JEE Main 2026 — 23 Jan Shift 2 (3:00 PM - 6:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
72
Total marks
288
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

Let A={0,1,2,…,9}A=\{0, 1, 2, \ldots, 9\}. Let R be a relation on A defined by (x,y)∈R(x, y)\in R if and only if ∣x−y∣|x-y| is a multiple of 3. Given below are two statements: Statement I: n(R)=36n(R)=36. Statement II: R is an equivalence relation. 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 correct
  2. B.Both Statement I and Statement II are incorrect
  3. C.Statement I is incorrect but Statement II is correct
  4. D.Statement I is correct but Statement II is incorrect

Correct answer: C

Explanation: Classes {0,3,6,9}, {1,4,7}, {2,5,8}: 16+9+9=3416+9+9=34 pairs, so I is false; R is an equivalence relation.

Question 2

Mathematics · 4 marks · −1

Consider two sets A={x∈Z:∣(∣x−3∣−3)∣≤1}A=\{x\in\mathbb{Z}: |(|x-3|-3)|\le1\} and B={x∈R−{1,2}:(x−2)(x−4)x−1log⁡e(∣x−2∣)=0}B=\left\{x\in\mathbb{R}-\{1, 2\}: \frac{(x-2)(x-4)}{x-1}\log_e(|x-2|)=0\right\}. Then the number of onto functions f:A→Bf: A\to B is equal to

  1. A.32
  2. B.62
  3. C.81
  4. D.79

Correct answer: B

Explanation: A = {-1, 0, 1, 5, 6, 7} (6 elements); B = {3, 4} (2 elements): 26−2=622^6-2=62.

Question 3

Mathematics · 4 marks · −1

The sum of all the real solutions of the equation log⁡(x+3)(6x2+28x+30)=5−2log⁡(6x+10)(x2+6x+9)\log_{(x+3)}(6x^2+28x+30)=5-2\log_{(6x+10)}(x^2+6x+9) is equal to

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

Correct answer: B

Question 4

Mathematics · 4 marks · −1

If z=32+i2z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=−1i=\sqrt{-1}, then (z201−i)8(z^{201}-i)^8 is equal to

  1. A.0
  2. B.256
  3. C.-1
  4. D.1

Correct answer: B

Explanation: z201=ei201π/6=ei⋅3π/2=−iz^{201}=e^{i201\pi/6}=e^{i\cdot3\pi/2}=-i; (−2i)8=256(-2i)^8=256.

Question 5

Mathematics · 4 marks · −1

The system of linear equations x+y+z=6x+y+z=6, 2x+5y+az=362x+5y+az=36, x+2y+3z=bx+2y+3z=b has

  1. A.infinitely many solutions for a=8a=8 and b=14b=14
  2. B.infinitely many solutions for a=8a=8 and b=16b=16
  3. C.unique solution for a=8a=8 and b=16b=16
  4. D.unique solution for a=8a=8 and b=14b=14

Correct answer: A

Question 6

Mathematics · 4 marks · −1

Let ∑k=1nak=αn2+βn\sum_{k=1}^{n}a_k=\alpha n^2+\beta n. If a10=59a_{10}=59 and a6=7a1a_6=7a_1, then α+β\alpha+\beta is equal to

  1. A.3
  2. B.5
  3. C.7
  4. D.12

Correct answer: B

Explanation: an=α(2n−1)+βa_n=\alpha(2n-1)+\beta: 19α+β=5919\alpha+\beta=59, 11α+β=7(α+β)⇒β=2α311\alpha+\beta=7(\alpha+\beta) \Rightarrow \beta=\frac{2\alpha}{3}; α=3,β=2\alpha=3,\beta=2.

Question 7

Mathematics · 4 marks · −1

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

  1. A.403
  2. B.384
  3. C.429
  4. D.455

Correct answer: D

Explanation: (153)=455\binom{15}{3}=455.

Question 8

Mathematics · 4 marks · −1

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is pq\frac{p}{q}, gcd⁡(p,q)=1\gcd(p, q)=1, then p+qp+q is equal to

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

Correct answer: A

Explanation: 35⋅1018+25⋅918=4890=815\frac35\cdot\frac{10}{18}+\frac25\cdot\frac9{18}=\frac{48}{90}=\frac{8}{15}; 23.

Question 9

Mathematics · 4 marks · −1

If the mean and the variance of the data: Class 4$-8,88, 8-12,1212, 12-16,1616, 16-$20 with Frequency 3, λ\lambda, 4, 7 are μ\mu and 19 respectively, then the value of λ+μ\lambda+\mu is

  1. A.21
  2. B.18
  3. C.19
  4. D.20

Correct answer: C

Question 10

Mathematics · 4 marks · −1

If the points of intersection of the ellipses x2+2y2−6x−12y+23=0x^2+2y^2-6x-12y+23=0 and 4x2+2y2−20x−12y+35=04x^2+2y^2-20x-12y+35=0 lie on a circle of radius r and centre (a,b)(a, b), then the value of ab+18r2ab+18r^2 is

  1. A.53
  2. B.51
  3. C.55
  4. D.52

Correct answer: C

Question 11

Mathematics · 4 marks · −1

An equilateral triangle OAB is inscribed in the parabola y2=4xy^2=4x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is

  1. A.4(6+3)4(6+\sqrt{3})
  2. B.4(3−3)4(3-\sqrt{3})
  3. C.2(8−33)2(8-3\sqrt{3})
  4. D.2(3+3)2(3+\sqrt{3})

Correct answer: B

Explanation: A, B =(12,±43)=(12,\pm4\sqrt3); circle centre (12,0), radius 434\sqrt3; minimum distance 12−43=4(3−3)12-4\sqrt3=4(3-\sqrt3).

