JEE Main 2025 — 2 Apr Shift 1 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

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

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If the system of linear equations 3x+y+βz=33x+y+\beta z=3, 2x+αy−z=−32x+\alpha y-z=-3, x+2y+z=4x+2y+z=4 has infinitely many solutions, then the value of 22β−9α22\beta-9\alpha is :

  1. A.37
  2. B.43
  3. C.49
  4. D.31

Correct answer: D

Question 2

Mathematics · 4 marks · −1

Let z be a complex number such that ∣z∣=1|z|=1. If 2+k2zk+zˉ=kz\frac{2+k^2z}{k+\bar{z}}=kz, k∈Rk\in\mathbf{R}, then the maximum distance of k+ik2k+ik^2 from the circle ∣z−(1+2i)∣=1|z-(1+2i)|=1 is :

  1. A.3+1\sqrt{3}+1
  2. B.2
  3. C.5+1\sqrt{5}+1
  4. D.3

Correct answer: C

Question 3

Mathematics · 4 marks · −1

If S and S' are the foci of the ellipse x218+y29=1\frac{x^2}{18}+\frac{y^2}{9}=1 and P be a point on the ellipse, then min⁡(SP⋅S′P)+max⁡(SP⋅S′P)\min(SP\cdot S'P)+\max(SP\cdot S'P) is equal to :

  1. A.3(6+2)3(6+\sqrt{2})
  2. B.9
  3. C.27
  4. D.3(1+2)3(1+\sqrt{2})

Correct answer: C

Explanation: SP⋅SPˊ=a2−e2x2SP\cdot S\'P=a^2-e^2x^2 ranges from b2=9b^2=9 to a2=18a^2=18; sum 27.

Question 4

Mathematics · 4 marks · −1

Let A=[α−16β]A=\begin{bmatrix}\alpha & -1\\ 6 & \beta\end{bmatrix}, α>0\alpha>0, such that det⁡(A)=0\det(A)=0 and α+β=1\alpha+\beta=1. If I denotes 2×22\times2 identity matrix, then the matrix (I+A)8(I+A)^8 is :

  1. A.[1025−5112024−1024]\begin{bmatrix}1025 & -511\\ 2024 & -1024\end{bmatrix}
  2. B.[257−64514−127]\begin{bmatrix}257 & -64\\ 514 & -127\end{bmatrix}
  3. C.[4−16−1]\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}
  4. D.[766−2551530−509]\begin{bmatrix}766 & -255\\ 1530 & -509\end{bmatrix}

Correct answer: D

Explanation: αβ+6=0, α+β=1⇒α=3,β=−2\alpha\beta+6=0,\ \alpha+\beta=1\Rightarrow\alpha=3,\beta=-2; A2=AA^2=A (trace 1, det 0), so (I+A)8=I+(28−1)A=I+255A(I+A)^8=I+(2^8-1)A=I+255A.

Question 5

Mathematics · 4 marks · −1

If θ∈[−2π,2π]\theta\in[-2\pi, 2\pi], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02\sqrt{2}\cos^2\theta+(2-\sqrt{6})\cos\theta-\sqrt{3}=0, is equal to :

  1. A.8
  2. B.10
  3. C.6
  4. D.12

Correct answer: A

Explanation: Roots cos⁡θ=32\cos\theta=\frac{\sqrt3}{2} and −12-\frac{1}{\sqrt2}; each gives 4 solutions in [−2π,2π][-2\pi,2\pi]: total 8.

Question 6

Mathematics · 4 marks · −1

If the function f(x)=2x3−9ax2+12a2x+1f(x)=2x^3-9ax^2+12a^2x+1, where a>0a>0, attains its local maximum and local minimum values at p and q, respectively, such that p2=qp^2=q, then f(3)f(3) is equal to :

  1. A.37
  2. B.23
  3. C.55
  4. D.10

Correct answer: A

Explanation: p=a,q=2ap=a, q=2a, a2=2a⇒a=2a^2=2a\Rightarrow a=2; f(3)=54−162+144+1=37f(3)=54-162+144+1=37.

Question 7

Mathematics · 4 marks · −1

If a⃗\vec{a} is a nonzero vector such that its projections on the vectors 2i^−j^+2k^2\hat{i}-\hat{j}+2\hat{k}, i^+2j^−2k^\hat{i}+2\hat{j}-2\hat{k} and k^\hat{k} are equal, then a unit vector along a⃗\vec{a} is :

  1. A.1155(7i^+9j^+5k^)\frac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}+5\hat{k})
  2. B.1155(7i^+9j^−5k^)\frac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}-5\hat{k})
  3. C.1155(−7i^+9j^+5k^)\frac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}+5\hat{k})
  4. D.1155(−7i^+9j^−5k^)\frac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}-5\hat{k})

Correct answer: A

Question 8

Mathematics · 4 marks · −1

The term independent of xx in the expansion of ((x+1)(x2/3+1−x1/3)−(x−1)(x−x1/2))10\left(\frac{(x+1)}{\left(x^{2/3}+1-x^{1/3}\right)}-\frac{(x-1)}{\left(x-x^{1/2}\right)}\right)^{10}, x>1x>1, is :

  1. A.240
  2. B.210
  3. C.150
  4. D.120

Correct answer: B

Explanation: Expression reduces to (x1/3−x−1/2)10(x^{1/3}-x^{-1/2})^{10}; general term exponent 10−r3−r2=0⇒r=4\frac{10-r}{3}-\frac r2=0\Rightarrow r=4: 10C4=210{}^{10}C_4=210.

