JEE Main 2025 — 8 Apr 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

The sum of the squares of the roots of ∣x−2∣2+∣x−2∣−2=0|x-2|^2+|x-2|-2=0 and the squares of the roots of x2−2∣x−3∣−5=0x^2-2|x-3|-5=0, is

  1. A.36
  2. B.26
  3. C.30
  4. D.24

Correct answer: A

Explanation: First: |x−2| = 1 ⇒ x = 1, 3 (squares 10). Second: x ≥ 3: x2−2x+1=0x^2-2x+1=0 (x = 1 rejected); x < 3: x2+2x−11=0x^2+2x-11=0 ⇒ x=−1±23x=-1\pm2\sqrt3 (both < 3), squares sum (−2)2+22=26(−2)^2+22=26; total 36.

Question 2

Mathematics · 4 marks · −1

The integral ∫−13/2(∣π2xsin⁡(πx)∣)dx\int_{-1}^{3/2}\left(\left|\pi^2x\sin(\pi x)\right|\right)dx is equal to :

  1. A.3+2π3+2\pi
  2. B.1+3π1+3\pi
  3. C.2+3π2+3\pi
  4. D.4+π4+\pi

Correct answer: B

Question 3

Mathematics · 4 marks · −1

If 114+124+134+…∞=π490\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots\infty=\frac{\pi^4}{90}, 114+134+154+…∞=α\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots\infty=\alpha, 124+144+164+…∞=β\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots\infty=\beta, then αβ\frac{\alpha}{\beta} is equal to

  1. A.14
  2. B.23
  3. C.18
  4. D.15

Correct answer: D

Explanation: β = S/16, α = 15S/16: ratio 15.

Question 4

Mathematics · 4 marks · −1

Let A={θ∈[0,2π]:1+10 Re(2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=0}A=\left\{\theta\in[0, 2\pi]: 1+10\,\mathrm{Re}\left(\frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right)=0\right\}. Then ∑θ∈Aθ2\sum_{\theta\in A}\theta^2 is equal to

  1. A.6π26\pi^2
  2. B.274π2\frac{27}{4}\pi^2
  3. C.214π2\frac{21}{4}\pi^2
  4. D.8π28\pi^2

Correct answer: C

Explanation: Re =2cos⁡2θ−3sin⁡2θcos⁡2θ+9sin⁡2θ=\frac{2\cos^2\theta-3\sin^2\theta}{\cos^2\theta+9\sin^2\theta}; equation gives tan⁡2θ=1\tan^2\theta=1: θ = π/4, 3π/4, 5π/4, 7π/4; sum of squares 84π216=21π24\frac{84\pi^2}{16}=\frac{21\pi^2}{4}.

Question 5

Mathematics · 4 marks · −1

Let a⃗=i^+2j^+k^\vec{a}=\hat{i}+2\hat{j}+\hat{k} and b⃗=2i^+j^−k^\vec{b}=2\hat{i}+\hat{j}-\hat{k}. Let c^\hat{c} be a unit vector in the plane of the vectors a⃗\vec{a} and b⃗\vec{b} and be perpendicular to a⃗\vec{a}. Then such a vector c^\hat{c} is :

  1. A.13(i^−j^+k^)\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})
  2. B.15(j^−2k^)\frac{1}{\sqrt{5}}(\hat{j}-2\hat{k})
  3. C.12(−i^+k^)\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})
  4. D.13(−i^+j^−k^)\frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}-\hat{k})

Correct answer: C

Explanation: a⃗×(a⃗×b⃗)=(1,2,1)×(−3,3,−3)=(−9,0,9)\vec a\times(\vec a\times\vec b)=(1,2,1)\times(-3,3,-3)=(-9,0,9), which lies in the plane of a, b and is ⟂ a: c^=12(−i^+k^)\hat c=\frac{1}{\sqrt2}(-\hat i+\hat k).

Question 6

Mathematics · 4 marks · −1

The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1(1+tan⁡2(12)+1tan⁡(12))\cot^{-1}\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right)-\cot^{-1}\left(\frac{\sqrt{1+\tan^2\left(\frac{1}{2}\right)}+1}{\tan\left(\frac{1}{2}\right)}\right) is equal to

  1. A.π+52\pi+\frac{5}{2}
  2. B.π−52\pi-\frac{5}{2}
  3. C.π+32\pi+\frac{3}{2}
  4. D.π−32\pi-\frac{3}{2}

Correct answer: B

Question 7

Mathematics · 4 marks · −1

Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle α with the positive x-axis and the equations of its diagonals are (3+1)x+(3−1)y=0(\sqrt{3}+1)x+(\sqrt{3}-1)y=0 and (3−1)x−(3+1)y+83=0(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0. Then a2a^2 is equal to

  1. A.24
  2. B.48
  3. C.32
  4. D.16

Correct answer: B

Explanation: Distance of O from the other diagonal =838=26=\frac{8\sqrt3}{\sqrt8}=2\sqrt6 = half-diagonal; a2=2(26)2=48a^2=2(2\sqrt6)^2=48.

