JEE Main 2026 — 21 Jan Shift 2 (3:00 PM - 6:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

Let A={2,3,5,7,9}A=\{2, 3, 5, 7, 9\}. Let R be the relation on A defined by xRyxRy if and only if 2x≤3y2x\le3y. 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 symmetric relation. Then l+ml+m is equal to :

  1. A.23
  2. B.21
  3. C.25
  4. D.27

Correct answer: C

Question 2

Mathematics · 4 marks · −1

Let zz be the complex number satisfying ∣z−5∣≤3|z-5|\le3 and having maximum positive principal argument. Then 34∣5z−125iz+16∣234\left|\frac{5z-12}{5iz+16}\right|^2 is equal to :

  1. A.12
  2. B.16
  3. C.26
  4. D.20

Correct answer: D

Question 3

Mathematics · 4 marks · −1

The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+⋯+(x+2n−2)(x+2n)=8n3x(x+2)+(x+2)(x+4)+\cdots+(x+2n-2)(x+2n)=\frac{8n}{3} are two consecutive even integers, is :

  1. A.9
  2. B.3
  3. C.12
  4. D.6

Correct answer: B

Question 4

Mathematics · 4 marks · −1

Let α\alpha and β\beta be the roots of the equation x2+2ax+(3a+10)=0x^2+2ax+(3a+10)=0 such that α<1<β\alpha<1<\beta. Then the set of all possible values of a is :

  1. A.(−∞,−115)∪(5,∞)\left(-\infty, \frac{-11}{5}\right)\cup(5, \infty)
  2. B.(−∞,−3)(-\infty, -3)
  3. C.(−∞,−2)∪(5,∞)(-\infty, -2)\cup(5, \infty)
  4. D.(−∞,−115)\left(-\infty, \frac{-11}{5}\right)

Correct answer: D

Explanation: f(1)<0f(1)<0: 1+2a+3a+10<0⇒a<−1151+2a+3a+10<0 \Rightarrow a<-\frac{11}{5}.

Question 5

Mathematics · 4 marks · −1

For the matrices A=[3−41−1]A=\begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} and B=[−2949−1318]B=\begin{bmatrix}-29 & 49\\ -13 & 18\end{bmatrix}, if (A15+B)[xy]=[00](A^{15}+B)\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}, then among the following which one is **true** ?

  1. A.x=16,y=3x=16, y=3
  2. B.x=18,y=11x=18, y=11
  3. C.x=5,y=7x=5, y=7
  4. D.x=11,y=2x=11, y=2

Correct answer: D

Explanation: A=I+NA=I+N with N2=0N^2=0, so A15=I+15N=[31−6015−29]A^{15}=I+15N=\begin{bmatrix}31&-60\\15&-29\end{bmatrix}; A15+B=[2−112−11]A^{15}+B=\begin{bmatrix}2&-11\\2&-11\end{bmatrix}, so 2x=11y2x=11y: x=11,y=2x=11, y=2.

Question 6

Mathematics · 4 marks · −1

If the system of equations 3x+y+4z=33x+y+4z=3, 2x+αy−z=−32x+\alpha y-z=-3, x+2y+z=4x+2y+z=4 has no solution, then the value of α\alpha is equal to :

  1. A.19
  2. B.13
  3. C.4
  4. D.23

Correct answer: A

Question 7

Mathematics · 4 marks · −1

Let a1,a22,a322,…,a1029a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9} be a G.P. of common ratio 12\frac{1}{\sqrt{2}}. If a1+a2+…+a10=62a_1+a_2+\ldots+a_{10}=62, then a1a_1 is equal to :

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

Correct answer: D

Explanation: ak=a1(2)k−1a_k=a_1(\sqrt2)^{k-1}; sum =a1(2)10−12−1=31a1(2+1)=62⇒a1=2(2−1)=a_1\frac{(\sqrt2)^{10}-1}{\sqrt2-1}=31a_1(\sqrt2+1)=62 \Rightarrow a_1=2(\sqrt2-1).

Question 8

Mathematics · 4 marks · −1

Let A={x:∣x2−10∣≤6}A=\{x: |x^2-10|\le6\} and B={x:∣x−2∣>1}B=\{x: |x-2|>1\}. Then

  1. A.A−B=[2,3)A-B=[2, 3)
  2. B.A∩B=[−4,−2]∪[3,4]A\cap B=[-4, -2]\cup[3, 4]
  3. C.B−A=(−∞,−4)∪(−2,1)∪(4,∞)B-A=(-\infty, -4)\cup(-2, 1)\cup(4, \infty)
  4. D.A∪B=(−∞,1]∪(2,∞)A\cup B=(-\infty, 1]\cup(2, \infty)

Correct answer: C

Question 9

Mathematics · 4 marks · −1

The largest n∈Nn\in\mathbb{N}, for which 7n7^n divides 101!101!, is :

  1. A.15
  2. B.19
  3. C.16
  4. D.18

Correct answer: C

Explanation: ⌊101/7⌋+⌊101/49⌋=14+2=16\lfloor101/7\rfloor+\lfloor101/49\rfloor=14+2=16.

