JEE Main 2025 — 28 Jan Shift 2 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
74
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If f(x)=∫1x1/4(1+x1/4)dxf(x)=\int\frac{1}{x^{1/4}(1+x^{1/4})}dx, f(0) = −6, then f(1) is equal to :

  1. A.4(log⁡e2−2)4(\log_e2-2)
  2. B.4(log⁡e2+2)4(\log_e2+2)
  3. C.log⁡e2+2\log_e2+2
  4. D.2−log⁡e22-\log_e2

Correct answer: A

Explanation: x=t4x=t^4: f=4(t22−t+ln⁡(1+t))+Cf=4\left(\frac{t^2}{2}-t+\ln(1+t)\right)+C, C = −6; f(1)=4(12−1+ln⁡2)−6=4(ln⁡2−2)f(1)=4(\frac12-1+\ln2)-6=4(\ln2-2).

Question 2

Mathematics · 4 marks · −1

Let A=[12−201]A=\begin{bmatrix}\frac{1}{\sqrt{2}} & -2\\ 0 & 1\end{bmatrix} and P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]P=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}, θ > 0. If B=PAPTB=PAP^T, C=PTB10PC=P^TB^{10}P and the sum of the diagonal elements of C is mn\frac{m}{n}, where gcd(m, n) = 1, then m + n is :

  1. A.65
  2. B.2049
  3. C.258
  4. D.127

Correct answer: A

Explanation: C=A10C=A^{10}; trace =(1/2)10+1=3332=(1/\sqrt2)^{10}+1=\frac{33}{32}: 65.

Question 3

Mathematics · 4 marks · −1

Bag B1B_1 contains 6 white and 4 blue balls, Bag B2B_2 contains 4 white and 6 blue balls, and Bag B3B_3 contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag B2B_2, is :

  1. A.13\frac{1}{3}
  2. B.25\frac{2}{5}
  3. C.23\frac{2}{3}
  4. D.415\frac{4}{15}

Correct answer: D

Explanation: 46+4+5=415\frac{4}{6+4+5}=\frac{4}{15}.

Question 4

Mathematics · 4 marks · −1

Let the coefficients of three consecutive terms TrT_r, Tr+1T_{r+1} and Tr+2T_{r+2} in the binomial expansion of (a+b)12(a+b)^{12} be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (34+43)12\left(\sqrt[4]{3}+\sqrt[3]{4}\right)^{12}. Then p + q is equal to :

  1. A.287
  2. B.283
  3. C.295
  4. D.299

Correct answer: B

Explanation: No three consecutive binomial coefficients of (a+b)12(a+b)^{12} are in G.P., so p = 0. Rational terms of (31/4+41/3)12(3^{1/4}+4^{1/3})^{12}: r = 0 gives 44=2564^4=256, r = 12 gives 33=273^3=27; q = 283. Hence p + q = 283.

Question 5

Mathematics · 4 marks · −1

Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is :

  1. A.13\frac{1}{3}
  2. B.14\frac{1}{4}
  3. C.23\frac{2}{3}
  4. D.12\frac{1}{2}

Correct answer: D

Question 6

Mathematics · 4 marks · −1

Let f:[0,3]→Af:[0, 3]\to A be defined by f(x)=2x3−15x2+36x+7f(x)=2x^3-15x^2+36x+7 and g:[0,∞)→Bg:[0, \infty)\to B be defined by g(x)=x2025x2025+1g(x)=\frac{x^{2025}}{x^{2025}+1}. If both the functions are onto and S={x∈Z:x∈AS=\{x\in\mathbf{Z}: x\in A or x∈B}x\in B\}, then n(S) is equal to :

  1. A.30
  2. B.31
  3. C.29
  4. D.36

Correct answer: A

Explanation: A = [7, 35] (f(0) = 7, max f(2) = 35); B = [0, 1): integers 0 and 7..35 → 30.

Question 7

Mathematics · 4 marks · −1

Let A, B, C be three points in xy-plane, whose position vector are given by 3i^+j^\sqrt{3}\hat{i}+\hat{j}, i^+3j^\hat{i}+\sqrt{3}\hat{j} and ai^+(1−a)j^a\hat{i}+(1-a)\hat{j} respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors OA⃗\vec{OA} and OB⃗\vec{OB} is 92\frac{9}{\sqrt{2}}, then the sum of all the possible values of a is :

  1. A.2
  2. B.1
  3. C.9/2
  4. D.0

Correct answer: B

Explanation: Bisector y = x; ∣2a−1∣2=92\frac{|2a-1|}{\sqrt2}=\frac{9}{\sqrt2} ⇒ a = 5 or −4: sum 1.