Question 8

Mathematics · 4 marks · −1

Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A defined by (x, y) ∈ R if and only if max{x, y} ∈ {3, 4}. Then among the statements (S₁) : The number of elements in R is 18, and (S₂) : The relation R is symmetric but neither reflexive nor transitive

  1. A.only (S1S_1) is true
  2. B.only (S2S_2) is true
  3. C.both are true
  4. D.both are false

Correct answer: B

Explanation: Pairs with max 3: 16 − 9 = 7; max 4: 25 − 16 = 9; total 16, so S₁ is false. R is symmetric, not reflexive ((0,0) ∉ R) and not transitive ((0,3),(3,0) ∈ R but (0,0) ∉ R), so S₂ is true.

Question 9

Mathematics · 4 marks · −1

Let the values of λ for which 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 x−λ3=y−44=z−55\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5} is 16\frac{1}{\sqrt{6}} be λ1\lambda_1 and λ2\lambda_2. Then the radius of the circle passing through the points (0, 0), (λ1,λ2)(\lambda_1, \lambda_2) and (λ2,λ1)(\lambda_2, \lambda_1) is

  1. A.3
  2. B.4
  3. C.523\frac{5\sqrt{2}}{3}
  4. D.23\frac{\sqrt{2}}{3}

Correct answer: C

Question 10

Mathematics · 4 marks · −1

Given below are two statements: Statement I : lim⁡x→0(tan⁡−1x+log⁡e1+x1−x−2xx5)=25\lim_{x\to0}\left(\frac{\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x}{x^5}\right)=\frac{2}{5}. Statement II : lim⁡x→1(x21−x)=1e2\lim_{x\to1}\left(x^{\frac{2}{1-x}}\right)=\frac{1}{e^2}. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: C

Explanation: tan⁡−1x+tanh⁡−1x=2x+2x55+…\tan^{-1}x+\tanh^{-1}x=2x+\frac{2x^5}{5}+\ldots ⇒ I true; x2/(1−x)→e−2x^{2/(1-x)}\to e^{-2} ⇒ II true.

Question 11

Mathematics · 4 marks · −1

Let the ellipse 3x2+py2=43x^2+py^2=4 pass through the centre C of the circle x2+y2−2x−4y−11=0x^2+y^2-2x-4y-11=0 of radius r. Let f1f_1, f2f_2 be the focal distances of the point C on the ellipse. Then 6f1f2−r6f_1f_2-r is equal to

  1. A.74
  2. B.70
  3. C.78
  4. D.68

Correct answer: B

Explanation: C = (1, 2), r = 4; 3+4p=43+4p=4 ⇒ p=14p=\frac14: x24/3+y216=1\frac{x^2}{4/3}+\frac{y^2}{16}=1 (major along y, a = 4, e2=1112e^2=\frac{11}{12}); f1f2=a2−e2y2=16−113=373f_1f_2=a^2-e^2y^2=16-\frac{11}{3}=\frac{37}{3}; 6⋅373−4=706\cdot\frac{37}{3}-4=70.

Question 12

Mathematics · 4 marks · −1

Let A=[22+p2+p+q46+2p8+3p+2q612+3p20+6p+3q]A=\begin{bmatrix}2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q\end{bmatrix}. If det (adj (adj (3A))) = 2m⋅3n2^m\cdot3^n, m, n ∈ N, then m + n is equal to

  1. A.24
  2. B.22
  3. C.20
  4. D.26

Correct answer: A

Explanation: Column operations reduce A to a triangular form with |A| = 8; ∣3A∣=27⋅8=2333|3A|=27\cdot8=2^3 3^3 and det(adj(adj M)) =∣M∣4=212312=|M|^4=2^{12}3^{12}: m + n = 24.

Question 13

Mathematics · 4 marks · −1

If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P(A∩Bˉ)=0.5P(A\cap\bar{B})=0.5, where Bˉ\bar{B} denotes the complement of B, then P(B∣(A∪Bˉ))P(B|(A\cup\bar{B})) is equal to

  1. A.12\frac{1}{2}
  2. B.16\frac{1}{6}
  3. C.14\frac{1}{4}
  4. D.13\frac{1}{3}

Correct answer: C

Explanation: P(A∩B)=0.2P(A\cap B)=0.2, P(A∪Bˉ)=0.7+0.6−0.5=0.8P(A\cup\bar B)=0.7+0.6-0.5=0.8; 0.20.8=14\frac{0.2}{0.8}=\frac14.

Question 14

Mathematics · 4 marks · −1

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

  1. A.230
  2. B.220
  3. C.210
  4. D.200

Correct answer: C

Explanation: 220−10=210220-10=210.

Question 15

Mathematics · 4 marks · −1

A line passing through the point P(a, 0) makes an acute angle α with the positive x-axis. Let this line be rotated about the point P through an angle α2\frac{\alpha}{2} in the clock-wise direction. If in the new position, the slope of the line is 2−32-\sqrt{3} and its distance from the origin is 12\frac{1}{\sqrt{2}}, then the value of 3a2tan⁡2α−233a^2\tan^2\alpha-2\sqrt{3} is

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

Correct answer: D

Explanation: α/2=15∘\alpha/2=15^\circ ⇒ α = 30°; distance asin⁡15∘=12a\sin15^\circ=\frac{1}{\sqrt2} ⇒ a2=12sin⁡215∘=11−cos⁡30a^2=\frac{1}{2\sin^215^\circ}=\frac{1}{1-\cos30}; 3a2⋅13−23=22−3−23=43a^2\cdot\frac13-2\sqrt3=\frac{2}{2-\sqrt3}-2\sqrt3=4.

