JEE Main 2025 — 7 Apr Shift 1 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

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

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The integral ∫0π(x+3)sin⁡x1+3cos⁡2xdx\int_0^{\pi}\frac{(x+3)\sin x}{1+3\cos^2x}dx is equal to

  1. A.π23(π+4)\frac{\pi}{2\sqrt{3}}(\pi+4)
  2. B.π3(π+2)\frac{\pi}{\sqrt{3}}(\pi+2)
  3. C.π3(π+1)\frac{\pi}{\sqrt{3}}(\pi+1)
  4. D.π33(π+6)\frac{\pi}{3\sqrt{3}}(\pi+6)

Correct answer: D

Explanation: ∫0πsin⁡x1+3cos⁡2xdx=2π33\int_0^\pi\frac{\sin x}{1+3\cos^2x}dx=\frac{2\pi}{3\sqrt3}; ∫x(…)=π2⋅2π33\int x(\ldots)=\frac\pi2\cdot\frac{2\pi}{3\sqrt3}; total 2π33(π2+3)=π33(π+6)\frac{2\pi}{3\sqrt3}(\frac\pi2+3)=\frac{\pi}{3\sqrt3}(\pi+6).

Question 2

Mathematics · 4 marks · −1

Let the set of all values of p∈Rp\in\mathbf{R}, for which both the roots of the equation x2−(p+2)x+(2p+9)=0x^2-(p+2)x+(2p+9)=0 are negative real numbers, be the interval (α,β](\alpha, \beta]. Then β−2α\beta-2\alpha is equal to

  1. A.20
  2. B.9
  3. C.0
  4. D.5

Correct answer: D

Explanation: Sum p+2<0p+2<0, product 2p+9>02p+9>0, D≥0D\ge0: (p+2)2≥4(2p+9)(p+2)^2\ge4(2p+9) ⇒ p≤−4p\le-4 or p≥8p\ge8. So p∈(−92,−4]p\in(-\frac92,-4]: −4+9=5-4+9=5.

Question 3

Mathematics · 4 marks · −1

If for θ∈[−π3,0]\theta\in\left[-\frac{\pi}{3}, 0\right], the points (x,y)=(3tan⁡(θ+π3),2tan⁡(θ+π6))(x, y)=\left(3\tan\left(\theta+\frac{\pi}{3}\right), 2\tan\left(\theta+\frac{\pi}{6}\right)\right) lie on xy+αx+βy+γ=0xy+\alpha x+\beta y+\gamma=0, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is equal to

  1. A.96
  2. B.75
  3. C.72
  4. D.80

Correct answer: B

Question 4

Mathematics · 4 marks · −1

Let A be a 3 × 3 matrix such that |adj (adj (adj A))| = 81. If S={n∈Z:(∣adj(adjA)∣)(n−1)22=∣A∣(3n2−5n−4)}S=\left\{n\in\mathbf{Z}: \left(|\mathrm{adj}(\mathrm{adj}A)|\right)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\right\}, then ∑n∈S∣A(n2+n)∣\sum_{n\in S}\left|A^{(n^2+n)}\right| is equal to

  1. A.820
  2. B.866
  3. C.732
  4. D.750

Correct answer: C

Question 5

Mathematics · 4 marks · −1

lim⁡x→0+tan⁡(5(x)1/3)log⁡e(1+3x2)(tan⁡−13x)2(e5(x)4/3−1)\lim_{x\to0^+}\frac{\tan\left(5(x)^{1/3}\right)\log_e(1+3x^2)}{\left(\tan^{-1}3\sqrt{x}\right)^2\left(e^{5(x)^{4/3}}-1\right)} is equal to

  1. A.1
  2. B.53\frac{5}{3}
  3. C.13\frac{1}{3}
  4. D.115\frac{1}{15}

Correct answer: C

Explanation: 5x1/3⋅3x29x⋅5x4/3=13\frac{5x^{1/3}\cdot3x^2}{9x\cdot5x^{4/3}}=\frac13.

Question 6

Mathematics · 4 marks · −1

Let x = −1 and x = 2 be the critical points of the function f(x)=x3+ax2+blog⁡e∣x∣+1f(x)=x^3+ax^2+b\log_e|x|+1, x≠0x\neq0. Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval [−2,−12]\left[-2, -\frac{1}{2}\right]. Then ∣M+m∣|M+m| is equal to (Take log⁡e2=0.7\log_e2=0.7) :

  1. A.21.1
  2. B.20.9
  3. C.19.8
  4. D.22.1

Correct answer: A

Question 7

Mathematics · 4 marks · −1

Among the statements (S1) : The set {z∈C−{−i}:∣z∣=1\{z\in\mathbf{C}-\{-i\}: |z|=1 and z−iz+i\frac{z-i}{z+i} is purely real}\} contains exactly two elements, and (S2) : The set {z∈C−{−1}:∣z∣=1\{z\in\mathbf{C}-\{-1\}: |z|=1 and z−1z+1\frac{z-1}{z+1} is purely imaginary}\} contains infinitely many elements.