Question 8

Mathematics · 4 marks · −1

The square of the distance of the point (157,327,7)\left(\frac{15}{7}, \frac{32}{7}, 7\right) from the line x+13=y+35=z+57\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} in the direction of the vector i^+4j^+7k^\hat{i}+4\hat{j}+7\hat{k} is :

  1. A.44
  2. B.54
  3. C.66
  4. D.41

Correct answer: C

Question 9

Mathematics · 4 marks · −1

Two equal sides of an isosceles triangle are along −x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is :

  1. A.−210-2\sqrt{10}
  2. B.6
  3. C.−6
  4. D.12

Correct answer: B

Explanation: The third side makes equal angles with the two given sides (slopes 1/2 and −1): ∣m−121+m2∣=∣m+11−m∣\left|\frac{m-\frac12}{1+\frac m2}\right|=\left|\frac{m+1}{1-m}\right|, giving m2−6m+1=0m^2-6m+1=0 (the other branch adds no new value); sum of slopes 6.

Question 10

Mathematics · 4 marks · −1

If the components of a⃗=αi^+βj^+γk^\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} along and perpendicular to b⃗=3i^+j^−k^\vec{b}=3\hat{i}+\hat{j}-\hat{k} respectively, are 1611(3i^+j^−k^)\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) and 111(−4i^−5j^−17k^)\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k}), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is equal to :

  1. A.16
  2. B.18
  3. C.26
  4. D.23

Correct answer: C

Explanation: a⃗=111(44,11,−33)=(4,1,−3)\vec a=\frac{1}{11}(44,11,-33)=(4,1,-3): 26.

Question 11

Mathematics · 4 marks · −1

Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0xtf(t)dtg(x)=\int_0^xtf(t)dt. If g(x3)=x6+x7g(x^3)=x^6+x^7, then value of ∑r=115f(r3)\sum_{r=1}^{15}f(r^3) is :

  1. A.310
  2. B.270
  3. C.340
  4. D.320

Correct answer: A

Explanation: 3x2⋅x3f(x3)=6x5+7x63x^2\cdot x^3f(x^3)=6x^5+7x^6 ⇒ f(x3)=2+73xf(x^3)=2+\frac73x; ∑115=30+73⋅120=310\sum_{1}^{15}=30+\frac73\cdot120=310.

Question 12

Mathematics · 4 marks · −1

If A and B are the points of intersection of the circle x2+y2−8x=0x^2+y^2-8x=0 and the hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 and a point P moves on the line 2x − 3y + 4 = 0, then the centroid of △PAB\triangle PAB lies on the line :

  1. A.4x − 9y = 12
  2. B.9x − 9y = 32
  3. C.6x − 9y = 20
  4. D.x + 9y = 36

Correct answer: C

Explanation: Intersections in x ≥ 3: x = 6, y=±23y=\pm2\sqrt3. Centroid h=12+xP3h=\frac{12+x_P}{3}, k=yP3k=\frac{y_P}{3} with 2xP−3yP+4=02x_P-3y_P+4=0 ⇒ 6h−9k=206h-9k=20.

Question 13

Mathematics · 4 marks · −1

If ∑r=113{1sin⁡(π4+(r−1)π6)sin⁡(π4+rπ6)}=a3+b\sum_{r=1}^{13}\left\{\frac{1}{\sin\left(\frac{\pi}{4}+(r-1)\frac{\pi}{6}\right)\sin\left(\frac{\pi}{4}+\frac{r\pi}{6}\right)}\right\}=a\sqrt{3}+b, a, b ∈ Z, then a2+b2a^2+b^2 is equal to :

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

Correct answer: A

Question 14

Mathematics · 4 marks · −1

Let f : R→R be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all x∈R, ∫02xF′(x)dx=6\int_0^2xF'(x)dx=6 and ∫02x2F′′(x)dx=40\int_0^2x^2F''(x)dx=40, then F′(2)+∫02F(x)dxF'(2)+\int_0^2F(x)dx is equal to :

  1. A.15
  2. B.13
  3. C.11
  4. D.9

Correct answer: C

Explanation: F(2) = 2. \int xF\' = 2F(2)-\int F=6 ⇒ ∫F=−2\int F=-2. \int x^2F\'\'=4F\'(2)-2\int xF\'=4F\'(2)-12=40 ⇒ F\'(2) = 13; answer 11.

Question 15

Mathematics · 4 marks · −1

For positive integers n, if 4an=(n2+5n+6)4a_n=(n^2+5n+6) and Sn=∑k=1n(1ak)S_n=\sum_{k=1}^{n}\left(\frac{1}{a_k}\right), then the value of 507S2025507S_{2025} is :

  1. A.540
  2. B.1350
  3. C.675
  4. D.135

Correct answer: C

Explanation: 1ak=4(1k+2−1k+3)\frac{1}{a_k}=4\left(\frac1{k+2}-\frac1{k+3}\right): S=4(13−12028)=675507S=4(\frac13-\frac1{2028})=\frac{675}{507}.

