JEE Main 2025 — 24 Jan 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 area of the region {(x,y):x2+4x+2≤y≤∣x+2∣}\{(x, y): x^2+4x+2\le y\le|x+2|\} is equal to

  1. A.7
  2. B.20/3
  3. C.24/5
  4. D.5

Correct answer: B

Explanation: With t=x+2t=x+2: t2−2≤y≤∣t∣t^2-2\le y\le|t|, region ∣t∣≤2|t|\le2; area 2∫02(t−t2+2)dt=2(2−83+4)=2032\int_0^2(t-t^2+2)dt=2(2-\frac83+4)=\frac{20}{3}.

Question 2

Mathematics · 4 marks · −1

For some n≠10n\neq10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1+x)n+4(1+x)^{n+4} be in A.P. Then the largest coefficient in the expansion of (1+x)n+4(1+x)^{n+4} is:

  1. A.35
  2. B.70
  3. C.10
  4. D.20

Correct answer: A

Explanation: NC4,NC5,NC6{}^NC_4, {}^NC_5, {}^NC_6 in AP ⇒ N = 7 or 14; n≠10n\neq10 ⇒ N = 7: largest 7C3=35{}^7C_3=35.

Question 3

Mathematics · 4 marks · −1

If I(m,n)=∫01xm−1(1−x)n−1dxI(m, n)=\int_0^1x^{m-1}(1-x)^{n-1}dx, m, n > 0, then I(9,14)+I(10,13)I(9, 14)+I(10, 13) is

  1. A.I(9, 13)
  2. B.I(9, 1)
  3. C.I(19, 27)
  4. D.I(1, 13)

Correct answer: A

Explanation: x8(1−x)13+x9(1−x)12=x8(1−x)12x^{8}(1-x)^{13}+x^9(1-x)^{12}=x^8(1-x)^{12}.

Question 4

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0\left(xy-5x^2\sqrt{1+x^2}\right)dx+(1+x^2)dy=0, y(0)=0y(0)=0. Then y(3)y(\sqrt{3}) is equal to

  1. A.222\sqrt{2}
  2. B.152\sqrt{\frac{15}{2}}
  3. C.532\frac{5\sqrt{3}}{2}
  4. D.143\sqrt{\frac{14}{3}}

Correct answer: C

Explanation: IF 1+x2\sqrt{1+x^2}: y1+x2=5x33y\sqrt{1+x^2}=\frac{5x^3}{3}; y(3)=5⋅333⋅2=532y(\sqrt3)=\frac{5\cdot3\sqrt3}{3\cdot2}=\frac{5\sqrt3}{2}.

Question 5

Mathematics · 4 marks · −1

Let the lines 3x−4y−α=03x-4y-\alpha=0, 8x−11y−33=08x-11y-33=0, and 2x−3y+λ=02x-3y+\lambda=0 be concurrent. If the image of the point (1, 2) in the line 2x−3y+λ=02x-3y+\lambda=0 is (5713,−4013)\left(\frac{57}{13}, \frac{-40}{13}\right), then ∣αλ∣|\alpha\lambda| is equal to

  1. A.91
  2. B.101
  3. C.113
  4. D.84

Correct answer: A

Explanation: Midpoint (3513,−713)(\frac{35}{13},-\frac{7}{13}) lies on the mirror ⇒ λ = −7; lines 2 and 3 meet at (11, 5), so α = 33 − 20 = 13; |αλ| = 91.

Question 6

Mathematics · 4 marks · −1

If α\alpha and β\beta are the roots of the equation 2z2−3z−2i=02z^2-3z-2i=0, where i=−1i=\sqrt{-1}, then 16⋅Re(α19+β19+α11+β11α15+β15)⋅Im(α19+β19+α11+β11α15+β15)16\cdot\mathrm{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\mathrm{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) is equal to

  1. A.441
  2. B.312
  3. C.398
  4. D.409

Correct answer: A

Question 7

Mathematics · 4 marks · −1

If the system of equations 2x−y+z=42x-y+z=4, 5x+λy+3z=125x+\lambda y+3z=12, 100x−47y+μz=212100x-47y+\mu z=212, has infinitely many solutions, then μ−2λ\mu-2\lambda is equal to

  1. A.55
  2. B.59
  3. C.57
  4. D.56

Correct answer: C

Question 8

Mathematics · 4 marks · −1

Let a⃗=i^+2j^+3k^\vec{a}=\hat{i}+2\hat{j}+3\hat{k}, b⃗=3i^+j^−k^\vec{b}=3\hat{i}+\hat{j}-\hat{k} and c⃗\vec{c} be three vectors such that c⃗\vec{c} is coplanar with a⃗\vec{a} and b⃗\vec{b}. If the vector c⃗\vec{c} is perpendicular to b⃗\vec{b} and a⃗⋅c⃗=5\vec{a}\cdot\vec{c}=5, then ∣c⃗∣|\vec{c}| is equal to

  1. A.16
  2. B.116\sqrt{\frac{11}{6}}
  3. C.18
  4. D.132\frac{1}{3\sqrt{2}}

Correct answer: B

Question 9

Mathematics · 4 marks · −1

For a statistical data x1,x2,…,x10x_1, x_2, \ldots, x_{10} of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\sum_{i=1}^{10}x_i^2=371. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is

  1. A.9
  2. B.5
  3. C.4
  4. D.7

Correct answer: D

Explanation: Sum 55 → 60 (mean 6); ∑x2=371−41+100=430\sum x^2=371-41+100=430; variance 43−36=743-36=7.

