JEE Main 2022 — 25 Jul Shift 1 (Previous Year Paper)

JEE Main · 2022 · with answers and explanations

questions
87
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The total number of functions, f:{1,2,3,4}→{1,2,3,4,5,6}f:\{1,2,3,4\}\to\{1,2,3,4,5,6\} such that f(1)+f(2)=f(3)f(1)+f(2)=f(3), is equal to:

  1. A.60
  2. B.90
  3. C.108
  4. D.126

Correct answer: B

Explanation: Pairs (f(1),f(2))(f(1),f(2)) with sum ≤6\le6: 15; f(4)f(4) free: 15×6=9015\times6=90.

Question 2

Mathematics · 4 marks · −1

If α,β,γ,δ\alpha,\beta,\gamma,\delta are the roots of the equation x4+x3+x2+x+1=0x^4+x^3+x^2+x+1=0, then α2021+β2021+γ2021+δ2021\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} is equal to:

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

Correct answer: B

Explanation: Roots are the non-real fifth roots of unity; 2021≡1(mod5)2021\equiv1\pmod5, so the sum equals ∑α=−1\sum\alpha=-1.

Question 3

Mathematics · 4 marks · −1

The number of θ∈(0,4π)\theta\in(0,4\pi) for which the system of linear equations 3(sin⁡3θ)x−y+z=23(\sin3\theta)x-y+z=2, 3(cos⁡2θ)x+4y+3z=33(\cos2\theta)x+4y+3z=3, 6x+7y+7z=96x+7y+7z=9 has no solution, is:

  1. A.6
  2. B.7
  3. C.8
  4. D.9

Correct answer: B

Question 4

Mathematics · 4 marks · −1

If lim⁡n→∞(n2−n−1+nα+β)=0\lim_{n\to\infty}\left(\sqrt{n^2-n-1}+n\alpha+\beta\right)=0, then 8(α+β)8(\alpha+\beta) is equal to:

  1. A.4
  2. B.−8-8
  3. C.−4-4
  4. D.8

Correct answer: C

Explanation: α=−1\alpha=-1 and n2−n−1−n→−12\sqrt{n^2-n-1}-n\to-\frac12, so β=12\beta=\frac12; 8(α+β)=−48(\alpha+\beta)=-4.

Question 5

Mathematics · 4 marks · −1

If the absolute maximum value of the function f(x)=(x2−2x+7)e(4x3−12x2−180x+31)f(x)=(x^2-2x+7)e^{(4x^3-12x^2-180x+31)} in the interval [−3,0][-3,0] is f(α)f(\alpha), then:

  1. A.α=0\alpha=0
  2. B.α=−3\alpha=-3
  3. C.α∈(−1,0)\alpha\in(-1,0)
  4. D.α∈(−3,−1]\alpha\in(-3,-1]

Correct answer: B

Explanation: On [−3,0][-3,0] the exponent g(x)=4x3−12x2−180x+31g(x)=4x^3-12x^2-180x+31 has derivative 12(x−5)(x+3)≤012(x-5)(x+3)\le0, so gg is decreasing from x=−3x=-3, and x2−2x+7x^2-2x+7 also decreases there; the maximum is at α=−3\alpha=-3.

Question 6

Mathematics · 4 marks · −1

The curve y(x)=ax3+bx2+cx+5y(x)=ax^3+bx^2+cx+5 touches the xx-axis at the point P(−2,0)P(-2,0) and cuts the yy-axis at the point Q, where y′y' is equal to 3. Then the local maximum value of y(x)y(x) is:

  1. A.274\frac{27}{4}
  2. B.294\frac{29}{4}
  3. C.374\frac{37}{4}
  4. D.92\frac92

Correct answer: A

Explanation: c=3c=3; y(−2)=0y(-2)=0, y(ˊ−2)=0y\'(-2)=0 give a=−12,b=−34a=-\frac12,b=-\frac34. y=ˊ−32x2−32x+3=0⇒x=1y\'=-\frac32x^2-\frac32x+3=0\Rightarrow x=1 (max): y(1)=274y(1)=\frac{27}{4}.

Question 7

Mathematics · 4 marks · −1

The area of the region given by A={(x,y):x2≤y≤min⁡{x+2,4−3x}}A=\{(x,y): x^2\le y\le\min\{x+2,4-3x\}\} is:

  1. A.318\frac{31}{8}
  2. B.176\frac{17}{6}
  3. C.196\frac{19}{6}
  4. D.278\frac{27}{8}

Correct answer: B

Explanation: Lines meet at x=12x=\frac12. ∫−11/2(x+2−x2)dx+∫1/21(4−3x−x2)dx=218+1724=176\int_{-1}^{1/2}(x+2-x^2)dx+\int_{1/2}^{1}(4-3x-x^2)dx=\frac{21}{8}+\frac{17}{24}=\frac{17}{6}.

Question 8

Mathematics · 4 marks · −1

For any real number xx, let [x][x] denote the largest integer less than equal to xx. Let ff be a real valued function defined on the interval [−10,10][-10,10] by f(x)={x−[x],if [x] is odd1+[x]−x,if [x] is evenf(x)=\begin{cases}x-[x], & \text{if }[x]\text{ is odd}\\ 1+[x]-x, & \text{if }[x]\text{ is even}\end{cases}. Then the value of π210∫−1010f(x)cos⁡πx dx\frac{\pi^2}{10}\int_{-10}^{10}f(x)\cos\pi x\,dx is:

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

Correct answer: A

Explanation: ff has period 2 and f(x)cos⁡πxf(x)\cos\pi x is even; on [0,1)[0,1), f=1−xf=1-x and on [1,2)[1,2), f=x−1f=x-1. ∫02fcos⁡πx dx=4π2\int_0^2 f\cos\pi x\,dx=\frac{4}{\pi^2}, so ∫−1010=40π2\int_{-10}^{10}=\frac{40}{\pi^2} and the value is 4.