Question 9

Mathematics · 4 marks · −1

Let the focal chord PQ of the parabola y2=4xy^2=4x make an angle of 60° with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point (0,α)(0, \alpha), then 5α25\alpha^2 is equal to :

  1. A.25
  2. B.15
  3. C.30
  4. D.20

Correct answer: B

Question 10

Mathematics · 4 marks · −1

Let a∈Ra\in\mathbf{R} and A be a matrix of order 3×33\times3 such that det⁡(A)=−4\det(A)=-4 and A+I=[1a1210a12]A+I=\begin{bmatrix}1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2\end{bmatrix}, where I is the identity matrix of order 3×33\times3. If det⁡((a+1) adj((a−1)A))\det((a+1)\,\mathrm{adj}((a-1)A)) is 2m3n2^m3^n, m,n∈{0,1,2,…,20}m, n\in\{0, 1, 2, \ldots, 20\}, then m+nm+n is equal to :

  1. A.15
  2. B.17
  3. C.14
  4. D.16

Correct answer: D

Question 11

Mathematics · 4 marks · −1

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the △BCD\triangle BCD is equal to :

  1. A.737\sqrt{3}
  2. B.12
  3. C.110\sqrt{110}
  4. D.340\sqrt{340}

Correct answer: C

Explanation: [BCD]2=52+62+72=110[BCD]^2=5^2+6^2+7^2=110.

Question 12

Mathematics · 4 marks · −1

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :

  1. A.1820
  2. B.45
  3. C.360
  4. D.2520

Correct answer: D

Explanation: 10!5! 3! 2!=2520\frac{10!}{5!\,3!\,2!}=2520.

Question 13

Mathematics · 4 marks · −1

Let A be the set of all functions f:Z→Zf:\mathbf{Z}\to\mathbf{Z} and R be a relation on A such that R={(f,g):f(0)=g(1) and f(1)=g(0)}R=\{(f, g): f(0)=g(1) \text{ and } f(1)=g(0)\}. Then R is :

  1. A.Reflexive but neither symmetric nor transitive
  2. B.Symmetric and transitive but not reflective
  3. C.Symmetric but neither reflective nor transitive
  4. D.Transitive but neither reflexive nor symmetric

Correct answer: C

Question 14

Mathematics · 4 marks · −1

The largest n∈Nn\in\mathbf{N} such that 3n3^n divides 50! is :

  1. A.21
  2. B.22
  3. C.23
  4. D.20

Correct answer: B

Explanation: ⌊50/3⌋+⌊50/9⌋+⌊50/27⌋=16+5+1=22\lfloor50/3\rfloor+\lfloor50/9\rfloor+\lfloor50/27\rfloor=16+5+1=22.

Question 15

Mathematics · 4 marks · −1

Let Pn=αn+βnP_n=\alpha^n+\beta^n, n∈Nn\in\mathbf{N}. If P10=123P_{10}=123, P9=76P_9=76, P8=47P_8=47 and P1=1P_1=1, then the quadratic equation having roots 1α\frac{1}{\alpha} and 1β\frac{1}{\beta} is :

  1. A.x2−x+1=0x^2-x+1=0
  2. B.x2−x−1=0x^2-x-1=0
  3. C.x2+x+1=0x^2+x+1=0
  4. D.x2+x−1=0x^2+x-1=0

Correct answer: D

Explanation: P10=P9−αβP8P_{10}=P_9-\alpha\beta P_8 gives αβ=−1\alpha\beta=-1, α+β=1\alpha+\beta=1; reciprocals: sum −1-1, product −1-1: x2+x−1=0x^2+x-1=0.

Question 16

Mathematics · 4 marks · −1

Let the vertices Q and R of the triangle PQR lie on the line x+35=y−12=z+43\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is mn\frac{m}{n} then :

  1. A.5m−221n=05m-2\sqrt{21}n=0
  2. B.5m−212n=05m-21\sqrt{2}n=0
  3. C.2m−521n=02m-5\sqrt{21}n=0
  4. D.m−521n=0m-5\sqrt{21}n=0

Correct answer: C

Question 17

Mathematics · 4 marks · −1

For α,β,γ∈R\alpha, \beta, \gamma\in\mathbf{R}, if lim⁡x→0x2sin⁡αx+(γ−1)ex2sin⁡2x−βx=3\lim_{x\to0}\frac{x^2\sin\alpha x+(\gamma-1)e^{x^2}}{\sin2x-\beta x}=3, then β+γ−α\beta+\gamma-\alpha is equal to :

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

Correct answer: D

Question 18

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be in an A.P. such that ∑k=112a2k−1=−725a1\sum_{k=1}^{12}a_{2k-1}=-\frac{72}{5}a_1, a1≠0a_1\neq0. If ∑k=1nak=0\sum_{k=1}^{n}a_k=0, then n is :

  1. A.10
  2. B.18
  3. C.17
  4. D.11

Correct answer: D

Question 19

Mathematics · 4 marks · −1

Let one focus of the hyperbola H: x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 be at (10,0)(\sqrt{10}, 0) and the corresponding directrix be x=910x=\frac{9}{\sqrt{10}}. If e and ll respectively are the eccentricity and the length of the latus rectum of H, then 9(e2+l)9(e^2+l) is equal to :

  1. A.16
  2. B.15
  3. C.12
  4. D.14

Correct answer: A

Explanation: ae=10ae=\sqrt{10}, a/e=9/10a/e=9/\sqrt{10} give a=3a=3, e2=10/9e^2=10/9, b2=1b^2=1, l=2/3l=2/3; 9(10/9+2/3)=169(10/9+2/3)=16.