Question 12

Mathematics · 4 marks · −1

Let PQ be a chord of the hyperbola x24−y2b2=1\frac{x^2}{4}-\frac{y^2}{b^2}=1, perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is 3\sqrt{3}, then the area of the triangle OPQ is

  1. A.115\frac{11}{5}
  2. B.95\frac{9}{5}
  3. C.835\frac{8\sqrt{3}}{5}
  4. D.232\sqrt{3}

Correct answer: C

Explanation: b2=8b^2=8; P =(h,h/3)=(h,h/\sqrt3): h24−h224=1⇒h2=245\frac{h^2}{4}-\frac{h^2}{24}=1 \Rightarrow h^2=\frac{24}{5}; area =h23=835=\frac{h^2}{\sqrt3}=\frac{8\sqrt3}{5}.

Question 13

Mathematics · 4 marks · −1

Let A(1, 2) and C($-$3, $-$6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line 7x−y=147x-y=14. If B(α,β)(\alpha, \beta) and D(γ,δ)(\gamma, \delta) are the other two vertices, then ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta| is equal to

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

Correct answer: A

Explanation: B + D = A + C = (-2, -4), so the sum is 6.

Question 14

Mathematics · 4 marks · −1

Let π2<θ<π\frac{\pi}{2}<\theta<\pi and cot⁡θ=−122\cot\theta=-\frac{1}{2\sqrt{2}}. Then the value of sin⁡(15θ2)(cos⁡8θ+sin⁡8θ)+cos⁡(15θ2)(cos⁡8θ−sin⁡8θ)\sin\left(\frac{15\theta}{2}\right)(\cos8\theta+\sin8\theta)+\cos\left(\frac{15\theta}{2}\right)(\cos8\theta-\sin8\theta) is equal to

  1. A.2−13\frac{\sqrt{2}-1}{\sqrt{3}}
  2. B.−23-\frac{\sqrt{2}}{\sqrt{3}}
  3. C.1−23\frac{1-\sqrt{2}}{\sqrt{3}}
  4. D.23\frac{\sqrt{2}}{\sqrt{3}}

Correct answer: C

Question 15

Mathematics · 4 marks · −1

Let a⃗=i^−2j^+3k^\vec{a}=\hat{i}-2\hat{j}+3\hat{k}, b⃗=2i^+j^−k^\vec{b}=2\hat{i}+\hat{j}-\hat{k}, c⃗=λi^+j^+k^\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k} and v⃗=a⃗×b⃗\vec{v}=\vec{a}\times\vec{b}. If v⃗⋅c⃗=11\vec{v}\cdot\vec{c}=11 and the length of the projection of b⃗\vec{b} on c⃗\vec{c} is p, then 9p29p^2 is equal to

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

Correct answer: A

Question 16

Mathematics · 4 marks · −1

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three vectors such that a⃗×b⃗=2(a⃗×c⃗)\vec{a}\times\vec{b}=2(\vec{a}\times\vec{c}). If ∣a⃗∣=1|\vec{a}|=1, ∣b⃗∣=4|\vec{b}|=4, ∣c⃗∣=2|\vec{c}|=2, and the angle between b⃗\vec{b} and c⃗\vec{c} is 60∘60^\circ, then ∣a⃗⋅c⃗∣|\vec{a}\cdot\vec{c}| is equal to

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

Correct answer: D

Question 17

Mathematics · 4 marks · −1

If f(x)={a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x,x≠0b,x=0f(x)=\begin{cases}\frac{a|x|+x^2-2(\sin|x|)(\cos|x|)}{x}, & x\ne0\\ b, & x=0\end{cases} is continuous at x=0x=0, then a+ba+b is equal to

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

Correct answer: D

Explanation: Near 0 the numerator is (a−2)∣x∣+O(x2)(a-2)|x|+O(x^2); continuity needs a=2a=2, and then the limit is 0, so b=0b=0: a+b=2a+b=2.

Question 18

Mathematics · 4 marks · −1

The least value of (cos⁡2θ−6sin⁡θcos⁡θ+3sin⁡2θ+2)(\cos^2\theta-6\sin\theta\cos\theta+3\sin^2\theta+2) is

  1. A.-1
  2. B.4+104+\sqrt{10}
  3. C.4−104-\sqrt{10}
  4. D.1

Correct answer: C

Explanation: =4−cos⁡2θ−3sin⁡2θ=4-\cos2\theta-3\sin2\theta, minimum 4−104-\sqrt{10}.