Question 10

Mathematics · 4 marks · −1

Let the line passing through the points (−1-1, 2, 1) and parallel to the line x−12=y+13=z4\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} intersect the line x+23=y−32=z−41\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1} at the point P. Then the distance of P from the point Q(4, −5-5, 1) is

  1. A.10
  2. B.5
  3. C.565\sqrt{6}
  4. D.555\sqrt{5}

Correct answer: D

Question 11

Mathematics · 4 marks · −1

Let f:R−{0}→Rf:\mathbf{R}-\{0\}\to\mathbf{R} be a function such that f(x)−6f(1x)=353x−52f(x)-6f\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}. If the lim⁡x→0(1αx+f(x))=β\lim_{x\to0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta; α,β∈R\alpha, \beta\in\mathbf{R}, then α+2β\alpha+2\beta is equal to

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

Correct answer: B

Explanation: Solving the pair of equations: f(x)=−2x−13x+12f(x)=-2x-\frac{1}{3x}+\frac12; the limit is finite only for α = 3, then β = 1/2: α + 2β = 4.

Question 12

Mathematics · 4 marks · −1

Let f(x)=2x+2+1622x+1+2x+4+32f(x)=\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}. Then the value of 8(f(115)+f(215)+…+f(5915))8\left(f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)\right) is equal to

  1. A.102
  2. B.92
  3. C.118
  4. D.108

Correct answer: C

Explanation: f(x)=22x+4f(x)=\frac{2}{2^x+4} and f(x)+f(4−x)=12f(x)+f(4-x)=\frac12; 29 pairs + f(2) = 29/2 + 1/4; × 8 = 118.

Question 13

Mathematics · 4 marks · −1

lim⁡x→0cosec x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)\lim_{x\to0}\mathrm{cosec}\,x\left(\sqrt{2\cos^2x+3\cos x}-\sqrt{\cos^2x+\sin x+4}\right) is:

  1. A.125\frac{1}{2\sqrt{5}}
  2. B.−125-\frac{1}{2\sqrt{5}}
  3. C.115\frac{1}{\sqrt{15}}
  4. D.0

Correct answer: B

Explanation: Rationalise: numerator ≈−sin⁡x\approx -\sin x to first order, denominator 252\sqrt5: −125-\frac{1}{2\sqrt5}.

Question 14

Mathematics · 4 marks · −1

Let circle C be the image of x2+y2−2x+4y−4=0x^2+y^2-2x+4y-4=0 in the line 2x−3y+5=02x-3y+5=0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α\alpha, β\beta), with β<4\beta<4, lies on C such that the length of the arc AB is (1/6)th(1/6)^{th} of the perimeter of C, then β−3α\beta-\sqrt{3}\alpha is equal to

  1. A.4−34-\sqrt{3}
  2. B.3
  3. C.4
  4. D.3+33+\sqrt{3}

Correct answer: C

Question 15

Mathematics · 4 marks · −1

Consider the region R={(x,y):x≤y≤9−113x2,x≥0}R=\left\{(x, y): x\le y\le9-\frac{11}{3}x^2, x\ge0\right\}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is:

  1. A.567121\frac{567}{121}
  2. B.730119\frac{730}{119}
  3. C.625111\frac{625}{111}
  4. D.821123\frac{821}{123}

Correct answer: A

Question 16

Mathematics · 4 marks · −1

Let Sn=12+16+112+120+…S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots upto n terms. If the sum of the first six terms of an A.P. with first term −p and common difference p is 2026S2025\sqrt{2026S_{2025}}, then the absolute difference between 20th20^{th} and 15th15^{th} terms of the A.P. is

  1. A.20
  2. B.25
  3. C.45
  4. D.90

Correct answer: B

Explanation: S2025=20252026S_{2025}=\frac{2025}{2026}, so the sum is 45 =3(−2p+5p)=9p=3(-2p+5p)=9p ⇒ p = 5; difference 5p=255p=25.

Question 17

Mathematics · 4 marks · −1

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

  1. A.917\frac{9}{17}
  2. B.819\frac{8}{19}
  3. C.817\frac{8}{17}
  4. D.919\frac{9}{19}

Correct answer: D

Explanation: p=19p=\frac19, q=536q=\frac{5}{36}: P=p1−(1−p)(1−q)=1/976/324=919P=\frac{p}{1-(1-p)(1-q)}=\frac{1/9}{76/324}=\frac{9}{19}.

Question 18

Mathematics · 4 marks · −1

Let in a △ABC\triangle ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}. Then the area (in sq. units) of △ABC\triangle ABC is:

  1. A.42
  2. B.21
  3. C.17
  4. D.56

Correct answer: B

Explanation: Distance of B from the line = 7; area 12⋅6⋅7=21\frac12\cdot6\cdot7=21.

