JEE Main 2026 — 22 Jan Shift 1 (9:00 AM - 12:00 PM) (Previous Year

JEE Main · 2026 · 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 the relation R on the set M={1,2,3,…,16}M=\{1, 2, 3, \ldots, 16\} be given by R={(x,y):4y=5x−3,x,y∈M}R=\{(x, y): 4y=5x-3, x, y\in M\}. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to

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

Correct answer: C

Explanation: R = {(3,3), (7,8), (11,13)}; add (8,7) and (13,11): 2.

Question 2

Mathematics · 4 marks · −1

If the domain of the function f(x)=sin⁡−1(5−x3+2x)+1log⁡e(10−x)f(x)=\sin^{-1}\left(\frac{5-x}{3+2x}\right)+\frac{1}{\log_e(10-x)} is (−∞,α]∪[β,γ)−{δ}(-\infty, \alpha]\cup[\beta, \gamma)-\{\delta\}, then 6(α+β+γ+δ)6(\alpha+\beta+\gamma+\delta) is equal to

  1. A.68
  2. B.66
  3. C.70
  4. D.67

Correct answer: C

Question 3

Mathematics · 4 marks · −1

The number of distinct real solutions of the equation x∣x+4∣+3∣x+2∣+10=0x|x+4|+3|x+2|+10=0 is

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

Correct answer: D

Question 4

Mathematics · 4 marks · −1

If A=[2335]A=\begin{bmatrix}2 & 3\\ 3 & 5\end{bmatrix}, then the determinant of the matrix (A2025−3A2024+A2023)(A^{2025}-3A^{2024}+A^{2023}) is

  1. A.28
  2. B.16
  3. C.24
  4. D.12

Correct answer: B

Explanation: det⁡A=1\det A=1; the matrix is A2023(A2−3A+I)A^{2023}(A^2-3A+I); by Cayley-Hamilton A2=7A−IA^2=7A-I, so A2−3A+I=4AA^2-3A+I=4A, det⁡(4A)=16\det(4A)=16.

Question 5

Mathematics · 4 marks · −1

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

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

Correct answer: A

Explanation: 2a+3d=32a+3d=3, 2a+5d=43⇒d=−56,a=1142a+5d=\frac43 \Rightarrow d=-\frac56, a=\frac{11}{4}; S12=6(2a+11d)=−22S_{12}=6(2a+11d)=-22.

Question 6

Mathematics · 4 marks · −1

The coefficient of x48x^{48} in (1+x)+2(1+x)2+3(1+x)3+…+100(1+x)100(1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100} is equal to

  1. A.100⋅101C49−101C50100\cdot{}^{101}C_{49}-{}^{101}C_{50}
  2. B.100⋅100C49−100C48100\cdot{}^{100}C_{49}-{}^{100}C_{48}
  3. C.100⋅100C49−100C50100\cdot{}^{100}C_{49}-{}^{100}C_{50}
  4. D.100C50+101C49{}^{100}C_{50}+{}^{101}C_{49}

Correct answer: A

Question 7

Mathematics · 4 marks · −1

Two distinct numbers aa and bb are selected at random from 1, 2, 3, ..., 50. The probability, that their product abab is divisible by 3, is

  1. A.825\frac{8}{25}
  2. B.5611225\frac{561}{1225}
  3. C.6641225\frac{664}{1225}
  4. D.2721225\frac{272}{1225}

Correct answer: C

Explanation: 16 multiples of 3: 1−(342)(502)=1−5611225=66412251-\frac{\binom{34}{2}}{\binom{50}{2}}=1-\frac{561}{1225}=\frac{664}{1225}.

Question 8

Mathematics · 4 marks · −1

If a random variable xx has the probability distribution: xx = 0, 1, 2, 3, 4, 5, 6, 7 with P(xx) = 0, 2k, k, 3k, 2k2^2, 2k, k2^2+k, 7k2^2 respectively, then P(3<x≤6)P(3<x\le6) is equal to

  1. A.0.22
  2. B.0.33
  3. C.0.34
  4. D.0.64

Correct answer: B

Explanation: 10k2+9k=1⇒k=11010k^2+9k=1 \Rightarrow k=\frac1{10}; P(4)+P(5)+P(6)=3k2+3k=0.33P(4)+P(5)+P(6)=3k^2+3k=0.33.

Question 9

Mathematics · 4 marks · −1

If the line αx+2y=1\alpha x+2y=1, where α∈R\alpha\in\mathbb{R}, does not meet the hyperbola x2−9y2=9x^2-9y^2=9, then a possible value of α\alpha is:

  1. A.0.5
  2. B.0.6
  3. C.0.8
  4. D.0.7

Correct answer: C

Question 10

Mathematics · 4 marks · −1

If the chord joining the points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) on the parabola y2=12xy^2=12x subtends a right angle at the vertex of the parabola, then x1x2−y1y2x_1x_2-y_1y_2 is equal to

  1. A.292
  2. B.288
  3. C.284
  4. D.280

Correct answer: B

Explanation: x1x2+y1y2=0x_1x_2+y_1y_2=0 and y1y2=−16a2=−144y_1y_2=-16a^2=-144, so x1x2=144x_1x_2=144; x1x2−y1y2=288x_1x_2-y_1y_2=288.