Question 19

Mathematics · 4 marks · −1

Let I(x)=∫3dx(4x+6)(4x2+8x+3)I(x)=\int\frac{3dx}{(4x+6)\left(\sqrt{4x^2+8x+3}\right)} and I(0)=34+20I(0)=\frac{\sqrt{3}}{4}+20. If I(12)=a2b+cI\left(\frac{1}{2}\right)=\frac{a\sqrt{2}}{b}+c, where a,b,c∈Na, b, c\in\mathbb{N}, gcd⁡(a,b)=1\gcd(a, b)=1, then a+b+ca+b+c is equal to

  1. A.29
  2. B.28
  3. C.30
  4. D.31

Correct answer: D

Question 20

Mathematics · 4 marks · −1

The area of the region enclosed between the circles x2+y2=4x^2+y^2=4 and x2+(y−2)2=4x^2+(y-2)^2=4 is:

  1. A.43(2π−33)\frac{4}{3}(2\pi-3\sqrt{3})
  2. B.23(4π−33)\frac{2}{3}(4\pi-3\sqrt{3})
  3. C.43(2π−3)\frac{4}{3}(2\pi-\sqrt{3})
  4. D.23(2π−33)\frac{2}{3}(2\pi-3\sqrt{3})

Correct answer: B

Explanation: Common chord y=1y=1; lens area =2⋅r22(2π3−sin⁡2π3)=4(2π3−32)=8π3−23=23(4π−33)=2\cdot\frac{r^2}{2}\left(\frac{2\pi}{3}-\sin\frac{2\pi}{3}\right)=4\left(\frac{2\pi}{3}-\frac{\sqrt3}{2}\right)=\frac{8\pi}{3}-2\sqrt3=\frac23(4\pi-3\sqrt3).

Question 21

Mathematics · 4 marks · −1

Let A=[02−3−2013−10]A=\begin{bmatrix}0 & 2 & -3\\ -2 & 0 & 1\\ 3 & -1 & 0\end{bmatrix} and B be a matrix such that B(I−A)=I+AB(I-A)=I+A. Then the sum of the diagonal elements of BTBB^TB is equal to ______

Correct answer: 3

Explanation: A is skew-symmetric, so B=(I+A)(I−A)−1B=(I+A)(I-A)^{-1} is orthogonal: BTB=IB^TB=I, trace 3.

Question 22

Mathematics · 4 marks · −1

Let S denote the set of 4-digit numbers abcd such that a>b>c>da>b>c>d and P denote the set of 5-digit numbers having product of its digits equal to 20. Then n(S)+n(P)n(S)+n(P) is equal to ______

Correct answer: 260

Explanation: n(S)=(104)=210n(S)=\binom{10}{4}=210. Digits {4,5,1,1,1}: 20; {2,2,5,1,1}: 30; total 50. Sum 260.

Question 23

Mathematics · 4 marks · −1

If the image of the point P(a,2,a)(a, 2, a) in the line x2=y+a1=z1\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1} is Q and the image of Q in the line x−2b2=y−a1=z+2b−5\frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5} is P, then a+ba+b is equal to ______.

Correct answer: 3

Question 24

Mathematics · 4 marks · −1

The number of elements in the set S={x:x∈[0,100] and ∫0xt2sin⁡(x−t)dt=x2}S=\left\{x: x\in[0, 100]\text{ and }\int_0^x t^2\sin(x-t)dt=x^2\right\} is ______

Correct answer: 16

Explanation: The integral equals x2−2+2cos⁡xx^2-2+2\cos x, so cos⁡x=1\cos x=1: x=2kπ≤100x=2k\pi\le100, k=0,…,15k=0,\ldots,15: 16.

Question 25

Mathematics · 4 marks · −1

If the solution curve y=f(x)y=f(x) of the differential equation (x2−4)y′−2xy+2x(4−x2)2=0(x^2-4)y'-2xy+2x(4-x^2)^2=0, x>2x>2, passes through the point (3,15)(3, 15), then the local maximum value of ff is ______

Correct answer: 16

Question 26

Physics · 4 marks · −1

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s2^2. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the airplane is ______ m. (g=10g=10 m/s2^2)

  1. A.82.5
  2. B.92.5
  3. C.62.5
  4. D.20

Correct answer: B

Explanation: Free fall 20 m, speed 20 m/s; decelerating to 5 m/s: 400−256=62.5\frac{400-25}{6}=62.5 m; total 20+62.5+10=92.520+62.5+10=92.5 m.

Question 27

Physics · 4 marks · −1

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t=0t=0 and at this instant the horizontal component of its velocity is vv. Another bead Q of the same mass as P is ejected from point A at t=0t=0 along the horizontal string AB, with the speed vv, friction between the beads and the respective strings may be neglected in both cases. Let tPt_P and tQt_Q be the respective times taken by beads P and Q to reach the point B, then the relation between tPt_P and tQt_Q is

Figure for question 27
  1. A.tP>tQt_P>t_Q
  2. B.tP>1.25tQt_P>1.25t_Q
  3. C.tP=tQt_P=t_Q
  4. D.tP<tQt_P<t_Q

Correct answer: D

Explanation: P gains speed as it descends, so its horizontal velocity component stays above vv through most of the path; it reaches B first.