  1. A.only (S2) is correct
  2. B.only (S1) is correct
  3. C.both are correct
  4. D.both are incorrect

Correct answer: A

Explanation: On |z| = 1, z−1z+1=itan⁡θ2\frac{z-1}{z+1}=i\tan\frac{\theta}{2} is purely imaginary for every admissible z (S2 true); z−iz+i\frac{z-i}{z+i} is real only when z is on the imaginary axis, i.e. z = i alone (S1 false).

Question 8

Mathematics · 4 marks · −1

Let the line L pass through (1, 1, 1) and intersect the lines x−12=y+13=z−14\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} and x−31=y−42=z1\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}. Then, which of the following points lies on the line L?

  1. A.(7, 15, 13)
  2. B.(10, −29, −50)
  3. C.(5, 4, 3)
  4. D.(4, 22, 7)

Correct answer: A

Question 9

Mathematics · 4 marks · −1

Let y = y (x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0x(x^2+e^x)dy+(e^x(x-2)y-x^3)dx=0, x > 0, passing through the point (1, 0). Then y (2) is equal to

  1. A.44+e2\frac{4}{4+e^2}
  2. B.22−e2\frac{2}{2-e^2}
  3. C.44−e2\frac{4}{4-e^2}
  4. D.22+e2\frac{2}{2+e^2}

Correct answer: A

Question 10

Mathematics · 4 marks · −1

Let P be the parabola, whose focus is (−2, 1) and directrix is 2x + y + 2 = 0. Then the sum of the ordinates of the points on P, whose abscissa is −2, is

  1. A.14\frac{1}{4}
  2. B.52\frac{5}{2}
  3. C.34\frac{3}{4}
  4. D.32\frac{3}{2}

Correct answer: D

Explanation: x = −2: 5(y−1)2=(y−2)25(y-1)^2=(y-2)^2 ⇒ 4y2−6y+1=04y^2-6y+1=0: sum 3/2.

Question 11

Mathematics · 4 marks · −1

The remainder when ((64)(64))(64)\left((64)^{(64)}\right)^{(64)} is divided by 7 is equal to

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

Correct answer: B

Explanation: 64 ≡ 1 (mod 7).

Question 12

Mathematics · 4 marks · −1

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are μ\mu and σ\sigma respectively, then 10(μ+σ)10(\mu+\sigma) is equal to

  1. A.451
  2. B.445
  3. C.447
  4. D.449

Correct answer: D

Explanation: μ = 39.9; ∑x2=100(26.01+1600)−2500+1600=161701\sum x^2=100(26.01+1600)-2500+1600=161701; σ2=1617.01−1592.01=25\sigma^2=1617.01-1592.01=25: σ = 5; 10(44.9) = 449.

Question 13

Mathematics · 4 marks · −1

Let C1C_1 be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let C2C_2 be the circle with centre (1, 3) that touches C1C_1 externally at the point (α,β)(\alpha, \beta). If (β−α)2=mn(\beta-\alpha)^2=\frac{m}{n}, gcd(m, n) = 1, then m + n is equal to

  1. A.9
  2. B.31
  3. C.22
  4. D.13

Correct answer: C

Explanation: C1C_1: centre (−3, −3), r = 3; distance to (1, 3) is 52\sqrt{52}; contact point =(−3,−3)+352(4,6)=(-3,-3)+\frac{3}{\sqrt{52}}(4,6), so β−α=652\beta-\alpha=\frac{6}{\sqrt{52}} and (β−α)2=913(\beta-\alpha)^2=\frac{9}{13}: 22.

Question 14

Mathematics · 4 marks · −1

Let the system of equations : 2x + 3y + 5z = 9, 7x + 3y − 2z = 8, 12x + 3y − (4 + λ\lambda)z = 16 − μ\mu, have infinitely many solutions. Then the radius of the circle centred at (λ\lambda, μ\mu) and touching the line 4x = 3y is

  1. A.215\frac{21}{5}
  2. B.75\frac{7}{5}
  3. C.7
  4. D.175\frac{17}{5}

Correct answer: B

Question 15

Mathematics · 4 marks · −1

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

  1. A.165
  2. B.135
  3. C.155
  4. D.145

Correct answer: C

Question 16

Mathematics · 4 marks · −1

If the shortest distance between the lines x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and x1=yα=z−51\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} is 56\frac{5}{\sqrt{6}}, then the sum of all possible values of α\alpha is

  1. A.−3
  2. B.−32-\frac{3}{2}
  3. C.3
  4. D.32\frac{3}{2}

Correct answer: A

Question 17

Mathematics · 4 marks · −1

Let the angle θ\theta, 0<θ<π20<\theta<\frac{\pi}{2} between two unit vectors a^\hat{a} and b^\hat{b} be sin⁡−1(659)\sin^{-1}\left(\frac{\sqrt{65}}{9}\right). If the vector c⃗=3a^+6b^+9(a^×b^)\vec{c}=3\hat{a}+6\hat{b}+9(\hat{a}\times\hat{b}), then the value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c}\cdot\hat{a})-3(\vec{c}\cdot\hat{b}) is

  1. A.27
  2. B.29
  3. C.24
  4. D.31

Correct answer: B

Explanation: cos⁡θ=49\cos\theta=\frac49: 9(3+6⋅49)−3(3⋅49+6)=51−22=299(3+6\cdot\frac49)-3(3\cdot\frac49+6)=51-22=29.