Question 20

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin y)(f(2x+2y)+f(2x-2y)), for all x,y∈Rx, y\in\mathbf{R}. If f′(0)=12f'(0)=\frac{1}{2}, then the value of 24f′′(5π3)24f''\left(\frac{5\pi}{3}\right) is :

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

Correct answer: B

Question 21

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a thrice differentiable odd function satisfying f′(x)≥0f'(x)\ge0, f′′(x)=f(x)f''(x)=f(x), f(0)=0f(0)=0, f′(0)=3f'(0)=3. Then 9f(log⁡e3)9f(\log_e3) is equal to ______.

Correct answer: 36

Explanation: f=3sinh⁡xf=3\sinh x; f(ln⁡3)=32(3−13)=4f(\ln3)=\frac32(3-\frac13)=4; answer 36.

Question 22

Mathematics · 4 marks · −1

Three distinct numbers are selected randomly from the set {1,2,3,…,40}\{1, 2, 3, \ldots, 40\}. If the probability, that the selected numbers are in an increasing G.P., is mn\frac{m}{n}, gcd⁡(m,n)=1\gcd(m, n)=1, then m+n is equal to ______.

Correct answer: 2477

Question 23

Mathematics · 4 marks · −1

Let [⋅][\cdot] denote the greatest integer function. If ∫0e3[1ex−1]dx=α−log⁡e2\int_0^{e^3}\left[\frac{1}{e^{x-1}}\right]dx=\alpha-\log_e2, then α3\alpha^3 is equal to ______.

Correct answer: 8

Question 24

Mathematics · 4 marks · −1

If the area of the region {(x,y):∣4−x2∣≤y≤x2,y≤4,x≥0}\{(x, y): |4-x^2|\le y\le x^2, y\le4, x\ge0\} is (802α−β)\left(\frac{80\sqrt{2}}{\alpha}-\beta\right), α,β∈N\alpha, \beta\in\mathbf{N}, then α+β\alpha+\beta is equal to ______.

Correct answer: 22

Question 25

Mathematics · 4 marks · −1

The absolute difference between the squares of the radii of the two circles passing through the point (−9,4)(-9, 4) and touching the lines x+y=3x+y=3 and x−y=3x-y=3, is equal to ______.

Correct answer: 768

Question 26

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Coefficient of viscosity (B) Intensity of wave (C) Pressure gradient (D) Compressibility. List - II: (I) [ML0T−3][ML^0T^{-3}] (II) [ML−2T−2][ML^{-2}T^{-2}] (III) [M−1LT2][M^{-1}LT^2] (IV) [ML−1T−1][ML^{-1}T^{-1}]. Choose the correct answer from the options given below :

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

Correct answer: D

Question 27

Physics · 4 marks · −1

A particle is subjected to two simple harmonic motions as : x1=7sin⁡5tx_1=\sqrt{7}\sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right) cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x×10−2 ms−2x\times10^{-2}\ ms^{-2}. The value of x is :

  1. A.175
  2. B.125
  3. C.25725\sqrt{7}
  4. D.575\sqrt{7}

Correct answer: A

Explanation: A2=7+28+2⋅7⋅27⋅12=49A^2=7+28+2\cdot\sqrt7\cdot2\sqrt7\cdot\frac12=49, A=7A=7 cm; amax=25×7=175a_{max}=25\times7=175 cm/s2^2.

Question 28

Physics · 4 marks · −1

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is : (C is the center of curvature of the spherical surface and R is the radius of curvature)

Figure for question 28
  1. A.0.24 m left to the spherical surface
  2. B.0.4 m left to the spherical surface
  3. C.0.24 m right to the spherical surface
  4. D.0.4 m right to the spherical surface

Correct answer: B

Explanation: 1.5v−1−0.2=0.50.4⇒1.5v=−3.75⇒v=−0.4\frac{1.5}{v}-\frac{1}{-0.2}=\frac{0.5}{0.4}\Rightarrow\frac{1.5}{v}=-3.75\Rightarrow v=-0.4 m (virtual, left).

Question 29

Physics · 4 marks · −1

The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged ?

  1. A.43.8 kJ
  2. B.87.7 kJ
  3. C.24.4 kJ
  4. D.48.7 kJ

Correct answer: B

Explanation: E=4.2×5.8×3600≈87.7E=4.2\times5.8\times3600\approx87.7 kJ.

Question 30

Physics · 4 marks · −1

A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be :

Figure for question 30
  1. A.0 rad/s
  2. B.30 rad/s
  3. C.10 rad/s
  4. D.20 rad/s

Correct answer: D

Explanation: 20×1=12(10)(0.1)2ω2⇒ω=2020\times1=\frac12(10)(0.1)^2\omega^2\Rightarrow\omega=20 rad/s.

