JEE Main 2025 — 23 Jan Shift 1 (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

Let I(x)=∫dx(x−11)1113(x+15)1513I(x)=\int\frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}. If I(37)−I(24)=14(1b113−1c113)I(37)-I(24)=\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right), b,c∈Nb, c\in\mathbf{N}, then 3(b+c)3(b+c) is equal to

  1. A.26
  2. B.40
  3. C.39
  4. D.22

Correct answer: C

Explanation: With t=x−11x+15t=\frac{x-11}{x+15}, dt=26(x+15)2dxdt=\frac{26}{(x+15)^2}dx and I=14t2/13I=\frac14t^{2/13}; I(37)−I(24)=14(141/13−191/13)I(37)-I(24)=\frac14\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right), so b = 4, c = 9 and 3(b+c)=393(b+c)=39.

Question 2

Mathematics · 4 marks · −1

Let f(x)=log⁡exf(x)=\log_ex and g(x)=x4−2x3+3x2−2x+22x2−2x+1g(x)=\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}. Then the domain of f∘gf\circ g is

  1. A.R\mathbf{R}
  2. B.[0,∞)[0, \infty)
  3. C.(0,∞)(0, \infty)
  4. D.[1,∞)[1, \infty)

Correct answer: A

Explanation: Numerator =(x2−x+1)2+1>0=(x^2-x+1)^2+1>0 and denominator >0>0, so g>0g>0 for all real x.

Question 3

Mathematics · 4 marks · −1

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

  1. A.49\frac{4}{9}
  2. B.23\frac{2}{3}
  3. C.12\frac{1}{2}
  4. D.35\frac{3}{5}

Correct answer: C

Explanation: Sum 4: 4+4+1=94+4+1=9 ways; sum 5: 2+4+2+1=92+4+2+1=9 ways (out of 36): probability 1/2.

Question 4

Mathematics · 4 marks · −1

Let ∣zˉ−i2zˉ+i∣=13\left|\frac{\bar{z}-i}{2\bar{z}+i}\right|=\frac{1}{3}, z∈Cz\in\mathbf{C}, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0, 0), C and (α\alpha, 0) is 11 square units, then α2\alpha^2 equals:

  1. A.8125\frac{81}{25}
  2. B.50
  3. C.12125\frac{121}{25}
  4. D.100

Correct answer: D

Question 5

Mathematics · 4 marks · −1

Let P be the foot of the perpendicular from the point Q(10, −3-3, −1-1) on the line x−37=y−2−1=z+1−2\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}. Then the area of the right angled triangle PQR, where R is the point (3, −2-2, 1), is

  1. A.30\sqrt{30}
  2. B.3303\sqrt{30}
  3. C.8158\sqrt{15}
  4. D.9159\sqrt{15}

Correct answer: B

Question 6

Mathematics · 4 marks · −1

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to

  1. A.-1020
  2. B.-1200
  3. C.-1080
  4. D.-120

Correct answer: C

Explanation: 5(12+6d)=12+22d⇒d=−65(12+6d)=12+22d\Rightarrow d=-6; S20=10(6−114)=−1080S_{20}=10(6-114)=-1080.

Question 7

Mathematics · 4 marks · −1

The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is

  1. A.36000
  2. B.35000
  3. C.34000
  4. D.37000

Correct answer: A

Explanation: 8!−6!⋅3!=40320−4320=360008!-6!\cdot3!=40320-4320=36000.

Question 8

Mathematics · 4 marks · −1

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that length of arc ABlength of arc BC=15\frac{\text{length of arc AB}}{\text{length of arc BC}}=\frac{1}{5}, and OC⃗=αOA⃗+βOB⃗\vec{OC}=\alpha\vec{OA}+\beta\vec{OB}, then α+2(3−1)β\alpha+\sqrt{2}(\sqrt{3}-1)\beta is equal to

  1. A.535\sqrt{3}
  2. B.232\sqrt{3}
  3. C.2−32-\sqrt{3}
  4. D.2+32+\sqrt{3}

Correct answer: C

Question 9

Mathematics · 4 marks · −1

The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\int_{e^2}^{e^4}\frac{1}{x}\left(\frac{e^{((\log_ex)^2+1)^{-1}}}{e^{((\log_ex)^2+1)^{-1}}+e^{((6-\log_ex)^2+1)^{-1}}}\right)dx is

  1. A.log⁡e2\log_e2
  2. B.1
  3. C.2
  4. D.e2e^2

Correct answer: B

Explanation: With t=ln⁡xt=\ln x over [2,4][2,4], King property (t→6−tt\to6-t) gives I=12(4−2)=1I=\frac12(4-2)=1.