Question 9

Mathematics · 4 marks · −1

The slope of the tangent to a curve C:y=y(x)C: y=y(x) at any point (x,y)(x,y) on it is 2e2x−6e−x+92+9e−2x\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}. If C passes through the points (0,12+π22)\left(0,\frac12+\frac{\pi}{2\sqrt2}\right) and (α,12e2α)\left(\alpha,\frac12e^{2\alpha}\right), then eαe^\alpha is equal to:

  1. A.3+23−2\frac{3+\sqrt2}{3-\sqrt2}
  2. B.32(3+23−2)\frac{3}{\sqrt2}\left(\frac{3+\sqrt2}{3-\sqrt2}\right)
  3. C.12(2+12−1)\frac{1}{\sqrt2}\left(\frac{\sqrt2+1}{\sqrt2-1}\right)
  4. D.2+12−1\frac{\sqrt2+1}{\sqrt2-1}

Correct answer: B

Question 10

Mathematics · 4 marks · −1

The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0(x-y^2)dx+y(5x+y^2)dy=0 is:

  1. A.(y2+x)4=C∣(y2+2x)3∣(y^2+x)^4=C|(y^2+2x)^3|
  2. B.(y2+2x)4=C∣(y2+x)3∣(y^2+2x)^4=C|(y^2+x)^3|
  3. C.∣(y2+x)3∣=C(2y2+x)4|(y^2+x)^3|=C(2y^2+x)^4
  4. D.∣(y2+2x)3∣=C(2y2+x)4|(y^2+2x)^3|=C(2y^2+x)^4

Correct answer: A

Question 11

Mathematics · 4 marks · −1

A line, with the slope greater than one, passes through the point A(4,3)A(4,3) and intersects the line x−y−2=0x-y-2=0 at the point B. If the length of the line segment AB is 293\frac{\sqrt{29}}{3}, then B also lies on the line:

  1. A.2x+y=92x+y=9
  2. B.3x−2y=73x-2y=7
  3. C.x+2y=6x+2y=6
  4. D.2x−3y=32x-3y=3

Correct answer: C

Question 12

Mathematics · 4 marks · −1

Let the locus of the centre (α,β)(\alpha,\beta), β>0\beta>0, of the circle which touches the circle x2+(y−1)2=1x^2+(y-1)^2=1 externally and also touches the xx-axis be L. Then the area bounded by L and the line y=4y=4 is:

  1. A.3223\frac{32\sqrt2}{3}
  2. B.4023\frac{40\sqrt2}{3}
  3. C.643\frac{64}{3}
  4. D.323\frac{32}{3}

Correct answer: C

Explanation: α2+(β−1)2=β+1⇒x2=4y\sqrt{\alpha^2+(\beta-1)^2}=\beta+1\Rightarrow x^2=4y. Area with y=4y=4: 2∫04(4−x24)dx=6432\int_0^4\left(4-\frac{x^2}{4}\right)dx=\frac{64}{3}.

Question 13

Mathematics · 4 marks · −1

Let P be the plane containing the straight line x−39=y+4−1=z−7−5\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5} and perpendicular to the plane containing the straight lines x2=y3=z5\frac x2=\frac y3=\frac z5 and x3=y7=z8\frac x3=\frac y7=\frac z8. If d is the distance of P from the point (2,−5,11)(2,-5,11), then d2d^2 is equal to:

  1. A.1472\frac{147}{2}
  2. B.96
  3. C.323\frac{32}{3}
  4. D.54

Correct answer: C

Question 14

Mathematics · 4 marks · −1

Let ABC be a triangle such that BC→=a⃗\overrightarrow{BC}=\vec a, CA→=b⃗\overrightarrow{CA}=\vec b, AB→=c⃗\overrightarrow{AB}=\vec c, ∣a⃗∣=62|\vec a|=6\sqrt2, ∣b⃗∣=23|\vec b|=2\sqrt3 and b⃗⋅c⃗=12\vec b\cdot\vec c=12. Consider the statements: (S1): ∣(a⃗×b⃗)+(c⃗×b⃗)∣−∣c⃗∣=6(22−1)|(\vec a\times\vec b)+(\vec c\times\vec b)|-|\vec c|=6(2\sqrt2-1) (S2): ∠ACB=cos⁡−1(23)\angle ACB=\cos^{-1}\left(\sqrt{\frac23}\right). Then

  1. A.both (S1) and (S2) are true
  2. B.only (S1) is true
  3. C.only (S2) is true
  4. D.both (S1) and (S2) are false

Correct answer: C

Question 15

Mathematics · 4 marks · −1

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is:

  1. A.33232\frac{33}{2^{32}}
  2. B.33229\frac{33}{2^{29}}
  3. C.33228\frac{33}{2^{28}}
  4. D.33227\frac{33}{2^{27}}

Correct answer: C

Explanation: Mean 16, variance 8: n=32,p=12n=32,p=\frac12. 32+496232=33228\frac{32+496}{2^{32}}=\frac{33}{2^{28}}.

Question 16

Mathematics · 4 marks · −1

If the numbers appeared on the two throws of a fair six faced die are α\alpha and β\beta, then the probability that x2+αx+β>0x^2+\alpha x+\beta>0, for all x∈Rx\in\mathbb{R}, is:

  1. A.1736\frac{17}{36}
  2. B.49\frac49
  3. C.12\frac12
  4. D.1936\frac{19}{36}

Correct answer: A

Explanation: Need α2<4β\alpha^2<4\beta: α=1\alpha=1: 6; α=2\alpha=2: 5; α=3\alpha=3: 4; α=4\alpha=4: 2; total 17.