Question 10

Mathematics · 4 marks · −1

A random variable X takes values 0, 1, 2, 3 with probabilities 2a+130,8a−130,4a+130\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b respectively, where a,b∈Ra, b\in\mathbb{R}. Let μ\mu and σ\sigma respectively be the mean and standard deviation of X such that σ2+μ2=2\sigma^2+\mu^2=2. Then ab\frac{a}{b} is equal to :

  1. A.12
  2. B.3
  3. C.60
  4. D.30

Correct answer: C

Question 11

Mathematics · 4 marks · −1

Let one end of a focal chord of the parabola y2=16xy^2=16x be (16,16)(16, 16). If P(α,β)P(\alpha, \beta) divides this focal chord internally in the ratio 5:25:2, then the minimum value of α+β\alpha+\beta is equal to :

  1. A.5
  2. B.7
  3. C.16
  4. D.22

Correct answer: B

Question 12

Mathematics · 4 marks · −1

Let y2=12xy^2=12x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the xx-axis such that ∠OPA=90∘\angle OPA=90^\circ. Then the locus of the centroid of such triangles OPA is :

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

Correct answer: D

Question 13

Mathematics · 4 marks · −1

If the line αx+4y=7\alpha x+4y=\sqrt{7}, where α∈R\alpha\in\mathbb{R}, touches the ellipse 3x2+4y2=13x^2+4y^2=1 at the point P in the first quadrant, then one of the focal distances of P is :

  1. A.13+125\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{5}}
  2. B.13+127\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}
  3. C.13−125\frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{5}}
  4. D.13−1211\frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{11}}

Correct answer: B

Question 14

Mathematics · 4 marks · −1

Let the line L pass through the point (−3,5,2)(-3, 5, 2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1)(-2, r, 1) is 143\sqrt{\frac{14}{3}}, then the sum of all possible values of r is :

  1. A.16
  2. B.12
  3. C.10
  4. D.6

Correct answer: C

Question 15

Mathematics · 4 marks · −1

Let the line L1L_1 be parallel to the vector −3i^+2j^+4k^-3\hat{i}+2\hat{j}+4\hat{k} and pass through the point (2,6,7)(2, 6, 7), and the line L2L_2 be parallel to the vector 2i^+j^+3k^2\hat{i}+\hat{j}+3\hat{k} and pass through the point (4,3,5)(4, 3, 5). If the line L3L_3 is parallel to the vector −3i^+5j^+16k^-3\hat{i}+5\hat{j}+16\hat{k} and intersects the lines L1L_1 and L2L_2 at the points C and D, respectively, then ∣CD→∣2|\overrightarrow{CD}|^2 is equal to :

  1. A.290
  2. B.89
  3. C.312
  4. D.171

Correct answer: A

Question 16

Mathematics · 4 marks · −1

For a triangle ABC, let p⃗=BC→\vec{p}=\overrightarrow{BC}, q⃗=CA→\vec{q}=\overrightarrow{CA} and r⃗=BA→\vec{r}=\overrightarrow{BA}. If ∣p⃗∣=23|\vec{p}|=2\sqrt{3}, ∣q⃗∣=2|\vec{q}|=2 and cos⁡θ=13\cos\theta=\frac{1}{\sqrt{3}}, where θ\theta is the angle between p⃗\vec{p} and q⃗\vec{q}, then ∣p⃗×(q⃗−3r⃗)∣2+3∣r⃗∣2|\vec{p}\times(\vec{q}-3\vec{r})|^2+3|\vec{r}|^2 is equal to :

  1. A.340
  2. B.220
  3. C.200
  4. D.410

Correct answer: C

Question 17

Mathematics · 4 marks · −1

Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x)=x^3+x^2f'(1)+2xf''(2)+f'''(3), x∈Rx\in\mathbb{R}. Then the value of f′(5)f'(5) is :

  1. A.625\frac{62}{5}
  2. B.6575\frac{657}{5}
  3. C.25\frac{2}{5}
  4. D.1175\frac{117}{5}

Correct answer: D

Question 18

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a twice differentiable function such that f′′(x)>0f''(x)>0 for all x∈Rx\in\mathbb{R} and f′(a−1)=0f'(a-1)=0, where a is a real number. Let g(x)=f(tan⁡2x−2tan⁡x+a)g(x)=f(\tan^2x-2\tan x+a), 0<x<π20<x<\frac{\pi}{2}. Consider the following two statements : (I) g is increasing in (0,π4)\left(0, \frac{\pi}{4}\right) (II) g is deceasing in (π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right). Then,

  1. A.Only (II) is True
  2. B.Only (I) is True
  3. C.Both (I) and (II) are True
  4. D.Neither (I) nor (II) is True

Correct answer: D

Explanation: g′=f′(u)⋅2sec⁡2x(tan⁡x−1)g'=f'(u)\cdot2\sec^2x(\tan x-1) with u=(tan⁡x−1)2+a−1≥a−1u=(\tan x-1)^2+a-1\ge a-1, so f′(u)>0f'(u)>0 except at x=π/4x=\pi/4: g decreases on (0,π/4)(0,\pi/4) and increases on (π/4,π/2)(\pi/4,\pi/2); neither statement holds.