Question 16

Mathematics · 4 marks · −1

If the midpoint of a chord of the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 is (2,4/3)(\sqrt{2}, 4/3), and the length of the chord is 2α3\frac{2\sqrt{\alpha}}{3}, then α is :

  1. A.20
  2. B.22
  3. C.18
  4. D.26

Correct answer: B

Question 17

Mathematics · 4 marks · −1

Let f : R − {0} → (−∞, 1) be a polynomial of degree 2, satisfying f(x)f(1x)=f(x)+f(1x)f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right). If f(K) = −2K, then the sum of squares of all possible values of K is :

  1. A.7
  2. B.1
  3. C.9
  4. D.6

Correct answer: D

Explanation: f=1−x2f=1-x^2 (range ≤ 1): 1−K2=−2K1-K^2=-2K ⇒ K=1±2K=1\pm\sqrt2; sum of squares 6.

Question 18

Mathematics · 4 marks · −1

If α + iβ and γ + iδ are the roots of x2−(3−2i)x−(2i−2)=0x^2-(3-2i)x-(2i-2)=0, i=−1i=\sqrt{-1}, then αγ + βδ is equal to :

  1. A.2
  2. B.−6
  3. C.6
  4. D.−2

Correct answer: A

Explanation: Roots 2 and 1 − 2i: αγ + βδ = 2·1 + 0·(−2) = 2.

Question 19

Mathematics · 4 marks · −1

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x)=sec⁡−1(2[x]+1)f(x)=\sec^{-1}(2[x]+1) is :

  1. A.(−∞, −1]∪[0, ∞)
  2. B.(−∞, −1]∪[1, ∞)
  3. C.(−∞, ∞) − {0}
  4. D.(−∞, ∞)

Correct answer: D

Explanation: ∣2[x]+1∣≥1|2[x]+1|\ge1 always holds for integer [x].

Question 20

Mathematics · 4 marks · −1

The area of the region bounded by the curves x(1+y2)=1x(1+y^2)=1 and y2=2xy^2=2x is:

  1. A.π4−13\frac{\pi}{4}-\frac{1}{3}
  2. B.π2−13\frac{\pi}{2}-\frac{1}{3}
  3. C.2(π2−13)2\left(\frac{\pi}{2}-\frac{1}{3}\right)
  4. D.12(π2−13)\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{3}\right)

Correct answer: B

Explanation: Meet at y=±1y=\pm1: ∫−11(11+y2−y22)dy=π2−13\int_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy=\frac\pi2-\frac13.

Question 21

Mathematics · 4 marks · −1

The interior angles of a polygon with n sides, are in an A.P. with common difference 6°. If the largest interior angle of the polygon is 219°, then n is equal to ______.

Correct answer: 20

Question 22

Mathematics · 4 marks · −1

Let A and B be the two points of intersection of the line y + 5 = 0 and the mirror image of the parabola y2=4xy^2=4x with respect to the line x + y + 4 = 0. If d denotes the distance between A and B, and a denotes the area of △SAB\triangle SAB, where S is the focus of the parabola y2=4xy^2=4x, then the value of (a + d) is ______.

Correct answer: 14

Question 23

Mathematics · 4 marks · −1

If y = y(x) is the solution of the differential equation, 4−x2dydx=((sin⁡−1(x2))2−y)sin⁡−1(x2)\sqrt{4-x^2}\frac{dy}{dx}=\left(\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2-y\right)\sin^{-1}\left(\frac{x}{2}\right), −2≤x≤2-2\le x\le2, y(2)=π2−84y(2)=\frac{\pi^2-8}{4}, then y2(0)y^2(0) is equal to ______.

Correct answer: 4

Question 24

Mathematics · 4 marks · −1

Let f(x)=lim⁡n→∞∑r=0n(tan⁡(x/2r+1)+tan⁡3(x/2r+1)1−tan⁡2(x/2r+1))f(x)=\lim_{n\to\infty}\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1})+\tan^3(x/2^{r+1})}{1-\tan^2(x/2^{r+1})}\right). Then lim⁡x→0ex−ef(x)(x−f(x))\lim_{x\to0}\frac{e^x-e^{f(x)}}{(x-f(x))} is equal to ______.

Correct answer: 1

Explanation: Each term =tan⁡x2r−tan⁡x2r+1=\tan\frac{x}{2^r}-\tan\frac{x}{2^{r+1}}, so f(x) = tan x; the limit → e0=1e^0=1.

Question 25

Mathematics · 4 marks · −1

The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is ______.