Question 17

Mathematics · 4 marks · −1

The number of solutions of ∣cos⁡x∣=sin⁡x|\cos x|=\sin x, such that −4π≤x≤4π-4\pi\le x\le4\pi is:

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

Correct answer: C

Explanation: Solutions x=π4,3π4x=\frac\pi4,\frac{3\pi}4 in each period of 2π2\pi where sin⁡x≥0\sin x\ge0: 4 periods give 8.

Question 18

Mathematics · 4 marks · −1

A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is:

  1. A.5(23+3)5(2\sqrt3+3) m
  2. B.5(3+3)5(\sqrt3+3) m
  3. C.10(3+1)10(\sqrt3+1) m
  4. D.10(23+1)10(2\sqrt3+1) m

Correct answer: A

Explanation: AQ=53AQ=5\sqrt3; PQ=AQtan⁡75∘=53(2+3)=5(23+3)PQ=AQ\tan75^\circ=5\sqrt3(2+\sqrt3)=5(2\sqrt3+3).

Question 19

Mathematics · 4 marks · −1

Which of the following statements is a tautology?

  1. A.((∼p)∨q)⇒p((\sim p)\vee q)\Rightarrow p
  2. B.p⇒((∼p)∨q)p\Rightarrow((\sim p)\vee q)
  3. C.((∼p)∨q)⇒q((\sim p)\vee q)\Rightarrow q
  4. D.q⇒((∼p)∨q)q\Rightarrow((\sim p)\vee q)

Correct answer: D

Explanation: If qq is true then (∼p)∨q(\sim p)\vee q is true.

Question 20

Mathematics · 4 marks · −1

Let A=(2−1−110−11−10)A=\begin{pmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{pmatrix} and B=A−IB=A-I. If ω=3i−12\omega=\frac{\sqrt3i-1}{2}, then the number of elements in the set {n∈{1,2,…,100}:An+(ωB)n=A+B}\{n\in\{1,2,\dots,100\}: A^n+(\omega B)^n=A+B\} is equal to ________.

Correct answer: 17

Question 21

Mathematics · 4 marks · −1

The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is ________.

Correct answer: 1492

Explanation: Alphabetical letters A, D, I, K, M, N, N. Words starting A/D/I/K: 4⋅6!2!=14404\cdot\frac{6!}{2!}=1440; MA then D/I/K: 3⋅4!2!=363\cdot\frac{4!}{2!}=36; MAN then D/I: 2⋅3!=122\cdot3!=12; MANK then D: 2; MANKI then D: 1; then MANKIND: rank 1440+36+12+2+1+1=14921440+36+12+2+1+1=1492.

Question 22

Mathematics · 4 marks · −1

If the maximum value of the term independent of tt in the expansion of (t2x15+(1−x)110t)15\left(t^2x^{\frac15}+\frac{(1-x)^{\frac1{10}}}{t}\right)^{15}, x≥0x\ge0, is K, then 8K is equal to ________.

Correct answer: 6006

Question 23

Mathematics · 4 marks · −1

Let a, b be two non-zero real numbers. If p and r are the roots of the equation x2−8ax+2a=0x^2-8ax+2a=0 and q and s are the roots of the equation x2+12bx+6b=0x^2+12bx+6b=0, such that 1p,1q,1r,1s\frac1p,\frac1q,\frac1r,\frac1s are in A.P., then a−1−b−1a^{-1}-b^{-1} is equal to ________.

Correct answer: 38

Question 24

Mathematics · 4 marks · −1

Let a1=b1=1a_1=b_1=1, an=an−1+2a_n=a_{n-1}+2 and bn=an+bn−1b_n=a_n+b_{n-1} for every natural number n≥2n\ge2. Then ∑n=115an⋅bn\sum_{n=1}^{15}a_n\cdot b_n is equal to ________.

Correct answer: 27560

Explanation: an=2n−1a_n=2n-1, bn=n2b_n=n^2: ∑(2n3−n2)=2(14400)−1240=27560\sum(2n^3-n^2)=2(14400)-1240=27560.

Question 25

Mathematics · 4 marks · −1

Let f(x)={∣4x2−8x+5∣,if 8x2−6x+1≥0[4x2−8x+5],if 8x2−6x+1<0f(x)=\begin{cases}|4x^2-8x+5|, & \text{if }8x^2-6x+1\ge0\\ [4x^2-8x+5], & \text{if }8x^2-6x+1<0\end{cases} where [α][\alpha] denotes the greatest integer less than or equal to α\alpha. Then the number of points in R\mathbb{R} where ff is not differentiable is ________.

Correct answer: 3

Question 26

Mathematics · 4 marks · −1

If lim⁡n→∞(n+1)k−1nk+1[(nk+1)+(nk+2)+⋯+(nk+n)]=33⋅lim⁡n→∞1nk+1⋅[1k+2k+3k+⋯+nk]\lim_{n\to\infty}\frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\dots+(nk+n)]=33\cdot\lim_{n\to\infty}\frac{1}{n^{k+1}}\cdot[1^k+2^k+3^k+\dots+n^k], then the integral value of k is equal to ________.

Correct answer: 5

Explanation: LHS =k+12=k+\frac12, RHS =33k+1=\frac{33}{k+1}: (2k+1)(k+1)=66⇒k=5(2k+1)(k+1)=66\Rightarrow k=5.

Question 27

Mathematics · 4 marks · −1

Let the equation of two diameters of a circle x2+y2−2x+2fy+1=0x^2+y^2-2x+2fy+1=0 be 2px−y=12px-y=1 and 2x+py=4p2x+py=4p. Then the slope m∈(0,∞)m\in(0,\infty) of the tangent to the hyperbola 3x2−y2=33x^2-y^2=3 passing through the centre of the circle is equal to ________.