Question 18

Mathematics · 4 marks · −1

Let x1,x2,x3,x4x_1, x_2, x_3, x_4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1,x2,x3,x4x_1, x_2, x_3, x_4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4)\frac{1}{24}(x_1x_2x_3x_4) is:

  1. A.72
  2. B.18
  3. C.36
  4. D.216

Correct answer: D

Question 19

Mathematics · 4 marks · −1

If the area of the region bounded by the curves y=4−x24y=4-\frac{x^2}{4} and y=x−42y=\frac{x-4}{2} is equal to α\alpha, then 6α6\alpha equals

  1. A.250
  2. B.210
  3. C.240
  4. D.220

Correct answer: A

Explanation: Intersections x = −6, 4; area ∫−64(6−x2−x24)dx=1253\int_{-6}^4\left(6-\frac x2-\frac{x^2}{4}\right)dx=\frac{125}{3}; 6α = 250.

Question 20

Mathematics · 4 marks · −1

Let ABC be the triangle such that the equations of lines AB and AC be 3y − x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to

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

Correct answer: C

Explanation: A(1,1), B(−2,0), C(2,0); orthocentre (1, 3); area 12⋅4⋅3=6\frac12\cdot4\cdot3=6.

Question 21

Mathematics · 4 marks · −1

Consider the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 having one of its focus at P(−3, 0). If the latus ractum through its other focus subtends a right angle at P and a2b2=α2−βa^2b^2=\alpha\sqrt{2}-\beta, α,β∈N\alpha, \beta\in\mathbf{N}, then α+β\alpha+\beta is ______.

Correct answer: 1944

Question 22

Mathematics · 4 marks · −1

For n≥2n\ge2, let SnS_n denote the set of all subsets of {1, 2, …, n} with no two consecutive numbers. For example {1, 3, 5} ∈ S6S_6, but {1, 2, 4} ∉ S6S_6. Then n(S5)n(S_5) is equal to ______

Correct answer: 13

Explanation: Fibonacci: F(7) = 13 (including the empty set).

Question 23

Mathematics · 4 marks · −1

The number of points of discontinuity of the function f(x)=[x22]−[x]f(x)=\left[\frac{x^2}{2}\right]-\left[\sqrt{x}\right], x∈[0,4]x\in[0, 4], where [·] denotes the greatest integer function, is ______

Correct answer: 8

Question 24

Mathematics · 4 marks · −1

The number of relations on the set A = {1, 2, 3}, containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is ______

Correct answer: 6

Question 25

Mathematics · 4 marks · −1

The number of singular matrices of order 2, whose elements are from the set {2, 3, 6, 9}, is ______.

Correct answer: 36

Explanation: Counting quadruples (a, b, c, d) from {2, 3, 6, 9} with ad = bc gives 36.

Question 26

Physics · 4 marks · −1

A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

  1. A.655 mm
  2. B.355 mm
  3. C.555 mm
  4. D.288 mm

Correct answer: D

Explanation: fwfa=μ−1μ/μw−1=0.60.25=2.4\frac{f_w}{f_a}=\frac{\mu-1}{\mu/\mu_w-1}=\frac{0.6}{0.25}=2.4: fw=28.8f_w=28.8 cm = 288 mm.

Question 27

Physics · 4 marks · −1

Uniform magnetic fields of different strengths (B1B_1 and B2B_2), both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface ? (Consider the velocity of the particle to be normal to the magnetic field and B2>B1B_2>B_1)

Figure for question 27
  1. A.mvqB1(1−B2B1)×2\frac{mv}{qB_1}\left(1-\frac{B_2}{B_1}\right)\times2
  2. B.mvqB1(1−B1B2)×2\frac{mv}{qB_1}\left(1-\frac{B_1}{B_2}\right)\times2
  3. C.mvqB1(1−B2B1)\frac{mv}{qB_1}\left(1-\frac{B_2}{B_1}\right)
  4. D.mvqB1(1−B1B2)\frac{mv}{qB_1}\left(1-\frac{B_1}{B_2}\right)

Correct answer: B

Explanation: Semicircles of diameters 2r22r_2 and 2r12r_1 in opposite directions: 2mvq(1B1−1B2)2\frac{mv}{q}\left(\frac1{B_1}-\frac1{B_2}\right).