Question 10

Mathematics · 4 marks · −1

Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be i^+2j^+k^\hat{i}+2\hat{j}+\hat{k}, i^+3j^−2k^\hat{i}+3\hat{j}-2\hat{k} and 2i^+j^−k^2\hat{i}+\hat{j}-\hat{k} respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is 1103\frac{\sqrt{110}}{3} and the volume of the tetrahedron is 80562\frac{\sqrt{805}}{6\sqrt{2}}, then the position vector of E is

  1. A.16(12i^+12j^+k^)\frac{1}{6}(12\hat{i}+12\hat{j}+\hat{k})
  2. B.12(i^+4j^+7k^)\frac{1}{2}(\hat{i}+4\hat{j}+7\hat{k})
  3. C.16(7i^+12j^+k^)\frac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})
  4. D.112(7i^+4j^+3k^)\frac{1}{12}(7\hat{i}+4\hat{j}+3\hat{k})

Correct answer: C

Question 11

Mathematics · 4 marks · −1

If the system of equations (λ−1)x+(λ−4)y+λz=5(\lambda-1)x+(\lambda-4)y+\lambda z=5, λx+(λ−1)y+(λ−4)z=7\lambda x+(\lambda-1)y+(\lambda-4)z=7, (λ+1)x+(λ+2)y−(λ+2)z=9(\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9 has infinitely many solutions, then λ2+λ\lambda^2+\lambda is equal to

  1. A.10
  2. B.12
  3. C.6
  4. D.20

Correct answer: B

Question 12

Mathematics · 4 marks · −1

If the line 3x−2y+12=03x-2y+12=0 intersects the parabola 4y=3x24y=3x^2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

  1. A.tan⁡−1(45)\tan^{-1}\left(\frac{4}{5}\right)
  2. B.tan⁡−1(119)\tan^{-1}\left(\frac{11}{9}\right)
  3. C.tan⁡−1(97)\tan^{-1}\left(\frac{9}{7}\right)
  4. D.π2−tan⁡−1(32)\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{2}\right)

Correct answer: C

Explanation: 3x2−6x−24=0⇒x=4,−23x^2-6x-24=0\Rightarrow x=4,-2: A(4,12), B(−2,3); slopes 3 and −3/2; tan⁡θ=3+3/21−9/2\tan\theta=\frac{3+3/2}{1-9/2} → ∣ ⋅ ∣=97|\,\cdot\,|=\frac{9}{7} (obtuse angle π−tan⁡−197\pi-\tan^{-1}\frac97).

Question 13

Mathematics · 4 marks · −1

If A, B, and (adj(A−1)+adj(B−1))(\mathrm{adj}(A^{-1})+\mathrm{adj}(B^{-1})) are non-singular matrices of same order, then the inverse of A(adj(A−1)+adj(B−1))−1BA(\mathrm{adj}(A^{-1})+\mathrm{adj}(B^{-1}))^{-1}B, is equal to

  1. A.1∣AB∣(adj(B)+adj(A))\frac{1}{|AB|}(\mathrm{adj}(B)+\mathrm{adj}(A))
  2. B.AB−1+A−1BAB^{-1}+A^{-1}B
  3. C.AB−1∣A∣+BA−1∣B∣\frac{AB^{-1}}{|A|}+\frac{BA^{-1}}{|B|}
  4. D.adj(B−1)+adj(A−1)\mathrm{adj}(B^{-1})+\mathrm{adj}(A^{-1})

Correct answer: A

Question 14

Mathematics · 4 marks · −1

If π2≤x≤3π4\frac{\pi}{2}\le x\le\frac{3\pi}{4}, then cos⁡−1(1213cos⁡x+513sin⁡x)\cos^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right) is equal to

  1. A.x−tan⁡−143x-\tan^{-1}\frac{4}{3}
  2. B.x+tan⁡−145x+\tan^{-1}\frac{4}{5}
  3. C.x−tan⁡−1512x-\tan^{-1}\frac{5}{12}
  4. D.x+tan⁡−1512x+\tan^{-1}\frac{5}{12}

Correct answer: C

Explanation: =cos⁡−1cos⁡(x−α)=\cos^{-1}\cos(x-\alpha) with tan⁡α=5/12\tan\alpha=5/12; x−α∈[0,π]x-\alpha\in[0,\pi], so it equals x−tan⁡−1512x-\tan^{-1}\frac{5}{12}.

Question 15

Mathematics · 4 marks · −1

Let a curve y=f(x)y=f(x) pass through the points (0, 5) and (log⁡e2\log_e2, k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=02(3+y)e^{2x}dx-(7+e^{2x})dy=0, then k is equal to

  1. A.4
  2. B.8
  3. C.32
  4. D.16

Correct answer: B

Explanation: ln⁡(3+y)=ln⁡(7+e2x)+c\ln(3+y)=\ln(7+e^{2x})+c, and (0, 5) gives c = 0; at x=ln⁡2x=\ln2: 3+y=113+y=11, k = 8.

Question 16

Mathematics · 4 marks · −1

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is

  1. A.40
  2. B.44
  3. C.48
  4. D.52

Correct answer: B

Explanation: 14=12+N/2−1812⋅6⇒N/2=2214=12+\frac{N/2-18}{12}\cdot6\Rightarrow N/2=22, N = 44.