Correct answer: 64

Question 26

Physics · 4 marks · −1

A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is '−3', then the magnitude of the radius of curvature of the mirror is :

  1. A.3.75 cm
  2. B.30 cm
  3. C.7.5 cm
  4. D.15 cm

Correct answer: D

Explanation: v = 3u, |v| − |u| = 20 ⇒ u = −10, v = −30; f = −7.5, R = 15 cm.

Question 27

Physics · 4 marks · −1

A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between plates of capacitor is.

  1. A.2×102 J/m32\times10^2\ J/m^3
  2. B.1.8×103 J/m31.8\times10^3\ J/m^3
  3. C.2×10−4 J/m32\times10^{-4}\ J/m^3
  4. D.1.8×105 J/m31.8\times10^5\ J/m^3

Correct answer: B

Explanation: E=2×107E=2\times10^7 V/m; u=12ϵ0E2=1.77×103u=\frac12\epsilon_0E^2=1.77\times10^3 J/m³.

Question 28

Physics · 4 marks · −1

In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VABV_{AB} is correctly represented by :

Figure for question 28
  1. A.Positive half-cycles only (half-wave rectified above the axis)
  2. B.Full sine wave
  3. C.VABV_{AB} would be zero at all times
  4. D.Negative half-cycles only (below the axis)

Correct answer: D

Explanation: The diode conducts only on the half-cycle that makes A negative with respect to B, so VABV_{AB} shows only negative half-cycles.

Question 29

Physics · 4 marks · −1

An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by :

Figure for question 29
  1. A.μ02πIa[π2+2]\frac{\mu_0}{2\pi}\frac{I}{a}\left[\frac{\pi}{2}+2\right]
  2. B.μ04πIa[3π2+2]\frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{3\pi}{2}+2\right]
  3. C.μ04πIa[π2+1]\frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{\pi}{2}+1\right]
  4. D.μ04πIa[3π2+1]\frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{3\pi}{2}+1\right]

Correct answer: D

Explanation: Arc (3/4 circle): 3μ0I8a\frac{3\mu_0I}{8a}; lower semi-infinite wire with foot at its end: μ0I4πa\frac{\mu_0I}{4\pi a}; the upper wire passes through O’s line (no field). Total μ0I4πa(3π2+1)\frac{\mu_0I}{4\pi a}\left(\frac{3\pi}{2}+1\right).

Question 30

Physics · 4 marks · −1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Knowing initial position x0x_0 and initial momentum p0p_0 is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω. Reason (R) : The amplitude and phase can be expressed in terms of x0x_0 and p0p_0. In the light of the above statements, choose the correct answer from the options given below :

  1. A.(A) is true but (R) is false
  2. B.Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  3. C.Both (A) and (R) are true and (R) is the correct explanation of (A)
  4. D.(A) is false but (R) is true

Correct answer: C

Question 31

Physics · 4 marks · −1

The ratio of vapour densities of two gases at the same temperature is 425\frac{4}{25}, then the ratio of r.m.s. velocities will be :

  1. A.52\frac{5}{2}
  2. B.25\frac{2}{5}
  3. C.254\frac{25}{4}
  4. D.425\frac{4}{25}

Correct answer: A

Explanation: v∝1/Mv\propto1/\sqrt M: 5/2.

Question 32

Physics · 4 marks · −1

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be :

  1. A.5.6
  2. B.11.2
  3. C.2.8
  4. D.8.4

Correct answer: A

Explanation: v∝M/Rv\propto\sqrt{M/R}: factor 4=2\sqrt{4}=2 ⇒ 5.6.

Question 33

Physics · 4 marks · −1

The kinetic energy of translation of the molecules in 50 g of CO2CO_2 gas at 17°C is

  1. A.4102.8 J
  2. B.3986.3 J
  3. C.4205.5 J
  4. D.3582.7 J

Correct answer: A

Explanation: 32nRT=32⋅5044⋅8.314⋅290=4102.8\frac32nRT=\frac32\cdot\frac{50}{44}\cdot8.314\cdot290=4102.8 J.

Question 34

Physics · 4 marks · −1

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 π rad s−1s^{-1} about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim ? (π = 3.14)

  1. A.0.5024 V
  2. B.0.1256 V
  3. C.0.0628 V
  4. D.0.2512 V

Correct answer: D

Explanation: 12Bωr2=0.5(0.4)(31.4)(0.04)=0.2512\frac12B\omega r^2=0.5(0.4)(31.4)(0.04)=0.2512 V.

Question 35

Physics · 4 marks · −1

A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force F⃗=(2i^+3j^)\vec{F}=(2\hat{i}+3\hat{j})N, it moves to a new point Q having coordinates (6, 10)m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of :

  1. A.13 : 6
  2. B.4 : 3
  3. C.1 : 2
  4. D.6 : 13

Correct answer: D

Explanation: W = F·Δr = 6 + 18 = 24 J, so Pavg=6P_{avg}=6 W. Taking the body to start from rest, v⃗(4)=F⃗mt=(2,3)\vec v(4)=\frac{\vec F}{m}t=(2,3) m/s and P=F⃗⋅v⃗=13P=\vec F\cdot\vec v=13 W: ratio 6 : 13.