Question 28

Physics · 4 marks · −1

The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility χMg=1.2×10−5\chi_{Mg}=1.2\times10^{-5}) is :

  1. A.56×10−5\frac{5}{6}\times10^{-5}%
  2. B.53×10−5\frac{5}{3}\times10^{-5}%
  3. C.56×10−4\frac{5}{6}\times10^{-4}%
  4. D.65×10−3\frac{6}{5}\times10^{-3}%

Correct answer: D

Explanation: Increase =χ×100%=1.2×10−3%=\chi\times100\%=1.2\times10^{-3}\%.

Question 29

Physics · 4 marks · −1

Two projectiles are fired from ground with same initial speeds from same point at angles (45° + α\alpha) and (45° − α\alpha) with horizontal direction. The ratio of their times of flights is

  1. A.1−tan⁡α1+tan⁡α\frac{1-\tan\alpha}{1+\tan\alpha}
  2. B.1
  3. C.1+tan⁡α1−tan⁡α\frac{1+\tan\alpha}{1-\tan\alpha}
  4. D.1+sin⁡2α1−sin⁡2α\frac{1+\sin2\alpha}{1-\sin2\alpha}

Correct answer: C

Explanation: sin⁡(45+α)sin⁡(45−α)=1+tan⁡α1−tan⁡α\frac{\sin(45+\alpha)}{\sin(45-\alpha)}=\frac{1+\tan\alpha}{1-\tan\alpha}.

Question 30

Physics · 4 marks · −1

If ϵ0\epsilon_0 denotes the permittivity of free space and ΦE\Phi_E is the flux of the electric field through the area bounded by the closed surface, then dimensions of (ϵ0dϕEdt)\left(\epsilon_0\frac{d\phi_E}{dt}\right) are that of :

  1. A.electric potential
  2. B.electric field
  3. C.electric current
  4. D.electric charge

Correct answer: C

Question 31

Physics · 4 marks · −1

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

  1. A.3 : 4
  2. B.27 : 5
  3. C.5 : 36
  4. D.5 : 27

Correct answer: D

Explanation: λLλB=5/363/4=527\frac{\lambda_L}{\lambda_B}=\frac{5/36}{3/4}=\frac{5}{27}.

Question 32

Physics · 4 marks · −1

An ac current is represented as i=52+10cos⁡(650πt+π6)i=5\sqrt{2}+10\cos\left(650\pi t+\frac{\pi}{6}\right) Amp. The r.m.s value of the current is

  1. A.50 Amp
  2. B.10 Amp
  3. C.525\sqrt{2} Amp
  4. D.100 Amp

Correct answer: B

Explanation: 50+50=10\sqrt{50+50}=10 A.

Question 33

Physics · 4 marks · −1

A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is :

Figure for question 33
  1. A.16
  2. B.14
  3. C.10
  4. D.12

Correct answer: D

Explanation: Each edge R/6; resistance across an edge of a tetrahedron = r/2 = R/12.

Question 34

Physics · 4 marks · −1

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V)

Figure for question 34
  1. A.80 mA
  2. B.10 mA
  3. C.24 mA
  4. D.60 mA

Correct answer: B

Explanation: Load current = 4/400 = 10 mA.

Question 35

Physics · 4 marks · −1

A particle of charge q, mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius (r). Which of the following curves represents the variation of r with E?

Figure for question 35
  1. A.r increases as a concave-down curve from the origin (√E shape)
  2. B.r increases as a concave-up curve
  3. C.r increases linearly with E
  4. D.r decreases with E

Correct answer: A

Explanation: r=2mEqB∝Er=\frac{\sqrt{2mE}}{qB}\propto\sqrt E.

Question 36

Physics · 4 marks · −1

Two wires A and B are made of same material having ratio of lengths LALB=13\frac{L_A}{L_B}=\frac{1}{3} and their diameters ratio dAdB=2\frac{d_A}{d_B}=2. If both the wires are stretched using same force, what would be the ratio of their respective elongations ?

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

Correct answer: A

Explanation: ΔL∝L/d2\Delta L\propto L/d^2: 13⋅14=112\frac13\cdot\frac14=\frac1{12}.

Question 37

Physics · 4 marks · −1

In a hydrogen like ion, the energy difference between the 2nd2^{nd} excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

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

Correct answer: C

Explanation: 13.6Z2(1−19)=108.813.6Z^2(1-\frac19)=108.8 ⇒ Z2=9Z^2=9.

Question 38

Physics · 4 marks · −1

A rod of length 5 L is bent right angle keeping one side length as 2 L. The position of the centre of mass of the system: (Consider L = 10 cm)

Figure for question 38
  1. A.4i^+9j^4\hat{i}+9\hat{j}
  2. B.5i^+8j^5\hat{i}+8\hat{j}
  3. C.3i^+7j^3\hat{i}+7\hat{j}
  4. D.2i^+3j^2\hat{i}+3\hat{j}

Correct answer: A

Explanation: x=2L⋅L5L=4x=\frac{2L\cdot L}{5L}=4 cm, y=3L⋅1.5L5L=9y=\frac{3L\cdot1.5L}{5L}=9 cm.