Question 17

Mathematics · 4 marks · −1

Let the area of a △PQR\triangle PQR with vertices P(5, 4), Q(−2-2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are O(2,145)\left(2, \frac{14}{5}\right) and C(c, d) respectively, then c + 2d is equal to

  1. A.3
  2. B.2
  3. C.73\frac{7}{3}
  4. D.83\frac{8}{3}

Correct answer: A

Question 18

Mathematics · 4 marks · −1

The value of (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin70^\circ)(\cot10^\circ\cot70^\circ-1) is

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

Correct answer: A

Explanation: =cos⁡10cos⁡70−sin⁡10sin⁡70sin⁡10=cos⁡80sin⁡10=1=\frac{\cos10\cos70-\sin10\sin70}{\sin10}=\frac{\cos80}{\sin10}=1.

Question 19

Mathematics · 4 marks · −1

If the function f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},x<04,x=02xlog⁡e(2+k1x2+k2x),x>0f(x)=\begin{cases}\frac{2}{x}\{\sin(k_1+1)x+\sin(k_2-1)x\}, & x<0\\ 4, & x=0\\ \frac{2}{x}\log_e\left(\frac{2+k_1x}{2+k_2x}\right), & x>0\end{cases} is continuous at x = 0, then k12+k22k_1^2+k_2^2 is equal to

  1. A.5
  2. B.8
  3. C.20
  4. D.10

Correct answer: D

Explanation: 2(k1+k2)=42(k_1+k_2)=4, k1−k2=4k_1-k_2=4: k1=3,k2=−1k_1=3,k_2=-1; sum of squares 10.

Question 20

Mathematics · 4 marks · −1

Let R = {(1, 2), (2, 3), (3, 3)} be a relation defined on the set {1, 2, 3, 4}. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:

  1. A.8
  2. B.10
  3. C.9
  4. D.7

Correct answer: D

Explanation: Need classes {1,2,3} and {4}: 9 + 1 = 10 pairs; 3 present → 7 to add.

Question 21

Mathematics · 4 marks · −1

The sum of all rational terms in the expansion of (1+21/3+31/2)6\left(1+2^{1/3}+3^{1/2}\right)^6 is equal to ______.

Correct answer: 612

Question 22

Mathematics · 4 marks · −1

If the area of the larger portion bounded between the curves x2+y2=25x^2+y^2=25 and y=∣x−1∣y=|x-1| is 14(bπ+c)\frac{1}{4}(b\pi+c), b,c∈Nb, c\in\mathbf{N}, then b + c is equal to ______.

Correct answer: 77

Question 23

Mathematics · 4 marks · −1

Let the circle C touch the line x−y+1=0x-y+1=0, have the centre on the positive x-axis, and cut off a chord of length 413\frac{4}{\sqrt{13}} along the line −3x+2y=1-3x+2y=1. Let H be the hyperbola x2α2−y2β2=1\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then 2α2+3β22\alpha^2+3\beta^2 is equal to ______.

Correct answer: 19

Question 24

Mathematics · 4 marks · −1

If the set of all values of a, for which the equation 5x3−15x−a=05x^3-15x-a=0 has three distinct real roots, is the interval (α\alpha, β\beta), then β−2α\beta-2\alpha is equal to ______.

Correct answer: 30

Explanation: g=5x3−15xg=5x^3-15x has local max 10 and min −10: a∈(−10,10)a\in(-10,10); 10+20=3010+20=30.

Question 25

Mathematics · 4 marks · −1

If the equation a(b−c)x2+b(c−a)x+c(a−b)=0a(b-c)x^2+b(c-a)x+c(a-b)=0 has equal roots, where a + c = 15 and b=365b=\frac{36}{5}, then a2+c2a^2+c^2 is equal to ______.

Correct answer: 117

Explanation: x = 1 is a root, so equal roots ⇒ product c(a−b)a(b−c)=1\frac{c(a-b)}{a(b-c)}=1 ⇒ b=2aca+cb=\frac{2ac}{a+c} ⇒ ac=54ac=54; a2+c2=225−108=117a^2+c^2=225-108=117.

Question 26

Physics · 4 marks · −1

A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ×10−2y\pi\times10^{-2} s, where the value of y is (Acceleration due to gravity, g = 10 m/s2m/s^2, density of water = 103 kg/m310^3\ kg/m^3)

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

Correct answer: B

Explanation: T=2πmρAg=2π0.01103⋅10−2⋅10=2π×10−2T=2\pi\sqrt{\frac{m}{\rho Ag}}=2\pi\sqrt{\frac{0.01}{10^3\cdot10^{-2}\cdot10}}=2\pi\times10^{-2}, y = 2.

Question 27

Physics · 4 marks · −1

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be

  1. A.1.0 cm
  2. B.2.0 cm
  3. C.1.5 cm
  4. D.0.5 cm

Correct answer: A

Explanation: x=(1/16)⋅1515/16=1x=\frac{(1/16)\cdot15}{15/16}=1 cm.