Question 16

Mathematics · 4 marks · −1

Let f(x) be a positive function and I1=∫−1/212x f(2x(1−2x))dxI_1=\int_{-1/2}^{1}2x\,f(2x(1-2x))dx and I2=∫−12f(x(1−x))dxI_2=\int_{-1}^{2}f(x(1-x))dx. Then the value of I2I1\frac{I_2}{I_1} is equal to ______

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

Correct answer: B

Question 17

Mathematics · 4 marks · −1

Let f(x) = x − 1 and g(x)=exg(x)=e^x for x ∈ R. If dydx=(e−2xg(f(f(x)))−yx)\frac{dy}{dx}=\left(e^{-2\sqrt{x}}g(f(f(x)))-\frac{y}{\sqrt{x}}\right), y(0) = 0, then y(1) is

  1. A.2e−1e3\frac{2e-1}{e^3}
  2. B.e−1e4\frac{e-1}{e^4}
  3. C.1−e3e4\frac{1-e^3}{e^4}
  4. D.1−e2e4\frac{1-e^2}{e^4}

Correct answer: B

Question 18

Mathematics · 4 marks · −1

The number of integral terms in the expansion of (512+718)1016\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016} is

  1. A.127
  2. B.128
  3. C.129
  4. D.130

Correct answer: B

Explanation: A term is integral when r ≡ 0 (mod 8) and 1016 − r is even: r = 0, 8, 16, …, 1016, i.e. 128 terms.

Question 19

Mathematics · 4 marks · −1

Let α be a solution of x2+x+1=0x^2+x+1=0, and for some a and b in R, [4 a b][11613−1−12−2−14−8]=[0 0 0][4\ a\ b]\begin{bmatrix}1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8\end{bmatrix}=[0\ 0\ 0]. If 4α4+mαa+nαb=3\frac{4}{\alpha^4}+\frac{m}{\alpha^a}+\frac{n}{\alpha^b}=3, then m + n is equal to ______

  1. A.3
  2. B.8
  3. C.7
  4. D.11

Correct answer: D

Question 20

Mathematics · 4 marks · −1

Let the function f(x)=x3+3x+3f(x)=\frac{x}{3}+\frac{3}{x}+3, x ≠ 0 be strictly increasing in (−∞,α1)∪(α2,∞)(-\infty, \alpha_1)\cup(\alpha_2, \infty) and strictly decreasing in (α3,α4)∪(α4,α5)(\alpha_3, \alpha_4)\cup(\alpha_4, \alpha_5). Then ∑i=15αi2\sum_{i=1}^{5}\alpha_i^2 is equal to

  1. A.48
  2. B.36
  3. C.28
  4. D.40

Correct answer: B

Explanation: α₁ = −3, α₂ = 3, α₃ = −3, α₄ = 0, α₅ = 3: sum 36.

Question 21

Mathematics · 4 marks · −1

Let the area of the triangle formed by the lines x + 2 = y − 1 = z, x−35=y−1=z−11\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1} and x−3=y−33=z−21\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1} be A. Then A2A^2 is equal to ______

Correct answer: 56

Question 22

Mathematics · 4 marks · −1

Let the area of the bounded region {(x,y):0≤9x≤y2,y≥3x−6}\{(x, y): 0\le9x\le y^2, y\ge3x-6\} be A. Then 6A is equal to ______

Correct answer: 15

Question 23

Mathematics · 4 marks · −1

The product of the last two digits of (1919)1919(1919)^{1919} is ______

Correct answer: 63

Explanation: 19191919≡1919191919^{1919}\equiv19^{1919} (mod 100); 1910≡119^{10}\equiv1, 199≡7919^{9}\equiv79: last two digits 79, product 63.

Question 24

Mathematics · 4 marks · −1

Let r be the radius of the circle, which touches x- axis at point (a, 0), a < 0 and the parabola y2=9xy^2=9x at the point (4, 6). Then r is equal to ______

Correct answer: 30

Question 25

Mathematics · 4 marks · −1

Let the domain of the function f(x)=cos⁡−1(4x+53x−7)f(x)=\cos^{-1}\left(\frac{4x+5}{3x-7}\right) be [α, β] and the domain of g(x)=log⁡2(2−6log⁡27(2x+5))g(x)=\log_2(2-6\log_{27}(2x+5)) be (γ, δ). Then |7(α + β) + 4(γ + δ)| is equal to ______

Correct answer: 96

Question 26

Physics · 4 marks · −1

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take g = 10 m/s², specific heat of water = 4200 J/(kg K))

  1. A.0.48 K
  2. B.0.36 K
  3. C.0.23 K
  4. D.0.14 K

Correct answer: A

Explanation: ΔT = gh/c = 2000/4200 = 0.48 K.

Question 27

Physics · 4 marks · −1

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be

  1. A.10[1−(2−x)2]3/210\left[1-(2-x)^2\right]^{3/2} m/s
  2. B.10[1−(2−x)2]1/210\left[1-(2-x)^2\right]^{1/2} m/s
  3. C.10[1−(2−x)2]210\left[1-(2-x)^2\right]^2 m/s
  4. D.10[1−(2−x)2]10\left[1-(2-x)^2\right] m/s

Correct answer: B

Explanation: ω=10\omega=10, amplitude 1: v=101−(2−x)2v=10\sqrt{1-(2-x)^2}.