Question 39

Physics · 4 marks · −1

Two charges q1q_1 and q2q_2 are separated by a distance of 30 cm. A third charge q3q_3 initially at 'C' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of q3q_3 from C to D is given by q3K4πϵ0\frac{q_3K}{4\pi\epsilon_0}, the value of K is:

Figure for question 39
  1. A.8q1q_1
  2. B.8q2q_2
  3. C.6q2q_2
  4. D.6q1q_1

Correct answer: B

Explanation: Only q2q_2 matters: ΔU=kq2q3(10.1−10.5)=8kq2q3\Delta U=kq_2q_3\left(\frac{1}{0.1}-\frac{1}{0.5}\right)=8kq_2q_3.

Question 40

Physics · 4 marks · −1

Two plane polarized light waves combine at a certain point whose electric field components are E1=E0sin⁡ωtE_1=E_0\sin\omega t, E2=E0sin⁡(ωt+π3)E_2=E_0\sin\left(\omega t+\frac{\pi}{3}\right). Find the amplitude of the resultant wave.

  1. A.1.7 E0E_0
  2. B.0.9 E0E_0
  3. C.3.4 E0E_0
  4. D.E0E_0

Correct answer: A

Explanation: 2E0cos⁡30∘=3E0≈1.7E02E_0\cos30^\circ=\sqrt3E_0\approx1.7E_0.

Question 41

Physics · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. Triatomic rigid gas B. Diatomic non-rigid gas C. Monoatomic gas D. Diatomic rigid gas. LIST-II: I. CpCv=53\frac{C_p}{C_v}=\frac{5}{3} II. CpCv=75\frac{C_p}{C_v}=\frac{7}{5} III. CpCv=43\frac{C_p}{C_v}=\frac{4}{3} IV. CpCv=97\frac{C_p}{C_v}=\frac{9}{7}. Choose the correct answer from the options given below:

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

Correct answer: A

Question 42

Physics · 4 marks · −1

A cubic block of mass m is sliding down on an inclined plane at 60° with an acceleration of g2\frac{g}{2}, the value of coefficient of kinetic friction is

  1. A.32\frac{\sqrt{3}}{2}
  2. B.3−1\sqrt{3}-1
  3. C.23\frac{\sqrt{2}}{3}
  4. D.1−321-\frac{\sqrt{3}}{2}

Correct answer: B

Explanation: 32−μ2=12\frac{\sqrt3}{2}-\frac{\mu}{2}=\frac12 ⇒ μ=3−1\mu=\sqrt3-1.

Question 43

Physics · 4 marks · −1

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is:

  1. A.10 D
  2. B.5 D
  3. C.1 D
  4. D.20 D

Correct answer: A

Explanation: 1F=130+110−10300=110\frac1F=\frac1{30}+\frac1{10}-\frac{10}{300}=\frac{1}{10} cm⁻¹ = 10 D.

Question 44

Physics · 4 marks · −1

Two harmonic waves moving in the same direction superimpose to form a wave x = a cos (1.5t) cos (50.5t) where t is in seconds. Find the period with which they beat. (close to nearest integer)

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

Correct answer: A

Explanation: Beat angular frequency 2×1.5=32\times1.5=3: T=2π/3≈2T=2\pi/3\approx2 s.

Question 45

Physics · 4 marks · −1

An object of mass 1000 g experiences a time dependent force F⃗=(2ti^+3t2j^)\vec{F}=(2t\hat{i}+3t^2\hat{j}) N. The power generated by the force at time t is:

  1. A.(3t3+5t53t^3+5t^5) W
  2. B.(2t3+3t52t^3+3t^5) W
  3. C.(2t2+18t32t^2+18t^3) W
  4. D.(2t2+3t32t^2+3t^3) W

Correct answer: B

Explanation: v⃗=(t2,t3)\vec v=(t^2,t^3) (from rest); P=2t3+3t5P=2t^3+3t^5.

Question 46

Physics · 4 marks · −1

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is ______ ×10−1\times10^{-1} J. (Take π\pi = 3.14)

Figure for question 46

Correct answer: 314

Explanation: Area =π(100×103)(100×10−6)=31.4=\pi(100\times10^3)(100\times10^{-6})=31.4 J = 314 × 10⁻¹ J.

Question 47

Physics · 4 marks · −1

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is x8\frac{x}{8} where x = ______. (Given σ=6.0×10−8\sigma=6.0\times10^{-8} W m−2K−4m^{-2}K^{-4}, π\pi = 3.14 and assume that the emissivity of wire material is same at all wavelength.)

Correct answer: 5

Explanation: A=πdL=1.57×10−4A=\pi dL=1.57\times10^{-4}; e=94.26×10−8⋅1.57×10−4⋅1.6×1013=0.625=58e=\frac{94.2}{6\times10^{-8}\cdot1.57\times10^{-4}\cdot1.6\times10^{13}}=0.625=\frac58.