Question 19

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation sec⁡xdydx−2y=2+3sin⁡x\sec x\frac{dy}{dx}-2y=2+3\sin x, x∈(−π2,π2)x\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), y(0)=−74y(0)=-\frac{7}{4}. Then y(π6)y\left(\frac{\pi}{6}\right) is equal to :

  1. A.−52-\frac{5}{2}
  2. B.−54-\frac{5}{4}
  3. C.−32−7-3\sqrt{2}-7
  4. D.−33−7-3\sqrt{3}-7

Correct answer: A

Question 20

Mathematics · 4 marks · −1

If the area of the region {(x,y):1−2x≤y≤4−x2,x≥0,y≥0}\{(x, y): 1-2x\le y\le4-x^2, x\ge0, y\ge0\} is αβ\frac{\alpha}{\beta}, α,β∈N\alpha, \beta\in\mathbb{N}, gcd⁡(α,β)=1\gcd(\alpha, \beta)=1, then the value of (α+β)(\alpha+\beta) is :

  1. A.67
  2. B.85
  3. C.91
  4. D.73

Correct answer: D

Explanation: ∫02(4−x2)dx−∫01/2(1−2x)dx=163−14=6112\int_0^2(4-x^2)dx-\int_0^{1/2}(1-2x)dx=\frac{16}{3}-\frac14=\frac{61}{12}; 61+12=7361+12=73.

Question 21

Mathematics · 4 marks · −1

If (115C0+115C1)(115C1+115C2)⋯(115C12+115C13)=α1314C0 14C1⋯ 14C12\left(\frac{1}{^{15}C_0}+\frac{1}{^{15}C_1}\right)\left(\frac{1}{^{15}C_1}+\frac{1}{^{15}C_2}\right)\cdots\left(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}\right)=\frac{\alpha^{13}}{^{14}C_0\,^{14}C_1\cdots\,^{14}C_{12}}, then 30α30\alpha is equal to ______.

Correct answer: 32

Explanation: 115Cr+115Cr+1=1615⋅114Cr\frac1{^{15}C_r}+\frac1{^{15}C_{r+1}}=\frac{16}{15}\cdot\frac{1}{^{14}C_r}, so α=1615\alpha=\frac{16}{15} and 30α=3230\alpha=32.

Question 22

Mathematics · 4 marks · −1

If P is a point on the circle x2+y2=4x^2+y^2=4, Q is a point on the straight line 5x+y+2=05x+y+2=0 and x−y+1=0x-y+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is ______.

Correct answer: 2

Explanation: P is the reflection of Q in y=x+1y=x+1: Q(a,b)↦(b−1,a+1)(a,b)\mapsto(b-1,a+1), so P lies on the reflected line x+5y−2=0x+5y-2=0. With x2+y2=4x^2+y^2=4: 26y2−20y=026y^2-20y=0, giving P =(2,0)=(2,0) and (−2413,1013)\left(-\frac{24}{13},\frac{10}{13}\right); sum of abscissae 213\frac{2}{13}, times 13 = 2.

Question 23

Mathematics · 4 marks · −1

Let the maximum value of (sin⁡−1x)2+(cos⁡−1x)2(\sin^{-1}x)^2+(\cos^{-1}x)^2 for x∈[−32,12]x\in\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right] be mnπ2\frac{m}{n}\pi^2, where gcd⁡(m,n)=1\gcd(m, n)=1. Then m+nm+n is equal to ______.

Correct answer: 65

Explanation: Maximum at x=−32x=-\frac{\sqrt3}{2}: π29+25π236=2936π2\frac{\pi^2}{9}+\frac{25\pi^2}{36}=\frac{29}{36}\pi^2, so 65.

Question 24

Mathematics · 4 marks · −1

Let [⋅][\cdot] denote the greatest integer function and f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x)=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^{n}\left[\frac{k^2}{3^x}\right]. Then 12∑j=1∞f(j)12\sum_{j=1}^{\infty}f(j) is equal to ______.

Correct answer: 2

Explanation: f(x)=13⋅3xf(x)=\frac{1}{3\cdot3^x}; 12∑j≥113j+1=12⋅16=212\sum_{j\ge1}\frac{1}{3^{j+1}}=12\cdot\frac16=2.

Question 25

Mathematics · 4 marks · −1

If ∫014cot⁡−1(1−2x+4x2) dx=atan⁡−1(2)−blog⁡e(5)\int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx=a\tan^{-1}(2)-b\log_e(5), where a,b∈Na, b\in\mathbb{N}, then (2a+b)(2a+b) is equal to ______.

Correct answer: 9

Question 26

Physics · 4 marks · −1

Keeping the significant figures in view, the sum of the physical quantities 52.01 m, 153.2 m and 0.123 m is :

  1. A.205.33 m
  2. B.205.333 m
  3. C.205 m
  4. D.205.3 m

Correct answer: D

Explanation: Sum 205.333 rounded to the least number of decimal places (one): 205.3 m.