Question 28

Physics · 4 marks · −1

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ______ m/s.

  1. A.12212\sqrt{2}
  2. B.12
  3. C.10210\sqrt{2}
  4. D.24224\sqrt{2}

Correct answer: A

Explanation: Masses 4, 4, 6 kg: 6v=4×182⇒v=1226v=4\times18\sqrt2 \Rightarrow v=12\sqrt2.

Question 29

Physics · 4 marks · −1

A small metallic sphere of diameter 2 mm and density 10.5 g/cm3^3 is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm3^3 respectively. The terminal velocity attained by the sphere is ______ cm/s. (π=227\pi=\frac{22}{7} and g=10g=10 m/s2^2)

  1. A.1.5
  2. B.2.0
  3. C.3.0
  4. D.1.0

Correct answer: B

Explanation: v=2r2(ρ−σ)g9η=2(10−3)2(9000)(10)9×1=0.02v=\frac{2r^2(\rho-\sigma)g}{9\eta}=\frac{2(10^{-3})^2(9000)(10)}{9\times1}=0.02 m/s = 2 cm/s.

Question 30

Physics · 4 marks · −1

An air bubble of volume 2.9 cm3^3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17∘17^\circC. The volume of the bubble when it reaches the surface, where the water temperature is 27∘27^\circC, is ______ cm3^3. (g=10g=10 m/s2^2, density of water =103=10^3 kg/m3^3, and 1 atm pressure is 10510^5 Pa)

  1. A.2.0
  2. B.3.0
  3. C.4.5
  4. D.4.2

Correct answer: C

Explanation: V=2.9×1.5×300290=4.5V=2.9\times1.5\times\frac{300}{290}=4.5 cm3^3.

Question 31

Physics · 4 marks · −1

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27∘27^\circC and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27∘27^\circC doing the same amount of work W, then its final temperature will be (close to) ______ ∘^\circC. (log⁡e2=0.693\log_e2=0.693)

  1. A.-56
  2. B.-30
  3. C.-189
  4. D.-117

Correct answer: A

Explanation: RTln⁡2=52RΔT⇒ΔT=83RT\ln2=\frac52R\Delta T \Rightarrow \Delta T=83 K; final ≈217\approx217 K =−56∘=-56^\circC.

Question 32

Physics · 4 marks · −1

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 cm2^2 and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4∘4^\circC, then the piston will move ______ cm. (atmospheric pressure =105=10^5 Pa)

  1. A.15.5
  2. B.1.55
  3. C.1.45
  4. D.14.5

Correct answer: A

Explanation: ΔU=3R×4=99.8\Delta U=3R\times4=99.8 J; work =26.2=26.2 J =105×17×10−4×x⇒x=0.155=10^5\times17\times10^{-4}\times x \Rightarrow x=0.155 m = 15.5 cm.

Question 33

Physics · 4 marks · −1

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25% more charge from the battery. The dielectric constant of mica is ______.

  1. A.2.0
  2. B.1.0
  3. C.1.5
  4. D.2.5

Correct answer: A

Explanation: 53+2/K=1.25⇒K=2\frac{5}{3+2/K}=1.25 \Rightarrow K=2.

Question 34

Physics · 4 marks · −1

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I=10sin⁡(ωt)I=10\sin(\omega t) A, where ω=1000\omega=1000 rad/s. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v=1v=1 cm/s. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/2\alpha/\sqrt{2} μ\muA. The value of α\alpha is ______. [Resistance of the loop = 10 Ω\Omega]

  1. A.197
  2. B.100
  3. C.80
  4. D.280

Correct answer: A

Explanation: ε0=μ0nI0ωπr2=4π×10−7×500×10×1000×π×10−4=1.97×10−3\varepsilon_0=\mu_0n I_0\omega\pi r^2=4\pi\times10^{-7}\times500\times10\times1000\times\pi\times10^{-4}=1.97\times10^{-3} V; I0=197 μI_0=197\ \muA.

Question 35

Physics · 4 marks · −1

The current passing through a conducting loop in the form of equilateral triangle of side 434\sqrt{3} cm is 2 A. The magnetic field at its centroid is α×10−5\alpha\times10^{-5} T. The value of α\alpha is ______. (Given : μ0=4π×10−7\mu_0=4\pi\times10^{-7} SI units)

  1. A.333\sqrt{3}
  2. B.232\sqrt{3}
  3. C.3\sqrt{3}
  4. D.32\frac{\sqrt{3}}{2}

Correct answer: A

Explanation: Distance to each side 2 cm; each side gives μ0I4πd(2sin⁡60∘)=3×10−5\frac{\mu_0I}{4\pi d}(2\sin60^\circ)=\sqrt3\times10^{-5}; total 33×10−53\sqrt3\times10^{-5} T.