Question 28

Physics · 4 marks · −1

A body of mass 2 kg moving with velocity of v⃗in=3i^+4j^\vec{v}_{in}=3\hat{i}+4\hat{j} ms−1s^{-1} enters into a constant force field of 6N directed along positive z–axis. If the body remains in the field for a period of 53\frac{5}{3} seconds, then velocity of the body when it emerges from force field is.

  1. A.3i^+4j^+5k^3\hat{i}+4\hat{j}+5\hat{k}
  2. B.3i^+4j^−5k^3\hat{i}+4\hat{j}-5\hat{k}
  3. C.3i^+4j^+5k^3\hat{i}+4\hat{j}+\sqrt{5}\hat{k}
  4. D.4i^+3j^+5k^4\hat{i}+3\hat{j}+5\hat{k}

Correct answer: A

Explanation: Δvz=3×53=5\Delta v_z=3\times\frac53=5.

Question 29

Physics · 4 marks · −1

A quantity Q is formulated as X−2Y+32Z−25X^{-2}Y^{+\frac{3}{2}}Z^{-\frac{2}{5}}. X, Y and Z are independent parameters which have fractional errors of 0.1, 0.2 and 0.5, respectively in measurement. The maximum fractional error of Q is

  1. A.0.6
  2. B.0.1
  3. C.0.8
  4. D.0.7

Correct answer: D

Explanation: 0.2 + 0.3 + 0.2 = 0.7.

Question 30

Physics · 4 marks · −1

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

  1. A.λL38π2\frac{\lambda L^3}{8\pi^2}
  2. B.λL316π2\frac{\lambda L^3}{16\pi^2}
  3. C.λL34π2\frac{\lambda L^3}{4\pi^2}
  4. D.λL312\frac{\lambda L^3}{12}

Correct answer: A

Explanation: 12(λL)(L2π)2=λL38π2\frac12(\lambda L)\left(\frac{L}{2\pi}\right)^2=\frac{\lambda L^3}{8\pi^2}.

Question 31

Physics · 4 marks · −1

A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ______.

  1. A.607\frac{60}{7}
  2. B.30
  3. C.15
  4. D.45

Correct answer: C

Explanation: F = −60 cm; 1v=−160−120\frac1v=-\frac1{60}-\frac1{20} ⇒ |v| = 15 cm.

Question 32

Physics · 4 marks · −1

For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as

  1. A.A3A^3
  2. B.Independent of A
  3. C.A2/3A^{2/3}
  4. D.A1/3A^{1/3}

Correct answer: B

Question 33

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 : Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R : Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. 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 34

Physics · 4 marks · −1

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1T_1 and T2T_2 are the total flying times of first and second ball, respectively, then the ratio of T1T_1 and T2T_2 is

  1. A.2 : 1
  2. B.2\sqrt{2} :1
  3. C.222\sqrt{2} :1
  4. D.4 : 1

Correct answer: C

Explanation: H∝T2H\propto T^2: 8=22\sqrt8=2\sqrt2.

Question 35

Physics · 4 marks · −1

The output voltage in the following circuit is (Consider ideal diode case)

Figure for question 35
  1. A.10 V
  2. B.0 V
  3. C.−5 V
  4. D.+5 V

Correct answer: B

Explanation: D2 conducts to the grounded input, clamping the output to 0 V.

Question 36

Physics · 4 marks · −1

Two metal spheres of radius R and 3R have same surface charge density σ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ1\sigma_1 and σ2\sigma_2, respectively. The ratio σ1σ2\frac{\sigma_1}{\sigma_2} is

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

Correct answer: A

Explanation: Equal potentials: σ ∝ 1/R ⇒ 3.

Question 37

Physics · 4 marks · −1

Electric charge is transferred to an irregular metallic disk as shown in figure. If σ1\sigma_1, σ2\sigma_2, σ3\sigma_3 and σ4\sigma_4 are charge densities at given points then, choose the correct answer from the options given below: A. σ1>σ3\sigma_1>\sigma_3; σ2=σ4\sigma_2=\sigma_4 B. σ1>σ2\sigma_1>\sigma_2; σ3>σ4\sigma_3>\sigma_4 C. σ1>σ3>σ2=σ4\sigma_1>\sigma_3>\sigma_2=\sigma_4 D. σ1<σ3<σ2=σ4\sigma_1<\sigma_3<\sigma_2=\sigma_4 E. σ1=σ2=σ3=σ4\sigma_1=\sigma_2=\sigma_3=\sigma_4

Figure for question 37
  1. A.A, B and C Only
  2. B.A and C Only
  3. C.D and E Only
  4. D.B and C Only

Correct answer: A

Question 38

Physics · 4 marks · −1

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid–glass combination will be

  1. A.70011\frac{700}{11} cm
  2. B.80011\frac{800}{11} cm
  3. C.50011\frac{500}{11} cm
  4. D.60011\frac{600}{11} cm

Correct answer: D

Explanation: Glass meniscus: 1f=0.5(120−130)=1120\frac1f=0.5\left(\frac1{20}-\frac1{30}\right)=\frac1{120}; liquid plano-convex (R = 30): 0.330=1100\frac{0.3}{30}=\frac1{100}; total 11600\frac{11}{600} ⇒ f = 600/11 cm.