Question 48

Physics · 4 marks · −1

For ac circuit shown in figure, R = 100 kΩ and C = 100 pF and the phase difference between VinV_{in} and (VB−VA)(V_B-V_A) is 90°. The input signal frequency is 10x10^x rad/sec, where 'x' is ______.

Figure for question 48

Correct answer: 5

Explanation: VB−VAV_B-V_A is 90° out of phase when ωRC=1\omega RC=1: ω=105\omega=10^{5} rad/s.

Question 49

Physics · 4 marks · −1

A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure. The moment of inertia of the given system about PQ axis is x15\frac{x}{15}I, where I is the moment of inertia of the disc about its diameter. The value of x is ______.

Figure for question 49

Correct answer: 199

Explanation: IA=MR24=II_A=\frac{MR^2}{4}=I; IB=25MR2+MR2I_B=\frac25MR^2+MR^2, IC=23MR2+MR2I_C=\frac23MR^2+MR^2; total =MR2(14+75+53)=19960MR2=19915I=MR^2(\frac14+\frac75+\frac53)=\frac{199}{60}MR^2=\frac{199}{15}I.

Question 50

Physics · 4 marks · −1

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm. The value of H is ______ cm. (Consider liquids are immisible)

Correct answer: 80

Explanation: 60(1−11.2)+H(1−11.6)=4060(1-\frac{1}{1.2})+H(1-\frac{1}{1.6})=40 ⇒ 10+0.375H=4010+0.375H=40, H = 80.

Question 51

Chemistry · 4 marks · −1

Which of the following amine (s) show (s) positive carbylamine test? A. aniline B. (CH3)2NH(CH_3)_2NH C. CH3NH2CH_3NH_2 D. (CH3)3N(CH_3)_3N E. N-methylaniline. Choose the correct answer from the options given below:

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

Correct answer: D

Question 52

Chemistry · 4 marks · −1

The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol−1mol^{-1} respectively. The above values are lowest among their group members. The nature of their ions A2+A^{2+} and B4+B^{4+} respectively is

  1. A.oxidising and reducing
  2. B.both reducing
  3. C.both oxidising
  4. D.reducing and oxidising

Correct answer: D

Explanation: A = Sn: Sn2+Sn^{2+} reducing; B = Pb: Pb4+Pb^{4+} oxidising.

Question 53

Chemistry · 4 marks · −1

Given below are two statements: Statement I: D-(+)-glucose + D-(+) fructose →−H2O\xrightarrow{-H_2O} Sucrose; sucrose →hydrolysis\xrightarrow{\text{hydrolysis}} D-(+) glucose + D-(+) fructose. Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: D

Explanation: Fructose is D-(−) (laevorotatory), so Statement I as written is false.

Question 54

Chemistry · 4 marks · −1

Which of the following compounds is least likely to give effervescence of CO2CO_2 in presence of aq. NaHCO3NaHCO_3 ?

Figure for question 54
  1. A.3-Nitrobenzoic acid
  2. B.2,4,6-Trinitrophenol (picric acid)
  3. C.3-Nitrophenol
  4. D.Anilinium chloride (PhNH3+Cl−PhNH_3^+Cl^-)

Correct answer: C

Question 55

Chemistry · 4 marks · −1

The reactions which cannot be applied to prepare an alkene by elimination, are A. 1-bromo-2-methylcyclohexane + NaOEt B. CH3−CH2−CH(Br)−CH3CH_3-CH_2-CH(Br)-CH_3 + KOH (aq) C. (CH3)3C−Br(CH_3)_3C-Br + NaOMe D. Cyclohexanol + Na2Cr2O7/H2SO4Na_2Cr_2O_7/H_2SO_4 E. (CH3)3C−OH(CH_3)_3C-OH + Cu, 573 K. Choose the correct answer from the options given below:

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

Correct answer: C

Explanation: Aqueous KOH gives substitution (alcohol); dichromate oxidises the alcohol to a ketone. (Cu at 573 K dehydrates a 3° alcohol to an alkene.)

Question 56

Chemistry · 4 marks · −1

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would (Given log 2 = 0.30)

  1. A.reduce to 0.5
  2. B.remain same
  3. C.increase to 1.3
  4. D.increase to 2

Correct answer: C

Question 57

Chemistry · 4 marks · −1

Total enthalpy change for freezing of 1 mol of water at 10°C to ice at −10°C is ______ (Given: ΔfusH\Delta_{fus}H = x kJ/mol, Cp[H2O(l)]C_p[H_2O(l)] = y J mol−1K−1mol^{-1}K^{-1}, Cp[H2O(s)]C_p[H_2O(s)] = z J mol−1K−1mol^{-1}K^{-1})

  1. A.10(100x + y + z)
  2. B.x − 10y − 10z
  3. C.−10(100x + y + z)
  4. D.−x − 10y − 10z

Correct answer: C

Explanation: −10y−1000x−10z-10y-1000x-10z J =−10(100x+y+z)=-10(100x+y+z) J.