Question 27

Physics · 4 marks · −1

A spherical body of radius rr and density σ\sigma falls freely through a viscous liquid having density ρ\rho and viscosity η\eta and attains a terminal velocity v0v_0. Estimated maximum error in the quantity η\eta is : (Ignore errors associated with σ\sigma, ρ\rho and gg, gravitational acceleration)

  1. A.2[Δrr−Δv0v0]2\left[\frac{\Delta r}{r}-\frac{\Delta v_0}{v_0}\right]
  2. B.2[Δrr+Δv0v0]2\left[\frac{\Delta r}{r}+\frac{\Delta v_0}{v_0}\right]
  3. C.2Δrr+Δv0v0\frac{2\Delta r}{r}+\frac{\Delta v_0}{v_0}
  4. D.2Δrr−Δv0v02\frac{\Delta r}{r}-\frac{\Delta v_0}{v_0}

Correct answer: C

Explanation: η=2r2(σ−ρ)g9v0\eta=\frac{2r^2(\sigma-\rho)g}{9v_0}, so Δηη=2Δrr+Δv0v0\frac{\Delta\eta}{\eta}=\frac{2\Delta r}{r}+\frac{\Delta v_0}{v_0}.

Question 28

Physics · 4 marks · −1

A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.

  1. A.20 s and 100 m
  2. B.40 s and 100 m
  3. C.40 s and 0 m
  4. D.40 s and 200 m

Correct answer: D

Explanation: Heading perpendicular each way: 20 s per crossing, 40 s total; drift 5×40=2005\times40=200 m downstream.

Question 29

Physics · 4 marks · −1

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is ______ (assume no slipping of string on pulley).

Figure for question 29
  1. A.(M−m)g[(136)M+m]\frac{(M-m)g}{\left[\left(\frac{13}{6}\right)M+m\right]}
  2. B.(M−m)g[(83)M+m]\frac{(M-m)g}{\left[\left(\frac{8}{3}\right)M+m\right]}
  3. C.(M−m)g2M+m\frac{(M-m)g}{2M+m}
  4. D.(M−m)gM+m\frac{(M-m)g}{M+m}

Correct answer: B

Explanation: I=MR2+2⋅M(2R)212=53MR2I=MR^2+2\cdot\frac{M(2R)^2}{12}=\frac53MR^2; a=(M−m)gM+m+53M=(M−m)g83M+ma=\frac{(M-m)g}{M+m+\frac53M}=\frac{(M-m)g}{\frac83M+m}.

Question 30

Physics · 4 marks · −1

Two cars A and B each of mass 10310^3 kg are moving on parallel tracks separated by a distance of 10 m, in same direction with speeds 72 km/h and 36 km/h. The magnitude of angular momentum of car A with respect to car B is ______ J.s.

  1. A.3×1053\times10^5
  2. B.10510^5
  3. C.3.6×1053.6\times10^5
  4. D.2×1052\times10^5

Correct answer: B

Explanation: L=mvreld=103×10×10=105L=m v_{rel} d=10^3\times10\times10=10^5 J s.

Question 31

Physics · 4 marks · −1

A body of mass 2 kg is moving along xx-direction such that its displacement as function of time is given by x(t)=αt2+βt+γx(t)=\alpha t^2+\beta t+\gamma m, where α=1\alpha=1 m/s2^2, β=1\beta=1 m/s and γ=1\gamma=1 m. The work done on the body during the time interval t=2t=2 s to t=3t=3 s, is ______ J.

  1. A.42
  2. B.24
  3. C.12
  4. D.49

Correct answer: B

Explanation: v=2t+1v=2t+1: 5 to 7 m/s; W=12⋅2(49−25)=24W=\frac12\cdot2(49-25)=24 J.

Question 32

Physics · 4 marks · −1

A large drum having radius R is spinning around its axis with angular velocity ω\omega, as shown in figure. The minimum value of ω\omega so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M as μ\mu, is :

Figure for question 32
  1. A.μgR\sqrt{\frac{\mu g}{R}}
  2. B.gμR\sqrt{\frac{g}{\mu R}}
  3. C.2gμR\sqrt{\frac{2g}{\mu R}}
  4. D.g2μR\sqrt{\frac{g}{2\mu R}}

Correct answer: B

Explanation: μMω2R≥Mg\mu M\omega^2R\ge Mg.

Question 33

Physics · 4 marks · −1

Surface tension of two liquids (having same densities), T1T_1 and T2T_2, are measured using capillary rise method utilizing two tubes with inner radii of r1r_1 and r2r_2 where r1>r2r_1>r_2. The measured liquid heights in these tubes are h1h_1 and h2h_2 respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h1h_1 and h2h_2 and surfaces tensions T1T_1 and T2T_2 satisfy the relation :

  1. A.h1>h2h_1>h_2 and T1<T2T_1<T_2
  2. B.h1=h2h_1=h_2 and T1=T2T_1=T_2
  3. C.h1<h2h_1<h_2 and T1=T2T_1=T_2
  4. D.h1>h2h_1>h_2 and T1=T2T_1=T_2

Correct answer: C

Question 34

Physics · 4 marks · −1

The r.m.s. speed of oxygen molecules at 47∘47^\circC is equal to that of the hydrogen molecules kept at ______ ∘^\circC. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)

  1. A.-20
  2. B.-253
  3. C.-235
  4. D.-100

Correct answer: B

Explanation: TH=320/16=20T_H=320/16=20 K =−253∘=-253^\circC.