Question 11

Mathematics · 4 marks · −1

Let the set of all values of rr, for which the circles (x+1)2+(y+4)2=r2(x+1)^2+(y+4)^2=r^2 and x2+y2−4x−2y−4=0x^2+y^2-4x-2y-4=0 intersect at two distinct points be the interval (α,β)(\alpha, \beta). Then αβ\alpha\beta is equal to

  1. A.25
  2. B.21
  3. C.24
  4. D.20

Correct answer: A

Explanation: Second circle: centre (2,1), radius 3; distance between (−1,−4)(-1,-4) and (2,1)(2,1) is 34\sqrt{34}; ∣r−3∣<34<r+3|r-3|<\sqrt{34}<r+3 gives (34−3,34+3)(\sqrt{34}-3,\sqrt{34}+3), product 34−9=2534-9=25.

Question 12

Mathematics · 4 marks · −1

The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{6}, where −126<x<126-\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}, is equal to

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

Correct answer: A

Question 13

Mathematics · 4 marks · −1

Let AB→=2i^+4j^−5k^\overrightarrow{AB}=2\hat{i}+4\hat{j}-5\hat{k} and AD→=i^+2j^+λk^\overrightarrow{AD}=\hat{i}+2\hat{j}+\lambda\hat{k}, λ∈R\lambda\in\mathbb{R}. Let the projection of the vector v⃗=i^+j^+k^\vec{v}=\hat{i}+\hat{j}+\hat{k} on the diagonal AC→\overrightarrow{AC} of the parallelogram ABCD be of length one unit. If α,β\alpha, \beta, where α>β\alpha>\beta, be the roots of the equation λ2x2−6λx+5=0\lambda^2x^2-6\lambda x+5=0, then 2α−β2\alpha-\beta is equal to

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

Correct answer: C

Question 14

Mathematics · 4 marks · −1

Let P(α,β,γ)P(\alpha, \beta, \gamma) be the point on the line x−12=y+1−3=z\frac{x-1}{2}=\frac{y+1}{-3}=z at a distance 4144\sqrt{14} from the point (1,−1,0)(1, -1, 0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3} and x+52=y−101=z−31\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}, is equal to

  1. A.7547\sqrt{\frac{5}{4}}
  2. B.4574\sqrt{\frac{5}{7}}
  3. C.2742\sqrt{\frac{7}{4}}
  4. D.4754\sqrt{\frac{7}{5}}

Correct answer: D

Question 15

Mathematics · 4 marks · −1

If the image of the point P(1,2,a)P(1, 2, a) in the line x−63=y−72=7−z2\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2} is Q(5,b,c)Q(5, b, c), then a2+b2+c2a^2+b^2+c^2 is equal to

  1. A.293
  2. B.298
  3. C.264
  4. D.283

Correct answer: B

Question 16

Mathematics · 4 marks · −1

Let f(x)=x2025−x2000f(x)=x^{2025}-x^{2000}, x∈[0,1]x\in[0, 1] and the minimum value of the function f(x)f(x) in the interval [0,1][0, 1] be (80)80(n)−81(80)^{80}(n)^{-81}. Then nn is equal to

  1. A.-40
  2. B.-81
  3. C.-80
  4. D.-41

Correct answer: B

Question 17

Mathematics · 4 marks · −1

Let f:[1,∞)→Rf:[1, \infty)\to\mathbb{R} be a differentiable function. If 6∫1xf(t) dt=3xf(x)+x3−46\int_1^x f(t)\,dt=3xf(x)+x^3-4 for all x≥1x\ge1, then the value of f(2)−f(3)f(2)-f(3) is

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

Correct answer: A

Question 18

Mathematics · 4 marks · −1

Let the line x=−1x=-1 divide the area of the region {(x,y):1+x2≤y≤3−x}\{(x, y): 1+x^2\le y\le3-x\} in the ratio m:nm:n, gcd⁡(m,n)=1\gcd(m, n)=1. Then m+nm+n is equal to

  1. A.27
  2. B.26
  3. C.25
  4. D.28

Correct answer: A

Explanation: Region for −2≤x≤1-2\le x\le1: left part ∫−2−1(2−x−x2)dx=76\int_{-2}^{-1}(2-x-x^2)dx=\frac76, right part ∫−11=103\int_{-1}^{1}=\frac{10}{3}; ratio 7:207:20, sum 27.

Question 19

Mathematics · 4 marks · −1

Let the solution curve of the differential equation x dy−y dx=x2+y2 dxx\,dy-y\,dx=\sqrt{x^2+y^2}\,dx, x>0x>0, y(1)=0y(1)=0, be y=y(x)y=y(x). Then y(3)y(3) is equal to

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

Correct answer: A

Explanation: Homogeneous: y+x2+y2=x2y+\sqrt{x^2+y^2}=x^2; at x=3x=3: 9+y2=9−y⇒y=4\sqrt{9+y^2}=9-y \Rightarrow y=4.