Question 58

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent. Statement II: Sodium metal can be used to dry diethyl ether and not ethyl alcohol. In the light of given statements, choose the correct answer from the options given below

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

Correct answer: B

Question 59

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Ozonolysis followed by treatment with Zn, H2OH_2O of cis-2-butene gives ethanal. Statement II: The product obtained by ozonolysis followed by treatment with Zn, H2OH_2O of 3, 6-dimethyloct-4-ene has no chiral carbon atom. 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 true but Statement II is false
  3. C.Both Statement I and Statement II are False
  4. D.Statement I is false but Statement II are true

Correct answer: B

Explanation: 3,6-Dimethyloct-4-ene gives 2-methylbutanal, which has a chiral carbon.

Question 60

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulfate. Statement II: If the molar conductance at infinite dilution for ferrous, ammonium and sulfate ions are x1x_1, x2x_2 and x3x_3 S cm2cm^2 mol−1mol^{-1}, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by x1+x2+2x3x_1+x_2+2x_3. In the light of the given statements, choose the correct answer from the options given below:

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

Correct answer: B

Explanation: FeSO4(NH4)2SO4⋅6H2OFeSO_4(NH_4)_2SO_4\cdot6H_2O: Λ∞=x1+2x2+2x3\Lambda^\infty=x_1+2x_2+2x_3.

Question 61

Chemistry · 4 marks · −1

Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A. t/min = 10: pressure of system 160 mm Hg; t = ∞: 240 mm Hg. Which of the following option is incorrect?

  1. A.The reaction never goes to completion
  2. B.Rate constant of the reaction is 1.693 min−1min^{-1}
  3. C.Initial pressure of A is 80 mm Hg
  4. D.Partial pressure of A after 10 minute is 40 mm Hg

Correct answer: B

Explanation: P0=80P_0=80; at 10 min 80+2x=16080+2x=160 ⇒ x = 40, pA=40p_A=40 (one half-life) ⇒ k = 0.0693 min⁻¹, so the stated 1.693 is incorrect.

Question 62

Chemistry · 4 marks · −1

The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is

  1. A.6
  2. B.9
  3. C.8
  4. D.10

Correct answer: A

Explanation: Cr has the lowest (397 kJ/mol) among these: 3d⁵4s¹ → 6.

Question 63

Chemistry · 4 marks · −1

An octahedral complex having molecular composition Co⋅5NH3⋅Cl⋅SO4Co\cdot5NH_3\cdot Cl\cdot SO_4 has two isomers A and B. The solution of A gives a white precipitate with AgNO3AgNO_3 solution and the solution of B gives white precipitate with BaCl2BaCl_2 solution. The type of isomerism exhibited by the complex is,

  1. A.Ionisation isomerism
  2. B.Co-ordinate isomerism
  3. C.Linkage isomerism
  4. D.Geometrical isomerism

Correct answer: A

Question 64

Chemistry · 4 marks · −1

At the sea level, the dry air mass percentage composition is given as nitrogen gas: 70.0, oxygen gas: 27.0 and argon gas: 3.0. If total pressure is 1.15 atm, then calculate the ratio of following respectively: (i) partial pressure of nitrogen gas to partial pressure of oxygen gas (ii) partial pressure of oxygen gas to partial pressure of argon gas (Given: Molar mass of N, O and Ar are 14, 16 and 40 g mol−1mol^{-1} respectively.)

  1. A.2.59, 11.85
  2. B.5.46, 17.8
  3. C.2.96, 11.2
  4. D.4.26, 19.3

Correct answer: C

Explanation: Moles: N2N_2 2.5, O2O_2 0.844, Ar 0.075 → 2.96 and 11.25.

Question 65

Chemistry · 4 marks · −1

Which of the following is the correct IUPAC name of given organic compound (X)? [structure: CH3−CH=C(CH3)−CH2BrCH_3-CH=C(CH_3)-CH_2Br]

Figure for question 65
  1. A.2-Bromo-2-methylbut-2-ene
  2. B.3-Bromo-3-methylprop-2-ene
  3. C.4-Bromo-3-methylbut-2-ene
  4. D.1-Bromo-2-methylbut-2-ene

Correct answer: D

Question 66

Chemistry · 4 marks · −1

[Visible spectrum 400–750 nm shown.] Which of the following statements are correct, if the threshold frequency of caesium is 5.16×10145.16\times10^{14} Hz? A. When Cs is placed inside a vacuum chamber with an ammeter connected to it and yellow light is focused on Cs, the ammeter shows the presence of current. B. When the brightness of the yellow light is dimmed, the value of the current in the ammeter is reduced. C. When a red light is used instead of the yellow light, the current produced is higher with respect to the yellow light. D. When a blue light is used, the ammeter shows the formation of current. E. When a white light is used. the ammeter shows formation of current. Choose the correct answer from the options given below:

Figure for question 66
  1. A.A, D and E Only
  2. B.A, C, D and E Only
  3. C.A, B, D and E Only
  4. D.B, C and D Only

Correct answer: C

Explanation: Threshold λ ≈ 581 nm: yellow (<581) works, red does not.