Question 35

Physics · 4 marks · −1

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [take π=227\pi=\frac{22}{7}]

  1. A.28
  2. B.176
  3. C.14
  4. D.88

Correct answer: C

Explanation: KE oscillates at 2ω2\omega: ω=88\omega=88 rad/s, f=882π=14f=\frac{88}{2\pi}=14 Hz.

Question 36

Physics · 4 marks · −1

A capacitor C is first charged fully with potential difference of V0V_0 and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In tt s 25% of the initial energy in the capacitor is transferred to the inductor. The value of tt is ______ s.

  1. A.πLC2\pi\sqrt{\frac{LC}{2}}
  2. B.πLC6\frac{\pi\sqrt{LC}}{6}
  3. C.πLC3\frac{\pi\sqrt{LC}}{3}
  4. D.πLC2\frac{\pi\sqrt{LC}}{2}

Correct answer: B

Explanation: UC=U0cos⁡2ωt=0.75U0⇒ωt=π/6U_C=U_0\cos^2\omega t=0.75U_0 \Rightarrow \omega t=\pi/6.

Question 37

Physics · 4 marks · −1

Two known resistances of R ΩR\ \Omega and 2R Ω2R\ \Omega and one unknown resistance X ΩX\ \Omega are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X ΩX\ \Omega, then the value of X is ______ Ω\Omega.

Figure for question 37
  1. A.R
  2. B.(3−1)R(\sqrt{3}-1)R
  3. C.2(3−1)R2(\sqrt{3}-1)R
  4. D.(3+1)R(\sqrt{3}+1)R

Correct answer: B

Explanation: R(2R+X)3R+X=X⇒X2+2RX−2R2=0⇒X=(3−1)R\frac{R(2R+X)}{3R+X}=X \Rightarrow X^2+2RX-2R^2=0 \Rightarrow X=(\sqrt3-1)R.

Question 38

Physics · 4 marks · −1

The total length of potentiometer wire AB is 50 cm in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ______ cm.

Figure for question 38
  1. A.30
  2. B.25
  3. C.15
  4. D.20

Correct answer: A

Explanation: APPB=64⇒AP=30\frac{AP}{PB}=\frac64 \Rightarrow AP=30 cm.

Question 39

Physics · 4 marks · −1

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is rr. The magnetic field at the centre O of the circular loop is :

Figure for question 39
  1. A.μ02πIr(π−1)i^\frac{\mu_0}{2\pi}\frac{I}{r}(\pi-1)\hat{i}
  2. B.μ02πIr(π+1)i^\frac{\mu_0}{2\pi}\frac{I}{r}(\pi+1)\hat{i}
  3. C.−μ02πIr(π−1)i^-\frac{\mu_0}{2\pi}\frac{I}{r}(\pi-1)\hat{i}
  4. D.−μ02πIr(π+1)i^-\frac{\mu_0}{2\pi}\frac{I}{r}(\pi+1)\hat{i}

Correct answer: C

Question 40

Physics · 4 marks · −1

Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of uu towards each other. The minimum value of uu, so that they can just touch each other is : (Take k=14πϵ0k=\frac{1}{4\pi\epsilon_0} and assume kQ2>Gm2kQ^2>Gm^2 where G is the Gravitational constant)

  1. A.kQ24mR(1−Gm2kQ2)\sqrt{\frac{kQ^2}{4mR}\left(1-\frac{Gm^2}{kQ^2}\right)}
  2. B.kQ22mR(1−Gm2kQ2)\sqrt{\frac{kQ^2}{2mR}\left(1-\frac{Gm^2}{kQ^2}\right)}
  3. C.kQ22mR(1−Gm22kQ2)\sqrt{\frac{kQ^2}{2mR}\left(1-\frac{Gm^2}{2kQ^2}\right)}
  4. D.kQ24mR(1+Gm2kQ2)\sqrt{\frac{kQ^2}{4mR}\left(1+\frac{Gm^2}{kQ^2}\right)}

Correct answer: A

Explanation: mu2=(kQ2−Gm2)(12R−14R)mu^2=(kQ^2-Gm^2)\left(\frac{1}{2R}-\frac{1}{4R}\right).

Question 41

Physics · 4 marks · −1

A battery with EMF E and internal resistance rr is connected across a resistance R. The power consumption in R will be maximum when :

  1. A.R=rR=r
  2. B.R=r/2R=r/2
  3. C.R=2rR=\sqrt{2}r
  4. D.R=2rR=2r

Correct answer: A

Question 42

Physics · 4 marks · −1

Given below are two statements : Statement I : In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits. Statement II : In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength. In the light of the above statements, choose the **correct answer** from the options given below :

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

Correct answer: D

Explanation: Angular separation λ/d\lambda/d is independent of D and grows with λ\lambda.

Question 43

Physics · 4 marks · −1

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface. If refractive index of the material of prism is 2\sqrt{2}, the angle θ\theta of prism is.