Question 28

Physics · 4 marks · −1

Refer to the circuit diagram given in the figure. which of the following observations are correct? A. Total resistance of circuit is 6 Ω\Omega B. Current in Ammeter is 1A C. Potential across AB is 4 Volts. D. Potential across CD is 4 Volts E. Total resistance of the circuit is 8 Ω\Omega. Choose the correct answer from the options given below:

Figure for question 28
  1. A.A, C and D Only
  2. B.B, C and E Only
  3. C.A, B and C Only
  4. D.A, B and D Only

Correct answer: D

Explanation: Parallel 2 Ω + 4 Ω = 6 Ω, I = 1 A; VCD=4V_{CD}=4 V, VAB=0.5×4=2V_{AB}=0.5\times4=2 V.

Question 29

Physics · 4 marks · −1

Given below are two statements: Statement I: The hot water flows faster than cold water Statement II: Soap water has higher surface tension as compared to fresh water. In the light above statements, choose the correct answer from the options given below

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

Correct answer: D

Question 30

Physics · 4 marks · −1

A radioactive nucleus n2n_2 has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1n_1. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2n_2 to the number of nuclei of n1n_1, after one half-life of n1n_1?

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

Correct answer: B

Explanation: After T1T_1, n1=N/2n_1=N/2 and n2=N/8n_2=N/8: ratio 1/4.

Question 31

Physics · 4 marks · −1

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness 'h' in terms of the angle of incidence 'i' and angle of refraction 'r', if the glass slab is placed in air medium?

  1. A.htan⁡(i−r)tan⁡r\frac{h\tan(i-r)}{\tan r}
  2. B.hcos⁡(i−r)sin⁡r\frac{h\cos(i-r)}{\sin r}
  3. C.h
  4. D.hsin⁡(i−r)cos⁡r\frac{h\sin(i-r)}{\cos r}

Correct answer: D

Question 32

Physics · 4 marks · −1

Identify the valid statements relevant to the given circuit at the instant when the key is closed. A. There will be no current through resistor R. B. There will be maximum current in the connecting wires. C. Potential difference between the capacitor plates A and B is minimum. D. Charge on the capacitor plates is minimum. Choose the correct answer from the options given below:

Figure for question 32
  1. A.A, B, D Only
  2. B.B, C, D Only
  3. C.C, D Only
  4. D.A, C Only

Correct answer: B

Explanation: At the instant of closing, the uncharged capacitor acts as a short: current is maximum, charge and p.d. are minimum (zero).

Question 33

Physics · 4 marks · −1

A solid sphere of mass 'm' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of length 'L' and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is v1v_1. If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2v_2. The ratio v12:v22v_1^2:v_2^2 is

  1. A.1 : 3
  2. B.1:21:\sqrt{2}
  3. C.1:31:\sqrt{3}
  4. D.1 : 2

Correct answer: B

Explanation: v2∝sin⁡θv^2\propto\sin\theta: 1/21/2=1:2\frac{1/2}{1/\sqrt2}=1:\sqrt2.

Question 34

Physics · 4 marks · −1

Given a thin convex lens (refractive index μ2\mu_2), kept in a liquid (refractive index μ1\mu_1, μ1<μ2\mu_1<\mu_2) having radii of curvatures ∣R1∣|R_1| and ∣R2∣|R_2|. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

  1. A.μ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R2∣\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}
  2. B.(μ2+μ1)∣R1∣(μ2−μ1)\frac{(\mu_2+\mu_1)|R_1|}{(\mu_2-\mu_1)}
  3. C.μ1∣R1∣⋅∣R2∣μ2(2∣R1∣+∣R2∣)−μ1∣R1∣⋅∣R2∣\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2(2|R_1|+|R_2|)-\mu_1\sqrt{|R_1|\cdot|R_2|}}
  4. D.μ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R1∣\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_1|}

Correct answer: A

Question 35

Physics · 4 marks · −1

The position of a particle moving on x-axis is given by x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A\sin t+B\cos^2t+Ct^2+D, where t is time. The dimension of ABCD\frac{ABC}{D} is

  1. A.L3T−2L^3T^{-2}
  2. B.L
  3. C.L2L^2
  4. D.L2T−2L^2T^{-2}

Correct answer: D

Explanation: A, B, D have dimension L; C has LT−2LT^{-2}: L⋅L⋅LT−2L=L2T−2\frac{L\cdot L\cdot LT^{-2}}{L}=L^2T^{-2}.

Question 36

Physics · 4 marks · −1

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is ______ grams. (Latent heat of fusion of lead = 2.5×104 JKg−12.5\times10^4\ JKg^{-1} and specific heat capacity of lead = 125 JKg−1K−1JKg^{-1}K^{-1})

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

Correct answer: A

Explanation: m(125×300+25000)=625⇒m=0.01m(125\times300+25000)=625\Rightarrow m=0.01 kg.