Question 36

Physics · 4 marks · −1

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '−2' then the radius of curvature of meniscus is :

  1. A.23\frac{2}{3} cm
  2. B.1 cm
  3. C.13\frac{1}{3} cm
  4. D.43\frac{4}{3} cm

Correct answer: A

Explanation: m=n1vn2u=−2m=\frac{n_1v}{n_2u}=-2 ⇒ v=2⋅1.4⋅131.3=28v=\frac{2\cdot1.4\cdot13}{1.3}=28 cm; 1.428+1.313=0.1R\frac{1.4}{28}+\frac{1.3}{13}=\frac{0.1}{R} ⇒ R=23R=\frac23 cm.

Question 37

Physics · 4 marks · −1

A bar magnet has total length 2l = 20 units and the field point P is at a distance d = 10 units from the centre of the magnet. If the relative uncertainty of length measurement is 1%, then uncertainty of the magnetic field at point P is :

Figure for question 37
  1. A.3%
  2. B.4%
  3. C.5%
  4. D.10%

Correct answer: B

Explanation: B=μ0m4π(d2+l2)3/2B=\frac{\mu_0m}{4\pi(d^2+l^2)^{3/2}} with m∝lm\propto l: ΔBB=Δll+32⋅2dΔd+2lΔld2+l2=1%+3%=4%\frac{\Delta B}{B}=\frac{\Delta l}{l}+\frac32\cdot\frac{2d\Delta d+2l\Delta l}{d^2+l^2}=1\%+3\%=4\%.

Question 38

Physics · 4 marks · −1

Which of the following phenomena can not be explained by wave theory of light ?

  1. A.Compton effect
  2. B.Diffraction of light
  3. C.Reflection of light
  4. D.Refraction of light

Correct answer: A

Question 39

Physics · 4 marks · −1

The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as

  1. A.1n\frac{1}{n}
  2. B.1n4\frac{1}{n^4}
  3. C.1n2\frac{1}{n^2}
  4. D.1n3\frac{1}{n^3}

Correct answer: D

Question 40

Physics · 4 marks · −1

The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4s ?

Figure for question 40
  1. A.13 m
  2. B.30 m
  3. C.11 m
  4. D.10 m

Correct answer: B

Explanation: 12(2)(10)+2(10)=30\frac12(2)(10)+2(10)=30 m.

Question 41

Physics · 4 marks · −1

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water ? (Given : density of water = 1000 kg m−3m^{-3})

  1. A.4000 cm3cm^3
  2. B.600 cm3cm^3
  3. C.400 cm3cm^3
  4. D.1400 cm3cm^3

Correct answer: B

Explanation: Submerged 400 cm³ of 1000 cm³: 600 cm³ outside.

Question 42

Physics · 4 marks · −1

The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]\vec{B}=\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)30\sin\left[\omega\left(t-\frac{z}{c}\right)\right] (S.I. Units). The corresponding electric field in S.I. units is :

  1. A.E⃗=(12i^−32j^)30csin⁡[ω(t−zc)]\vec{E}=\left(\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]
  2. B.E⃗=(34i^+14j^)30ccos⁡[ω(t−zc)]\vec{E}=\left(\frac{3}{4}\hat{i}+\frac{1}{4}\hat{j}\right)30c\cos\left[\omega\left(t-\frac{z}{c}\right)\right]
  3. C.E⃗=(12i^+32j^)30csin⁡[ω(t+zc)]\vec{E}=\left(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t+\frac{z}{c}\right)\right]
  4. D.E⃗=(32i^−12j^)30csin⁡[ω(t+zc)]\vec{E}=\left(\frac{\sqrt{3}}{2}\hat{i}-\frac{1}{2}\hat{j}\right)30c\sin\left[\omega\left(t+\frac{z}{c}\right)\right]

Correct answer: A

Explanation: Wave along +z; E⃗×B⃗\vec E\times\vec B must be along +z: E^=12i^−32j^\hat E=\frac12\hat i-\frac{\sqrt3}{2}\hat j.

Question 43

Physics · 4 marks · −1

A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be (Take 'g' as acceleration due to gravity)

  1. A.3Ma2a+g\frac{3Ma}{2a+g}
  2. B.2Ma3a+g\frac{2Ma}{3a+g}
  3. C.2Ma3a−g\frac{2Ma}{3a-g}
  4. D.3Ma2a−g\frac{3Ma}{2a-g}

Correct answer: B

Explanation: B=M(g+a)=(M−m)(g+3a)B=M(g+a)=(M-m)(g+3a) ⇒ m=2Mag+3am=\frac{2Ma}{g+3a}.