Question 31

Physics · 4 marks · −1

Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of He+He^+ ion in its first excited state. (B) the energy of H atom in ground state is same as that for Li++Li^{++} ion in its second excited state. (C) the energy of H atom in its ground state is same as that of He+He^+ ion for its ground state. (D) the energy of He+He^+ ion in its first excited state is same as that for Li++Li^{++} ion in its ground state. Choose the correct answer from the options given below :

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

Correct answer: B

Explanation: E∝Z2/n2E\propto Z^2/n^2: H(n=1)=1, He+He^+(n=2)=1, Li2+Li^{2+}(n=3)=1.

Question 32

Physics · 4 marks · −1

The relationship between the magnetic susceptibility (χ\chi) and the magnetic permeability (μ\mu) is given by : (μ0\mu_0 is the permeability of free space and μr\mu_r is relative permeability)

  1. A.χ=μr+1\chi=\mu_r+1
  2. B.χ=1−μμ0\chi=1-\frac{\mu}{\mu_0}
  3. C.χ=μμ0−1\chi=\frac{\mu}{\mu_0}-1
  4. D.χ=μrμ0+1\chi=\frac{\mu_r}{\mu_0}+1

Correct answer: C

Question 33

Physics · 4 marks · −1

A small bob of mass 100 mg and charge +10 μ\muC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density 'σ\sigma' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be. (Given, ϵ0=8.85×10−12 Fm\epsilon_0=8.85\times10^{-12}\ \frac{F}{m} and acceleration due to gravity, g = 10 ms2\frac{m}{s^2})

Figure for question 33
  1. A.17.7 nC/m2nC/m^2
  2. B.0.885 nC/m2nC/m^2
  3. C.885 nC/m2nC/m^2
  4. D.1.77 nC/m2nC/m^2

Correct answer: D

Explanation: qσ2ϵ0=mgtan⁡45∘⇒σ=2ϵ0mgq=2(8.85×10−12)(10−4)(10)10−5=1.77×10−9\frac{q\sigma}{2\epsilon_0}=mg\tan45^\circ\Rightarrow\sigma=\frac{2\epsilon_0mg}{q}=\frac{2(8.85\times10^{-12})(10^{-4})(10)}{10^{-5}}=1.77\times10^{-9}.

Question 34

Physics · 4 marks · −1

A river is flowing from west to east direction with speed of 9 km h−1h^{-1}. If a boat capable of moving at a maximum speed of 27 km h−1h^{-1} in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150° to direction of river flow, then the width of the river is :

  1. A.112.5 m
  2. B.112.5×3112.5\times\sqrt{3} m
  3. C.75 m
  4. D.300 m

Correct answer: A

Explanation: Cross-stream speed 27sin⁡150∘=13.527\sin150^\circ=13.5 km/h =3.75=3.75 m/s; width =3.75×30=112.5=3.75\times30=112.5 m.

Question 35

Physics · 4 marks · −1

A monochromatic light is incident on a metallic plate having work function ϕ\phi. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is : (Given : The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

  1. A.2m(hcλ−ϕ)/eB\sqrt{2m\left(\frac{hc}{\lambda}-\phi\right)}/eB
  2. B.2m(hcλ−ϕ)/eB2\sqrt{m\left(\frac{hc}{\lambda}-\phi\right)}/eB
  3. C.m(hcλ−ϕ)/eB\sqrt{m\left(\frac{hc}{\lambda}-\phi\right)}/eB
  4. D.8m(hcλ−ϕ)/eB\sqrt{8m\left(\frac{hc}{\lambda}-\phi\right)}/eB

Correct answer: D

Explanation: AB = diameter =2peB=22mKeB=8mKeB=\frac{2p}{eB}=\frac{2\sqrt{2mK}}{eB}=\frac{\sqrt{8mK}}{eB}.

Question 36

Physics · 4 marks · −1

Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α\alpha'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

  1. A.α/4\alpha/4
  2. B.α/8\alpha/8
  3. C.α/2\alpha/2
  4. D.α\alpha

Correct answer: A

Explanation: Each half: (M/2)(L/2)212\frac{(M/2)(L/2)^2}{12}; total =ML248=α/4=\frac{ML^2}{48}=\alpha/4.

Question 37

Physics · 4 marks · −1

In an adiabatic process, which of the following statements is true ?

  1. A.Work done by the gas equals the increase in internal energy
  2. B.The molar heat capacity is infinite
  3. C.The internal energy of the gas decreases as the temperature increases
  4. D.The molar heat capacity is zero

Correct answer: D

Question 38

Physics · 4 marks · −1

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is :

  1. A.(4d/3, 0, 0)
  2. B.(d/3, 0, 0)
  3. C.(3d/4, 0, 0)
  4. D.(d/4, 0, 0)

Correct answer: D

Explanation: qx2=9q(d−x)2⇒x=d/4\frac{q}{x^2}=\frac{9q}{(d-x)^2}\Rightarrow x=d/4.

Question 39

Physics · 4 marks · −1

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :

Figure for question 39
  1. A.+45∘+45^\circ
  2. B.−45∘-45^\circ
  3. C.−α-\alpha
  4. D.−α2-\frac{\alpha}{2}

Correct answer: B

Explanation: A: u=−30⇒v=60u=-30\Rightarrow v=60. B: u=−29⇒v=580/9≈64.44u=-29\Rightarrow v=580/9\approx64.44, height =2×(−64.44/29)=−4.44=2\times(-64.44/29)=-4.44 cm. Image displacement (4.44, −4.44) → −45∘-45^\circ.