Question 39

Physics · 4 marks · −1

A monoatomic gas having γ=53\gamma=\frac{5}{3} is stored in a thermally insulated container and the gas is suddenly compressed to (18)th\left(\frac{1}{8}\right)^{th} of its initial volume. The ratio of final pressure and initial pressure is: (γ is the ratio of specific heats of the gas at constant pressure and at constant volume)

  1. A.16
  2. B.28
  3. C.32
  4. D.40

Correct answer: C

Explanation: 85/3=328^{5/3}=32.

Question 40

Physics · 4 marks · −1

In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case

  1. A.there shall be an interference pattern for red distinct from that for green.
  2. B.there shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.
  3. C.there shall be no interference fringes.
  4. D.there shall be alternate interference fringes of red and green.

Correct answer: C

Explanation: Red and green light are not coherent (different frequencies): no interference.

Question 41

Physics · 4 marks · −1

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y1(x,t)=4sin⁡(kx−ωt)y_1(x, t)=4\sin(kx-\omega t) and y2(x,t)=2sin⁡(kx−ωt+2π3)y_2(x, t)=2\sin\left(kx-\omega t+\frac{2\pi}{3}\right), are: (Take the angular frequency of initial waves same as ω)

  1. A.[6,π/3][6, \pi/3]
  2. B.[3,π/6][\sqrt{3}, \pi/6]
  3. C.[6,2π/3][6, 2\pi/3]
  4. D.[23,π/6][2\sqrt{3}, \pi/6]

Correct answer: D

Explanation: A2=16+4−8=12A^2=16+4-8=12 ⇒ 232\sqrt3; tan⁡ϕ=2sin⁡1204+2cos⁡120=33\tan\phi=\frac{2\sin120}{4+2\cos120}=\frac{\sqrt3}{3} ⇒ π/6.

Question 42

Physics · 4 marks · −1

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is P × 101110^{11} Nm−2m^{-2}, where the value of P is: (Take g = 3π m/s²)

  1. A.10
  2. B.25
  3. C.5
  4. D.2.5

Correct answer: C

Explanation: Y=50⋅3π⋅3π(9×10−6)(10−4)=5×1011Y=\frac{50\cdot3\pi\cdot3}{\pi(9\times10^{-6})(10^{-4})}=5\times10^{11}.

Question 43

Physics · 4 marks · −1

An infinitely long wire has uniform linear charge density λ = 2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt{3} cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm2C−1m^2C^{-1}, where x is: [Neglect any edge effects and use 14πϵ0\frac{1}{4\pi\epsilon_0} = 9 × 10910^9 SI units]

  1. A.0.72π0.72\pi
  2. B.1.44π1.44\pi
  3. C.2.16π2.16\pi
  4. D.6.48π6.48\pi

Correct answer: C

Explanation: Body diagonal = 3 cm; ϕ=λ⋅0.03ϵ0=2×10−9×0.03×36π×109=2.16π\phi=\frac{\lambda\cdot0.03}{\epsilon_0}=2\times10^{-9}\times0.03\times36\pi\times10^9=2.16\pi.

Question 44

Physics · 4 marks · −1

Figure shows a current carrying square loop ABCD of edge length is 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is

Figure for question 44
  1. A.3πμ0I2a\frac{3\pi\mu_0I}{\sqrt{2}a}
  2. B.2μ0I3πa\frac{\sqrt{2}\mu_0I}{3\pi a}
  3. C.2μ0I3πa\frac{2\mu_0I}{3\pi a}
  4. D.μ0I2πa\frac{\mu_0I}{2\pi a}

Correct answer: B

Explanation: i1=2I3i_1=\frac{2I}{3} (via B), i2=I3i_2=\frac{I}{3} (via D). Each side contributes 2μ0i4π(a/2)\frac{\sqrt2\mu_0 i}{4\pi(a/2)} at the centre; the two paths give opposite fields, net 22μ02πa⋅I3=2μ0I3πa\frac{2\sqrt2\mu_0}{2\pi a}\cdot\frac{I}{3}=\frac{\sqrt2\mu_0I}{3\pi a}.

Question 45

Physics · 4 marks · −1

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is v1v_1, and that in the other string having radius of cross section R/2 is v2v_2. Then v2v1\frac{v_2}{v_1} =

  1. A.4
  2. B.2\sqrt{2}
  3. C.2
  4. D.8

Correct answer: C

Explanation: v∝1/rv\propto1/r: 2.

Question 46

Physics · 4 marks · −1

A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to change of volume of 0.8 cm3cm^3. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was ______ litre. (Take 1 atm = 10510^5 Pa)

Correct answer: 4

Explanation: V=BΔVΔP=2×109×0.8×10−64×105=4×10−3V=\frac{B\Delta V}{\Delta P}=\frac{2\times10^9\times0.8\times10^{-6}}{4\times10^5}=4\times10^{-3} m³ = 4 L.

Question 47

Physics · 4 marks · −1

An electron is released from rest near an infinite non-conducting sheet of uniform charge density '−σ'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nthn^{th} power of time. The numerical value of n is ______.

