

Chemistry · 4 marks · −1

Given below are two statements: Statement I : The N - N single bond is weaker and longer than that of P - P single bond. Statement II : Compounds of group 15 elements in + 3 oxidation states readily undergo disproportionation reactions. 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.Both Statement I and Statement II are true
  3. C.Statement I is true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: A

Question 55

Chemistry · 4 marks · −1

Given below are two statements: Statement I : A catalyst cannot alter the equilibrium constant (KcK_c) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant. 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.Both Statement I and Statement II are true
  4. D.Statement I is true but Statement II is false

Correct answer: D

Question 56

Chemistry · 4 marks · −1

Among 10−910^{-9} g (each) of the following elements, which one will have the highest number of atoms? Element: Pb, Po, Pr and Pt

  1. A.Pb
  2. B.Pr
  3. C.Po
  4. D.Pt

Correct answer: B

Explanation: Lowest atomic mass gives most atoms: Pr (140.9).

Question 57

Chemistry · 4 marks · −1

Number of molecules from below which cannot give iodoform reaction is: Ethanol, Isopropyl alcohol, Bromoacetone, 2−Butanol, 2−Butanone, Butanal, 2−Pentanone, 3−Pentanone, Pentanal and 3−Pentanol.

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

Correct answer: C

Explanation: Cannot: butanal, 3-pentanone, pentanal, 3-pentanol = 4.

Question 58

Chemistry · 4 marks · −1

In the following system, PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g) at equilibrium, upon addition of xenon gas at constant T & p, the concentration of

  1. A.PCl5PCl_5, PCl3PCl_3 & Cl2Cl_2 remain constant
  2. B.PCl5PCl_5 will increase
  3. C.Cl2Cl_2 will decrease
  4. D.PCl3PCl_3 will increase

Correct answer: D

Explanation: Adding inert gas at constant p increases the volume, so the equilibrium shifts forward (more moles of PCl3 and Cl2).

Question 59

Chemistry · 4 marks · −1

Which of the following statements are correct? A. The process of adding an electron to a neutral gaseous atom is always exothermic. B. The process of removing an electron from an isolated gaseous atom is always endothermic. C. The 1st1^{st} ionization energy of boron is less than that of beryllium. D. The electronegativity of C is 2.5 in CH4CH_4 and CCl4CCl_4 E. Li is the most electropositive among elements of group I. Choose the correct answer from the options given below:

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

Correct answer: A

Question 60

Chemistry · 4 marks · −1

The correct order of the complexes [Co(NH3)5(H2O)]3+[Co(NH_3)_5(H_2O)]^{3+} (A), [Co(NH3)6]3+[Co(NH_3)_6]^{3+} (B), [Co(CN)6]3−[Co(CN)_6]^{3-} (C) and [CoCl(NH3)5]2+[CoCl(NH_3)_5]^{2+} (D) in terms of wavelength of light absorbed is

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

Correct answer: B

Explanation: Larger Δo\Delta_o → shorter λ: Δ\Delta: C > B > A > D, so λ: D > A > B > C.

Question 61

Chemistry · 4 marks · −1

The least acidic compound, among the following is: (A) 3-ethoxyphenol (B) benzenesulfonic acid (C) butanoic acid (CH3CH2CH2COOHCH_3CH_2CH_2COOH) (D) ethyl propiolate (EtO2C−C≡C−HEtO_2C-C\equiv C-H)

Figure for question 61
  1. A.C
  2. B.A
  3. C.B
  4. D.D

Correct answer: D

Explanation: The terminal alkyne C–H (pKa ≈ 20+) is far less acidic than a phenol, carboxylic or sulfonic acid.

Question 62

Chemistry · 4 marks · −1

Which of the following is the correct structure of L−Fructose? (Fischer projections, top to bottom: C1 CH2OHCH_2OH, C2 C=O, then C3, C4, C5, then C6 CH2OHCH_2OH; the side on which each OH is drawn is given.)