Question 40

Physics · 4 marks · −1

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 Ω\Omega resistor connected in series, the zener current is found to be 4 times load current. The load current (ILI_L) and load resistance (RLR_L) are :

  1. A.IL=0.02I_L=0.02 mA; RL=250 ΩR_L=250\ \Omega
  2. B.IL=10I_L=10 mA; RL=500 ΩR_L=500\ \Omega
  3. C.IL=10I_L=10 A; RL=0.5 ΩR_L=0.5\ \Omega
  4. D.IL=20I_L=20 mA; RL=250 ΩR_L=250\ \Omega

Correct answer: B

Explanation: I=20/400=50I=20/400=50 mA =5IL⇒IL=10=5I_L\Rightarrow I_L=10 mA; RL=5/0.01=500 ΩR_L=5/0.01=500\ \Omega.

Question 41

Physics · 4 marks · −1

A square Lamina OABC of length 10 cm is pivoted at 'O'. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :

Figure for question 41
  1. A.10 N
  2. B.0 (zero)
  3. C.20 N
  4. D.10210\sqrt{2} N

Correct answer: A

Explanation: Torques about O: F (at C, leftward) gives +0.1F+0.1F; at B the two 10 N forces cancel; at A the downward 10 N gives −1-1 N m. So F = 10 N.

Question 42

Physics · 4 marks · −1

The equation for real gas is given by (P+aV2)(V−b)=RT\left(P+\frac{a}{V^2}\right)(V-b)=RT, where P, V, T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2ab^{-2} is equivalent to that of :

  1. A.Energy density
  2. B.Planck's constant
  3. C.Strain
  4. D.Compressibility

Correct answer: A

Explanation: a/b2a/b^2 has dimensions of pressure = energy density.

Question 43

Physics · 4 marks · −1

Let B1B_1 be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B2B_2 be the magnitude of magnetic field at an axial distance 'x' from the center. For x:R=3:4x:R=3:4, B2B1\frac{B_2}{B_1} is :

  1. A.25 : 16
  2. B.64 : 125
  3. C.4 : 5
  4. D.16 : 25

Correct answer: B

Explanation: B2B1=(R2R2+x2)3/2=(16/25)3/2=64/125\frac{B_2}{B_1}=\left(\frac{R^2}{R^2+x^2}\right)^{3/2}=(16/25)^{3/2}=64/125.

Question 44

Physics · 4 marks · −1

A light wave is propagating with plane wave fronts of the type x+y+z=x+y+z= constant. The angle made by the direction of wave propagation with the x-axis is :

  1. A.cos⁡−1(23)\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)
  2. B.cos⁡−1(2/3)\cos^{-1}(2/3)
  3. C.cos⁡−1(13)\cos^{-1}\left(\frac{1}{3}\right)
  4. D.cos⁡−1(13)\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)

Correct answer: D

Question 45

Physics · 4 marks · −1

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ+\sigma and −2σ-2\sigma. The force experienced by a point charge +q placed at the mid point between two plates will be :

Figure for question 45
  1. A.3σq4ϵ0\frac{3\sigma q}{4\epsilon_0}
  2. B.3σq2ϵ0\frac{3\sigma q}{2\epsilon_0}
  3. C.σq2ϵ0\frac{\sigma q}{2\epsilon_0}
  4. D.σq4ϵ0\frac{\sigma q}{4\epsilon_0}

Correct answer: B

Explanation: Both fields point from left to right: σ2ϵ0+2σ2ϵ0=3σ2ϵ0\frac{\sigma}{2\epsilon_0}+\frac{2\sigma}{2\epsilon_0}=\frac{3\sigma}{2\epsilon_0}.

Question 46

Physics · 4 marks · −1

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ______ μ\mum.

Correct answer: 6

Explanation: Small-angle: 5λ/a=π/6⇒a=30λπ=6 μ5\lambda/a=\pi/6\Rightarrow a=\frac{30\lambda}{\pi}=6\ \mum.

Question 47

Physics · 4 marks · −1

A steel wire of length 2 m and Young\'s modulus 2.0×1011 N m−22.0\times10^{11}\ N\,m^{-2} is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−310^{-3} respectively, then the elastic potential energy density of the wire is ______ ×105\times10^5 (in SI units).

Correct answer: 25

Explanation: Longitudinal strain =10−3/0.2=5×10−3=10^{-3}/0.2=5\times10^{-3}; u=12Yϵ2=12(2×1011)(25×10−6)=25×105u=\frac12Y\epsilon^2=\frac12(2\times10^{11})(25\times10^{-6})=25\times10^5.

Question 48

Physics · 4 marks · −1

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm2cm^2 with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5×103 kg/m31.5\times10^3\ kg/m^3. What force one needs to apply on the hinged door so that it does not get opened ? (Acceleration due to gravity = 10 m/s2m/s^2)

Correct answer: 150

Explanation: ΔP=(1500−1000)(10)(3)=15000\Delta P=(1500-1000)(10)(3)=15000 Pa; F=15000×0.01=150F=15000\times0.01=150 N.