Question 36

Physics · 4 marks · −1

Two shorts dipoles (A, B), A having charges ±2\pm2 μ\muC and length 1 cm and B having charges ±4\pm4 μ\muC and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point P, equi-distant from the centres of both dipoles is ______ N/C.

Figure for question 36
  1. A.4.52×1044.5\sqrt{2}\times10^4
  2. B.92×1049\sqrt{2}\times10^4
  3. C.9162×105\frac{9}{16}\sqrt{2}\times10^5
  4. D.9162×104\frac{9}{16}\sqrt{2}\times10^4

Correct answer: D

Explanation: A (axial, p=2×10−8p=2\times10^{-8}): 2kpr3=2⋅9×109⋅2×10−80.064=5.625×103\frac{2kp}{r^3}=\frac{2\cdot9\times10^9\cdot2\times10^{-8}}{0.064}=5.625\times10^3; B (equatorial, p=4×10−8p=4\times10^{-8}): kpr3=5.625×103\frac{kp}{r^3}=5.625\times10^3; perpendicular, resultant 9162×104\frac{9}{16}\sqrt2\times10^4 N/C.

Question 37

Physics · 4 marks · −1

Two charges 7 μ\muC and −2-2 μ\muC are placed at (−9,0,0)(-9, 0, 0) cm and (9,0,0)(9, 0, 0) cm respectively in an external field E=Ar2r^E=\frac{A}{r^2}\hat{r}, where A=9×105A=9\times10^5 N/C.m2^2. Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______ J.

  1. A.24.3
  2. B.49.3
  3. C.-90.7
  4. D.1.4

Correct answer: B

Explanation: External potential V=A/r=107V=A/r=10^7 V at r=0.09r=0.09 m: (7−2)×10−6×107=50(7-2)\times10^{-6}\times10^7=50 J; mutual 9×109(7)(−2)×10−120.18=−0.7\frac{9\times10^9(7)(-2)\times10^{-12}}{0.18}=-0.7 J; total 49.3 J.

Question 38

Physics · 4 marks · −1

A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is ______ mV.

  1. A.13.2
  2. B.8.25
  3. C.6.6
  4. D.1.32

Correct answer: D

Explanation: Perimeter 44 cm; areas 154 and 121 cm2^2; ε=0.2×33×10−40.5=1.32\varepsilon=\frac{0.2\times33\times10^{-4}}{0.5}=1.32 mV.

Question 39

Physics · 4 marks · −1

The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant k=3k=3 and permeability of μ=2μ0\mu=2\mu_0, is (μ0\mu_0 = permeability of vacuum)

  1. A.6:1\sqrt{6}:1
  2. B.6:1
  3. C.36:1
  4. D.3:2

Correct answer: A

Question 40

Physics · 4 marks · −1

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is ______. (tan⁡−1(1.52)=57.7∘\tan^{-1}(1.52)=57.7^\circ, refractive indices of air and glass are 1.00 and 1.52, respectively.)

  1. A.39.6∘39.6^\circ
  2. B.32.3∘32.3^\circ
  3. C.42.6∘42.6^\circ
  4. D.36.3∘36.3^\circ

Correct answer: B

Explanation: r=90∘−57.7∘=32.3∘r=90^\circ-57.7^\circ=32.3^\circ.

Question 41

Physics · 4 marks · −1

Which of the following pair of nuclei are isobars of the element?

  1. A.92236^{236}_{92}U and 92238^{238}_{92}U
  2. B.12^{2}_{1}H and 13^{3}_{1}H
  3. C.13^{3}_{1}H and 23^{3}_{2}He
  4. D.80198^{198}_{80}Hg and 79197^{197}_{79}Au

Correct answer: C

Question 42

Physics · 4 marks · −1

For the given logic gate circuit, which of the following is the correct truth table ?

Figure for question 42
  1. A.n m z: 0 0 0; 0 1 1; 1 1 0; 1 0 1
  2. B.0 0 1; 0 1 1; 1 1 0; 1 0 0
  3. C.0 0 1; 0 1 0; 1 1 1; 1 0 0
  4. D.0 0 1; 0 1 0; 1 1 0; 1 0 0

Correct answer: B

Explanation: n(n+m)=nn(n+m)=n, so z=nˉz=\bar n.

Question 43

Physics · 4 marks · −1

Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x256Mr2\frac{x}{256}Mr^2. The value of xx is ______.

Figure for question 43

Correct answer: 109

Explanation: Each hole: mass M/16M/16, I=M16(r232+9r216)=19512Mr2I=\frac{M}{16}\left(\frac{r^2}{32}+\frac{9r^2}{16}\right)=\frac{19}{512}Mr^2; remainder 128−19256Mr2\frac{128-19}{256}Mr^2, x=109x=109.

Question 44

Physics · 4 marks · −1

A ball of radius r and density ρ\rho dropped through a viscous liquid of density σ\sigma and viscosity η\eta attains its terminal velocity at time t, given by t=Aρarbηcσdt=A\rho^ar^b\eta^c\sigma^d, where A is a constant and a, b, c and d are integers. The value of b+ca+d\frac{b+c}{a+d} is ______.