Question 19

Mathematics · 4 marks · −1

The product of all the rational roots of the equation (x2−9x+11)2−(x−4)(x−5)=3(x^2-9x+11)^2-(x-4)(x-5)=3, is equal to

  1. A.14
  2. B.28
  3. C.7
  4. D.21

Correct answer: A

Explanation: With t=x2−9xt=x^2-9x: (t+11)2−(t+20)=3(t+11)^2-(t+20)=3 ⇒ t2+21t+98=0t^2+21t+98=0 ⇒ t = −7, −14; x2−9x+14=0x^2-9x+14=0 gives 2, 7 (rational); product 14.

Question 20

Mathematics · 4 marks · −1

Let the product of the focal distances of the point (3,12)\left(\sqrt{3}, \frac{1}{2}\right) on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, (a > b), be 74\frac{7}{4}. Then the absolute difference of the eccentricities of two such ellipses is

  1. A.1−32\frac{1-\sqrt{3}}{\sqrt{2}}
  2. B.1−223\frac{1-2\sqrt{2}}{\sqrt{3}}
  3. C.3−2223\frac{3-2\sqrt{2}}{2\sqrt{3}}
  4. D.3−2232\frac{3-2\sqrt{2}}{3\sqrt{2}}

Correct answer: C

Question 21

Mathematics · 4 marks · −1

Let A be a 3 × 3 matrix such that XTAX=OX^TAX=O for all nonzero 3 × 1 matrices X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}. If A[111]=[14−5]A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}, A[121]=[04−8]A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}, and det⁡(adj(2(A+I)))=2α3β5γ\det(\mathrm{adj}(2(A+I)))=2^\alpha3^\beta5^\gamma, α,β,γ∈N\alpha, \beta, \gamma\in\mathbf{N}, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is ______.

Correct answer: 44

Question 22

Mathematics · 4 marks · −1

Let f be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt2(x+2)^2f(x)-3(x+2)^2=10\int_0^x(t+2)f(t)dt, x≥0x\ge0. Then f(2) is equal to ______.

Correct answer: 19

Question 23

Mathematics · 4 marks · −1

If for some α,β\alpha, \beta; α≤β\alpha\le\beta, α+β=8\alpha+\beta=8 and sec⁡2(tan⁡−1α)+cosec2(cot⁡−1β)=36\sec^2(\tan^{-1}\alpha)+\mathrm{cosec}^2(\cot^{-1}\beta)=36, then α2+β\alpha^2+\beta is ______.

Correct answer: 14

Explanation: 2+α2+β2=362+\alpha^2+\beta^2=36, α+β=8\alpha+\beta=8 ⇒ αβ=15\alpha\beta=15: α = 3, β = 5; 9 + 5 = 14.

Question 24

Mathematics · 4 marks · −1

Let S={p1,p2,…,p10}S=\{p_1, p_2, \ldots, p_{10}\} be the set of first ten prime numbers. Let A=S∪PA=S\cup P, where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs (x, y), x∈Sx\in S, y∈Ay\in A, such that x divides y, is ______.

Correct answer: 5120

Explanation: Each prime divides 29=5122^9=512 of the non-empty subsets-products (A = all 1023 products): 10×512=512010\times512=5120.

Question 25

Mathematics · 4 marks · −1

The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is ______.

Correct answer: 125

Explanation: Multiples of 6 among 3-digit numbers: 150; those divisible by both 4 and 9 (i.e. by 36): 25; 150 − 25 = 125.

Question 26

Physics · 4 marks · −1

For an experimental expression y=32.3×112527.4y=\frac{32.3\times1125}{27.4}, where all the digits are significant. Then to report the value of y we should write

  1. A.y = 1326.19
  2. B.y = 1326.2
  3. C.y = 1326.186
  4. D.y = 1330

Correct answer: D

Question 27

Physics · 4 marks · −1

A force F=α+βx2F=\alpha+\beta x^2 acts on an object in the x–direction. The work done by the force is 5 J when the object is displaced by 1 m. If the constant α\alpha = 1N then β\beta will be

  1. A.12 N/m2m^2
  2. B.15 N/m2m^2
  3. C.8 N/m2m^2
  4. D.10 N/m2m^2

Correct answer: A

Explanation: 1+β/3=51+\beta/3=5 ⇒ β = 12.

Question 28

Physics · 4 marks · −1

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process). Choose the correct answer from the options given below:

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

Correct answer: D

Question 29

Physics · 4 marks · −1

A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

  1. A.132g\frac{1}{3\sqrt{2}}g
  2. B.12g\frac{1}{\sqrt{2}}g
  3. C.2g\sqrt{2}g
  4. D.2g3\frac{\sqrt{2}g}{3}

Correct answer: D

Explanation: a=23gsin⁡45∘=2g3a=\frac{2}{3}g\sin45^\circ=\frac{\sqrt2g}{3}.

Question 30

Physics · 4 marks · −1

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3m^3. If the pressure inside the bubble is 2100 N/m2m^2 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10m/s2s^2)

  1. A.0.05
  2. B.0.25
  3. C.0.1
  4. D.0.02

Correct answer: A

Explanation: 2100−2000=2T10−32100-2000=\frac{2T}{10^{-3}} ⇒ T = 0.05.