Question 20

Mathematics · 4 marks · −1

The value of ∫−π/2π/2(1[x]+4)dx\int_{-\pi/2}^{\pi/2}\left(\frac{1}{[x]+4}\right)dx, where [⋅][\cdot] denotes the greatest integer function, is

  1. A.160(π−7)\frac{1}{60}(\pi-7)
  2. B.160(21π−1)\frac{1}{60}(21\pi-1)
  3. C.760(3π−1)\frac{7}{60}(3\pi-1)
  4. D.760(π−3)\frac{7}{60}(\pi-3)

Correct answer: C

Question 21

Mathematics · 4 marks · −1

Let α=−1+i32\alpha=\frac{-1+i\sqrt{3}}{2} and β=−1−i32\beta=\frac{-1-i\sqrt{3}}{2}, i=−1i=\sqrt{-1}. If (7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10(7-7\alpha+9\beta)^{20}+(9+7\alpha-7\beta)^{20}+(-7+9\alpha+7\beta)^{20}+(14+7\alpha+7\beta)^{20}=m^{10}, then mm is ______

Correct answer: 49

Question 22

Mathematics · 4 marks · −1

Let A be a 3×33\times3 matrix such that A+AT=OA+A^T=O. If A[1−10]=[332]A\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}3\\3\\2\end{bmatrix}, A2[1−10]=[−319−24]A^2\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}-3\\19\\-24\end{bmatrix} and det⁡(adj⁡(2adj⁡(A+I)))=(2)α⋅(3)β⋅(11)γ\det(\operatorname{adj}(2\operatorname{adj}(A+I)))=(2)^\alpha\cdot(3)^\beta\cdot(11)^\gamma, α,β,γ\alpha, \beta, \gamma are non-negative integers, then α+β+γ\alpha+\beta+\gamma is equal to ______

Correct answer: 18

Question 23

Mathematics · 4 marks · −1

Let ABC be a triangle. Consider four points p1,p2,p3,p4p_1, p_2, p_3, p_4 on the side AB, five points p5,p6,p7,p8,p9p_5, p_6, p_7, p_8, p_9 on the side BC, and four points p10,p11,p12,p13p_{10}, p_{11}, p_{12}, p_{13} on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1,p2,…,p13p_1, p_2, \ldots, p_{13}, is ______

Correct answer: 660

Explanation: Choose 5 points with at most 2 on any side: distributions (2,2,1) of sides with 4,5,4 points: (42)(52)(41)+(42)(51)(42)+(41)(52)(42)=240+180+240=660\binom42\binom52\binom41+\binom42\binom51\binom42+\binom41\binom52\binom42=240+180+240=660.

Question 24

Mathematics · 4 marks · −1

If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\frac{\cos^2 48^\circ-\sin^2 12^\circ}{\sin^2 24^\circ-\sin^2 6^\circ}=\frac{\alpha+\beta\sqrt{5}}{2}, where α,β∈N\alpha, \beta\in\mathbb{N}, then α+β\alpha+\beta is equal to ______

Correct answer: 4

Explanation: Numerator =cos⁡60∘cos⁡36∘=\cos60^\circ\cos36^\circ, denominator =sin⁡30∘sin⁡18∘=\sin30^\circ\sin18^\circ; ratio 5+15−1=3+52\frac{\sqrt5+1}{\sqrt5-1}=\frac{3+\sqrt5}{2}, so 4.

Question 25

Mathematics · 4 marks · −1

If ∫(sin⁡x)−112(cos⁡x)−52dx=−p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C\int(\sin x)^{-\frac{11}{2}}(\cos x)^{-\frac{5}{2}}dx=-\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}}-\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}}-\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}}+\frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}}+C, where pip_i and qiq_i are positive integers with gcd⁡(pi,qi)=1\gcd(p_i, q_i)=1 for i=1,2,3,4i=1, 2, 3, 4 and C is the constant of integration, then 15p1p2p3p4q1q2q3q4\frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} is equal to ______

Correct answer: 16

Question 26

Physics · 4 marks · −1

Match the LIST-I with LIST-II. List-I: A. Spring constant; B. Thermal conductivity; C. Boltzmann constant; D. Inductive reactance. List-II: I. ML2T−2K−1ML^2T^{-2}K^{-1}; II. ML0T−2ML^0T^{-2}; III. ML2T−3A−2ML^2T^{-3}A^{-2}; IV. MLT−3K−1MLT^{-3}K^{-1}. Choose the correct answer from the options given below:

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

Correct answer: A

Question 27

Physics · 4 marks · −1

A meter bridge with two resistances R1R_1 and R2R_2 as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 Ω\Omega resistance is connected in parallel to R2R_2. The values of resistances R1R_1 and R2R_2 are ______.

Figure for question 27
  1. A.R2=4 ΩR_2=4\ \Omega, R1=43 ΩR_1=\frac{4}{3}\ \Omega
  2. B.R2=16 ΩR_2=16\ \Omega, R1=163 ΩR_1=\frac{16}{3}\ \Omega
  3. C.R2=8 ΩR_2=8\ \Omega, R1=163 ΩR_1=\frac{16}{3}\ \Omega
  4. D.R2=12 ΩR_2=12\ \Omega, R1=123 ΩR_1=\frac{12}{3}\ \Omega

Correct answer: C

Explanation: R1=23R2R_1=\frac23R_2 and then R1=16R216+R2⇒R2=8R_1=\frac{16R_2}{16+R_2} \Rightarrow R_2=8, R1=163R_1=\frac{16}{3}.

Question 28

Physics · 4 marks · −1

The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is ______ m/s.

  1. A.100021000\sqrt{2}
  2. B.1000
  3. C.2005200\sqrt{5}
  4. D.10010100\sqrt{10}

Correct answer: D

Explanation: ve∝Rρv_e\propto R\sqrt\rho: 104×0.1×0.1=1001010^4\times0.1\times\sqrt{0.1}=100\sqrt{10} m/s.

Question 29

Physics · 4 marks · −1

A projectile is thrown upward at an angle 60∘60^\circ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45∘45^\circ with the horizontal. The initial speed of the projectile is ______ m/s.