Figure for question 62
  1. A.C3 OH left, C4 OH left, C5 OH left
  2. B.C3 OH right, C4 OH right, C5 OH left
  3. C.C3 OH left, C4 OH right, C5 OH left
  4. D.C3 OH right, C4 OH left, C5 OH left

Correct answer: D

Explanation: D-fructose is C3 left, C4 right, C5 right; its mirror image L-fructose is C3 right, C4 left, C5 left.

Question 63

Chemistry · 4 marks · −1

Which of the following postulate of Bohr's model of hydrogen atom is not in agreement with quantum mechanical model of an atom?

  1. A.An atom in a stationary state does not emit electromagnetic radiation as long as it stays in the same state.
  2. B.An atom can take only certain distinct energies E1E_1, E2E_2, E3E_3, etc. These allowed states of constant energy are called the stationary states of atom.
  3. C.The electron in a H atom's stationary state moves in a circle around the nucleus.
  4. D.When an electron makes a transition from a higher energy stationary state to a lower energy stationary state, then it emits a photon of light.

Correct answer: C

Question 64

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Molecules/ion): A. PF5PF_5 B. SF6SF_6 C. Ni(CO)4Ni(CO)_4 D. [PtCl4]2−[PtCl_4]^{2-}. LIST-II (Hybridisation of central atom): I. dsp2dsp^2 II. sp3dsp^3d III. sp3d2sp^3d^2 IV. sp3sp^3. Choose the correct answer from the options given below:

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

Correct answer: B

Question 65

Chemistry · 4 marks · −1

Identify the correct statements from the following. A. Hexan-3-one and hexan-2-one (drawn as skeletal ketones) are metamers. B. Propyl cyanide (CH3CH2CH2CNCH_3CH_2CH_2CN) and propyl isocyanide (CH3CH2CH2NCCH_3CH_2CH_2NC) are functional isomers. C. Pentan-1-ol and 3-methylbutan-1-ol are position isomers. D. Pentan-1-amine and N-ethylpropan-1-amine (skeletal: CH3CH2CH2−NH−CH2CH3CH_3CH_2CH_2-NH-CH_2CH_3) are homologous. Choose the correct answer from the options given below:

Figure for question 65
  1. A.A & B Only
  2. B.B & C Only
  3. C.C & D Only
  4. D.A,B & C Only

Correct answer: A

Question 66

Chemistry · 4 marks · −1

Which compound would give 3-methyl-6-oxoheptanal upon ozonolysis?

Figure for question 66
  1. A.1,3-Dimethylcyclohex-1-ene (second methyl on the ring carbon next to the =CH)
  2. B.1,4-Dimethylcyclohex-1-ene (second methyl two carbons away from the =CH)
  3. C.1,6-Dimethylcyclohex-1-ene (second methyl on the sp3 carbon next to the methyl-bearing alkene carbon)
  4. D.1,5-Dimethylcyclohex-1-ene (second methyl two carbons away from the methyl-bearing alkene carbon on the saturated side)

Correct answer: B

Explanation: 3-Methyl-6-oxoheptanal OHC−CH2−CH(CH3)−CH2−CH2−COCH3OHC-CH_2-CH(CH_3)-CH_2-CH_2-COCH_3 comes from cleaving the ring C=C of 1,4-dimethylcyclohexene.

Question 67

Chemistry · 4 marks · −1

Identify [A], [B] and [C], respectively in the following reaction sequence: [A] →NaNO2,HCl,273−278K\xrightarrow{NaNO_2, HCl, 273-278K} benzenediazonium chloride →KI\xrightarrow{KI} [B] →2Na, Dry ether\xrightarrow{2Na,\ \text{Dry ether}} [C]

Figure for question 67
  1. A.[A] aniline, [B] iodobenzene, [C] biphenyl
  2. B.[A] aniline, [B] chlorobenzene, [C] iodobenzene
  3. C.[A] aniline, [B] chlorobenzene, [C] benzene
  4. D.[A] nitrobenzene, [B] iodobenzene, [C] biphenyl

Correct answer: A

Question 68

Chemistry · 4 marks · −1

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water = 0.52 K kg mol−1mol^{-1})

  1. A.375.3 K
  2. B.377.3 K
  3. C.379.2 K
  4. D.277.3 K

Correct answer: B

Explanation: ΔTb=0.52×4/0.5=4.16\Delta T_b=0.52\times4/0.5=4.16 K → 377.3 K.