Question 67

Chemistry · 4 marks · −1

When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown coloured precipitate is formed. X and Y respectively, are

  1. A.X = NH3NH_3 and Y = K2HgI4K_2HgI_4 + KOH
  2. B.X = HCl and Y = NH4ClNH_4Cl
  3. C.X = NH3NH_3 and Y= HgO
  4. D.X = NH4ClNH_4Cl and Y = KOH

Correct answer: A

Question 68

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Molecule / ion): A. ICl2−ICl_2^- B. H2OH_2O C. SO2SO_2 D. XeF4XeF_4. LIST-II (Bond pair : lone pair on the central atom): I. 4 : 2 II. 4 : 1 III. 2 : 3 IV. 2 : 2. Choose the correct answer from the options given below:

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

Correct answer: C

Explanation: SO2SO_2 counted as 4 bond pairs (two double bonds) : 1 lone pair.

Question 69

Chemistry · 4 marks · −1

The first transition series metal 'M' has the highest enthalpy of atomisation in its series. One of its aquated ion (Mn+M^{n+}) exists in green colour. The nature of the oxide formed by the above Mn+M^{n+} ion is:

  1. A.basic
  2. B.amphoteric
  3. C.neutral
  4. D.acidic

Correct answer: A

Explanation: M = V; V3+V^{3+}(aq) is green; V2O3V_2O_3 is basic.

Question 70

Chemistry · 4 marks · −1

A person's wound was exposed to some bacteria and then bacterial growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay(r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine? [Given: N = No. of bacteria, t = time, bacterial growth follows 1st1^{st} order kinetics.]

Figure for question 70
  1. A.Before: N/N₀ rises exponentially from 1; After: r decreases with N (hyperbola-like)
  2. B.Before: N/N₀ rises exponentially from 1; After: r increases linearly with N
  3. C.Before: N/N₀ rises exponentially from 1; After: r increases parabolically (∝ N²) from the origin
  4. D.Before: N/N₀ rises from 0; After: r rises from a non-zero intercept

Correct answer: C

Question 71

Chemistry · 4 marks · −1

An organic compound weighing 500 mg, produced 220 mg of CO2CO_2, on complete combustion. The percentage composition of carbon in the compound is ______%. (nearest integer) (Given molar mass in g mol−1mol^{-1} of C:12, O:16)

Correct answer: 12

Explanation: 60 mg C → 12%.

Question 72

Chemistry · 4 marks · −1

Thyroxine, the hormone has given below structure [3,5,3′,5′-tetraiodothyronine: HO–C₆H₂I₂–O–C₆H₂I₂–CH₂–CH(NH₂)–COOH]. The percentage of iodine in thyroxine is ______%. (nearest integer) (Given molar mass in g mol−1mol^{-1} C:12, H:1, O:16, N:14, I:127)

Correct answer: 65

Explanation: C15H11I4NO4=777C_{15}H_{11}I_4NO_4=777; 508/777=65.4%508/777=65.4\% → 65.

Question 73

Chemistry · 4 marks · −1

1 Faraday electricity was passed through Cu2+Cu^{2+} (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag+Ag^+ (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______ mV (nearest integer) Given : ECu2+/Cu∘E^\circ_{Cu^{2+}/Cu} = 0.34 V, EAg+/Ag∘E^\circ_{Ag^+/Ag} = 0.8 V, 2.303RTF\frac{2.303RT}{F} = 0.06V

Figure for question 73

Correct answer: 400

Explanation: [Cu2+]=1.0[Cu^{2+}]=1.0 M, [Ag+]=0.1[Ag^+]=0.1 M; E=0.46−0.03log⁡10.01=0.40E=0.46-0.03\log\frac{1}{0.01}=0.40 V = 400 mV.

Question 74

Chemistry · 4 marks · −1

The percentage dissociation of a salt (MX3MX_3) solution at given temperature (van't Hoff factor i = 2) is ______ %(Nearest integer)

Correct answer: 33

Explanation: i=1+3α=2i=1+3\alpha=2 ⇒ α = 33%.

Question 75

Chemistry · 4 marks · −1

The number of paramagnetic complexes among [FeF6]3−[FeF_6]^{3-}, [Fe(CN)6]3−[Fe(CN)_6]^{3-}, [Mn(CN)6]3−[Mn(CN)_6]^{3-}, [Co(C2O4)3]3−[Co(C_2O_4)_3]^{3-}, [MnCl6]3−[MnCl_6]^{3-}, and [CoF6]3−[CoF_6]^{3-}, which involved d2sp3d^2sp^3 hybridization is ______

Correct answer: 2

Explanation: d2sp3d^2sp^3 and paramagnetic: [Fe(CN)6]3−[Fe(CN)_6]^{3-} (1 unpaired), [Mn(CN)6]3−[Mn(CN)_6]^{3-} (2 unpaired) = 2.

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