Question 37

Physics · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I: A. Pressure varies inversely with volume of an ideal gas. B. Heat absorbed goes partly to increase internal energy and partly to do work. C. Heat is neither absorbed nor released by a system. D. No work is done on or by a gas. LIST-II: I. Adiabatic process II. Isochoric process III. Isothermal process IV. Isobaric process. Choose the correct answer from the options given below:

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

Correct answer: C

Question 38

Physics · 4 marks · −1

The electric field of an electromagnetic wave in free space is E⃗=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)\vec{E}=57\cos[7.5\times10^6t-5\times10^{-3}(3x+4y)](4\hat{i}-3\hat{j}) N/C. The associated magnetic field in Tesla is

  1. A.B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec{B}=-\frac{57}{3\times10^8}\cos[7.5\times10^6t-5\times10^{-3}(3x+4y)](\hat{k})
  2. B.B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec{B}=\frac{57}{3\times10^8}\cos[7.5\times10^6t-5\times10^{-3}(3x+4y)](\hat{k})
  3. C.B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B}=\frac{57}{3\times10^8}\cos[7.5\times10^6t-5\times10^{-3}(3x+4y)](5\hat{k})
  4. D.B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B}=-\frac{57}{3\times10^8}\cos[7.5\times10^6t-5\times10^{-3}(3x+4y)](5\hat{k})

Correct answer: D

Question 39

Physics · 4 marks · −1

The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ______ km.

Figure for question 39
  1. A.6
  2. B.3
  3. C.12
  4. D.9

Correct answer: C

Explanation: 200+4002×2+400×28.5=600+11400=12000\frac{200+400}{2}\times2+400\times28.5=600+11400=12000 m.

Question 40

Physics · 4 marks · −1

The electric flux is ϕ=ασ+βλ\phi=\alpha\sigma+\beta\lambda where λ\lambda and σ\sigma are linear and surface charge density, respectively. (αβ)\left(\frac{\alpha}{\beta}\right) represents

  1. A.area
  2. B.displacement
  3. C.electric field
  4. D.charge

Correct answer: B

Explanation: ασ\alpha\sigma and βλ\beta\lambda share dimensions, so α/β=λ/σ\alpha/\beta=\lambda/\sigma = length.

Question 41

Physics · 4 marks · −1

Regarding self-inductance: A. The self-inductance of the coil depends on its geometry. B. Self-inductance does not depend on the permeability of the medium. C. Self-induced e.m.f. opposes any change in the current in a circuit. D. Self-inductance is electromagnetic analogue of mass in mechanics. E. Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below:

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

Correct answer: B

Question 42

Physics · 4 marks · −1

A sub-atomic particle of mass 10−3010^{-30} kg is moving with a velocity 2.21×1062.21\times10^6 m/s. Under the matter wave consideration, the particle will behave closely like ______. (h = 6.63×10−346.63\times10^{-34} J.s)

  1. A.Gamma rays
  2. B.X-rays
  3. C.Visible radiation
  4. D.Infra-red radiation

Correct answer: B

Explanation: λ=6.63×10−342.21×10−24=3×10−10\lambda=\frac{6.63\times10^{-34}}{2.21\times10^{-24}}=3\times10^{-10} m (X-ray range).

Question 43

Physics · 4 marks · −1

A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5). The centre of curvature is in the glass medium. A point object 'O' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to

  1. A.1.5 R
  2. B.3 R
  3. C.2 R
  4. D.5 R

Correct answer: D

Explanation: 1.5x+1x=0.5R⇒x=5R\frac{1.5}{x}+\frac{1}{x}=\frac{0.5}{R}\Rightarrow x=5R.

Question 44

Physics · 4 marks · −1

A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charge q separated by a distance 2a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

Figure for question 44
  1. A.ar∼3\frac{a}{r}\sim3
  2. B.ar∼20\frac{a}{r}\sim20
  3. C.ar∼10\frac{a}{r}\sim10
  4. D.ar∼0.5\frac{a}{r}\sim0.5

Correct answer: A

Question 45

Physics · 4 marks · −1

Consider a moving coil galvanometer (MCG): A. The torsional constant in moving coil galvanometer has dimentions [ML2T−2][ML^2T^{-2}] B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C. If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles. D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer. E. Current sensitivity of MCG depends inversely on number of turns of coil. Choose the correct answer from the options given below:

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

Correct answer: D

Question 46

Physics · 4 marks · −1

A force f=x2yi^+y2j^f=x^2y\hat{i}+y^2\hat{j} acts on a particle in a plane x + y = 10. The work done by this force during a displacement from (0, 0) to (4m, 2m) is ______ Joule (round off to the nearest integer)

Correct answer: 152

Question 47

Physics · 4 marks · −1

Two particles are located at equal distance from origin. The position vectors of those are represented by A⃗=2i^+3nj^+2k^\vec{A}=2\hat{i}+3n\hat{j}+2\hat{k} and B⃗=2i^−2j^+4pk^\vec{B}=2\hat{i}-2\hat{j}+4p\hat{k}, respectively. If both the vectors are at right angle to each other, the value of n−1n^{-1} is ______.

Correct answer: 3

Question 48

Physics · 4 marks · −1

An ideal gas initially at 0°C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is 3/2, the change in temperature due to the thermodynamic process is ______ K.

Correct answer: 273

Explanation: T2=273×41/2=546T_2=273\times4^{1/2}=546 K; change 273 K.