Correct answer: 1

Explanation: t∼ρr2ηt\sim\frac{\rho r^2}{\eta}: a=1,b=2,c=−1,d=0a=1,b=2,c=-1,d=0; 11=1\frac{1}{1}=1.

Question 45

Physics · 4 marks · −1

The velocity of sound in air is doubled when the temperature is raised from 0∘0^\circC to α∘\alpha^\circC. The value of α\alpha is ______.

Correct answer: 819

Explanation: T=4×273=1092T=4\times273=1092 K, i.e. 819∘819^\circC.

Question 46

Physics · 4 marks · −1

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm.

Correct answer: 16

Explanation: ff−8=−ff−24⇒f=16\frac{f}{f-8}=-\frac{f}{f-24} \Rightarrow f=16 cm.

Question 47

Physics · 4 marks · −1

The average energy released per fission for the nucleus of 92235^{235}_{92}U is 190 MeV. When all the atoms of 47 g pure 92235^{235}_{92}U undergo fission process, the energy released is α×1023\alpha\times10^{23} MeV. The value of α\alpha is ______. (Avogadro Number =6×1023=6\times10^{23} per mole)

Correct answer: 228

Explanation: 0.2 mol ×6×1023×190=228×1023\times6\times10^{23}\times190=228\times10^{23} MeV.

Question 48

Chemistry · 4 marks · −1

The work functions of two metals (MA_A and MB_B) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA_A : MB_B is in the ratio of 2.642 : 1. The work functions (in eV) of MA_A and MB_B are respectively.

  1. A.2.3, 4.6
  2. B.3.1, 6.2
  3. C.1.4, 2.8
  4. D.1.5, 3.0

Correct answer: A

Explanation: 6−ϕ6−2ϕ=2.642⇒ϕ=2.3\frac{6-\phi}{6-2\phi}=2.642 \Rightarrow \phi=2.3 eV.

Question 49

Chemistry · 4 marks · −1

Identify the INCORRECT statements from the following: A. Notation 1224^{24}_{12}Mg represents 24 protons and 12 neutrons. B. Wavelength of a radiation of frequency 4.5×10154.5\times10^{15} s−1^{-1} is 6.7×10−86.7\times10^{-8} m. C. One radiation has wavelength = λ1\lambda_1 (900 nm) and energy = E1_1. Other radiation has wavelength = λ2\lambda_2 (300 nm) and energy = E2_2. E1_1 : E2_2 = 3 : 1. D. Number of photons of light of wavelength 2000 pm that provides 1 J of energy is 1.006×10161.006\times10^{16}. Choose the correct answer from the options given below:

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

Correct answer: B

Question 50

Chemistry · 4 marks · −1

Which statements are NOT TRUE about XeO2_2F2_2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO2_2F2_2. C. The O−-Xe−-O bond angle is close to 180∘180^\circ. D. The F−-Xe−-F bond angle is close to 180∘180^\circ. E. Xe has 16 valence electrons in XeO2_2F2_2. Choose the correct answer from the options given below:

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

Correct answer: B

Question 51

Chemistry · 4 marks · −1

It is noticed that Pb2+^{2+} is more stable than Pb4+^{4+} but Sn2+^{2+} is less stable than Sn4+^{4+}. Observe the following reactions. PbO2_2 + Pb →\to 2PbO; ΔrG∘(1)\Delta_rG^\circ(1). SnO2_2 + Sn →\to 2SnO; ΔrG∘(2)\Delta_rG^\circ(2). Identify the correct set from the following

  1. A.ΔrG∘(1)>0\Delta_rG^\circ(1)>0; ΔrG∘(2)<0\Delta_rG^\circ(2)<0
  2. B.ΔrG∘(1)<0\Delta_rG^\circ(1)<0; ΔrG∘(2)>0\Delta_rG^\circ(2)>0
  3. C.ΔrG∘(1)>0\Delta_rG^\circ(1)>0; ΔrG∘(2)>0\Delta_rG^\circ(2)>0
  4. D.ΔrG∘(1)<0\Delta_rG^\circ(1)<0; ΔrG∘(2)<0\Delta_rG^\circ(2)<0

Correct answer: B

Question 52

Chemistry · 4 marks · −1

Consider the electrochemical cell shown: two identical metal electrodes M dip into MX solutions separated by a semi permeable membrane; the left compartment has M+^+ at concentration c2c_2 and the right compartment M+^+ at concentration c1c_1. A metal electrode (M) is undergoing redox reaction by forming M+^+ (M →\to M+^+ + e−^-). The cation M+^+ is present in two different concentrations c1c_1 and c2c_2 as shown above. Which of the following statement is correct for generating a positive cell potential?

Figure for question 52
  1. A.If c1c_1 is present at anode, then c1>c2c_1>c_2.
  2. B.If c1c_1 is present at cathode, then c1>c2c_1>c_2.
  3. C.If c1c_1 is present at anode, then c1=c2c_1=c_2.
  4. D.If c1c_1 is present at cathode, then c1<c2c_1<c_2.