Correct answer: 2

Explanation: Constant force: p ∝ t, λ ∝ 1/t, dλ/dt ∝ 1/t²: n = 2.

Question 48

Physics · 4 marks · −1

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. (Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm3cm^3.) The unknown mass is ______ kg.

Correct answer: 3

Explanation: (m−0.5)×2=0.2×25(m-0.5)\times2=0.2\times25 (rod treated as light) ⇒ m = 3 kg.

Question 49

Physics · 4 marks · −1

Space between the plates of a parallel plate capacitor of plate area 4 cm2cm^2 and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is ______ pF. (Given ϵ0\epsilon_0 = 8.85 × 10−1210^{-12} F/m)

Figure for question 49

Correct answer: 15

Explanation: C=2ϵ0Ad158=2(8.85×10−12)(4×10−4)1.77×10−3⋅158=7.5C=\frac{2\epsilon_0A}{d}\frac{15}{8}=\frac{2(8.85\times10^{-12})(4\times10^{-4})}{1.77\times10^{-3}}\cdot\frac{15}{8}=7.5 pF; total 15 pF.

Question 50

Physics · 4 marks · −1

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is ______ m.

Correct answer: 40

Explanation: Δω = 10π rad/s; I = 25π·40/(10π) = 100 = MR²/4 ⇒ R = 20, D = 40 m.

Question 51

Chemistry · 4 marks · −1

Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?

  1. A.C6H5OH+C6H5NH2C_6H_5OH+C_6H_5NH_2
  2. B.H2O+CH3COC2H5H_2O+CH_3COC_2H_5
  3. C.CS2+CH3COCH3CS_2+CH_3COCH_3
  4. D.CH3OH+CHCl3CH_3OH+CHCl_3

Correct answer: A

Explanation: Phenol–aniline shows negative deviation (maximum boiling azeotrope).

Question 52

Chemistry · 4 marks · −1

HA(aq) ⇌ H+H^+(aq) + A−A^-(aq). The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given : Kf(H2O)K_f(H_2O) = 1.8 K kg mol−1mol^{-1}, molality ≡ molarity

  1. A.1.1×10−21.1\times10^{-2}
  2. B.1.89×10−11.89\times10^{-1}
  3. C.1.38×10−31.38\times10^{-3}
  4. D.1.90×10−31.90\times10^{-3}

Correct answer: C

Explanation: i = 1.111, α = 0.111; Ka=0.1α21−α=1.38×10−3K_a=\frac{0.1\alpha^2}{1-\alpha}=1.38\times10^{-3}.

Question 53

Chemistry · 4 marks · −1

Choose the correct set of reagents for the following conversion. Ethyl benzene → 4-bromostyrene (CH=CH2CH=CH_2 para to Br).

  1. A.Cl2Cl_2/Fe ; Br2Br_2/anhy.AlCl3AlCl_3 ; aq. KOH
  2. B.Br2Br_2/Fe ; Cl2Cl_2, Δ ; alc.KOH
  3. C.Br2Br_2/anhy.AlCl3AlCl_3 ; Cl2Cl_2, Δ ; aq. KOH
  4. D.Cl2Cl_2/anhy.AlCl3AlCl_3 ; Br2Br_2/Fe ; alc. KOH

Correct answer: B

Question 54

Chemistry · 4 marks · −1

On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H2OH_2O and 0.307 g CO2CO_2. The percentages of hydrogen and oxygen in the given organic compound respectively are:

  1. A.6.72, 39.87
  2. B.6.72, 53.41
  3. C.7.55, 43.85
  4. D.53.41, 39.6

Correct answer: B

Explanation: H = 0.0141 g (6.72%), C = 0.0837 g (39.87%), O = 53.41%.

Question 55

Chemistry · 4 marks · −1

1,2-dibromocyclooctane →(i) KOH (alc.) (ii) NaNH2 (iii) Hg2+/H+ (iv) Zn–Hg/H+\xrightarrow{\text{(i) KOH (alc.) (ii) } NaNH_2\ \text{(iii) } Hg^{2+}/H^+\ \text{(iv) Zn–Hg}/H^+} P (major product). 'P' is

  1. A.Cyclooctane
  2. B.Cyclooctyne
  3. C.Cyclooctanol
  4. D.Cyclooctanone

Correct answer: A

Explanation: Cyclooctyne → cyclooctanone → Clemmensen → cyclooctane.

Question 56

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Complex/ Species): A. [Ni(CO)4][Ni(CO)_4] B. [Ni(CN)4]2−[Ni(CN)_4]^{2-} C. [NiCl4]2−[NiCl_4]^{2-} D. [MnBr4]2−[MnBr_4]^{2-}. LIST-II (Shape & magnetic moment): I. Tetrahedral, 2.8 BM II. Square planar, 0 BM III. Tetrahedral, 0 BM IV. Tetrahedral, 5.9 BM. Choose the correct answer from the options given below:

  1. A.A-III, B-IV, C-II, D-I
  2. B.A-IV, B-I, 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 57

Chemistry · 4 marks · −1

The atomic number of the element from the following with lowest 1st1^{st} ionisation enthalpy is :

  1. A.35
  2. B.19
  3. C.32
  4. D.87

Correct answer: D

Question 58

Chemistry · 4 marks · −1

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1t_1 and t2t_2 (s), respectively. The ratio t1/t2t_1/t_2 will be:

  1. A.43\frac{4}{3}
  2. B.32\frac{3}{2}
  3. C.23\frac{2}{3}
  4. D.34\frac{3}{4}

Correct answer: C

Explanation: 2 half-lives vs 3 half-lives.