Question 69

Chemistry · 4 marks · −1

In the following reactions, which one is NOT correct?

  1. A.C6H5N2+Cl−→EtOHC6H5OEtC_6H_5N_2^+Cl^-\xrightarrow{EtOH}C_6H_5OEt
  2. B.C6H5N2+Cl−→CuCN/KCNC6H5CNC_6H_5N_2^+Cl^-\xrightarrow{CuCN/KCN}C_6H_5CN
  3. C.C6H5N2+Cl−→H3PO2/H2OC6H6C_6H_5N_2^+Cl^-\xrightarrow{H_3PO_2/H_2O}C_6H_6
  4. D.C6H5N2+Cl−→Potassium IodideC6H5IC_6H_5N_2^+Cl^-\xrightarrow{\text{Potassium Iodide}}C_6H_5I

Correct answer: A

Explanation: Ethanol reduces the diazonium salt to benzene (it is oxidised to ethanal); it does not give phenetole.

Question 70

Chemistry · 4 marks · −1

During estimation of nitrogen by Dumas' method of compound X (0.42 g) [X = piperazine, a six-membered ring with two NH groups at positions 1 and 4] ______ mL of N2N_2 gas will be liberated at STP. (nearest integer) (Given molar mass in g mol−1mol^{-1} : C : 12, H : 1, N : 14)

Figure for question 70

Correct answer: 109

Explanation: C4H10N2C_4H_{10}N_2 = 86 g/mol; 0.42/86 mol N2N_2 × 22400 = 109 mL.

Question 71

Chemistry · 4 marks · −1

Consider the following reactions: A + NaCl + H2SO4→CrO2Cl2H_2SO_4\to CrO_2Cl_2 + Side Products (A in little amount). CrO2Cl2CrO_2Cl_2(Vapour) + NaOH →\to B + NaCl + H2OH_2O. B + H+→H^+\to C + H2OH_2O. The number of terminal 'O' present in the compound 'C' is ______.

Correct answer: 6

Explanation: B = Na2CrO4Na_2CrO_4, C = Na2Cr2O7Na_2Cr_2O_7 (dichromate): 6 terminal O.

Question 72

Chemistry · 4 marks · −1

The number of optical isomers exhibited by the iron complex (A) obtained from the following reaction is ______. FeCl3+KOH+H2C2O4→AFeCl_3+KOH+H_2C_2O_4\to A

Correct answer: 2

Explanation: K3[Fe(C2O4)3]K_3[Fe(C_2O_4)_3] exists as a pair of enantiomers: 2.

Question 73

Chemistry · 4 marks · −1

0.5 g of an organic compound on combustion gave 1.46 g of CO2CO_2 and 0.9 g of H2OH_2O. The percentage of carbon in the compound is ______. (Nearest integer) [Given : Molar mass (in g mol−1mol^{-1}) C : 12, H : 1, O : 16]

Correct answer: 80

Explanation: C = 1.46 × 12/44 = 0.398 g → 79.6% ≈ 80.

Question 74

Chemistry · 4 marks · −1

Given : ΔHsub⊖\Delta H^\ominus_{sub}[C (graphite)] = 710 kJ mol−1mol^{-1}, ΔC−HH⊖\Delta_{C-H}H^\ominus = 414 kJ mol−1mol^{-1}, ΔH−HH⊖\Delta_{H-H}H^\ominus = 436 kJ mol−1mol^{-1}, ΔC=CH⊖\Delta_{C=C}H^\ominus = 611 kJ mol−1mol^{-1}. The ΔHf⊖\Delta H_f^\ominus for CH2=CH2CH_2=CH_2 is ______ kJ mol−1mol^{-1} (nearest integer value)

Correct answer: 25

Explanation: 2(710)+2(436)−[4(414)+611]=2292−2267=252(710)+2(436)-[4(414)+611]=2292-2267=25.

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