Figure for question 43
  1. A.90∘90^\circ
  2. B.45∘45^\circ
  3. C.75∘75^\circ
  4. D.60∘60^\circ

Correct answer: D

Question 44

Physics · 4 marks · −1

The energy of an electron in an orbit of the Bohr's atom is −0.04E0-0.04E_0 eV where E0E_0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2πLh\frac{2\pi L}{h} is ______ :

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

Correct answer: C

Explanation: 1/n2=0.04⇒n=51/n^2=0.04 \Rightarrow n=5.

Question 45

Physics · 4 marks · −1

The charge stored by the capacitor C in the given circuit in the steady state is ______ μ\muC.

Figure for question 45
  1. A.10
  2. B.7.5
  3. C.5
  4. D.12.5

Correct answer: A

Explanation: Current 2.5/(1+4)=0.52.5/(1+4)=0.5 A; voltage across the parallel branches 2.5−0.5=22.5-0.5=2 V; Q=5×2=10 μQ=5\times2=10\ \muC.

Question 46

Physics · 4 marks · −1

The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be ______ cm/s.

Correct answer: 5

Explanation: v∝r2v\propto r^2: 20/4=520/4=5.

Question 47

Physics · 4 marks · −1

A diatomic gas (γ=1.4)(\gamma=1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas ______ J.

Correct answer: 350

Explanation: Q=γγ−1W=3.5×100=350Q=\frac{\gamma}{\gamma-1}W=3.5\times100=350 J.

Question 48

Physics · 4 marks · −1

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, σ=10\sigma=10 mho/m. The ratio of maximum conduction current density to maximum displacement current density is ______. [Take 14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 Nm2^2/C2^2]

Correct answer: 1800

Explanation: σωϵ0=10×4π×9×1092π×108=1800\frac{\sigma}{\omega\epsilon_0}=\frac{10\times4\pi\times9\times10^9}{2\pi\times10^8}=1800.

Question 49

Physics · 4 marks · −1

In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 μ\mum is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by α10\frac{\alpha}{10}, where α\alpha is ______.

Correct answer: 14

Explanation: (μ−1)tDd=0.02⇒μ−1=0.4(\mu-1)t\frac{D}{d}=0.02 \Rightarrow \mu-1=0.4, μ=1.4\mu=1.4, α=14\alpha=14.

Question 50

Physics · 4 marks · −1

A particle having electric charge 3×10−193\times10^{-19} C and mass 6×10−276\times10^{-27} kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10−12\alpha\times10^{-12} m. The value of α\alpha is ______. (Take Planck's constant =6.6×10−34=6.6\times10^{-34} J.s)

Correct answer: 10

Explanation: λ=h2mqV=6.6×10−346.6×10−23=10−11\lambda=\frac{h}{\sqrt{2mqV}}=\frac{6.6\times10^{-34}}{6.6\times10^{-23}}=10^{-11} m, α=10\alpha=10.

Question 51

Chemistry · 4 marks · −1

Aqueous HCl reacts with MnO2_2(s) to form MnCl2_2(aq), Cl2_2(g) and H2_2O(ll). What is the weight (in g) of Cl2_2 liberated when 8.7 g of MnO2_2(s) is reacted with excess aqueous HCl solution ? (Given Molar mass in g mol−1^{-1} Mn = 55, Cl = 35.5, O = 16, H = 1)

  1. A.21.3
  2. B.71
  3. C.14.2
  4. D.7.1

Correct answer: D

Explanation: MnO2_2 + 4HCl →\to MnCl2_2 + Cl2_2 + 2H2_2O; 0.1 mol MnO2_2 gives 0.1 mol Cl2_2 = 7.1 g.

Question 52

Chemistry · 4 marks · −1

Consider the following spectral lines for atomic hydrogen : A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series. The correct arrangement of the above lines in ascending order of energy is :

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

Correct answer: D

Explanation: In units of R: A 19−116=0.049\frac19-\frac1{16}=0.049; B 14−116=0.1875\frac14-\frac1{16}=0.1875; C 19−136=0.083\frac19-\frac1{36}=0.083; D 116−164=0.047\frac1{16}-\frac1{64}=0.047. So D < A < C < B.

Question 53

Chemistry · 4 marks · −1

The correct increasing order of C−-H(A), C−-O(B), C=O(C) and C≡\equivN(D) bonds in terms of covalent bond length is :

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

Correct answer: A

Explanation: C−-H 107 pm < C≡\equivN 116 pm < C=O 121 pm < C−-O 143 pm.

Question 54

Chemistry · 4 marks · −1

Consider the following data : ΔfH⊖\Delta_fH^\ominus (methane, g) =−X=-X kJ mol−1^{-1}; Enthalpy of sublimation of graphite =Y=Y kJ mol−1^{-1}; Dissociation enthalpy of H2_2 =Z=Z kJ mol−1^{-1}. The bond enthalpy of C−-H bond is given by :

  1. A.X+Y+4Z2\frac{X+Y+4Z}{2}
  2. B.X+Y+2Z4\frac{X+Y+2Z}{4}
  3. C.−X+Y+Z4\frac{-X+Y+Z}{4}
  4. D.X+Y+ZX+Y+Z

Correct answer: B

Explanation: 4 ϵC−H=Y+2Z−(−X)4\,\epsilon_{C-H}=Y+2Z-(-X).

Question 55

Chemistry · 4 marks · −1

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature ?