Question 49

Physics · 4 marks · −1

A person travelling on a straight line moves with a uniform velocity v1v_1 for a distance x and with a uniform velocity v2v_2 for the next 32x\frac{3}{2}x distance. The average velocity in this motion is 507\frac{50}{7} m/s. If v1v_1 is 5 m/s then v2v_2 = ______ m/s.

Correct answer: 10

Explanation: 2.5xx/5+1.5x/v2=507⇒v2=10\frac{2.5x}{x/5+1.5x/v_2}=\frac{50}{7}\Rightarrow v_2=10.

Question 50

Physics · 4 marks · −1

γA\gamma_A is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γB\gamma_B is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If γAγB=(1+1n)\frac{\gamma_A}{\gamma_B}=\left(1+\frac{1}{n}\right), then the value of n is ______.

Correct answer: 3

Explanation: fB=3+3+2=8f_B=3+3+2=8: γB=10/8\gamma_B=10/8, γA=5/3\gamma_A=5/3; ratio =4/3=4/3, n = 3.

Question 51

Chemistry · 4 marks · −1

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2CO_2 and 0.567 g of H2OH_2O. The empirical formula mass of compound (X) is ______ g. (Given molar mass in g mol−1mol^{-1} C : 12, H : 1, O : 16)

  1. A.15
  2. B.45
  3. C.30
  4. D.60

Correct answer: C

Explanation: C = 0.398 g, H = 0.063 g, O = 0.539 g → C : H : O ≈ 1 : 2 : 1 (CH2OCH_2O = 30).

Question 52

Chemistry · 4 marks · −1

CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO_3(s)+2HCl(aq)\to CaCl_2(aq)+CO_2(g)+H_2O(l). Consider the above reaction, what mass of CaCl2CaCl_2 will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3CaCO_3 ? (Given : Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol−1mol^{-1}, respectively)

  1. A.2.636 g
  2. B.5.272 g
  3. C.3.908 g
  4. D.10.545 g

Correct answer: D

Explanation: HCl = 0.19 mol (limiting) → 0.095 mol CaCl2CaCl_2 × 111 = 10.545 g.

Question 53

Chemistry · 4 marks · −1

According to Bohr's model of hydrogen atom, which of the following statement is incorrect ?

  1. A.Radius of 4th4^{th} orbit is four times larger than that of 2nd2^{nd} orbit
  2. B.Radius of 3rd3^{rd} orbit is nine times larger than that of 1st1^{st} orbit
  3. C.Radius of 8th8^{th} orbit is four times larger than that of 4th4^{th} orbit
  4. D.Radius of 6th6^{th} orbit is three times larger than that of 4th4^{th} orbit

Correct answer: D

Explanation: r∝n2r\propto n^2: r6/r4=36/16=2.25r_6/r_4=36/16=2.25, not 3.

Question 54

Chemistry · 4 marks · −1

Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true ?

Figure for question 54
  1. A.dU≠0dU\neq0
  2. B.dw≠0dw\neq0
  3. C.The pressure in the vessel B before opening the stopcock is zero
  4. D.dq≠0dq\neq0

Correct answer: C

Explanation: Free expansion into an evacuated vessel: B was at zero pressure; q = w = ΔU = 0.

Question 55

Chemistry · 4 marks · −1

Consider the following compound (X): HI−C≡C−CH2II−CHIII(CH3)−CH3IV\overset{I}{H}-C\equiv C-\overset{II}{CH_2}-\overset{III}{CH}(CH_3)-\overset{IV}{CH_3}. The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C–H bond are :

Figure for question 55
  1. A.II, I
  2. B.II, IV
  3. C.I, IV
  4. D.III, II

Correct answer: A

Explanation: Propargylic radical (II) is resonance-stabilised; the sp (alkynyl) radical from I is least stable.

Question 56

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : In octahedral complexes, when Δo<P\Delta_o<P high spin complexes are formed. When Δo>P\Delta_o>P low spin complexes are formed. Statement (II) : In tetrahedral complexes because of Δt<P\Delta_t<P, low spin complexes are rarely formed. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: D

Question 57

Chemistry · 4 marks · −1

Among SO2SO_2, NF3NF_3, NH3NH_3, XeF2XeF_2, ClF3ClF_3 and SF4SF_4, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is :

  1. A.sp3d2sp^3d^2
  2. B.dsp2dsp^2
  3. C.sp3dsp^3d
  4. D.sp3sp^3

Correct answer: C

Explanation: XeF2XeF_2 has 3 lone pairs but zero dipole; ClF3ClF_3 (2 lone pairs, polar) is sp3dsp^3d.

Question 58

Chemistry · 4 marks · −1

Designate whether each of the following compounds is aromatic or not aromatic. (a) cyclopentadienyl anion (b) cyclopentadienyl cation (c) cyclobutadiene dication (cyclobutene ring carrying + charges on the two sp2^2 carbons) (d) cyclobutadiene dianion (cyclobutene ring carrying − charges on the two carbons) (e) cycloheptatrienyl (tropylium) cation (f) cyclooctatetraene (g) cyclobutadiene (h) cyclopropenyl cation

Figure for question 58
  1. A.a, b, c, d aromatic and e, f, g, h not aromatic
  2. B.b, e, f, g aromatic and a, c, d, h not aromatic
  3. C.e, g aromatic and a, b, c, d, f, h not aromatic
  4. D.a, c, d, e, h aromatic and b, f, g not aromatic

Correct answer: D

Explanation: (4n+2)π planar cyclic conjugated: a (6π), c (2π), d (6π), e (6π), h (2π).