Question 59

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Reagent): A. Sodium bicarbonate solution B. Neutral ferric chloride C. ceric ammonium nitrate D. alkaline KMnO4KMnO_4. LIST-II (Functional Group detected): I. double bond/ unsaturation II. carboxylic acid III. phenolic - OH IV. alcoholic - OH. Choose the correct answer from the options given below:

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

Correct answer: B

Question 60

Chemistry · 4 marks · −1

A →(i) NaOH; (ii) H3O+\xrightarrow{\text{(i) NaOH; (ii) } H_3O^+} B →(i) EtOH; (ii) H2SO4,Δ\xrightarrow{\text{(i) EtOH; (ii) } H_2SO_4, \Delta} C. 'A' shows positive Lassaign's test for N and its molar mass is 121. 'B' gives effervescence with aq NaHCO3NaHCO_3. 'C' gives fruity smell. Identify A, B and C from the following.

Figure for question 60
  1. A.A = 2-aminobenzaldehyde, B = 2-aminobenzoic acid, C = ethyl 2-(ethylamino)benzoate
  2. B.A = 4-aminobenzaldehyde, B = an aldol-type hydroxy-ketone dimer, C = a hemiacetal dimer
  3. C.A = benzamide (C6H5CONH2C_6H_5CONH_2), B = benzoic acid, C = ethyl benzoate
  4. D.A = benzylhydrazine (C6H5CH2NHNH2C_6H_5CH_2NHNH_2), B = benzoic acid, C = ethyl benzoate

Correct answer: C

Question 61

Chemistry · 4 marks · −1

Given below are two statements: Statement I : H2SeH_2Se is more acidic than H2TeH_2Te. Statement II : H2SeH_2Se has higher bond enthalpy for dissociation than H2TeH_2Te. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: C

Question 62

Chemistry · 4 marks · −1

Correct statements for an element with atomic number 9 are A. There can be 5 electrons for which ms=+12m_s=+\frac{1}{2} and 4 electrons for which ms=−12m_s=-\frac{1}{2}. B. There is only one electron in pzp_z orbital. C. The last electron goes to orbital with n = 2 and ll = 1. D. The sum of angular nodes of all the atomic orbitals is 1. Choose the correct answer from the options given below:

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

Correct answer: C

Question 63

Chemistry · 4 marks · −1

What is the correct IUPAC name of [cyclopentene ring bearing OH at one sp³ carbon and an ethyl group on the other sp³ carbon flanking the C=C]?

Figure for question 63
  1. A.1-Ethylcyclopent-2-en-3-ol
  2. B.4-Ethyl-1-hydroxycyclopent-2-ene
  3. C.1-Ethyl-3-hydroxycyclopent-2-ene
  4. D.4-Ethylcyclopent-2-en-1-ol

Correct answer: D

Question 64

Chemistry · 4 marks · −1

Given below are two statements: Statement I : A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II : cis- and trans- platin are heteroleptic complexes of Pd. 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.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are false

Correct answer: B

Question 65

Chemistry · 4 marks · −1

Valine, H2N−CH(CH(CH3)2)−COOHH_2N-CH(CH(CH_3)_2)-COOH, at pH = 2 gives A and at pH = 10 gives B. Choose the correct option for structures of A and B, respectively

  1. A.A: H2N−CH(iPr)−COO−H_2N-CH(iPr)-COO^-; B: H3N+−CH(iPr)−COOHH_3N^+-CH(iPr)-COOH
  2. B.A: H2N−CH(iPr)−COO−H_2N-CH(iPr)-COO^-; B: H3N+−CH(iPr)−COO−H_3N^+-CH(iPr)-COO^-
  3. C.A: H3N+−CH(iPr)−COO−H_3N^+-CH(iPr)-COO^-; B: H3N+−CH(iPr)−COOHH_3N^+-CH(iPr)-COOH
  4. D.A: H3N+−CH(iPr)−COOHH_3N^+-CH(iPr)-COOH; B: H2N−CH(iPr)−COO−H_2N-CH(iPr)-COO^-

Correct answer: D

Question 66

Chemistry · 4 marks · −1

Which one of the following reactions will not lead to the desired ether formation in major proportion ? (iso-Bu ⇒ isobutyl, sec-Bu ⇒ sec-butyl, nPr ⇒ n-propyl, ᵗBu ⇒ tert-butyl, Et ⇒ ethyl)

  1. A.tBuO−Na++EtBr→tBu−O−Et^tBuO^-Na^++EtBr\to{}^tBu-O-Et
  2. B.Cyclohexyl-O−Na+O^-Na^+ + CH3BrCH_3Br → cyclohexyl–O–CH3CH_3
  3. C.Na+−ONa^+{}^-O-Ph + n-PrBr → n-Pr–O–Ph
  4. D.iso-BuO−Na+O^-Na^+ + sec-BuBr → sec-Bu–O–iso-Bu

Correct answer: D

Explanation: A secondary halide with a strong alkoxide mainly undergoes elimination.