Figure for question 55
  1. A.EcellE_{cell} increases linearly with time
  2. B.EcellE_{cell} constant with time (horizontal line)
  3. C.EcellE_{cell} decreases with time (falling line)
  4. D.Vertical line at a fixed time

Correct answer: B

Explanation: NTA key: the plot of constant EcellE_{cell} (option 2). Strictly, EcellE_{cell} falls slowly as [Zn2+^{2+}]/[Cu2+^{2+}] changes, but over the ideal short-time picture it is treated as constant.

Question 56

Chemistry · 4 marks · −1

Decomposition of A is a first order reaction at T(K) and is given by A(g) →\to B(g) + C(g). In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min−1^{-1}) of the reaction ? (log 2 = 0.3)

  1. A.6.9×10−46.9\times10^{-4}
  2. B.6.9×10−16.9\times10^{-1}
  3. C.6.9×10−26.9\times10^{-2}
  4. D.6.9×10−36.9\times10^{-3}

Correct answer: D

Explanation: Half of A has decomposed, so t1/2=100t_{1/2}=100 min and k=0.693/100k=0.693/100.

Question 57

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The correct order in terms of atomic/ionic radii is Al > Mg > Mg2+^{2+} > Al3+^{3+}. Statement II : The correct order in terms of the magnitude of electron gain enthalpy is Cl > Br > S > O. 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 false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: C

Explanation: Mg > Al in atomic radius, so I is false; II is correct.

Question 58

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The correct order in terms of bond dissociation enthalpy is Cl2_2 > Br2_2 > F2_2 > I2_2. Statement II : The correct trend in the covalent character of the metal halides is [SnCl4_4 > SnCl2_2], [PbCl4_4 > PbCl2_2] and [UF4_4 > UF6_6]. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: C

Explanation: Higher oxidation state is more covalent, so UF6_6 > UF4_4: II is false.

Question 59

Chemistry · 4 marks · −1

Given below are some of the statements about Mn and Mn2_2O7_7. Identify the correct statements. A. Mn forms the oxide Mn2_2O7_7, in which Mn is in its highest oxidation state. B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn. C. Mn2_2O7_7 is an ionic oxide. D. The structure of Mn2_2O7_7 consists of one bridged oxygen. Choose the correct answer from the options given below :

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

Correct answer: A

Question 60

Chemistry · 4 marks · −1

Given below are two statements : Statement I : Crystal Field Stabilization Energy (CFSE) of [Cr(H2_2O)6_6]2+^{2+} is greater than that of [Mn(H2_2O)6_6]2+^{2+}. Statement II : Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. 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: C

Explanation: High-spin d4^4 CFSE −0.6Δo-0.6\Delta_o vs d5^5 zero; [Fe(CN)6_6]3−^{3-} has 1 unpaired electron, [Fe(CN)6_6]4−^{4-} none.

Question 61

Chemistry · 4 marks · −1

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : (nearest integer) (Given, molar mass in g mol−1^{-1} : O = 16, Mg = 24, P = 31)

  1. A.20
  2. B.40
  3. C.30
  4. D.50

Correct answer: D

Explanation: Mg2_2P2_2O7_7 = 222: 62222×1.79×100≈50%\frac{62}{222}\times1.79\times100\approx50\%.

Question 62

Chemistry · 4 marks · −1

The correct order of the rate of the reaction for the following reaction with respect to nucleophiles is : CH3_3Br + Nu⊖^\ominus →\to CH3_3Nu + Br⊖^\ominus

  1. A.CH3_3COO−^- > PhO−^- > −^-OH > ClO4−_4^-
  2. B.ClO4−_4^- > CH3_3COO−^- > −^-OH > PhO−^-
  3. C.−^-OH > PhO−^- > CH3_3COO−^- > ClO4−_4^-
  4. D.PhO−^- > −^-OH > CH3_3COO−^- > ClO4−_4^-

Correct answer: C

Question 63

Chemistry · 4 marks · −1

Given below are two statements : Statement I : Compound (X), shown below, dissolves in NaHCO3_3 solution and has two chiral carbon atoms. [X = 2-(4-bromobenzoyl)cyclopentane-1-carboxylic acid: a cyclopentane ring bearing a COOH group on one carbon and, on the adjacent carbon, a C(=O) group attached to a 4-bromophenyl ring.] Statement II : Compound (Y), shown below, has two carbons with sp3^3 hybridization, one carbon with sp2^2 and one carbon with sp hybridization. [Y = CH3_3CH2_2CH=C=O (ethylketene).] In the light of the above statements, choose the correct answer from the options given below :

Figure for question 63
  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: C

Question 64

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Pair of Compounds): A. 2-Methylpropene and but-1-ene; B. Cis-but-2-ene and trans-but-2-ene; C. 2-Butanol and diethyl ether; D. But-1-ene and but-2-ene. List - II (Type of Isomers): I. Stereoisomers; II. Position isomers; III. Chain isomers; IV. Functional group isomers. Choose the correct answer from the options given below :

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

Correct answer: D

Question 65

Chemistry · 4 marks · −1

Given below are four compounds : (a) n-propyl choride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride. Percentage of carbon in the one which exhibits optical isomerism is :

  1. A.56
  2. B.40
  3. C.46
  4. D.52

Correct answer: D

Explanation: sec-Butyl chloride C4_4H9_9Cl (92.5): 48/92.5=51.9%48/92.5=51.9\%.