Question 31

Physics · 4 marks · −1

Consider the following statements: A. The junction area of solar cell is made very narrow compared to a photo diode. B. Solar cells are not connected with any external bias. C. LED is made of lightly doped p-n junction. D. Increase of forward current results in continuous increase of LED light intensity. E. LEDs have to be connected in forward bias for emission of light. Choose the correct answer from the options given below:

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

Correct answer: C

Question 32

Physics · 4 marks · −1

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then Dd\frac{D}{d} is

  1. A.10
  2. B.25
  3. C.165\frac{16}{5}
  4. D.154\frac{15}{4}

Correct answer: B

Explanation: 12.5 s = 6.25 T: D = 6 × 4 cm + 1 cm = 25 cm; starting from mean position, displacement after T/4 is 1 cm: ratio 25.

Question 33

Physics · 4 marks · −1

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is

  1. A.0.15 m
  2. B.0.25 m
  3. C.0.20 m
  4. D.0.10 m

Correct answer: D

Explanation: P=2PL+PMP=2P_L+P_M, 1fL=0.31.2R=14R\frac{1}{f_L}=\frac{0.3}{1.2R}=\frac{1}{4R}, plane mirror P = 0: 10.2=24R\frac{1}{0.2}=\frac{2}{4R} ⇒ R = 0.1 m.

Question 34

Physics · 4 marks · −1

The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be

  1. A.5 J
  2. B.15 J
  3. C.20 J
  4. D.10 J

Correct answer: B

Explanation: W∝(n1/3−1)W\propto(n^{1/3}-1): 10 J for 2, so 15 J for 3.

Question 35

Physics · 4 marks · −1

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and ϵ0\epsilon_0 is the permittivity of free space between the plates, then potential energy stored in the capacitor is

  1. A.12ϵ0E2Ad\frac{1}{2}\epsilon_0E^2Ad
  2. B.14ϵ0E2Ad\frac{1}{4}\epsilon_0E^2Ad
  3. C.ϵ0E2Ad\epsilon_0E^2Ad
  4. D.34ϵ0E2Ad\frac{3}{4}\epsilon_0E^2Ad

Correct answer: A

Question 36

Physics · 4 marks · −1

An electron of mass 'm' with an initial velocity v⃗=v0i^\vec{v}=v_0\hat{i} (v0>0v_0>0) enters an electric field E⃗=−E0k^\vec{E}=-E_0\hat{k}. If the initial de Broglie wavelength is λ0\lambda_0, the value after time t would be

  1. A.λ0\lambda_0
  2. B.λ01−e2E02t2m2v02\frac{\lambda_0}{\sqrt{1-\frac{e^2E_0^2t^2}{m^2v_0^2}}}
  3. C.λ01+e2E02t2m2v02\frac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}
  4. D.λ01+e2E02t2m2v02\lambda_0\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}

Correct answer: C

Question 37

Physics · 4 marks · −1

A car of mass 'm' moves on a banked road having radius 'r' and banking angle θ\theta. To avoid slipping from banked road, the maximum permissible speed of the car is v0v_0. The coefficient of friction μ\mu between the wheels of the car and the banked road is

  1. A.μ=v02−rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta}
  2. B.μ=v02+rgtan⁡θrg−v02tan⁡θ\mu=\frac{v_0^2+rg\tan\theta}{rg-v_0^2\tan\theta}
  3. C.μ=v02+rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_0^2+rg\tan\theta}{rg+v_0^2\tan\theta}
  4. D.μ=v02−rgtan⁡θrg−v02tan⁡θ\mu=\frac{v_0^2-rg\tan\theta}{rg-v_0^2\tan\theta}

Correct answer: A

Explanation: v02=rgμ+tan⁡θ1−μtan⁡θv_0^2=rg\frac{\mu+\tan\theta}{1-\mu\tan\theta} ⇒ μ=v02−rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta}.

Question 38

Physics · 4 marks · −1

An alternating current is given by I=IAsin⁡ωt+IBcos⁡ωtI=I_A\sin\omega t+I_B\cos\omega t. The r.m.s current will be

  1. A.∣IA+IB∣2\frac{|I_A+I_B|}{\sqrt{2}}
  2. B.IA2+IB2\sqrt{I_A^2+I_B^2}
  3. C.IA2+IB22\frac{\sqrt{I_A^2+I_B^2}}{2}
  4. D.IA2+IB22\sqrt{\frac{I_A^2+I_B^2}{2}}

Correct answer: D

Question 39

Physics · 4 marks · −1

A parallel plate capacitor was made with two rectangular plates, each with a length of ll = 3 cm and breath of b = 1 cm. The distance between the plates is 3 μ\mum. Out of the following, which are the ways to increase the capacitance by a factor of 10? A. ll = 30 cm, b = 1 cm, d = 1 μ\mum B. ll = 3 cm, b = 1 cm, d = 30 μ\mum C. ll = 6 cm, b = 5 cm, d = 3 μ\mum D. ll = 1 cm, b = 1 cm, d = 10 μ\mum E. ll = 5 cm, b = 2 cm, d = 1 μ\mum. Choose the correct answer from the options given below:

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

Correct answer: C

Explanation: Need A/d = 10 × (3/3) = 10 cm²/μm: C gives 30/3 = 10 ✓, E gives 10/1 = 10 ✓.