Question 44

Physics · 4 marks · −1

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

  1. A.290 g
  2. B.200 g
  3. C.190 g
  4. D.300 g

Correct answer: C

Explanation: 400(30)=250(10)+m(50)400(30)=250(10)+m(50) ⇒ m = 190 g.

Question 45

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Angular Impulse (B) Latent Heat (C) Electrical resistivity (D) Electromotive force. List - II: (I) [M0L2T−2][M^0L^2T^{-2}] (II) [ML2T−3A−1][ML^2T^{-3}A^{-1}] (III) [ML2T−1][ML^2T^{-1}] (IV) [ML3T−3A−2][ML^3T^{-3}A^{-2}]. Choose the correct answer from the options given below :

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

Correct answer: A

Question 46

Physics · 4 marks · −1

An electric dipole of dipole moment 6×10−66\times10^{-6} Cm is placed in uniform electric field of magnitude 10610^6 V/m. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.

Correct answer: 12

Explanation: 2pE=122pE=12 J.

Question 47

Physics · 4 marks · −1

The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ______ A.

Figure for question 47

Correct answer: 2

Explanation: Balanced: R = 20 Ω. Arms 30 Ω and 60 Ω in parallel = 20 Ω; I = 40/20 = 2 A.

Question 48

Physics · 4 marks · −1

The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of 7×1067\times10^6 Pa, would be ______ mm3mm^3. (Given bulk modulus of copper = 1.4×10111.4\times10^{11} N m−2m^{-2})

Correct answer: 50

Explanation: ΔV=PVB=7×106×10−31.4×1011=5×10−8\Delta V=\frac{PV}{B}=\frac{7\times10^6\times10^{-3}}{1.4\times10^{11}}=5\times10^{-8} m³ = 50 mm³.

Question 49

Physics · 4 marks · −1

A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝ tnt^n, then value of n is ______.

Figure for question 49

Correct answer: 1

Explanation: Length of bar in the circuit ∝ distance ∝ t: E = Bℓv ∝ t, n = 1.

Question 50

Physics · 4 marks · −1

A thin transparent film with refractive index 1.4, is held on circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ______ π×10−13m3/s\pi\times10^{-13}m^3/s.

Correct answer: 54

Explanation: Thickness change per minimum =λ2μ=200=\frac{\lambda}{2\mu}=200 nm; rate =π(0.018)2×2×10−712=5.4π×10−12=54π×10−13=\frac{\pi(0.018)^2\times2\times10^{-7}}{12}=5.4\pi\times10^{-12}=54\pi\times10^{-13}.

Question 51

Chemistry · 4 marks · −1

Identify the inorganic sulphides that are yellow in colour : (A) (NH4)2S(NH_4)_2S (B) PbS (C) CuS (D) As2S3As_2S_3 (E) As2S5As_2S_5. Choose the correct answer from the options given below :

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

Correct answer: C

Question 52

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Complex): (A) [CoF6]3−[CoF_6]^{3-} (B) [NiCl4]2−[NiCl_4]^{2-} (C) [Co(NH3)6]3+[Co(NH_3)_6]^{3+} (D) [Ni(CN)4]2−[Ni(CN)_4]^{2-}. List - II (Hybridisation of central metal ion): (I) d2sp3d^2sp^3 (II) sp3sp^3 (III) sp3d2sp^3d^2 (IV) dsp2dsp^2. Choose the correct answer from the options given below :

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

Correct answer: C

Question 53

Chemistry · 4 marks · −1

The product B formed in the following reaction sequence is : 1-methyl-4-(prop-1-enyl)benzene →HCl\xrightarrow{HCl} (A) (Major) →AgCN\xrightarrow{AgCN} (B) (Major)

Figure for question 53
  1. A.p-Tolyl–CH2–CH(NC)–CH3 (isocyanide on C-2 of the side chain)
  2. B.p-Tolyl–CH2–CH(NC)–CH3 (same skeleton, drawn with NC below)
  3. C.p-Tolyl–CH(NC)–CH2CH3 (benzylic isocyanide)
  4. D.p-Tolyl–CH(CN)–CH3 (benzylic nitrile, one carbon shorter chain)

Correct answer: C

Explanation: HCl adds to give the benzylic chloride (stable benzylic cation); AgCN gives the isocyanide at the benzylic carbon.