Correct answer: B

Explanation: A concentration cell has E>0E>0 when the cathode compartment is the more concentrated one.

Question 53

Chemistry · 4 marks · −1

Given above is the concentration vs time plot for a dissociation reaction : A →\to nB. Based on the data of the initial phase of the reaction (initial 10 min), the value of n is ______.

Figure for question 53
  1. A.4
  2. B.5
  3. C.2
  4. D.3

Correct answer: D

Explanation: In the first 10 min Δ[A]≈−0.01\Delta[A]\approx-0.01 M while Δ[B]≈+0.03\Delta[B]\approx+0.03 M, so n = 3.

Question 54

Chemistry · 4 marks · −1

Observe the following reactions at T(K). I. A →\to products. II. 5Br−^-(aq) + BrO3−_3^-(aq) + 6H+^+(aq) →\to 3Br2_2(aq) + 3H2_2O(ll). Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of −Δ[Br−]Δt-\frac{\Delta[Br^-]}{\Delta t} at 10.10 am is 2×10−42\times10^{-4} mol L−1^{-1} min−1^{-1}. The concentration of A at 10.10 am is 10−210^{-2} mol L−1^{-1}. What is the first order rate constant (in min−1^{-1}) of reaction I?

  1. A.2×10−32\times10^{-3}
  2. B.4×10−34\times10^{-3}
  3. C.10−210^{-2}
  4. D.10−310^{-3}

Correct answer: B

Explanation: Rate of II =15×2×10−4=4×10−5=\frac15\times2\times10^{-4}=4\times10^{-5}; k=4×10−5/10−2=4×10−3k=4\times10^{-5}/10^{-2}=4\times10^{-3} min−1^{-1}.

Question 55

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II: The ionic radius of O2−^{2-} is larger than that of F−^-. 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

Question 56

Chemistry · 4 marks · −1

Elements X and Y belong to Group 15. The difference between the electronegativity values of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'. 'X' & 'Y' are respectively

  1. A.As & Bi
  2. B.Bi & N
  3. C.As & Sb
  4. D.N & As

Correct answer: D

Question 57

Chemistry · 4 marks · −1

The oxidation state of chromium in the final product formed in the reaction between KI and acidified K2_2Cr2_2O7_7 solution is:

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

Correct answer: A

Question 58

Chemistry · 4 marks · −1

Identify the CORRECT set of details from the following: A. [Co(NH3_3)6_6]3+^{3+} : Inner orbital complex; d2^2sp3^3 hybridized B. [MnCl6_6]3−^{3-} : Outer orbital complex; sp3^3d2^2 hybridized C. [CoF6_6]3−^{3-} : Outer orbital complex; d2^2sp3^3 hybridized D. [FeF6_6]3−^{3-} : Outer orbital complex; sp3^3d2^2 hybridized E. [Ni(CN)4_4]2−^{2-} : Inner orbital complex; sp3^3 hybridized. Choose the correct answer from the options given below:

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

Correct answer: C

Question 59

Chemistry · 4 marks · −1

In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is

  1. A.37.57%
  2. B.34.79%
  3. C.53.58%
  4. D.87.65%

Correct answer: C

Explanation: 35.5143.5×0.52530.2425×100=53.58%\frac{35.5}{143.5}\times\frac{0.5253}{0.2425}\times100=53.58\%.

Question 60

Chemistry · 4 marks · −1

Given below are two statements: Statement I: (CH3)3C+(CH_3)_3\overset{+}{C} is more stable than C+H3\overset{+}{C}H_3 as nine hyperconjugation interactions are possible in (CH3)3C+(CH_3)_3\overset{+}{C}. Statement II: C+H3\overset{+}{C}H_3 is less stable than (CH3)3C+(CH_3)_3\overset{+}{C} as only three hyperconjugation interactions are possible in C+H3\overset{+}{C}H_3. 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 false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: D

Explanation: CH3+_3^+ has no C−-H bond beta to the cationic centre, so no hyperconjugation: II is false.

Question 61

Chemistry · 4 marks · −1

A vicinal dibromide on a substituted cyclopentane (shown in figure) is treated with (i) Zn, Δ\Delta (ii) HBr to give the major product (P). Identify (P)

Figure for question 61
  1. A.Tertiary bromide: Br on the substituted ring carbon (see figure)
  2. B.Br on the side chain carbon next to the ring (see figure)
  3. C.Br on a secondary ring carbon (see figure)
  4. D.Br on the ring carbon between the two alkyl groups (see figure)

Correct answer: A

Explanation: Zn debromination regenerates the alkene; Markovnikov addition of HBr places Br on the more substituted (tertiary) carbon.