Correct answer: 2

Explanation: The centre (1,−f)(1,-f) lies on both diameters; the consistent solution gives centre (1,1)(1,1). Tangent y−1=m(x−1)y-1=m(x-1) to x2−y23=1x^2-\frac{y^2}{3}=1: (1−m)2=m2−3⇒m=2(1-m)^2=m^2-3\Rightarrow m=2.

Question 28

Mathematics · 4 marks · −1

The sum of diameters of the circles that touch (i) the parabola 75x2=64(5y−3)75x^2=64(5y-3) at the point (85,65)\left(\frac85,\frac65\right) and (ii) the yy-axis, is equal to ________.

Correct answer: 10

Question 29

Physics · 4 marks · −1

If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:

  1. A.[PA−1T0][PA^{-1}T^0]
  2. B.[PAT−1][PAT^{-1}]
  3. C.[PA−1T][PA^{-1}T]
  4. D.[PA−1T−1][PA^{-1}T^{-1}]

Correct answer: A

Explanation: η=FA(dv/dx)\eta=\frac{F}{A(dv/dx)} has dimensions of pressure ×\times time =P/TA⋅T=PA−1=\frac{P/T}{A}\cdot T=PA^{-1}.

Question 30

Physics · 4 marks · −1

Which of the following physical quantities have the same dimensions?

  1. A.Electric displacement (D⃗\vec D) and surface charge density
  2. B.Displacement current and electric field
  3. C.Current density and surface charge density
  4. D.Electric potential and energy

Correct answer: A

Question 31

Physics · 4 marks · −1

A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be: (Given cos⁡135∘=−0.7\cos135^\circ=-0.7)

Figure for question 31
  1. A.42 m
  2. B.47 m
  3. C.19 m
  4. D.40 m

Correct answer: B

Explanation: 3π4r=60⇒r≈25.5\frac{3\pi}{4}r=60\Rightarrow r\approx25.5 m; AB=r2−2cos⁡135∘=r3.4≈47AB=r\sqrt{2-2\cos135^\circ}=r\sqrt{3.4}\approx47 m.

Question 32

Physics · 4 marks · −1

A body of mass 0.5 kg travels on straight line path with velocity v=(3x2+4)v=(3x^2+4) m/s. The net workdone by the force during its displacement from x=0x=0 to x=2x=2 m is:

  1. A.64 J
  2. B.60 J
  3. C.120 J
  4. D.128 J

Correct answer: B

Explanation: vv: 4 to 16 m/s; ΔK=12(0.5)(256−16)=60\Delta K=\frac12(0.5)(256-16)=60 J.

Question 33

Physics · 4 marks · −1

A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be:

  1. A.53\sqrt{\frac53}
  2. B.45\sqrt{\frac45}
  3. C.35\sqrt{\frac35}
  4. D.1415\sqrt{\frac{14}{15}}

Correct answer: D

Explanation: v2=2gh1+k2/R2v^2=\frac{2gh}{1+k^2/R^2}: 1.41.5=1415\frac{1.4}{1.5}=\frac{14}{15}.

Question 34

Physics · 4 marks · −1

Three identical particles A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:

  1. A.21 G
  2. B.100 G
  3. C.59 G
  4. D.42 G

Correct answer: B

Explanation: F=Gm2132(1+2⋅12⋅12)=104G169(1.707)≈101 G≈100 GF=\frac{Gm^2}{13^2}\left(1+2\cdot\frac{1}{2}\cdot\frac{1}{\sqrt2}\right)=\frac{10^4G}{169}(1.707)\approx101\,G\approx100\,G.

Question 35

Physics · 4 marks · −1

A certain amount of gas of volume V at 27°C temperature and pressure 2×1072\times10^7 Nm−2^{-2} expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use γ=1.5\gamma=1.5):

  1. A.3.536×1053.536\times10^5 Pa
  2. B.3.536×1063.536\times10^6 Pa
  3. C.1.25×1061.25\times10^6 Pa
  4. D.1.25×1051.25\times10^5 Pa

Correct answer: B

Explanation: 107×2−1.5=3.536×10610^7\times2^{-1.5}=3.536\times10^6 Pa.

Question 36

Physics · 4 marks · −1

Following statements are given: (A) The average kinetic energy of a gas molecule decreases when the temperature is reduced. (B) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature. (C) The average kinetic energy of a gas molecule decreases with increase in volume. (D) Pressure of a gas increases with increase in temperature at constant pressure. (E) The volume of gas decreases with increase in temperature. Choose the correct answer from the options given below:

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

Correct answer: A

Question 37

Physics · 4 marks · −1

In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two springs of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2k'. If mass 'm' in (A) and in (B) are displaced by distance 'x' horizontally and then released, then time period T1T_1 and T2T_2 corresponding to (A) and (B) respectively follow the relation.

Figure for question 37
  1. A.T1T2=32\frac{T_1}{T_2}=\frac{3}{\sqrt2}
  2. B.T1T2=32\frac{T_1}{T_2}=\sqrt{\frac32}
  3. C.T1T2=23\frac{T_1}{T_2}=\sqrt{\frac23}
  4. D.T1T2=23\frac{T_1}{T_2}=\frac{\sqrt2}{3}

Correct answer: A

Explanation: T1=2π3m/2kT_1=2\pi\sqrt{3m/2k}, T2=2πm/3kT_2=2\pi\sqrt{m/3k}; ratio =9/2=32=\sqrt{9/2}=\frac{3}{\sqrt2}.

Question 38

Physics · 4 marks · −1

A condenser of 2 μ\muF capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across it\'s plates with respect to the charge (Q) on the condenser?