Question 54

Chemistry · 4 marks · −1

Arrange the following in increasing order of solubility product : Ca(OH)2Ca(OH)_2, AgBr, PbS, HgS

  1. A.HgS < PbS < AgBr < Ca(OH)2Ca(OH)_2
  2. B.HgS < AgBr < PbS < Ca(OH)2Ca(OH)_2
  3. C.Ca(OH)2Ca(OH)_2 < AgBr < HgS < PbS
  4. D.PbS < HgS < Ca(OH)2Ca(OH)_2 < AgBr

Correct answer: A

Question 55

Chemistry · 4 marks · −1

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth ? Where N - Number of Bacteria at any time, N0N_0 - Initial number of Bacteria.

Figure for question 55
  1. A.N/N₀ rises from 0 and saturates (concave down)
  2. B.N/N₀ decays (concave up, decreasing)
  3. C.N/N₀ rises exponentially from a positive intercept (concave up)
  4. D.N/N₀ decreases linearly

Correct answer: C

Question 56

Chemistry · 4 marks · −1

Identify correct conversion during acidic hydrolysis from the following : (A) starch gives galactose. (B) cane sugar gives equal amount of glucose and fructose. (C) milk sugar gives glucose and galactose. (D) amylopectin gives glucose and fructose. (E) amylose gives only glucose. Choose the correct answer from the options given below :

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

Correct answer: A

Question 57

Chemistry · 4 marks · −1

The purification method based on the following physical transformation is : Solid (X) →Heat\xrightarrow{\text{Heat}} Vapour (X) →Cool\xrightarrow{\text{Cool}} Solid (X)

  1. A.Sublimation
  2. B.Distillation
  3. C.Crystallization
  4. D.Extraction

Correct answer: A

Question 58

Chemistry · 4 marks · −1

The amphoteric oxide among V2O3V_2O_3, V2O4V_2O_4 and V2O5V_2O_5, upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

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

Correct answer: A

Explanation: NTA key: +5, i.e. the amphoteric oxide is taken as V2O5V_2O_5, which dissolves in alkali as vanadate VO43−VO_4^{3-} (V +5).

Question 59

Chemistry · 4 marks · −1

Consider an elementary reaction A(g) + B(g) → C(g) + D(g). If the volume of reaction mixture is suddenly reduced to 13\frac{1}{3} of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is :

  1. A.3
  2. B.13\frac{1}{3}
  3. C.9
  4. D.19\frac{1}{9}

Correct answer: C

Question 60

Chemistry · 4 marks · −1

Identify product [A], [B] and [C] in the following reaction sequence. CH3−C≡CH→H2Pd/CCH_3-C\equiv CH\xrightarrow[H_2]{Pd/C} [A] →(i) O3; (ii) Zn, H2O\xrightarrow{\text{(i) } O_3;\ \text{(ii) Zn, } H_2O} [B] + [C]

  1. A.[A] : CH2=CH2CH_2=CH_2, [B] : CH3COCH3CH_3COCH_3, [C] : HCHO
  2. B.[A] : CH3−CH=CH2CH_3-CH=CH_2, [B] : CH3CHOCH_3CHO, [C] : HCHO
  3. C.[A] : CH3CH2CH3CH_3CH_2CH_3, [B] : CH3CHOCH_3CHO, [C] : HCHO
  4. D.[A] : CH3−CH=CH2CH_3-CH=CH_2, [B] : CH3CHOCH_3CHO, [C] : CH3CH2OHCH_3CH_2OH

Correct answer: B

Question 61

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number. Statement (II) : Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers. 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 false but Statement II is true
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are true

Correct answer: A

Question 62

Chemistry · 4 marks · −1

The major product of the following reaction is : C6H5−CH(CH3)−CH(Br)−CH2−CH(Br)−CH2CH3C_6H_5-CH(CH_3)-CH(Br)-CH_2-CH(Br)-CH_2CH_3 →KOH/EtOH (excess),Δ\xrightarrow{\text{KOH/EtOH (excess)}, \Delta} Major product

Figure for question 62
  1. A.2-Phenylhepta-2,4-diene
  2. B.2-Phenylhepta-2,5-diene
  3. C.6-Phenylhepta-3,5-diene
  4. D.6-Phenylhepta-2,4-diene

Correct answer: A

Explanation: Eliminations give the conjugated diene also conjugated with the ring: Ph–C(CH3)=CH–CH=CH–CH2CH3 (2-phenylhepta-2,4-diene).

Question 63

Chemistry · 4 marks · −1

Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______. Given : Density of nitric acid solution is 1.25 g/mL.

  1. A.32
  2. B.55
  3. C.45
  4. D.40

Correct answer: A

Explanation: 40 g solution / 1.25 = 32 mL.

Question 64

Chemistry · 4 marks · −1

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases above. Choose the correct option regarding ΔU :

Figure for question 64
  1. A.ΔU(Case-III) > ΔU(Case-II) > ΔU(Case-I)
  2. B.ΔU(Case-I) > ΔU(Case-III) > ΔU(Case-II)
  3. C.ΔU(Case-I) > ΔU(Case-II) > ΔU(Case-III)
  4. D.ΔU(Case-I) = ΔU(Case-II) = ΔU(Case-III)

Correct answer: D

Explanation: ΔU = 0 for every cycle.