Question 62

Chemistry · 4 marks · −1

Which of the following statements are TRUE about Haloform reaction?: A. Sodium hypochlorite reacts with KI to give KOI. B. KOI is a reducing agent. C. α,β\alpha,\beta-unsaturated methylketone (CH3−_3-CH=CH−-CO−-CH3_3) will give iodoform reaction. D. Isopropyl alcohol will not give iodoform test. E. Methanoic acid will give positive iodoform test. Choose the correct answer from the options given below:

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

Correct answer: C

Question 63

Chemistry · 4 marks · −1

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

  1. A.CH3_3CH2−_2-O−-C(CH3_3)2_2CH2_2CH3_3
  2. B.CH3_3CH2−_2-O−-CH(CH3_3)CH(CH3_3)2_2 (ethyl 3-methylbutan-2-yl ether)
  3. C.CH3_3CH2−_2-O−-CH(Ph)CH(CH3_3)2_2
  4. D.PhCH2−_2-O−-CH(CH3_3)2_2

Correct answer: B

Explanation: Ethanol and 3-methylbutan-2-ol both give the iodoform test.

Question 64

Chemistry · 4 marks · −1

Given below are two statements: Statement I: benzene-1,2-diamine can be synthesized from 1,2-dipropylbenzene (C6_6H4_4(CH2_2CH2_2CH3_3)2_2, ortho) using simpler reagents in the order i) Acidic KMnO4_4, ii) Ammonia, iii) Bromine and alkali. Statement II: 4-methylaniline (pp-toluidine) can be converted into 3,5-dibromotoluene using reagents in the order i) Bromine-H2_2O ii) NaNO2_2/HCl (0 - 5 ∘^\circC) (iii) Aq. H3_3PO2_2. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: D

Question 65

Chemistry · 4 marks · −1

A student has been given a compound \"x\" of molecular formula- C6_6H7_7N. 'x' is sparingly soluble in water. However, on addition of dilute mineral acid, 'x' becomes soluble in water. 'x' when treated with CHCl3_3 and KOH(alc), 'y' is produced. 'y' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, 'x' gives a compound 'z' which is soluble in alkali. The number of different \"H\" atoms present in 'z' is:-

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

Correct answer: D

Explanation: z = C6_6H5_5SO2_2NHC6_6H5_5: 3 + 3 aromatic kinds + N−-H = 7.

Question 66

Chemistry · 4 marks · −1

Both human DNA and RNA are chiral molecules. The chirality in DNA and RNA arises due to the presence of

  1. A.Base unit
  2. B.Chiral phosphate unit
  3. C.L−-sugar component
  4. D.D−-sugar component

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Iodoform test can differentiate between A. Methanol and Ethanol B. CH3_3COOH and CH3_3CH2_2COOH C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone. Choose the correct answer from the options given below:

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

Correct answer: C

Question 68

Chemistry · 4 marks · −1

Total number of unpaired electrons present in the central metal atoms/ions of [Ni(CO)4_4], [NiCl4_4]2−^{2-}, [PtCl2_2(NH3_3)2_2], [Ni(CN)4_4]2−^{2-} and [Pt(CN)4_4]2−^{2-} is ______.

Correct answer: 2

Explanation: Only [NiCl4_4]2−^{2-} (2 unpaired): total 2.

Question 69

Chemistry · 4 marks · −1

Consider the following reaction of benzene: Benzene + H3_3C−-CO−-CH2−_2-CH2−_2-CO−-Cl →anhydrous AlCl3\xrightarrow{\text{anhydrous AlCl}_3} (P) →aq NaOH, Δ\xrightarrow{\text{aq NaOH, }\Delta} (Q). In compound (Q), the percentage of oxygen is ______%. (Nearest integer)

Correct answer: 10

Explanation: P = PhCOCH2_2CH2_2COCH3_3; intramolecular aldol condensation gives 3-phenylcyclopent-2-en-1-one C11_{11}H10_{10}O (158): %O = 10.

Question 70

Chemistry · 4 marks · −1

200 cc of x×10−3x\times10^{-3} M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here xx = ______.

Correct answer: 375

Explanation: 6×0.2x×10−3=0.45⇒x=3756\times0.2x\times10^{-3}=0.45 \Rightarrow x=375.

Question 71

Chemistry · 4 marks · −1

X2_2(g) + Y2_2(g) ⇌\rightleftharpoons 2Z(g). X2_2(g) and Y2_2(g) are added to a 1 L flask and it is found that the system attains the above equilibrium at T(K) with the number of moles of X2_2(g), Y2_2(g) and Z(g) being 3, 3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of Z(g) is added to the flask and the temperature is maintained at T(K). Then the number of moles of Z(g) in the flask when the new equilibrium is established is ______. (Nearest integer)

Correct answer: 15

Explanation: K=9K=9; (19−2x)2(3+x)2=9⇒x=2\frac{(19-2x)^2}{(3+x)^2}=9 \Rightarrow x=2; Z = 15 mol.

Question 72

Chemistry · 4 marks · −1

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is ______. (Nearest integer)

Correct answer: 200

Explanation: 34pA+14pB=500\frac34p_A+\frac14p_B=500, 45pA+15pB=520⇒pA=600\frac45p_A+\frac15p_B=520 \Rightarrow p_A=600, pB=200p_B=200.

View all papers — JEE Main

Practice this paper with a timer

Attempt it online, get scored instantly, and see which subjects are costing you marks.

Start practising free