  1. A.20220\sqrt{2}
  2. B.40
  3. C.20320\sqrt{3}
  4. D.40240\sqrt{2}

Correct answer: A

Explanation: ucos⁡60∘=20cos⁡45∘⇒u=202u\cos60^\circ=20\cos45^\circ \Rightarrow u=20\sqrt2.

Question 30

Physics · 4 marks · −1

Net gravitational force at the center of a square is found to be F1F_1 when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is F2F_2 when the positions of 3M and 4M are interchanged. The ratio F1F2\frac{F_1}{F_2} is α5\frac{\alpha}{\sqrt{5}}. The value of α\alpha is ______.

Figure for question 30
  1. A.1
  2. B.3
  3. C.252\sqrt{5}
  4. D.2

Correct answer: D

Question 31

Physics · 4 marks · −1

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ______ kg.m2^2.

  1. A.0.18
  2. B.0.63
  3. C.0.72
  4. D.0.36

Correct answer: B

Explanation: 75(5)(0.1)2+75(10)(0.2)2=0.07+0.56=0.63\frac75(5)(0.1)^2+\frac75(10)(0.2)^2=0.07+0.56=0.63.

Question 32

Physics · 4 marks · −1

Rods xx and yy of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points A and F are maintained at 100∘100^\circC and 40∘40^\circC respectively. Given the thermal conductivity of rod xx is three times of that of rod yy, the temperature at junction points B and E are (close to):

Figure for question 32
  1. A.60∘60^\circC and 45∘45^\circC respectively
  2. B.89∘89^\circC and 73∘73^\circC respectively
  3. C.80∘80^\circC and 70∘70^\circC respectively
  4. D.80∘80^\circC and 60∘60^\circC respectively

Correct answer: B

Explanation: Thermal conductances (units of y): AB 3, BE 12+32=2\frac12+\frac32=2, EF 1. H=601/3+1/2+1=32.7H=\frac{60}{1/3+1/2+1}=32.7; TB=100−H/3≈89∘T_B=100-H/3\approx89^\circC, TE=40+H≈73∘T_E=40+H\approx73^\circC.

Question 33

Physics · 4 marks · −1

Given below are two statements: Statement I: Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid. Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: D

Question 34

Physics · 4 marks · −1

A cylindrical tube AB of length ll, closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity ω\omega about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If PAP_A and PBP_B are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)

Figure for question 34
  1. A.PB=PAexp⁡(Mω2l2/RT)P_B=P_A\exp(M\omega^2l^2/RT)
  2. B.PB=PAP_B=P_A
  3. C.PB=PAexp⁡(Mω2l2/3RT)P_B=P_A\exp(M\omega^2l^2/3RT)
  4. D.PB=PAexp⁡(Mω2l2/2RT)P_B=P_A\exp(M\omega^2l^2/2RT)

Correct answer: D

Explanation: dP=ρω2r drdP=\rho\omega^2r\,dr with ρ=PMRT\rho=\frac{PM}{RT}: ln⁡PBPA=Mω2l22RT\ln\frac{P_B}{P_A}=\frac{M\omega^2l^2}{2RT}.

Question 35

Physics · 4 marks · −1

The volume of an ideal gas increases 8 times and temperature becomes (1/4)th(1/4)^{th} of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ=0)(\Delta Q=0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

  1. A.He
  2. B.O2_2
  3. C.CO2_2
  4. D.NH3_3

Correct answer: A

Explanation: TVγ−1TV^{\gamma-1} constant: 8γ−1=4⇒γ=538^{\gamma-1}=4 \Rightarrow \gamma=\frac53, monatomic.

Question 36

Physics · 4 marks · −1

XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is ll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L>l)(L>l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is ______ m/s. (gg = acceleration due to gravity)

Figure for question 36
  1. A.mgRB2l2\frac{mgR}{B^2l^2}
  2. B.2mgRB2L2\frac{2mgR}{B^2L^2}
  3. C.8mgRB2l2\frac{8mgR}{B^2l^2}
  4. D.2mgRB2l2\frac{2mgR}{B^2l^2}

Correct answer: A

Explanation: B2l2vR=mg\frac{B^2l^2v}{R}=mg.

Question 37

Physics · 4 marks · −1

Six point charges are kept 60∘60^\circ apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the center of the circle is ______. (ϵ0\epsilon_0 is permittivity of free space)

Figure for question 37
  1. A.Q4πϵ0R2(3i^−j^)\frac{Q}{4\pi\epsilon_0R^2}(\sqrt{3}\hat{i}-\hat{j})
  2. B.−Q4πϵ0R2(3i^−j^)-\frac{Q}{4\pi\epsilon_0R^2}(\sqrt{3}\hat{i}-\hat{j})
  3. C.−(5Q8πϵ0R2)(i^−3j^)-\left(\frac{5Q}{8\pi\epsilon_0R^2}\right)(\hat{i}-3\hat{j})
  4. D.−5Q8πϵ0R2(i^+3j^)-\frac{5Q}{8\pi\epsilon_0R^2}(\hat{i}+\sqrt{3}\hat{j})

Correct answer: B

Explanation: Equivalent to a single −2Q-2Q at 150∘150^\circ: field points toward it, 2kQR2(−32,12)\frac{2kQ}{R^2}\left(-\frac{\sqrt3}{2},\frac12\right).