Question 65

Chemistry · 4 marks · −1

The total number of compounds from below when treated with hot KMnO4KMnO_4 giving benzoic acid is : toluene, ethylbenzene, isopropylbenzene (cumene), tert-butylbenzene, isobutylbenzene, 2-phenylpropan-2-ol, propylbenzene, 4,4′-dimethylbiphenyl.

Figure for question 65
  1. A.6
  2. B.5
  3. C.4
  4. D.3

Correct answer: B

Explanation: Benzylic H present: toluene, ethylbenzene, cumene, isobutylbenzene, propylbenzene (5). tert-Butylbenzene and 2-phenylpropan-2-ol lack benzylic H; the biphenyl gives a dicarboxylic acid.

Question 66

Chemistry · 4 marks · −1

Identify correct statements : (A) Primary amines do not give diazonium salts when treated with NaNO2NaNO_2 in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl3CHCl_3 and ethanolic KOH form carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily. Choose the correct answer from the options given below :

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

Correct answer: B

Question 67

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : methylcyclopentane and hex-2-ene (drawn as skeletal structures) are isomeric compounds. Statement (II) : pentan-1-amine (CH3(CH2)4NH2CH_3(CH_2)_4NH_2) and N-ethylpropan-1-amine... drawn as CH3CH2CH2−NH−CH2CH3CH_3CH_2CH_2-NH-CH_2CH_3 are functional group isomers. 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 false but Statement II is true
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are true

Correct answer: D

Explanation: Both are C6H12C_6H_{12} (I true); primary and secondary amines of the same formula are treated as functional isomers (NCERT) (II true).

Question 68

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Saccharides): (A) Sucrose (B) Maltose (C) Lactose (D) Amylopectin. List - II (Glycosidic-linkages found): (I) α 1−4 (II) α 1−4 and α 1−6 (III) α 1−β 2 (IV) β 1−4. Choose the correct answer from the options given below :

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

Correct answer: C

Question 69

Chemistry · 4 marks · −1

Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom ? (A) 1s < 2p < 3d < 4s (B) 1s < 2s = 2p < 3s = 3p (C) 1s < 2s < 2p < 3s < 3p (D) 1s < 2s < 4s < 3d. Choose the correct answer from the options given below :

  1. A.(A) and (C) only
  2. B.(B) and (D) only
  3. C.(A) and (B) only
  4. D.(C) and (D) only

Correct answer: D

Question 70

Chemistry · 4 marks · −1

Consider the following data : Heat of formation of CO2CO_2(g) = −393.5 kJ mol−1mol^{-1}; Heat of formation of H2OH_2O(l) = −286.0 kJ mol−1mol^{-1}; Heat of combustion of benzene = −3267.0 kJ mol−1mol^{-1}. The heat of formation of benzene is ______ kJ mol−1mol^{-1}. (Nearest integer)

Correct answer: 48

Explanation: 6(−393.5)+3(−286)+3267=486(-393.5)+3(-286)+3267=48.

Question 71

Chemistry · 4 marks · −1

Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ______. (Nearest integer).

Correct answer: 2

Explanation: OH−OH^- = 0.01 × 0.6 = 6 mmol = 6 mmol e⁻; I=0.006×96500300=1.93I=\frac{0.006\times96500}{300}=1.93 ≈ 2 A.

Question 72

Chemistry · 4 marks · −1

A group 15 element forms dπ − dπ bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form dπ − dπ bond. The atomic number of the element is ______.

Correct answer: 15

Explanation: P (forms dπ–dπ; PH3PH_3 is more basic than AsH3AsH_3, SbH3SbH_3): 15.

Question 73

Chemistry · 4 marks · −1

Total number of molecules/species from following which will be paramagnetic is ______. O2O_2, O2+O_2^+, O2−O_2^-, NO, NO2NO_2, CO, K2[NiCl4]K_2[NiCl_4], [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3, K2[Ni(CN)4]K_2[Ni(CN)_4]

Correct answer: 6

Explanation: O2O_2, O2+O_2^+, O2−O_2^-, NO, NO2NO_2, [NiCl4]2−[NiCl_4]^{2-} = 6.

Question 74

Chemistry · 4 marks · −1

The spin only magnetic moment (μ) value (B.M.) of the compound with strongest oxidising power among Mn2O3Mn_2O_3, TiO and VO is ______ B.M. (Nearest integer).

Correct answer: 5

Explanation: Mn2O3Mn_2O_3 (Mn3+Mn^{3+} d4d^4): 24=4.9\sqrt{24}=4.9 ≈ 5.

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