Question 49

Physics · 4 marks · −1

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω\Omega, the value of the current in the circuit will be ______ A.

Figure for question 49

Correct answer: 3

Explanation: Pulling out increases R so current decreases: 12+3×8=12I⇒I=312+3\times8=12I\Rightarrow I=3 A.

Question 50

Chemistry · 4 marks · −1

Given below are two statements: Statement I: In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds. Statement II: The sodium fusion extract of an organic compound having N and S gives prussian blue colour with FeSO4FeSO_4 and Na4[Fe(CN)6]Na_4[Fe(CN)_6]. 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 true
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are false

Correct answer: C

Explanation: With both N and S present, NaSCN forms and gives blood-red colour with Fe3+, not Prussian blue.

Question 51

Chemistry · 4 marks · −1

FeO42−→+2.0 VFe3+→0.8 VFe2+→−0.5 VFe0FeO_4^{2-}\xrightarrow{+2.0\,V}Fe^{3+}\xrightarrow{0.8\,V}Fe^{2+}\xrightarrow{-0.5\,V}Fe^0. In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of EFeO42−/Fe2+⊖E^\ominus_{FeO_4^{2-}/Fe^{2+}} is

  1. A.2.1 V
  2. B.1.2 V
  3. C.1.4 V
  4. D.1.7 V

Correct answer: D

Explanation: E=3(2.0)+1(0.8)4=1.7E=\frac{3(2.0)+1(0.8)}{4}=1.7 V.

Question 52

Chemistry · 4 marks · −1

Which among the following react with Hinsberg's reagent? A. Aniline (C6H5NH2C_6H_5NH_2) B. N,N-Dimethylaniline (C6H5N(CH3)2C_6H_5N(CH_3)_2) C. CH3−NH2CH_3-NH_2 D. N(CH3)3N(CH_3)_3 E. Diphenylamine (C6H5−NH−C6H5C_6H_5-NH-C_6H_5). Choose the correct answer from the options given below:

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

Correct answer: D

Explanation: Primary and secondary amines react; tertiary amines do not.

Question 53

Chemistry · 4 marks · −1

The incorrect statement among the following is

  1. A.PH3PH_3 shows lower proton affinity than NH3NH_3.
  2. B.PF3PF_3 exists but NF5NF_5 does not.
  3. C.NO2NO_2 can dimerise easily.
  4. D.SO2SO_2 can act as an oxidizing agent, but not as a reducing agent.

Correct answer: D

Question 54

Chemistry · 4 marks · −1

What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in g mol−1mol^{-1} of C, H, O, Br are 12, 1, 16, 80 respectively)

  1. A.20.44 g
  2. B.6.0 g
  3. C.4.0 g
  4. D.10.22 g

Correct answer: D

Explanation: 2/94 mol phenol × 3 mol Br2Br_2 × 160 g = 10.21 g.

Question 55

Chemistry · 4 marks · −1

The major product of the following reaction is: CH3CH2CH=O→refluxexcess HCHO, alkaliCH_3CH_2CH=O\xrightarrow[\text{reflux}]{\text{excess HCHO, alkali}} ?

  1. A.CH3−C(=CH2)−CH=OCH_3-C(=CH_2)-CH=O
  2. B.CH3−C(CH2OH)3CH_3-C(CH_2OH)_3
  3. C.CH3−CH2−CH2−OHCH_3-CH_2-CH_2-OH
  4. D.CH3−CH(CH2OH)−CH=OCH_3-CH(CH_2OH)-CH=O

Correct answer: B

Explanation: Two aldol additions of HCHO at the α-carbon, then crossed Cannizzaro reduces CHO to CH2OHCH_2OH: CH3C(CH2OH)3CH_3C(CH_2OH)_3.

Question 56

Chemistry · 4 marks · −1

The complex that shows Facial - Meridional isomerism is:

  1. A.[Co(NH3)4Cl2]+[Co(NH_3)_4Cl_2]^+
  2. B.[Co(en)3]3+[Co(en)_3]^{3+}
  3. C.[Co(NH3)3Cl3][Co(NH_3)_3Cl_3]
  4. D.[Co(en)2Cl2]+[Co(en)_2Cl_2]^+

Correct answer: C

Question 57

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Classification of molecules based on octet rule): A. Molecules obeying octet rule B. Molecules with incomplete octet C. Molecules with incomplete octet with odd electron D. Molecules with expanded octet. LIST-II (Example): I. NO, NO2NO_2 II. BCl3BCl_3, AlCl3AlCl_3 III. H2SO4H_2SO_4, PCl5PCl_5 IV. CCl4CCl_4, CO2CO_2. Choose the correct answer from the options given below:

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

Correct answer: B

Question 58

Chemistry · 4 marks · −1

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this? Given : Rydberg constant RH=105 cm−1R_H=10^5\ cm^{-1}, h = 6.6×10−346.6\times10^{-34} J s, c = 3×1083\times10^8 m/s)

  1. A.Balmer series, ∞→2\infty\to2
  2. B.Paschen series, 5→35\to3
  3. C.Paschen series, ∞→3\infty\to3
  4. D.Lyman series, ∞→1\infty\to1

Correct answer: C

Explanation: 1λ=RH9\frac1\lambda=\frac{R_H}{9} gives λ=900\lambda=900 nm for Paschen ∞→3\infty\to3.