Question 38

Physics · 4 marks · −1

A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E} is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ______. (gg: acceleration due to gravity)

  1. A.m2g2−q2E2\sqrt{m^2g^2-q^2E^2}
  2. B.m2g2+q2E2\sqrt{m^2g^2+q^2E^2}
  3. C.mg+qEmg+qE
  4. D.mg−qEmg-qE

Correct answer: B

Question 39

Physics · 4 marks · −1

Electric field in a region is given by E⃗=Axi^+Byj^\vec{E}=Ax\hat{i}+By\hat{j}, where A=10A=10 V/m2^2 and B=5B=5 V/m2^2. If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is ______ V.

  1. A.1000
  2. B.500
  3. C.2000
  4. D.0

Correct answer: C

Explanation: V=−Ax22−By22+CV=-\frac{Ax^2}{2}-\frac{By^2}{2}+C; 500=−500−1000+C⇒C=2000500=-500-1000+C \Rightarrow C=2000.

Question 40

Physics · 4 marks · −1

Three identical coils C1C_1, C2C_2 and C3C_3 are closely placed such that they share a common axis. C2C_2 is exactly midway. C1C_1 carries current I in anti-clockwise direction while C3C_3 carries current I in clockwise direction. An induced current flows through C2C_2 will be in clockwise direction when

  1. A.C1C_1 and C3C_3 move with equal speeds away from C2C_2
  2. B.C1C_1 moves away from C2C_2 and C3C_3 moves towards C2C_2
  3. C.C1C_1 moves towards C2C_2 and C3C_3 moves away from C2C_2
  4. D.C1C_1 and C3C_3 move with equal speeds towards C2C_2

Correct answer: C

Question 41

Physics · 4 marks · −1

Consider an equilateral prism (refractive index 2\sqrt{2}). A ray of light is incident on its one surface at a certain angle ii. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to ______.

  1. A.15∘15^\circ
  2. B.40∘40^\circ
  3. C.20∘20^\circ
  4. D.30∘30^\circ

Correct answer: A

Explanation: Critical angle 45∘45^\circ at the second face, so r1=60∘−45∘=15∘r_1=60^\circ-45^\circ=15^\circ.

Question 42

Physics · 4 marks · −1

7.9 MeV α\alpha-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ______ m. [14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 Nm2^2/C2^2 and electron charge =1.6×10−19=1.6\times10^{-19} C]

  1. A.1.69×10−121.69\times10^{-12}
  2. B.1.44×10−131.44\times10^{-13}
  3. C.2.88×10−142.88\times10^{-14}
  4. D.5.76×10−145.76\times10^{-14}

Correct answer: D

Explanation: Distance of closest approach 9×109×2×79×1.6×10−197.9×106=2.88×10−14\frac{9\times10^9\times2\times79\times1.6\times10^{-19}}{7.9\times10^6}=2.88\times10^{-14} m taken as the radius; diameter 5.76×10−145.76\times10^{-14} m.

Question 43

Physics · 4 marks · −1

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 α\alpha particles is ______ kHz. [mass of He nucleus = 4.002 amu, 1 amu =1.66×10−27=1.66\times10^{-27} kg, h =6.6×10−34=6.6\times10^{-34} J.s and c =3×108=3\times10^8 m/s]

  1. A.9×10199\times10^{19}
  2. B.9×10209\times10^{20}
  3. C.14.94×102014.94\times10^{20}
  4. D.14.94×101914.94\times10^{19}

Correct answer: D

Explanation: Δm=16.008−15.348=0.66\Delta m=16.008-15.348=0.66 amu; ν=0.66×1.66×10−27×9×10166.6×10−34=1.494×1023\nu=\frac{0.66\times1.66\times10^{-27}\times9\times10^{16}}{6.6\times10^{-34}}=1.494\times10^{23} Hz =14.94×1019=14.94\times10^{19} kHz.

Question 44

Physics · 4 marks · −1

Find the correct combination of A, B, C and D inputs which can cause the LED to glow.

Figure for question 44
  1. A.0100
  2. B.1000
  3. C.0011
  4. D.1101

Correct answer: D

Explanation: NAND(Aˉ,Bˉ)=A+B(\bar A,\bar B)=A+B must be 1 and NOR(C,D)(C,D) must be 0: A=1, B=1, C=0, D=1 works.

Question 45

Physics · 4 marks · −1

A circular disc has radius R1R_1 and thickness T1T_1. Another circular disc made of the same material has radius R2R_2 and thickness T2T_2. If the moment of inertia of both discs are same and R1R2=2\frac{R_1}{R_2}=2 then T1T2=1α\frac{T_1}{T_2}=\frac{1}{\alpha}. The value of α\alpha is ______.

Correct answer: 16

Explanation: I∝R4TI\propto R^4T: T1/T2=(R2/R1)4=1/16T_1/T_2=(R_2/R_1)^4=1/16.

Question 46

Physics · 4 marks · −1

Two loudspeakers (L1L_1 and L2L_2) are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point A, at equidistance to both loud speakers, is moved by 25 m along the line AB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is ______ Hz (Speed of sound in air is 324 m/s and 5=2.23\sqrt{5}=2.23)

Figure for question 46

Correct answer: 600

Explanation: At B: ∣L2B∣=402+302=50|L_2B|=\sqrt{40^2+30^2}=50, ∣L1B∣=402+202=205=44.6|L_1B|=\sqrt{40^2+20^2}=20\sqrt5=44.6; path difference 5.4 m =10λ=10\lambda, λ=0.54\lambda=0.54 m, f=600f=600 Hz.