Figure for question 38
  1. A.Straight line from origin reaching V = 2.5 x 10^6 V at Q = 5 C
  2. B.Curve bending upward (convex) from origin reaching V = 5 x 10^6 V at Q = 5 C
  3. C.Straight line from origin reaching V = 5 x 10^6 V at Q = 5 C
  4. D.Curve rising and saturating near V = 2.5 x 10^6 V by Q = 5 C

Correct answer: A

Explanation: V=Q/CV=Q/C is linear, reaching 5/2×10−6=2.5×1065/2\times10^{-6}=2.5\times10^6 V.

Question 39

Physics · 4 marks · −1

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be:

  1. A.8 : 5
  2. B.5 : 4
  3. C.5 : 3
  4. D.8 : 7

Correct answer: B

Explanation: q∝mrq\propto\frac{\sqrt m}{r}: 3/26/5=54\frac{3/2}{6/5}=\frac54.

Question 40

Physics · 4 marks · −1

To increase the resonant frequency in series LCR circuit,

  1. A.source frequency should be increased.
  2. B.another resistance should be added in series with the first resistance.
  3. C.another capacitor should be added in series with the first capacitor.
  4. D.the source frequency should be decreased.

Correct answer: C

Explanation: Series capacitors reduce C, raising 12πLC\frac{1}{2\pi\sqrt{LC}}.

Question 41

Physics · 4 marks · −1

A small square loop of wire of side ll is placed inside a large square loop of wire L (L≫lL\gg l). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

Figure for question 41
  1. A.22μ0L2πl\frac{2\sqrt2\mu_0L^2}{\pi l}
  2. B.μ0l222πL\frac{\mu_0l^2}{2\sqrt2\pi L}
  3. C.22μ0l2πL\frac{2\sqrt2\mu_0l^2}{\pi L}
  4. D.μ0L222πl\frac{\mu_0L^2}{2\sqrt2\pi l}

Correct answer: C

Explanation: Field at centre of large square =22μ0IπL=\frac{2\sqrt2\mu_0I}{\pi L}; M=22μ0l2πLM=\frac{2\sqrt2\mu_0l^2}{\pi L}.

Question 42

Physics · 4 marks · −1

The rms value of conduction current in a parallel plate capacitor is 6.9 μ\muA. The capacity of this capacitor, if it is connected to 230 V ac supply with an angular frequency of 600 rad/s, will be:

  1. A.5 pF
  2. B.50 pF
  3. C.100 pF
  4. D.200 pF

Correct answer: B

Explanation: C=IVω=6.9×10−6230×600=50C=\frac{I}{V\omega}=\frac{6.9\times10^{-6}}{230\times600}=50 pF.

Question 43

Physics · 4 marks · −1

Which of the following statement is correct?

  1. A.In primary rainbow, observer sees red colour on the top and violet on the bottom
  2. B.In primary rainbow, observer sees violet colour on the top and red on the bottom
  3. C.In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops.
  4. D.Primary rainbow is less bright than secondary rainbow.

Correct answer: A

Question 44

Physics · 4 marks · −1

Time taken by light to travel in two different materials A and B of refractive indices μA\mu_A and μB\mu_B of same thickness is t1t_1 and t2t_2 respectively. If t2−t1=5×10−10t_2-t_1=5\times10^{-10} s and the ratio of μA\mu_A to μB\mu_B is 1 : 2. Then, the thickness of material, in meter is: (Given vAv_A and vBv_B are velocities of light in A and B materials respectively.)

  1. A.5×10−10vA5\times10^{-10}v_A m
  2. B.5×10−105\times10^{-10} m
  3. C.1.5×10−101.5\times10^{-10} m
  4. D.5×10−10vB5\times10^{-10}v_B m

Correct answer: A

Explanation: vA=2vBv_A=2v_B; dvB−dvA=dvA=5×10−10⇒d=5×10−10vA\frac{d}{v_B}-\frac{d}{v_A}=\frac{d}{v_A}=5\times10^{-10}\Rightarrow d=5\times10^{-10}v_A.

Question 45

Physics · 4 marks · −1

A metal exposed to light of wavelength 800 nm and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The workfunction of the metal is: (Take hc = 1230 eV-nm).

  1. A.1.537 eV
  2. B.2.46 eV
  3. C.0.615 eV
  4. D.1.23 eV

Correct answer: C

Explanation: 2(1.5375−ϕ)=2.46−ϕ⇒ϕ=0.6152(1.5375-\phi)=2.46-\phi\Rightarrow\phi=0.615 eV.

Question 46

Physics · 4 marks · −1

The momentum of an electron revolving in nthn^{th} orbit is given by: (Symbols have their usual meanings)

  1. A.nh2πr\frac{nh}{2\pi r}
  2. B.nh2r\frac{nh}{2r}
  3. C.nh2π\frac{nh}{2\pi}
  4. D.2πrnh\frac{2\pi r}{nh}

Correct answer: A

Explanation: mvr=nh2πmvr=\frac{nh}{2\pi}.

Question 47

Physics · 4 marks · −1

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

  1. A.μ⃗L=eL⃗2m\vec\mu_L=\frac{e\vec L}{2m}
  2. B.μ⃗L=−eL⃗2m\vec\mu_L=-\frac{e\vec L}{2m}
  3. C.μ⃗l=−eL⃗m\vec\mu_l=-\frac{e\vec L}{m}
  4. D.μ⃗l=2eL⃗m\vec\mu_l=\frac{2e\vec L}{m}

Correct answer: B

Question 48

Physics · 4 marks · −1

In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5 V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. The truth table of the given circuit will be:

Figure for question 48
  1. A.A B Y: 0 0 0; 1 0 0; 0 1 0; 1 1 1
  2. B.A B Y: 0 0 0; 1 0 1; 0 1 1; 1 1 1
  3. C.A B Y: 0 0 0; 1 0 0; 0 1 0; 1 1 0
  4. D.A B Y: 0 0 1; 1 0 1; 0 1 1; 1 1 0

Correct answer: A

Explanation: Y is high only when both diodes are reverse biased (A = B = 1): AND gate.