Question 59

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : The metallic radius of Al is less than that of Ga. Statement (II) : The ionic radius of Al3+Al^{3+} is less than that of Ga3+Ga^{3+}. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: A

Explanation: Metallic radius Al (143 pm) > Ga (135 pm); ionic radius Al3+Al^{3+} (53.5 pm) < Ga3+Ga^{3+} (62 pm).

Question 60

Chemistry · 4 marks · −1

A molecule with the formula AX4YAX_4Y has all it\'s elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is :

  1. A.Octahedral
  2. B.Pentagonal planar
  3. C.Trigonal bipyramidal
  4. D.Square pyramidal

Correct answer: D

Explanation: XeF4OXeF_4O: 5 bond pairs + 1 lone pair on Xe → square pyramidal.

Question 61

Chemistry · 4 marks · −1

The property/properties that show irregularity in first four elements of group-17 is/are : (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy. Choose the correct answer from the options given below :

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

Correct answer: C

Explanation: Electron gain enthalpy: Cl > F > Br > I (F anomalous); other properties vary regularly.

Question 62

Chemistry · 4 marks · −1

Consider the following molecules : (p) CH3−CH2−CO−ClCH_3-CH_2-CO-Cl (q) CH3−CH2−CO−O−CO−CH3CH_3-CH_2-CO-O-CO-CH_3 (r) CH3−CH2−CO−O−CH2−CH3CH_3-CH_2-CO-O-CH_2-CH_3 (s) CH3−CH2−CO−NH2CH_3-CH_2-CO-NH_2. The correct order of rate of hydrolysis is :

  1. A.p > q > r > s
  2. B.p > r > q > s
  3. C.r > q > p > s
  4. D.q > p > r > s

Correct answer: A

Explanation: Acyl chloride > anhydride > ester > amide.

Question 63

Chemistry · 4 marks · −1

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are :

  1. A.600 mm Hg, B
  2. B.600 mm Hg, A
  3. C.1400 mm Hg, A
  4. D.1400 mm Hg, B

Correct answer: B

Explanation: 500=14(200)+34pB0⇒pB0=600500=\frac14(200)+\frac34p_B^0\Rightarrow p_B^0=600; A (200 mm) is less volatile.

Question 64

Chemistry · 4 marks · −1

Which of the following graph correctly represents the plots of KHK_H at 1 bar for gases in water versus temperature?

Figure for question 64
  1. A.Graph: straight lines rising from the origin, steepest labelled O2O_2, then N2N_2, lowest He
  2. B.Graph: curves rising to a maximum then falling, highest He, then N2N_2, lowest CH4CH_4
  3. C.Graph: U-shaped curves (falling then rising steeply), top He, then N2N_2, lowest O2O_2
  4. D.Graph: curves rising to a maximum then falling, highest O2O_2, then N2N_2, lowest He

Correct answer: B

Question 65

Chemistry · 4 marks · −1

An optically active alkyl halide C4H9BrC_4H_9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2NaNH_2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :

  1. A.Butan-2-one
  2. B.Butan-2-ol
  3. C.Butan-1-al
  4. D.But-2-yne

Correct answer: A

Explanation: 2-Bromobutane → but-2-ene → 2,3-dibromobutane → but-2-yne → butan-2-one.

Question 66

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Vanillin (4-hydroxy-3-methoxybenzaldehyde) will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: B

Explanation: Phenolic OH reacts with NaOH and CHO reduces Tollens; no α-H, so no self aldol.

Question 67

Chemistry · 4 marks · −1

Choose the correct tests with respective observations. (A) CuSO4CuSO_4 (acidified with acetic acid) + K4[Fe(CN)6]→K_4[Fe(CN)_6]\to Chocolate brown precipitate. (B) FeCl3+K4[Fe(CN)6]→FeCl_3+K_4[Fe(CN)_6]\to Prussian blue precipitate. (C) ZnCl2+K4[Fe(CN)6]ZnCl_2+K_4[Fe(CN)_6], neutralised with NH4OH→NH_4OH\to White or bluish white precipitate. (D) MgCl2+K4[Fe(CN)6]→MgCl_2+K_4[Fe(CN)_6]\to Blue precipitate. (E) BaCl2+K4[Fe(CN)6]BaCl_2+K_4[Fe(CN)_6], neutralised with NaOH →\to White precipitate. Choose the correct answer from the options given below :

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

Correct answer: D

Question 68

Chemistry · 4 marks · −1

Identify the correct statement among the following :

  1. A.All naturally occurring amino acids are optically active.
  2. B.Amino acid, cysteine can easily undergo dimerisation due to the presence of free SH group.
  3. C.All naturally occurring amino acids except glycine contain one chiral centre.
  4. D.Glutamic acid is the only amino acid that contains a −COOH group at the side chain.