Question 40

Physics · 4 marks · −1

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1D from 2.5D ? ['D' stands for dioptre]

  1. A.0.1
  2. B.0.01
  3. C.0.04
  4. D.0.40

Correct answer: C

Explanation: Δf/f=ΔP/P=0.1/2.5=0.04\Delta f/f=\Delta P/P=0.1/2.5=0.04.

Question 41

Physics · 4 marks · −1

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is

  1. A.4000 Å
  2. B.3000 Å
  3. C.6000 Å
  4. D.2000 Å

Correct answer: B

Explanation: 12000−16000=13000\frac{1}{2000}-\frac{1}{6000}=\frac{1}{3000}.

Question 42

Physics · 4 marks · −1

An object of mass 'm' is projected from origin in a vertical xy plane at an angle 45° with the x–axis with an initial velocity v0v_0. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]

  1. A.mv0322g\frac{mv_0^3}{2\sqrt{2}g} along positive z–axis
  2. B.mv0342g\frac{mv_0^3}{4\sqrt{2}g} along positive z–axis
  3. C.mv0342g\frac{mv_0^3}{4\sqrt{2}g} along negative z–axis
  4. D.mv0322g\frac{mv_0^3}{2\sqrt{2}g} along negative z–axis

Correct answer: C

Explanation: L=mvxH=mv02⋅v024gL=mv_xH=m\frac{v_0}{\sqrt2}\cdot\frac{v_0^2}{4g}, direction r⃗×v⃗\vec r\times\vec v = −z.

Question 43

Physics · 4 marks · −1

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f1f_1 in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f2f_2 when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of f1f_1 and f2f_2 will be

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

Correct answer: D

Explanation: f1=4f_1=4 cm; 1f2=0.31.2⋅13=112\frac{1}{f_2}=\frac{0.3}{1.2}\cdot\frac13=\frac1{12}: ratio 1 : 3.

Question 44

Physics · 4 marks · −1

The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is

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

Correct answer: B

Explanation: 480n1=600n2480n_1=600n_2: n1=5n_1=5.

Question 45

Physics · 4 marks · −1

A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by

  1. A.9%
  2. B.4.5%
  3. C.3%
  4. D.2.5%

Correct answer: B

Explanation: 32×3%=4.5%\frac32\times3\%=4.5\%.

Question 46

Physics · 4 marks · −1

A current of 5A exists in a square loop of side 12\frac{1}{\sqrt{2}} m. Then the magnitude of the magnetic field B at the centre of the square loop will be p×10−6p\times10^{-6} T. where, value of p is ______. [Take μ0=4π×10−7\mu_0=4\pi\times10^{-7} T m A−1A^{-1}].

Correct answer: 8

Explanation: B=22μ0Iπa=22⋅4π×10−7⋅5π/2=8×10−6B=\frac{2\sqrt2\mu_0I}{\pi a}=\frac{2\sqrt2\cdot4\pi\times10^{-7}\cdot5}{\pi/\sqrt2}=8\times10^{-6} T.

Question 47

Physics · 4 marks · −1

A square loop of sides a = 1 m is held normally in front of a point charge q = 1C. The flux of the electric field through the shaded region is 5p×1ϵ0Nm2C\frac{5}{p}\times\frac{1}{\epsilon_0}\frac{Nm^2}{C}, where the value of p is ______.

Figure for question 47

Correct answer: 48

Explanation: The square subtends 1/6 of the total solid angle (face of a cube of side a centred on q): flux through square =q6ϵ0=\frac{q}{6\epsilon_0}; shaded 58⋅16ϵ0=548ϵ0\frac58\cdot\frac{1}{6\epsilon_0}=\frac{5}{48\epsilon_0}, p = 48.

Question 48

Physics · 4 marks · −1

A wire of resistance 9 Ω\Omega is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ______ ohm.

Correct answer: 2

Explanation: 3×69=2\frac{3\times6}{9}=2.

Question 49

Physics · 4 marks · −1

The temperature of 1 mole of an ideal monoatomic gas is increased by 50°C at constant pressure. The total heat added and change in internal energy are E1E_1 and E2E_2, respectively. If E1E2=x9\frac{E_1}{E_2}=\frac{x}{9} then the value of x is ______

Correct answer: 15

Explanation: CpCv=53=159\frac{C_p}{C_v}=\frac53=\frac{15}{9}, x = 15.

Question 50

Physics · 4 marks · −1

The least count of a screw guage is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ______ ×10−3\times10^{-3} mm

Correct answer: 35

Explanation: 0.01×1.75/0.5=0.0350.01\times1.75/0.5=0.035 mm = 35 × 10⁻³ mm.