Question 66

Chemistry · 4 marks · −1

The correct order of reactivity of the following benzyl halides towards reaction with KCN is : (a) 4-hydroxybenzyl bromide, (b) 4-aminobenzyl bromide, (c) 4-nitrobenzyl bromide, (d) 3-nitrobenzyl bromide

Figure for question 66
  1. A.b > a > c > d
  2. B.a > b > c > d
  3. C.a > b > d > c
  4. D.b > a > d > c

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Reagents): A. H2_2, Pd-BaSO4_4; B. SnCl2_2, HCl; C. CrO2_2Cl2_2, CS2_2; D. CO, HCl, Anhyd. AlCl3_3. List - II (Reaction Name (Involving aldehydes)): I. Etard Reaction; II. Rosenmund Reduction; III. Gatterman - Koch Reaction; IV. Stephen Reaction. Choose the correct answer from the options given below :

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

Correct answer: B

Question 68

Chemistry · 4 marks · −1

Nitrobenzene is subjected to the sequence: (1) Br2_2/FeBr3_3/Δ\Delta; (2) Sn/HCl/Δ\Delta; (3) pH neutralisation; (4) Br2_2/H2_2O; (5) NaNO2_2/HBr, 0$-55^\circ$C; (6) CuBr/NaBr →\to Major Product (P). Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

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

Correct answer: B

Explanation: m-bromonitrobenzene →\to 3-bromoaniline →\to 2,3,4,6-tetrabromoaniline (bromine water) →\to Sandmeyer replaces NH2_2 by Br: 5 Br.

Question 69

Chemistry · 4 marks · −1

The **correct** statements are : A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C. Nucleotides are joined together by glycosidic linkage between C1_1 and C4_4 carbons of the pentose sugar. D. Quaternary structure of proteins represents overall folding of the polypeptide chain. Choose the **correct** answer from the options given below :

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

Correct answer: B

Question 70

Chemistry · 4 marks · −1

On heating a mixture of common salt and K2_2Cr2_2O7_7 in equal amount along with concentrated H2_2SO4_4 in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are :

  1. A.CrO2_2Cl2_2 and +6
  2. B.CrO2_2Cl2_2 and +5
  3. C.Cr2_2O2_2Cl2_2 and +3
  4. D.Cr2_2O2_2Cl2_2 and +6

Correct answer: A

Question 71

Chemistry · 4 marks · −1

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol−1^{-1}) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is ______ ×10−2\times10^{-2}. (nearest integer) [Given : KbK_b of the solvent = 5.0 K kg mol−1^{-1}] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Correct answer: 3

Explanation: Molality 0.1, so 0.015 mol X; solvent 0.5 mol; relative lowering ≈0.015/0.5=0.03\approx0.015/0.5=0.03.

Question 72

Chemistry · 4 marks · −1

The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ______ g L−1^{-1}. (Nearest integer) Given : R = 0.08 L atm K−1^{-1} mol−1^{-1}. Assume complete dissociation of NaCl (Given : Molar mass of Na and Cl are 23 and 35.5 g mol−1^{-1} respectively.)

Correct answer: 15

Explanation: 12=2C(0.08)(300)⇒C=0.2512=2C(0.08)(300) \Rightarrow C=0.25 M =14.6≈15=14.6\approx15 g/L.

Question 73

Chemistry · 4 marks · −1

The first and second ionization constants of H2_2X are 2.5×10−82.5\times10^{-8} and 1.0×10−131.0\times10^{-13} respectively. The concentration of X2−^{2-} in 0.1 M H2_2X solution is ______ ×10−15\times10^{-15} M. (Nearest Integer)

Correct answer: 100

Explanation: [X2−^{2-}] ≈Ka2=10−13=100×10−15\approx K_{a2}=10^{-13}=100\times10^{-15}.

Question 74

Chemistry · 4 marks · −1

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX(s) ⇌\rightleftharpoons M+^+(aq) + X−^-(aq); Ksp=10−10K_{sp}=10^{-10}. If the standard reduction potential for M+^+(aq) →+e−\xrightarrow{+e^-} M(s) is EM+/M⊖=0.79E^\ominus_{M^+/M}=0.79 V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode EX−/MX(s)/M⊖E^\ominus_{X^-/MX(s)/M} is ______ mV. (nearest integer) [Given : 2.303RTF=0.059\frac{2.303RT}{F}=0.059 V]

Correct answer: 200

Explanation: 0.79+0.059log⁡10−10=0.200.79+0.059\log10^{-10}=0.20 V = 200 mV.

Question 75

Chemistry · 4 marks · −1

Identify the metal ions among Co2+^{2+}, Ni2+^{2+}, Fe2+^{2+}, V3+^{3+} and Ti2+^{2+} having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is ______.

Correct answer: 7

Explanation: Co2+^{2+} (3 unpaired, 3.87 BM) and Fe2+^{2+} (4, 4.90 BM) exceed 3.0 BM; 3 + 4 = 7.

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