Question 59

Chemistry · 4 marks · −1

2.8×10−32.8\times10^{-3} mol of CO2CO_2 is left after removing 102110^{21} molecules from its 'x' mg sample. The mass of CO2CO_2 taken initially is Given: NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ mol^{-1}

  1. A.196.2 mg
  2. B.98.3 mg
  3. C.48.2 mg
  4. D.150.4 mg

Correct answer: A

Explanation: 102110^{21} molecules =1.66×10−3=1.66\times10^{-3} mol; total 4.46×10−34.46\times10^{-3} mol × 44 = 196.2 mg.

Question 60

Chemistry · 4 marks · −1

The correct set of ions (aqueous solution) with same colour from the following is:

  1. A.Sc3+Sc^{3+}, Ti3+Ti^{3+}, Cr2+Cr^{2+}
  2. B.V2+V^{2+}, Cr3+Cr^{3+}, Mn3+Mn^{3+}
  3. C.Ti4+Ti^{4+}, V4+V^{4+}, Mn2+Mn^{2+}
  4. D.Zn2+Zn^{2+}, V3+V^{3+}, Fe3+Fe^{3+}

Correct answer: B

Explanation: V2+V^{2+}, Cr3+Cr^{3+} and Mn3+Mn^{3+} are all violet.

Question 61

Chemistry · 4 marks · −1

The correct stability order of the following species/molecules is: p = cyclopropenyl anion (three-membered ring with one C=C and a − charge); q = cyclopentadienyl anion (five-membered ring with two C=C, a − charge and an H on the charged carbon); r = cyclooctatetraene (eight-membered ring with four C=C).

Figure for question 61
  1. A.r > q > p
  2. B.p > q > r
  3. C.q > r > p
  4. D.q > p > r

Correct answer: C

Explanation: Aromatic q > non-aromatic r > antiaromatic p.

Question 62

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. LIST-I (Name reaction): A. Swarts reaction B. Sandmeyer's reaction C. Wurtz Fittig reaction D. Finkelstein reaction. LIST-II (Product obtainable): I. Ethyl benzene II. Ethyl iodide III. Cyanobenzene IV. Ethyl fluoride. Choose the correct answer from the options given below:

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

Correct answer: D

Question 63

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Fructose does not contain an aldehydic group but still reduces Tollen's reagent Statement II: In the presence of base, fructose undergoes rearrangement to give glucose. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: A

Question 64

Chemistry · 4 marks · −1

The element that does not belong to the same period of the remaining elements (modern periodic table) is:

  1. A.Osmium
  2. B.Palladium
  3. C.Iridium
  4. D.Platinum

Correct answer: B

Explanation: Pd is in period 5; Os, Ir, Pt are in period 6.

Question 65

Chemistry · 4 marks · −1

Which of the following happens when NH4OHNH_4OH is added gradually to the solution containing 1M A2+A^{2+} and 1M B3+B^{3+} ions? Given : Ksp[A(OH)2]=9×10−10K_{sp}[A(OH)_2]=9\times10^{-10} and Ksp[B(OH)3]=27×10−18K_{sp}[B(OH)_3]=27\times10^{-18} at 298 K.

  1. A.A(OH)2A(OH)_2 will precipitate before B(OH)3B(OH)_3
  2. B.A(OH)2A(OH)_2 and B(OH)3B(OH)_3 will precipitate together
  3. C.B(OH)3B(OH)_3 will precipitate before A(OH)2A(OH)_2
  4. D.Both A(OH)2A(OH)_2 and B(OH)3B(OH)_3 do not show precipitation with NH4OHNH_4OH

Correct answer: C

Explanation: [OH−][OH^-] needed: A: 3×10−53\times10^{-5}; B: 3×10−63\times10^{-6} → B precipitates first.

Question 66

Chemistry · 4 marks · −1

The d– electronic configuration of an octahedral Co(II) complex having magnetic moment of 3.95 BM is:

  1. A.e4t23e^4t_2^3
  2. B.t2g6eg1t_{2g}^6e_g^1
  3. C.t2g3eg0t_{2g}^3e_g^0
  4. D.t2g5eg2t_{2g}^5e_g^2

Correct answer: D

Explanation: 3 unpaired electrons (3.87 BM): high-spin octahedral d7d^7 = t2g5eg2t_{2g}^5e_g^2.