Question 51

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The conversion proceeds well in the less polar medium. CH3−CH2−CH2−CH2−Cl→HO−CH3−CH2−CH2−CH2−OH+Cl−CH_3-CH_2-CH_2-CH_2-Cl\xrightarrow{HO^-}CH_3-CH_2-CH_2-CH_2-OH+Cl^- Statement II: The conversion proceeds well in the more polar medium. CH3−CH2−CH2−CH2−Cl→R3N¨CH3−CH2−CH2−CH2−N+R3 Cl−CH_3-CH_2-CH_2-CH_2-Cl\xrightarrow{R_3\ddot{N}}CH_3-CH_2-CH_2-CH_2-\overset{+}{N}R_3\ Cl^-. 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.Both Statement I and Statement II are false
  4. D.Statement I is true but Statement II is false

Correct answer: A

Explanation: SN2 with a charged nucleophile giving neutral products is favoured by less polar media; neutral reactants giving charged products are favoured by more polar media.

Question 52

Chemistry · 4 marks · −1

Preparation of potassium permanganate from MnO2MnO_2 involves two step process in which the 1st step is a reaction with KOH and KNO3KNO_3 to produce

  1. A.K3MnO4K_3MnO_4
  2. B.KMnO4KMnO_4
  3. C.K2MnO4K_2MnO_4
  4. D.K4[Mn(OH)6]K_4[Mn(OH)_6]

Correct answer: C

Question 53

Chemistry · 4 marks · −1

Following are the four molecules "P", "Q", "R" and "S". Which one among the four molecules will react with H–Br(aq) at the fastest rate? P: 3,6-dihydro-2H-pyran (ring O not attached to the C=C); Q: 3,4-dihydro-2H-pyran (cyclic enol ether, ring O directly attached to the C=C); R: 1-methylcyclohexene; S: 1,2-dimethylcyclohexene.

Figure for question 53
  1. A.S
  2. B.R
  3. C.Q
  4. D.P

Correct answer: C

Explanation: Protonation of the enol ether Q gives an oxocarbenium ion stabilised by the ring-oxygen lone pair — fastest.

Question 54

Chemistry · 4 marks · −1

Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing ΔTf\Delta T_f, depression in the freezing point of a solvent in a solution?

Figure for question 54
  1. A.Plot 1: curves labelled "Solution" and "Liquid Solvent" rising with T and a dashed horizontal line; TfT_f and Tf∘T_f^\circ marked where each curve meets the dashed line
  2. B.Plot 2: curves labelled "Solution" and "Frozen Solvent" rising with T, with TfT_f and Tf∘T_f^\circ marked at their intersections with a dashed line
  3. C.Plot 3: a steep "Frozen solvent" curve crossing the "Liquid Solvent" curve at Tf∘T_f^\circ and the lower "Solution" curve at TfT_f (to the left)
  4. D.Plot 4: "Solution" curve above "Liquid Solvent" curve, with the frozen solvent curve meeting them; TfT_f to the left of Tf∘T_f^\circ

Correct answer: C

Explanation: Freezing points are where the solid-solvent VP curve meets the liquid-solvent (at Tf∘T_f^\circ) and the (lower) solution curve (at Tf<Tf∘T_f<T_f^\circ).

Question 55

Chemistry · 4 marks · −1

KspK_{sp} for Cr(OH)3Cr(OH)_3 is 1.6×10−301.6\times10^{-30}. What is the molar solubility of this salt in water?

  1. A.1.8×10−305\sqrt[5]{1.8\times10^{-30}}
  2. B.1.8×10−3027\frac{1.8\times10^{-30}}{27}
  3. C.1.6×10−302\sqrt[2]{1.6\times10^{-30}}
  4. D.1.6×10−30274\sqrt[4]{\frac{1.6\times10^{-30}}{27}}

Correct answer: D

Explanation: Ksp=27s4K_{sp}=27s^4.

Question 56

Chemistry · 4 marks · −1

One mole of the octahedral complex compound Co(NH3)5Cl3Co(NH_3)_5Cl_3 gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of AgNO3AgNO_3 solution to yield two moles of AgCl(s)AgCl_{(s)}. The structure of the complex is:

  1. A.[Co(NH3)3Cl3]⋅2NH3[Co(NH_3)_3Cl_3]\cdot2NH_3
  2. B.[Co(NH3)4Cl]⋅Cl2⋅NH3[Co(NH_3)_4Cl]\cdot Cl_2\cdot NH_3
  3. C.[Co(NH3)5Cl]Cl2[Co(NH_3)_5Cl]Cl_2
  4. D.[Co(NH3)4Cl2]⋅Cl⋅NH3[Co(NH_3)_4Cl_2]\cdot Cl\cdot NH_3

Correct answer: C

Question 57

Chemistry · 4 marks · −1

The product (A) formed in the following reaction sequence is CH3−C≡CH→i) Hg2+,H2SO4; ii) HCN; iii) H2/NiCH_3-C\equiv CH\xrightarrow{\text{i) } Hg^{2+}, H_2SO_4;\ \text{ii) HCN;\ iii) } H_2/Ni} (A) Product

  1. A.CH3−CH2−CH(NH2)−CH2−OHCH_3-CH_2-CH(NH_2)-CH_2-OH
  2. B.CH3−C(OH)(CH3)−CH2−NH2CH_3-C(OH)(CH_3)-CH_2-NH_2
  3. C.CH3−C(NH2)(CH3)−CH2−OHCH_3-C(NH_2)(CH_3)-CH_2-OH
  4. D.CH3−CH2−CH(OH)−CH2−NH2CH_3-CH_2-CH(OH)-CH_2-NH_2

Correct answer: B

Explanation: Propyne → acetone → acetone cyanohydrin → 1-amino-2-methylpropan-2-ol.