Question 49

Physics · 4 marks · −1

A car is moving with speed of 150 km/h and after applying the break it will move 27 m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ________ m distance.

Correct answer: 3

Explanation: s∝v2s\propto v^2: 27/9=327/9=3 m.

Question 50

Physics · 4 marks · −1

Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F1F_1 to F2F_2 is 1:x1:x where xx = ________.

Figure for question 50

Correct answer: 3

Explanation: Horizontal: F1=2−12=12F_1=\frac{2-1}{\sqrt2}=\frac{1}{\sqrt2}; vertical: F2=32F_2=\frac{3}{\sqrt2}; ratio 1:31:3.

Question 51

Physics · 4 marks · −1

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F under same conditions. The increase in length of this wire is ________ cm.

Correct answer: 5

Explanation: ΔL∝FLr2\Delta L\propto\frac{FL}{r^2}: 4⋅416=1\frac{4\cdot4}{16}=1, so 5 cm.

Question 52

Physics · 4 marks · −1

A unit scale is to be prepared whose length does not change with temperature and remains 20 cm, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is 40 cm and length of iron will be ________ cm. (αiron=1.2×10−5\alpha_{iron}=1.2\times10^{-5} K−1^{-1} and αbrass=1.8×10−5\alpha_{brass}=1.8\times10^{-5} K−1^{-1}).

Correct answer: 60

Explanation: 40×1.8=L×1.2⇒L=6040\times1.8=L\times1.2\Rightarrow L=60 cm.

Question 53

Physics · 4 marks · −1

An observer is riding on a bicycle and moving towards a hill at 18 kmh−1^{-1}. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 Hz and velocity of the sound in air is 320 m/s, the beat frequency between the two sounds heard by observer will be ________ Hz.

Correct answer: 20

Explanation: Direct: 640⋅315320=630640\cdot\frac{315}{320}=630; reflected: 640⋅325320=650640\cdot\frac{325}{320}=650; beats 20 Hz.

Question 54

Physics · 4 marks · −1

The volume charge density of a sphere of radius 6 m is 2 μ\muC cm−3^{-3}. The number of lines of force per unit surface area coming out from the surface of the sphere is ________ ×1010\times10^{10} NC−1^{-1}. [Given: Permittivity of vacuum ϵ0=8.85×10−12\epsilon_0=8.85\times10^{-12} C2^2 N−1^{-1}-m−2^{-2}]

Correct answer: 45

Explanation: E=ρR3ϵ0=2×63×8.85×10−12≈45×1010E=\frac{\rho R}{3\epsilon_0}=\frac{2\times6}{3\times8.85\times10^{-12}}\approx45\times10^{10}.

Question 55

Physics · 4 marks · −1

In the given figure, the value of VoV_o will be ________ V.

Figure for question 55

Correct answer: 4

Explanation: Millman: Vo=2+4+63=4V_o=\frac{2+4+6}{3}=4 V.

Question 56

Physics · 4 marks · −1

Eight copper wire of length ll and diameter d are joined in parallel to form a single composite conductor of resistance R. If a single copper wire of length 2l2l have the same resistance (R) then its diameter will be ________ d.

Correct answer: 4

Explanation: \frac{\rho l}{8A}=\frac{\rho2l}{A\'}\Rightarrow A\'=16A, diameter 4d4d.

Question 57

Physics · 4 marks · −1

The energy band gap of semiconducting material to produce violet (wavelength = 4000 Å) LED is ________ eV. (Round off to the nearest integer).

Correct answer: 3

Explanation: 12400/4000=3.1≈312400/4000=3.1\approx3 eV.

Question 58

Physics · 4 marks · −1

The required height of a TV tower which can cover the population of 6.03 lakh is h. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of h will be ________ m.

Correct answer: 150

Explanation: Area =6030=6030 km2=π(2Rh)⇒h≈0.15^2=\pi(2Rh)\Rightarrow h\approx0.15 km =150=150 m.

Question 59

Chemistry · 4 marks · −1

SO2_2Cl2_2 on reaction with excess of water results into acidic mixture SO2Cl2+2H2O→H2SO4+2HCl\mathrm{SO_2Cl_2+2H_2O\rightarrow H_2SO_4+2HCl}. 16 moles of NaOH is required for the complete neutralisation of the resultant acidic mixture. The number of moles of SO2_2Cl2_2 used is:

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

Correct answer: C

Explanation: Each mole gives 4 mol H+^+: 4 mol.

Question 60

Chemistry · 4 marks · −1

Which of the following sets of quantum numbers is not allowed?

  1. A.n=3,l=2,ml=0,s=+12n=3, l=2, m_l=0, s=+\frac12
  2. B.n=3,l=2,ml=−2,s=+12n=3, l=2, m_l=-2, s=+\frac12
  3. C.n=3,l=3,ml=−3,s=−12n=3, l=3, m_l=-3, s=-\frac12
  4. D.n=3,l=0,ml=0,s=−12n=3, l=0, m_l=0, s=-\frac12

Correct answer: C

Question 61

Chemistry · 4 marks · −1

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L−1^{-1} is 0.0405°C. Density of formic acid is 1.05 g mL−1^{-1}. The Van't Hoff factor of the formic acid solution is nearly: (Given for water kfk_f = 1.86 k kg mol−1^{-1})

  1. A.0.8
  2. B.1.1
  3. C.1.9
  4. D.2.4

Correct answer: C

Explanation: 0.525 g/L =0.0114=0.0114 m; i=0.04051.86×0.0114≈1.9i=\frac{0.0405}{1.86\times0.0114}\approx1.9.