Question 47

Physics · 4 marks · −1

Inductance of a coil with 10410^4 turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω\Omega. The energy density in the inductor when the current reaches (1e)\left(\frac{1}{e}\right) of its maximum value is απ×1e2\alpha\pi\times\frac{1}{e^2} J/m3^3. The value of α\alpha is ______. (μ0=4π×10−7\mu_0=4\pi\times10^{-7} Tm/A).

Correct answer: 20

Question 48

Physics · 4 marks · −1

The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, Ey=20sin⁡(3×106x−4.5×1014t)E_y=20\sin(3\times10^6x-4.5\times10^{14}t) V/m (where xx, tt and other values have S.I. units). The dielectric constant of the medium is ______ (speed of light in free space is 3×1083\times10^8 m/s)

Correct answer: 4

Explanation: v=ω/k=1.5×108v=\omega/k=1.5\times10^8 m/s, n=2n=2, K=n2=4K=n^2=4.

Question 49

Physics · 4 marks · −1

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance xx cm from the centre of the curvature of the spherical surface. The value of xx is ______ cm.

Correct answer: 100

Explanation: 1.5v=0.550⇒v=150\frac{1.5}{v}=\frac{0.5}{50} \Rightarrow v=150 cm from the surface, i.e. 100 cm from the centre.

Question 50

Chemistry · 4 marks · −1

In the reaction, 2Al(s) + 6HCl(aq) →\to 2Al3+^{3+}(aq) + 6Cl−^-(aq) + 3H2_2(g)

  1. A.11.2 L H2_2(g) at STP is produced for every mole of HCl consumed.
  2. B.12 L HCl(aq) is consumed for every 6L H2_2(g) produced.
  3. C.33.6 L H2_2(g) is produced regardless of temperature and pressure for every mole of Al that reacts.
  4. D.67.2 L H2_2(g) at STP is produced for every mole of Al that reacts.

Correct answer: A

Explanation: 6 mol HCl give 3 mol H2_2: 0.5 mol (11.2 L at STP) per mol HCl.

Question 51

Chemistry · 4 marks · −1

The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ______ J mol−1^{-1}. Given: RH=2.18×10−11R_H=2.18\times10^{-11} ergs.

  1. A.9.835×10129.835\times10^{12}
  2. B.9.835×1059.835\times10^{5}
  3. C.1.635×10−111.635\times10^{-11}
  4. D.1.635×10−181.635\times10^{-18}

Correct answer: B

Explanation: 34×2.18×10−18\frac34\times2.18\times10^{-18} J ×6.022×1023≈9.8×105\times6.022\times10^{23}\approx9.8\times10^5 J/mol.

Question 52

Chemistry · 4 marks · −1

The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNO3_3 molecule respectively are: H−-O(1)−^{(1)}-N(2)^{(2)}(=O(3)^{(3)})−-O(4)^{(4)} [O(1) bonded to H and N with two lone pairs; O(3) double-bonded to N with two lone pairs; O(4) single-bonded to N with three lone pairs]

  1. A.+1, 0, 0, -1
  2. B.0, -1, 0, +1
  3. C.0, +1, 0, -1
  4. D.0, 0, -1, +1

Correct answer: C

Question 53

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. List-I (Thermodynamic Process): A. Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2 dm3^3 to 20 dm3^3 at 300 K. B. Work done in irreversible isothermal expansion of 1 mol ideal gas from 1 m3^3 to 3 m3^3 at 300 K against a constant pressure of 3kPa. C. Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and CˉV=32R\bar{C}_V=\frac{3}{2}R. D. Change in enthalpy at constant pressure of 1 mol ideal gas with change of temperature = 337 K and Cˉp=52R\bar{C}_p=\frac{5}{2}R. List-II (Magnitude in kJ): I. 4; II. 11.5; III. 6; IV. 7. Choose the correct answer from the options given below:

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

Correct answer: D

Explanation: A: 2(8.314)(300)ln⁡10=11.52(8.314)(300)\ln10=11.5 kJ; B: 3×2=63\times2=6 kJ; C: 1.5×8.314×320=41.5\times8.314\times320=4 kJ; D: 2.5×8.314×337=72.5\times8.314\times337=7 kJ.

Question 54

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The Henry's law constant KHK_H is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute. Statement II: KHK_H does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: D

Question 55

Chemistry · 4 marks · −1

Consider a solution of CO2_2(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2_2 in vapour phase above water) [yy-axis] with log (mole fraction of CO2_2 in water) [xx-axis] at 25∘25^\circC?

Figure for question 55
  1. A.Straight line with positive slope and positive y-intercept
  2. B.Horizontal line
  3. C.Straight line with negative slope
  4. D.Straight line with positive slope passing through the origin

Correct answer: A

Explanation: log⁡p=log⁡KH+log⁡x\log p=\log K_H+\log x: slope 1, intercept log⁡KH\log K_H.

Question 56

Chemistry · 4 marks · −1

A 'p'-block element (E) and hydrogen form a binary cation (EHx_x)+^+, while EH3_3 on treatment with K2_2HgI4_4 in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ionisation enthalpy values (kJ mol−1^{-1}) for first element each from group 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element E.

  1. A.1402
  2. B.801
  3. C.1312
  4. D.1086

Correct answer: A

Explanation: E is nitrogen (NH4+_4^+; Nessler's reagent), IE1_1 = 1402 kJ/mol.