Question 58

Chemistry · 4 marks · −1

Which one of the carbocations from the following is most stable?

Figure for question 58
  1. A.C+H2−CH=CH−O−CH2−CO−CH3\overset{+}{C}H_2-CH=CH-O-CH_2-CO-CH_3
  2. B.C+H2−CH=CH−CH2−O−CH3\overset{+}{C}H_2-CH=CH-CH_2-O-CH_3
  3. C.C+H2−CH=CH−O−CH2CH3\overset{+}{C}H_2-CH=CH-O-CH_2CH_3
  4. D.C+H2−CH=CH−F\overset{+}{C}H_2-CH=CH-F

Correct answer: C

Explanation: In option 3 the O lone pair conjugates with the allyl cation and the ethyl group is purely donating (no −I carbonyl nearby).

Question 59

Chemistry · 4 marks · −1

Which of the following ions is the strongest oxidizing agent? (Atomic Number of Ce = 58, Eu = 63, Tb = 65, Lu = 71)

  1. A.Eu2+Eu^{2+}
  2. B.Lu3+Lu^{3+}
  3. C.Tb4+Tb^{4+}
  4. D.Ce3+Ce^{3+}

Correct answer: C

Question 60

Chemistry · 4 marks · −1

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

  1. A.Both ΔH and ΔS are (−ve)
  2. B.ΔH is (−ve) but ΔS is (+ve)
  3. C.Both ΔH and ΔS are (+ve)
  4. D.ΔH is (+ve) but ΔS is (−ve)

Correct answer: C

Question 61

Chemistry · 4 marks · −1

For a reaction, N2O5(g)→2NO2(g)+12O2(g)N_2O_{5(g)}\to2NO_{2(g)}+\frac{1}{2}O_{2(g)} in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is

  1. A.7/2 times of initial pressure
  2. B.5/2 times of initial pressure
  3. C.5 times of initial pressure
  4. D.7/4 times of initial pressure

Correct answer: D

Explanation: P0/2+P0+P0/4=74P0P_0/2+P_0+P_0/4=\frac74P_0.

Question 62

Chemistry · 4 marks · −1

For the given cell Fe(aq)2++Ag(aq)+→Fe(aq)3++Ag(s)Fe^{2+}_{(aq)}+Ag^+_{(aq)}\to Fe^{3+}_{(aq)}+Ag_{(s)}. The standard cell potential of the above reaction is Given: Ag++e−→AgAg^++e^-\to Ag E⊖=xE^\ominus=x V; Fe2++2e−→FeFe^{2+}+2e^-\to Fe E⊖=yE^\ominus=y V; Fe3++3e−→FeFe^{3+}+3e^-\to Fe E⊖=zE^\ominus=z V

  1. A.x + 2y
  2. B.x + y − z
  3. C.x + 2y − 3z
  4. D.y − 2x

Correct answer: C

Explanation: EFe3+/Fe2+∘=3z−2yE^\circ_{Fe^{3+}/Fe^{2+}}=3z-2y; Ecell=x−(3z−2y)=x+2y−3zE_{cell}=x-(3z-2y)=x+2y-3z.

Question 63

Chemistry · 4 marks · −1

Aman has been asked to synthesise the molecule 1-acetylcyclopentene (x). He thought of preparing the molecule using an aldol condensation reaction. He found a few cyclic alkenes in his laboratory. He thought of performing ozonolysis reaction on alkene to produce a dicarbonyl compound followed by aldol reaction to prepare "x". Predict the suitable alkene that can lead to the formation of "x".

Figure for question 63
  1. A.4-Methylcyclohexene (CH3 on the carbon para to the C=C)
  2. B.1-Methylcyclohexene (CH3 on a double-bond carbon)
  3. C.Methylenecyclohexane (exocyclic =CH2)
  4. D.3-Methylcyclohexene (CH3 on the allylic carbon)

Correct answer: B

Explanation: Ozonolysis of 1-methylcyclohexene gives 6-oxoheptanal, whose intramolecular aldol condensation gives 1-acetylcyclopentene.

Question 64

Chemistry · 4 marks · −1

The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of

  1. A.Atomicity
  2. B.Atomic size
  3. C.Electron gain enthalpy
  4. D.Electronegativity

Correct answer: A

Explanation: O2O_2 is diatomic, sulphur is S8S_8.