Question 67

Chemistry · 4 marks · −1

Ice at −5°C is heated to become vapor with temperature of 110°C at atmospheric pressure. The entropy change associated with this process can be obtained from

  1. A.∫268K273KCp,mTdT+ΔHm,fusionTf+ΔHm,vaporisationTb+∫273K373KCp,mdTT+∫373K383KCp,mdTT\int_{268K}^{273K}\frac{C_{p,m}}{T}dT+\frac{\Delta H_{m,fusion}}{T_f}+\frac{\Delta H_{m,vaporisation}}{T_b}+\int_{273K}^{373K}\frac{C_{p,m}dT}{T}+\int_{373K}^{383K}\frac{C_{p,m}dT}{T}
  2. B.∫268K383KCpdT+ΔHmelting273+ΔHboiling373\int_{268K}^{383K}C_pdT+\frac{\Delta H_{melting}}{273}+\frac{\Delta H_{boiling}}{373}
  3. C.∫268K273KCp,mdT+ΔHm,fusionTf+ΔHm,vaporisationTb+∫273K373KCp,mdT+∫373K383KCp,mdT\int_{268K}^{273K}C_{p,m}dT+\frac{\Delta H_{m,fusion}}{T_f}+\frac{\Delta H_{m,vaporisation}}{T_b}+\int_{273K}^{373K}C_{p,m}dT+\int_{373K}^{383K}C_{p,m}dT
  4. D.∫268K383KCpdT+qrevT\int_{268K}^{383K}C_pdT+\frac{q_{rev}}{T}

Correct answer: A

Question 68

Chemistry · 4 marks · −1

Propane molecule on chlorination under photochemical condition gives two di-chloro products, "x" and "y". Amongst "x" and "y", "x" is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from "x" when it is further treated with chlorine under the photochemical condition?

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

Correct answer: A

Explanation: x = 1,2-dichloropropane; further chlorination gives 1,1,2-, 1,2,2- and 1,2,3-trichloropropane = 3.

Question 69

Chemistry · 4 marks · −1

CrCl3⋅xNH3CrCl_3\cdot xNH_3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be (Given KfK_f = 1.86 K kg mol−1mol^{-1})

  1. A.[Cr(NH3)3Cl3][Cr(NH_3)_3Cl_3]
  2. B.[Cr(NH3)5Cl]Cl2[Cr(NH_3)_5Cl]Cl_2
  3. C.[Cr(NH3)4Cl2]Cl[Cr(NH_3)_4Cl_2]Cl
  4. D.[Cr(NH3)6]Cl3[Cr(NH_3)_6]Cl_3

Correct answer: B

Explanation: i=0.558/(1.86×0.1)=3i=0.558/(1.86\times0.1)=3.

Question 70

Chemistry · 4 marks · −1

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (KbK_b) of ethylamine is 10−x10^{-x}. The value of x is ______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

Correct answer: 7

Explanation: [OH−]=10−5[OH^-]=10^{-5}; Kb=10−10/10−3=10−7K_b=10^{-10}/10^{-3}=10^{-7}.

Question 71

Chemistry · 4 marks · −1

The standard enthalpy and standard entropy of decomposition of N2O4N_2O_4 to NO2NO_2 are 55.0 kJ mol−1mol^{-1} and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25°C in J mol−1mol^{-1} is ______ (Nearest integer)

Correct answer: 2850

Explanation: ΔG∘=55000−298×175=2850\Delta G^\circ=55000-298\times175=2850 J mol−1^{-1}.

Question 72

Chemistry · 4 marks · −1

During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %. (Given molar mass in g mol−1mol^{-1} of Ba : 137, S : 32, O : 16)

Correct answer: 40

Explanation: S = 466 × 32/233 = 64 mg; 40%.

Question 73

Chemistry · 4 marks · −1

Consider the following sequence of reactions to produce major product (A): 3-Nitrotoluene (m-nitrotoluene) →i) Br2,Fe; ii) Sn, HCl; iii) NaNO2,HCl,273K; iv) H3PO2,H2O\xrightarrow{\text{i) } Br_2, Fe;\ \text{ii) Sn, HCl;\ iii) } NaNO_2, HCl, 273K;\ \text{iv) } H_3PO_2, H_2O} (A) Major Product. Molar mass of product (A) is ______ g mol−1mol^{-1}. (Given molar mass in g mol−1mol^{-1} of C : 12, H: 1, O :16, Br : 80, N : 14, P:31)

Correct answer: 171

Explanation: Bromination, reduction of NO2NO_2, diazotisation and deamination leave a bromotoluene C7H7BrC_7H_7Br = 171.

Question 74

Chemistry · 4 marks · −1

For the thermal decomposition of N2O5(g)N_2O_5(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2N2O5(g)→2N2O4(g)+O2(g)2N_2O_5(g)\to2N_2O_4(g)+O_2(g). Sr. No. 1: Time 0 s, Total pressure 0.6 atm; Sr. No. 2: Time 100 s, Total pressure x. x = ______ ×10−3\times10^{-3} atm [nearest integer] Given : Rate constant for the reaction is 4.606×10−2 s−14.606\times10^{-2}\ s^{-1}.

Correct answer: 897

Explanation: pN2O5=0.6e−4.606=0.006p_{N_2O_5}=0.6e^{-4.606}=0.006 atm; total =0.6+12(0.6−0.006)=0.897=0.6+\frac12(0.6-0.006)=0.897 atm → 897.

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