Question 67

Chemistry · 4 marks · −1

The correct decreasing order of spin only magnetic moment values (BM) of Cu+Cu^+, Cu2+Cu^{2+}, Cr2+Cr^{2+} and Cr3+Cr^{3+} ions is :

  1. A.Cr2+>Cr3+>Cu2+>Cu+Cr^{2+}>Cr^{3+}>Cu^{2+}>Cu^+
  2. B.Cu2+>Cu+>Cr2+>Cr3+Cu^{2+}>Cu^+>Cr^{2+}>Cr^{3+}
  3. C.Cu+>Cu2+>Cr3+>Cr2+Cu^+>Cu^{2+}>Cr^{3+}>Cr^{2+}
  4. D.Cr3+>Cr2+>Cu+>Cu2+Cr^{3+}>Cr^{2+}>Cu^+>Cu^{2+}

Correct answer: A

Question 68

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. Carbocation B. C-Free radical C. Nucleophile D. Electrophile. LIST-II: I. Species that can supply a pair of electrons. II. Species that can receive a pair of electrons. III. sp2sp^2 hybridized carbon with empty p-orbital. IV. sp2/sp3sp^2/sp^3 hybridized carbon with one unpaired electron. Choose the correct answer from the options given below:

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

Correct answer: C

Question 69

Chemistry · 4 marks · −1

When 6-oxoheptanal (CH3CO−(CH2)4−CHOCH_3CO-(CH_2)_4-CHO) undergoes intramolecular aldol condensation, the major product formed is :

Figure for question 69
  1. A.Cyclohept-2-en-1-one
  2. B.2-Methylcyclopent-2-en-1-one
  3. C.1-Acetylcyclopentene
  4. D.Cyclopent-1-ene-1-carbaldehyde with 2-methyl (2-methylcyclopent-1-enecarbaldehyde)

Correct answer: C

Explanation: The methyl-ketone enolate attacks the aldehyde to form a five-membered ring: 1-acetylcyclopentene.

Question 70

Chemistry · 4 marks · −1

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is ______ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol−1mol^{-1})

Correct answer: 1

Explanation: AgI 0.02017 mol / 0.02 L ≈ 1 M.

Question 71

Chemistry · 4 marks · −1

The equilibrium constant for decomposition of H2OH_2O(g) H2O(g)⇌H2(g)+12O2(g)H_2O(g)\rightleftharpoons H_2(g)+\frac{1}{2}O_2(g) (ΔG° = 92.34 kJ mol−1mol^{-1}) is 8.0 × 10−310^{-3} at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α) of water is ______ × 10−210^{-2} (nearest integer value). [Assume α is negligible with respect to 1]

Correct answer: 5

Explanation: Kp=αα/2K_p=\alpha\sqrt{\alpha/2} ⇒ α3/2=82×10−3\alpha^{3/2}=8\sqrt2\times10^{-3} ⇒ α = 0.05 → 5.

Question 72

Chemistry · 4 marks · −1

The energy of an electron in first Bohr orbit of H-atom is −13.6 eV. The magnitude of energy value of electron in the first excited state of Be3+Be^{3+} is ______ eV (nearest integer value)

Correct answer: 54

Explanation: 13.6×16/4=54.413.6\times16/4=54.4 → 54.

Question 73

Chemistry · 4 marks · −1

Resonance in X2YX_2Y can be represented as X⊖=X⊕=Y↔X≡X⊕−Y⊖\overset{\ominus}{X}=\overset{\oplus}{X}=Y\leftrightarrow X\equiv\overset{\oplus}{X}-\overset{\ominus}{Y}. The enthalpy of formation of X2YX_2Y (X≡X(g)+12Y=Y(g)→X2Y(g)X\equiv X(g)+\frac{1}{2}Y=Y(g)\to X_2Y(g)) is 80 kJ mol−1mol^{-1}. The magnitude of resonance energy of X2YX_2Y is ______ kJ mol−1mol^{-1} (nearest integer value) Given : Bond energies of X≡X, X=X, Y=Y and X=Y are 940, 410, 500 and 602 kJ mol−1mol^{-1} respectively. valence X: 3, Y: 2

Correct answer: 98

Explanation: Calculated ΔHf\Delta H_f (structure X=X=Y) =940+250−(410+602)=178=940+250-(410+602)=178; resonance energy =178−80=98=178-80=98.

Question 74

Chemistry · 4 marks · −1

Consider the following half cell reaction Cr2O72−(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2O(l)Cr_2O_7^{2-}(aq)+6e^-+14H^+(aq)\to2Cr^{3+}(aq)+7H_2O(l). The reaction was conducted with the ratio of [Cr3+]2[Cr2O72−]=10−6\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-6}. The pH value at which the EMF of the half cell will become zero is ______. (nearest integer value) [Given : standard half cell reduction potential ECr2O72−,H+/Cr3+∘E^\circ_{Cr_2O_7^{2-}, H^+/Cr^{3+}} = 1.33V, 2.303RTF\frac{2.303RT}{F} = 0.059V.]

Correct answer: 10

Explanation: 0=1.33−0.0596(−6+14 pH)0=1.33-\frac{0.059}{6}(-6+14\,\mathrm{pH}) ⇒ pH ≈ 10.

View all papers — JEE Main

Practice this paper with a timer

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

Start practising free