Correct answer: B

Question 69

Chemistry · 4 marks · −1

The correct order of basic nature in aqueous solution for the bases NH3NH_3, H2N−NH2H_2N-NH_2, CH3CH2NH2CH_3CH_2NH_2, (CH3CH2)2NH(CH_3CH_2)_2NH and (CH3CH2)3N(CH_3CH_2)_3N is :

  1. A.NH3<H2N−NH2<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NHNH_3<H_2N-NH_2<(CH_3CH_2)_3N<CH_3CH_2NH_2<(CH_3CH_2)_2NH
  2. B.NH2−NH2<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NHNH_2-NH_2<NH_3<CH_3CH_2NH_2<(CH_3CH_2)_3N<(CH_3CH_2)_2NH
  3. C.NH3<H2N−NH2<CH3CH2NH2<(CH3CH2)2NH<(CH3CH2)3NNH_3<H_2N-NH_2<CH_3CH_2NH_2<(CH_3CH_2)_2NH<(CH_3CH_2)_3N
  4. D.H2N−NH2<NH3<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NHH_2N-NH_2<NH_3<(CH_3CH_2)_3N<CH_3CH_2NH_2<(CH_3CH_2)_2NH

Correct answer: B

Explanation: Aqueous order for ethyl amines: 2∘>3∘>1∘2^\circ>3^\circ>1^\circ; hydrazine is weaker than ammonia.

Question 70

Chemistry · 4 marks · −1

If equal volumes of AB2AB_2 and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2AY_2 at 300 K ? (Given KspK_{sp} (at 300 K) for AY2=5.2×10−7AY_2=5.2\times10^{-7})

  1. A.3.6×10−33.6\times10^{-3} M AB2AB_2, 5.0×10−45.0\times10^{-4} M XY
  2. B.2.0×10−42.0\times10^{-4} M AB2AB_2, 0.8×10−30.8\times10^{-3} M XY
  3. C.2.0×10−22.0\times10^{-2} M AB2AB_2, 2.0×10−22.0\times10^{-2} M XY
  4. D.1.5×10−41.5\times10^{-4} M AB2AB_2, 1.5×10−31.5\times10^{-3} M XY

Correct answer: C

Explanation: After mixing concentrations halve: (3) Q=(10−2)(10−2)2=10−6>KspQ=(10^{-2})(10^{-2})^2=10^{-6}>K_{sp}.

Question 71

Chemistry · 4 marks · −1

Consider the following equilibrium, CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g). 0.1 mol of CO along with a catalyst is present in a 2 dm3dm^3 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH3OHCH_3OH is formed. The KpθK_p^\theta is ______ ×10−3\times10^{-3} (nearest integer). Given : R = 0.08 dm3dm^3 bar K−1mol−1K^{-1}mol^{-1}. Assume only methanol is formed as the product and the system follows ideal gas behaviour.

Correct answer: 74

Question 72

Chemistry · 4 marks · −1

For the reaction A →\to products. [Graph: t1/2t_{1/2} (min) vs [A]0[A]_0 (mol L−1L^{-1}) is a straight line through the origin with slope = 76.92 (appropriate units)]. The concentration of A at 10 minutes is ______ ×10−3\times10^{-3} mol L−1L^{-1} (nearest integer). The reaction was started with 2.5 mol L−1L^{-1} of A.

Figure for question 72

Correct answer: 2435

Explanation: t1/2∝[A]0t_{1/2}\propto[A]_0 → zero order, slope =1/(2k)=1/(2k), k=0.0065k=0.0065 mol L−1^{-1}min−1^{-1}; [A]=2.5−0.065=2.435[A]=2.5-0.065=2.435 M → 2435.

Question 73

Chemistry · 4 marks · −1

Consider the following electrochemical cell at standard condition. Au(s)∣QH2,Q∣NH4X(0.01 M)∣∣Ag+(1 M)∣Ag(s)Au(s)|QH_2, Q|NH_4X(0.01\,M)||Ag^+(1\,M)|Ag(s), Ecell=+0.4E_{cell}=+0.4 V. The couple QH2/QQH_2/Q represents quinhydrone electrode, the half cell reaction is given below : Q+2e−+2H+→QH2Q + 2e^- + 2H^+ \to QH_2 (p-benzoquinone to hydroquinone), EQ/QH2∘=+0.7E^\circ_{Q/QH_2}=+0.7 V. [Given : EAg+/Ag∘=+0.8E^\circ_{Ag^+/Ag}=+0.8 V and 2.303RTF=0.06\frac{2.303RT}{F}=0.06 V]. The pKbpK_b value of the ammonium halide salt (NH4XNH_4X) used here is ______. (nearest integer)

Correct answer: 6

Question 74

Chemistry · 4 marks · −1

0.1 mol of the following given antiviral compound (P) will weigh ______ ×10−1\times10^{-1} g (nearest integer). (Given : molar mass in g mol−1mol^{-1} H : 1, C : 12, N : 14, O : 16, F : 19, I : 127)

Figure for question 74

Correct answer: 372

Explanation: C9H10FIN2O5=372C_9H_{10}FIN_2O_5=372 g/mol; 0.1 mol = 37.2 g = 372 ×10−1\times10^{-1} g.

Question 75

Chemistry · 4 marks · −1

A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (M3+/M2+M^{3+}/M^{2+}). It forms a metal complex of the type [M(CN)6]4−[M(CN)_6]^{4-}. The number of electrons present in the ege_g orbital of the complex is ______.

Correct answer: 1

Explanation: Co (E∘E^\circ = +1.8 V); [Co(CN)6]4−[Co(CN)_6]^{4-} is Co2+Co^{2+} d7d^7 low spin t2g6eg1t_{2g}^6e_g^1 → 1.

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