Question 62

Chemistry · 4 marks · −1

20 mL of 0.1 M NH4_4OH is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given: KbK_b(NH4_4OH) = 1×10−51\times10^{-5}, log 2 = 0.30, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84, log 11 = 1.04)

  1. A.3.2
  2. B.4.2
  3. C.5.2
  4. D.6.2

Correct answer: C

Explanation: Exact neutralisation: [NH4+]=130[\mathrm{NH_4^+}]=\frac{1}{30} M; pH =7−12(5)−12log⁡130≈5.2=7-\frac12(5)-\frac12\log\frac1{30}\approx5.2.

Question 63

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)} (B) CO(g)+3H2(g)→CH4(g)+H2O(g)\mathrm{CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)} (C) CO(g)+H2(g)→HCHO(g)\mathrm{CO(g)+H_2(g)\rightarrow HCHO(g)} (D) CO(g)+2H2(g)→CH3OH(g)\mathrm{CO(g)+2H_2(g)\rightarrow CH_3OH(g)}. List - II: (I) Cu (II) Cu/ZnO–Cr2_2O3_3 (III) Fex_xOy_y + K2_2O + Al2_2O3_3 (IV) Ni. Choose the correct answer from the options given below:

  1. A.(A) - (II), (B) - (IV), (C) - (I), (D) - (III)
  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: C

Question 64

Chemistry · 4 marks · −1

The IUPAC nomenclature of an element with electronic configuration [Rn] 5f$^{14}6d6d^17s7s^2$ is:

  1. A.Unnilbium
  2. B.Unnilunium
  3. C.Unnilquadium
  4. D.Unniltrium

Correct answer: D

Explanation: Z = 103: Unniltrium.

Question 65

Chemistry · 4 marks · −1

The compound(s) that is(are) removed as slag during the extraction of copper is: (A) CaO (B) FeO (C) Al2_2O3_3 (D) ZnO (E) NiO. Choose the correct answer from the options given below:

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

Correct answer: D

Question 66

Chemistry · 4 marks · −1

The reaction of H2_2O2_2 with potassium permanganate in acidic medium leads to the formation of mainly:

  1. A.Mn2+^{2+}
  2. B.Mn4+^{4+}
  3. C.Mn3+^{3+}
  4. D.Mn6+^{6+}

Correct answer: A

Question 67

Chemistry · 4 marks · −1

Choose the correct order of density of the alkali metals:

  1. A.Li < K < Na < Rb < Cs
  2. B.Li < Na < K < Rb < Cs
  3. C.Cs < Rb < K < Na < Li
  4. D.Li < Na < K < Cs < Rb

Correct answer: A

Question 68

Chemistry · 4 marks · −1

The geometry around boron in the product 'B' formed from the following reaction is BF3+NaH→450 KA+NaF\mathrm{BF_3+NaH\xrightarrow{450\,K}A+NaF}; A+NMe3→B\mathrm{A+NMe_3\rightarrow B}

  1. A.trigonal planar
  2. B.tetrahedral
  3. C.pyramidal
  4. D.square planar

Correct answer: B

Explanation: A = B2_2H6_6; B = Me3_3N·BH3_3 with tetrahedral boron.

Question 69

Chemistry · 4 marks · −1

The interhalogen compound formed from the reaction of bromine with excess of fluorine is a:

  1. A.hypohalite
  2. B.halate
  3. C.perhalate
  4. D.halite

Correct answer: B

Question 70

Chemistry · 4 marks · −1

The photochemical smog does not generally contain:

  1. A.NO
  2. B.NO2_2
  3. C.SO2_2
  4. D.HCHO

Correct answer: C

Question 71

Chemistry · 4 marks · −1

A compound 'A' on reaction with 'X' and 'Y' produces the same major product but different by product 'a' and 'b'. Oxidation of 'a' gives a substance produced by ants. Compound 'A' = H2C=C(CH3)−CH2−C(CH3)2−CH3\mathrm{H_2C=C(CH_3)-CH_2-C(CH_3)_2-CH_3}; with X: a + O=C(CH3)−CH2−C(CH3)2−CH3\mathrm{O=C(CH_3)-CH_2-C(CH_3)_2-CH_3}; with Y: b + O=C(CH3)−CH2−C(CH3)2−CH3\mathrm{O=C(CH_3)-CH_2-C(CH_3)_2-CH_3}. 'X' and 'Y' respectively are

  1. A.KMnO4_4/H+^+ and dil. KMnO4_4, 273 K
  2. B.KMnO4_4(dilute), 273 K and KMnO4_4/H+^+
  3. C.KMnO4_4/H+^+ and O3_3, H2_2O/Zn
  4. D.O3_3, H2_2O/Zn and KMnO4_4/H+^+

Correct answer: D

Explanation: By-product a oxidises to formic acid, so a = HCHO: X is ozonolysis (O3_3, H2_2O/Zn). With acidic KMnO4_4 the =CH2_2 end goes to CO2_2 (b). So X = O3_3/Zn and Y = KMnO4_4/H+^+.