Question 57

Chemistry · 4 marks · −1

A →\to product (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions: Run 1 ⇒\Rightarrow 100 mL of 10 M solution of reactant A; Run 2 ⇒\Rightarrow 200 mL of 10 M solution of reactant A; Run 3 ⇒\Rightarrow 100 mL of 10 M solution of reactant A + 100 mL of H2_2O added. The correct variation of rate of reaction is

  1. A.Run 3 < Run 1 = Run 2
  2. B.Run 1 = Run 2 = Run 3
  3. C.Run 1 < Run 2 < Run 3
  4. D.Run 3 < Run 1 < Run 2

Correct answer: A

Explanation: Rate depends on concentration: runs 1 and 2 are 10 M, run 3 is 5 M.

Question 58

Chemistry · 4 marks · −1

Two p-block elements X and Y form fluorides of the type EF3_3. The fluoride compound XF3_3 is a Lewis acid and YF3_3 is a Lewis base. The hybridizations of the central atoms of XF3_3 and YF3_3 respectively are

  1. A.Both sp2^2
  2. B.Both sp3^3
  3. C.sp2^2 and sp3^3
  4. D.sp3^3 and sp2^2

Correct answer: C

Question 59

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II: A group 15 element's hydride EH3_3 has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: A

Explanation: Longest H−-X bond is HI (largest radius), so I is false; lowest-boiling hydride is PH3_3 and P's maximum covalency is not 4, so II is false.

Question 60

Chemistry · 4 marks · −1

A first row transition metal (M) does not liberate H2_2 gas from dilute HCl. 1 mol of aqueous solution of MSO4_4 is treated with excess of aqueous KCN and then H2_2S(g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ______ mol.

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

Correct answer: B

Explanation: M = Cu; excess CN−^- forms the very stable [Cu(CN)4_4]3−^{3-}, so no CuS precipitates.

Question 61

Chemistry · 4 marks · −1

Consider the transition metal ions Mn3+^{3+}, Cr3+^{3+}, Fe3+^{3+} and Co3+^{3+} and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

  1. A.Cr3+^{3+} > Mn3+^{3+} > Fe3+^{3+} > Co3+^{3+}
  2. B.Fe3+^{3+} > Co3+^{3+} > Mn3+^{3+} > Cr3+^{3+}
  3. C.Mn3+^{3+} > Fe3+^{3+} > Co3+^{3+} > Cr3+^{3+}
  4. D.Cr3+^{3+} > Fe3+^{3+} > Co3+^{3+} > Mn3+^{3+}

Correct answer: A

Explanation: Low spin: d3^3 3, d4^4 2, d5^5 1, d6^6 0.

Question 62

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Phenol on treatment with CHCl3_3/aq. KOH under refluxing condition, followed by acidification produces pp-hydroxy benzaldehyde as the major product and oo-hydroxy benzaldehyde as the minor product. Statement II: The mixture of pp-hydroxybenzaldehyde and oo-hydroxybenzaldehyde can be easily separated through steam distillation. 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: A

Question 63

Chemistry · 4 marks · −1

The correct order of reactivity of CH3_3Br in methanol with the following nucleophiles is F−^-, I−^-, C2_2H5_5O−^- and C6_6H5_5O−^-

  1. A.I−^- > C2_2H5_5O−^- > F−^- > C6_6H5_5O−^-
  2. B.I−^- > C6_6H5_5O−^- > F−^- > C2_2H5_5O−^-
  3. C.I−^- > F−^- > C6_6H5_5O−^- > C2_2H5_5O−^-
  4. D.I−^- > C2_2H5_5O−^- > C6_6H5_5O−^- > F−^-

Correct answer: D

Question 64

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with CH3_3COCl/AlCl3_3 will give 3-nitroacetophenone (benzene ring bearing COCH3_3 and NO2_2 meta to each other). Statement II: −-NO2_2 group is a mm-directing, and deactivating group. In the light of the above statements, choose the most appropriate answer from the options given below

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

Correct answer: C

Explanation: Nitrobenzene is too deactivated for Friedel-Crafts acylation.

Question 65

Chemistry · 4 marks · −1

The correct order of the rate of reaction of the following reactants with nucleophile by S$_N1mechanismis:(Given:StructuresIandIIarerigid)(I)1−bromobicyclo[2.2.2]octane(bridgeheadBr);(II)1−bromobicyclo[2.2.1]heptane(bridgeheadBr);(III)tert−butylbromide;(IV)Ph1 mechanism is : (Given : Structures I and II are rigid) (I) 1-bromobicyclo[2.2.2]octane (bridgehead Br); (II) 1-bromobicyclo[2.2.1]heptane (bridgehead Br); (III) tert-butyl bromide; (IV) Ph_3CC-$Br

Figure for question 65
  1. A.III < I < II < IV
  2. B.I < II < III < IV
  3. C.II < I < III < IV
  4. D.IV < III < II < I

Correct answer: C

Question 66

Chemistry · 4 marks · −1

As compared with chlorocyclohexane, which of the following statements correctly apply to chlorobenzene? A. The magnitude of negative charge is more on chlorine atom. B. The C−-Cl bond has partial double bond character. C. C−-Cl bond is less polar. D. C−-Cl bond is longer due to repulsion between delocalised electrons of the aromatic ring and lone pairs of electrons of chlorine. E. The C−-Cl bond is formed using sp2^2 hybridised orbital of carbon. Choose the correct answer from the options given below:

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

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. List-I (Reagents): A. NH2−_2-NH2_2, KOH; B. Ag(NH3_3)2_2OH; C. Aq. CuSO4_4, Sodium Potassium tartarate, KOH; D. Zn - Hg, HCl. List-II (Name of Reaction involving carbonyl compounds): I. Tollen's Test; II. Clemmensen Reduction; III. Wolff - Kishner Reduction; IV. Fehling's Test. Choose the correct answer from the options given below:

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

Correct answer: D

Question 68

Chemistry · 4 marks · −1

'A' is a neutral organic compound (M. F : C8_8H9_9ON). On treatment with aqueous Br2_2/HO(−)^{(-)}, 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO2_2 / HCl (0-5 ∘^\circC) produces a compound 'C' which on treatment with CuCN/NaCN produces 'D'. Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified KMnO4_4 produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is

  1. A.4-Methylbenzamide (pp-CH3_3C6_6H4_4CONH2_2)
  2. B.N-Methylbenzamide (C6_6H5_5CONHCH3_3)
  3. C.3-Methylbenzamide (mm-CH3_3C6_6H4_4CONH2_2)
  4. D.2-Methylbenzamide (oo-CH3_3C6_6H4_4CONH2_2)

Correct answer: A

Explanation: E = 4-methylbenzoic acid; F = terephthalic acid, which has only two kinds of H (ring and COOH).

Question 69

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Sucrose is dextrorotatory. However, sucrose upon hydrolysis gives a solution having mixture of products. This solution shows laevorotation. Statement II: Hydrolysis of sucrose gives glucose and fructose. Since the laevorotation of glucose is more than the dextrorotation of fructose, the resulting solution becomes laevorotatory. 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.Statement I is true but Statement II is false
  3. C.Both Statement I and Statement II are false
  4. D.Both Statement I and Statement II are true

Correct answer: B

Question 70

Chemistry · 4 marks · −1

Dissociation of a gas A2_2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300K. A2_2(g) ⇌\rightleftharpoons 2A(g). The standard Gibbs energy of formation of the involved substances has been provided below: A2_2: −100.00-100.00 kJ mol−1^{-1}; A: −50.832-50.832 kJ mol−1^{-1}. The degree of dissociation of A2_2(g) is given by (x×10−2)1/2(x\times10^{-2})^{1/2} where xx = ______. (Nearest integer). [Given: R = 8 J mol−1^{-1} K−1^{-1}, log 2 = 0.3010, log 3 = 0.48] Assume degree of dissociation is not negligible.

Correct answer: 33

Explanation: ΔG∘=−101.664+100=−1.664\Delta G^\circ=-101.664+100=-1.664 kJ ⇒Kp=e1664/2400=2\Rightarrow K_p=e^{1664/2400}=2. 4α21−α2=2⇒α2=13=33×10−2\frac{4\alpha^2}{1-\alpha^2}=2 \Rightarrow \alpha^2=\frac13=33\times10^{-2}.

Question 71

Chemistry · 4 marks · −1

Consider the following electrochemical cell at 298K: Pt | HSnO2−_2^-(aq) | Sn(OH)62−_6^{2-}(aq) | OH−^-(aq) | Bi2_2O3_3(s) | Bi(s). If the reaction quotient at a given time is 10610^6, then the cell EMF (EcellE_{cell}) is ______ ×10−1\times10^{-1} V (Nearest integer). Given the standard half-cell reduction potential as EBi2O3/Bi,OH−∘=−0.44E^\circ_{Bi_2O_3/Bi,OH^-}=-0.44 V and ESn(OH)62−/HSnO2−,OH−∘=−0.90E^\circ_{Sn(OH)_6^{2-}/HSnO_2^-,OH^-}=-0.90 V

Correct answer: 4

Explanation: E∘=0.46E^\circ=0.46 V, n=6n=6: E=0.46−0.0596×6=0.40E=0.46-\frac{0.059}{6}\times6=0.40 V =4×10−1=4\times10^{-1} V.

Question 72

Chemistry · 4 marks · −1

The temperature at which the rate constants of the given below two gaseous reactions become equal is ______ K. (Nearest integer) X →\to Y, k1=106e−30000/Tk_1=10^6e^{-30000/T}; P →\to Q, k2=104e−24000/Tk_2=10^4e^{-24000/T}. Given: ln 10 = 2.303

Correct answer: 1303

Explanation: 100=e6000/T⇒T=60004.606≈1303100=e^{6000/T} \Rightarrow T=\frac{6000}{4.606}\approx1303 K.

Question 73

Chemistry · 4 marks · −1

The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C:Br ratio is 3:1. The percentage of bromine in the product (Y) is ______%. (Nearest integer) (Given : molar mass in g mol−1^{-1} H : 1, C : 12, O : 16, Br : 80)

Correct answer: 66

Explanation: Y = C6_6H10_{10}Br2_2 (242): 160/242=66%160/242=66\%.

Question 74

Chemistry · 4 marks · −1

Sodium fusion extract of an organic compound (Y) with CHCl3_3 and chlorine water gives violet color to the CHCl3_3 layer. 0.15g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is ______. (Nearest integer) (Given : molar mass g mol−1^{-1} C : 12, H : 1, Cl : 35.5, Br : 80, I : 127)

Correct answer: 43

Explanation: Violet layer: iodine. 127235×0.120.15×100=43%\frac{127}{235}\times\frac{0.12}{0.15}\times100=43\%.

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