Question 65

Chemistry · 4 marks · −1

Which of the following statement is true with respect to H2OH_2O, NH3NH_3 and CH4CH_4 ? A. The central atoms of all the molecules are sp3sp^3 hybridized. B. The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5°, respectively. C. The increasing order of dipole moment is CH4<NH3<H2OCH_4<NH_3<H_2O. D. Both H2OH_2O and NH3NH_3 are Lewis acids and CH4CH_4 is a Lewis base. E. A solution of NH3NH_3 in H2OH_2O is basic. In this solution NH3NH_3 and H2OH_2O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below:

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

Correct answer: C

Question 66

Chemistry · 4 marks · −1

Which of the following statements are NOT true about the periodic table? A. The properties of elements are function of atomic weights. B. The properties of elements are function of atomic numbers. C. Elements having similar outer electronic configurations are arranged in same period. D. An element's location reflects the quantum numbers of the last filled orbital. E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled. Choose the correct answer from the options given below:

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

Correct answer: A

Question 67

Chemistry · 4 marks · −1

Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z- direction] ? A. 2pz2p_z and 2px2p_x B. 2s and 2px2p_x C. 3dxy3d_{xy} and 3dx2−y23d_{x^2-y^2} D. 2s and 2pz2p_z E. 2pz2p_z and 3dx2−y23d_{x^2-y^2}. Choose the correct answer from the options given below:

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

Correct answer: A

Question 68

Chemistry · 4 marks · −1

Given below are two statements I and II. Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound. Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc H2SO4H_2SO_4. 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.Statement I is true but Statement II is false
  3. C.Both Statement I and Statement II are true
  4. D.Statement I is false but Statement II is true

Correct answer: B

Question 69

Chemistry · 4 marks · −1

Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?

  1. A.p- nitrobenzaldehyde < benzaldehyde < p-tolualdehyde < acetophenone
  2. B.acetophenone < benzaldehyde < p-tolualdehyde < p- nitrobenzaldehyde
  3. C.benzaldehyde < acetophenone < p- nitrobenzaldehyde < p-tolualdehyde
  4. D.acetophenone < p-tolualdehyde < benzaldehyde < p- nitrobenzaldehyde

Correct answer: D

Question 70

Chemistry · 4 marks · −1

The carbohydrate "Ribose" present in DNA, is A. A pentose sugar B. present in pyranose from C. in "D" configuration D. a reducing sugar, when free E. in α\alpha–anomeric form. Choose the correct answer from the options given below:

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

Correct answer: A

Question 71

Chemistry · 4 marks · −1

37.8 g N2O5N_2O_5 was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K 2N2O5(g)⇌2N2O4(g)+O2(g)2N_2O_{5(g)}\rightleftharpoons2N_2O_{4(g)}+O_{2(g)}. The total pressure at equilibrium was found to be 18.65 bar. Then, Kp = ______ ×10−2\times10^{-2} [nearest integer] Assume N2O5N_2O_5 to behave ideally under these conditions. Given: R = 0.082 bar L mol−1mol^{-1} K−1K^{-1}

Correct answer: 962

Question 72

Chemistry · 4 marks · −1

Consider the following reaction occurring in the blast furnace: Fe3O4(s)+4CO(g)→3Fe(l)+4CO2(g)Fe_3O_{4(s)}+4CO_{(g)}\to3Fe_{(l)}+4CO_{2(g)}. 'x' kg of iron is produced when 2.32×1032.32\times10^3 kg Fe3O4Fe_3O_4 and 2.8×1022.8\times10^2 kg CO are brought together in the furnace. The value of 'x' is ______. (nearest integer) {Given: molar mass of Fe3O4Fe_3O_4 = 232 g mol−1mol^{-1}, molar mass of CO = 28 g mol−1mol^{-1}, molar mass of Fe = 56 g mol−1mol^{-1}}

Correct answer: 420

Explanation: CO 10 kmol (limiting) → 7.5 kmol Fe = 420 kg.

Question 73

Chemistry · 4 marks · −1

Xg of benzoic acid on reaction with aq NaHCO3NaHCO_3 released CO2CO_2 that occupied 11.2 L volume at STP. X is ______ g.

Correct answer: 61

Explanation: 0.5 mol × 122 = 61 g.

Question 74

Chemistry · 4 marks · −1

Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with K4[Fe(CN)6]K_4[Fe(CN)_6] is ______. Cu2+Cu^{2+}, Fe3+Fe^{3+}, Ba2+Ba^{2+}, Ca2+Ca^{2+}, NH4+NH_4^+, Mg2+Mg^{2+}, Zn2+Zn^{2+}

Correct answer: 3

Explanation: Cu2+Cu^{2+} (chocolate brown), Fe3+Fe^{3+} (Prussian blue), Zn2+Zn^{2+} (white/bluish-white): 3.

Question 75

Chemistry · 4 marks · −1

Standard entropies of X2X_2, Y2Y_2 and XY5XY_5 are 70, 50 and 110 J K−1K^{-1} mol−1mol^{-1} respectively. The temperature in Kelvin at which the reaction 12X2+52Y2⇌XY5\frac{1}{2}X_2+\frac{5}{2}Y_2\rightleftharpoons XY_5 ΔH⊖=−35\Delta H^\ominus=-35 kJ mol−1mol^{-1} will be at equilibrium is ______ (Nearest integer)

Correct answer: 700

Explanation: ΔS=110−35−125=−50\Delta S=110-35-125=-50; T=35000/50=700T=35000/50=700 K.

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