Question 72

Chemistry · 4 marks · −1

Most stable product of the following reaction is: (3-methylpentan-2-ol drawn with wedge bonds at C-2 (OH) and C-3 (ethyl)) →(i) p-CH3C6H4SO2Cl, pyridine; (ii) NaCN, DMF\xrightarrow{\text{(i) }p\text{-CH}_3\text{C}_6\text{H}_4\text{SO}_2\text{Cl, pyridine; (ii) NaCN, DMF}}

Figure for question 72
  1. A.2-cyano-3-methylpentane with CN on a wedge at C-2 and ethyl on a wedge at C-3 (configuration retained at C-2)
  2. B.CN wedge at C-2 and the ethyl on a hashed bond at C-3
  3. C.CN wedge at C-2, ethyl wedge at C-3 drawn in the alternative skeleton orientation
  4. D.CN on a hashed bond at C-2 and ethyl on a hashed bond at C-3

Correct answer: B

Explanation: Tosylation retains configuration at C-2; S$_N2attackbyCN2 attack by CN^-$ then inverts C-2 while C-3 is untouched (answer per official key; the option drawings differ only in wedge/hash placement).

Question 73

Chemistry · 4 marks · −1

Which one of the following reactions does not represent correct combination of substrate and product under the given conditions?

  1. A.p-toluoyl chloride →H2, Pd.BaSO4\xrightarrow{\text{H}_2\text{, Pd.BaSO}_4} p-tolualdehyde
  2. B.p-tolunitrile →(i) DIBAL-H, (ii) H2O\xrightarrow{\text{(i) DIBAL-H, (ii) H}_2\text{O}} p-tolualdehyde
  3. C.ethyl p-toluate (CO2_2C2_2H5_5) →(i) AlH(iso-Bu)2, (ii) H2O\xrightarrow{\text{(i) AlH(iso-Bu)}_2\text{, (ii) H}_2\text{O}} p-tolualdehyde
  4. D.4-methylbenzyl alcohol →(i) Na2Cr2O7, (ii) H2SO4, H2O\xrightarrow{\text{(i) Na}_2\text{Cr}_2\text{O}_7\text{, (ii) H}_2\text{SO}_4\text{, H}_2\text{O}} p-tolualdehyde

Correct answer: D

Explanation: Acidified dichromate oxidises the primary alcohol on to the carboxylic acid.

Question 74

Chemistry · 4 marks · −1

An organic compound 'A' on reaction with NH3_3 followed by heating gives compound B. Which on further strong heating gives compound C (C8_8H5_5NO2_2). Compound C on sequential reaction with ethanolic KOH, alkyl chloride and hydrolysis with alkali gives a primary amine. The compound A is:

  1. A.benzene-1,2-dicarbaldehyde (phthalaldehyde)
  2. B.2-formylbenzoic acid
  3. C.benzene-1,2-dicarboxylic acid (phthalic acid)
  4. D.2-hydroxybenzaldehyde (salicylaldehyde)

Correct answer: C

Explanation: Phthalic acid → ammonium phthalate → phthalamide → phthalimide (C8H5NO2), used in Gabriel synthesis.

Question 75

Chemistry · 4 marks · −1

Melamine polymer is formed by the condensation of:

  1. A.2,4,6-triamino-1,3,5-triazine + HCHO
  2. B.2,4,6-triaminopyrimidine + HCHO
  3. C.benzene-1,3,5-triamine + HCHO
  4. D.pyridine-2,4-diamine + HCHO

Correct answer: A

Question 76

Chemistry · 4 marks · −1

During the denaturation of proteins, which of these structures will remain intact?

  1. A.Primary
  2. B.Secondary
  3. C.Tertiary
  4. D.Quaternary

Correct answer: A

Question 77

Chemistry · 4 marks · −1

Drugs used to bind to receptors, inhibiting its natural function and blocking a message are called:

  1. A.Agonists
  2. B.Antagonists
  3. C.Allosterists
  4. D.Anti histaminists

Correct answer: B

Question 78

Chemistry · 4 marks · −1

Given below are two statements: Statement I: On heating with KHSO4_4, glycerol is dehydrated and acrolein is formed. Statement II: Acrolein has fruity odour and can be used to test glycerol's presence. Choose the correct option.

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

Correct answer: C

Explanation: Acrolein has a pungent (not fruity) odour.

Question 79

Chemistry · 4 marks · −1

Among the following species N2_2, N2+_2^+, N2−_2^-, N22−_2^{2-}, O2_2, O2+_2^+, O2−_2^-, O22−_2^{2-} the number of species showing diamagnesim is ________.

Correct answer: 2

Explanation: N2_2 (14 e−^-) and O22−_2^{2-} (18 e−^-) are diamagnetic; N22−_2^{2-} is isoelectronic with O2_2 (paramagnetic). Answer 2.

Question 80

Chemistry · 4 marks · −1

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are −2220.0-2220.0 kJ mol−1^{-1}, −393.5-393.5 kJ mol−1^{-1} and −285.8-285.8 kJ mol−1^{-1} respectively. The magnitude of enthalpy of formation of propane (C3_3H8_8) is ________ kJ mol−1^{-1}. (Nearest integer)

Correct answer: 104

Explanation: 3(−393.5)+4(−285.8)+2220=−103.73(-393.5)+4(-285.8)+2220=-103.7: 104.

Question 81

Chemistry · 4 marks · −1

The pressure of a moist gas at 27°C is 4 atm. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is ________ ×10−1\times10^{-1} atm. (Nearest integer) (Given: The vapour pressure of water at 27°C is 0.4 atm.)

Correct answer: 22

Explanation: Dry gas 3.6→1.83.6\to1.8 atm; plus 0.4: 2.2 atm =22×10−1=22\times10^{-1}.

Question 82

Chemistry · 4 marks · −1

The cell potential for Zn|Zn2+^{2+}(aq)||Snx+^{x+}|Sn is 0.801 V at 298 K. The reaction quotient for the above reaction is 10−210^{-2}. The number of electrons involved in the given electrochemical cell reaction is ________. (Given: EZn2+∣Zn∘=−0.763E^\circ_{Zn^{2+}|Zn}=-0.763 V, ESnx+∣Sn∘=+0.008E^\circ_{Sn^{x+}|Sn}=+0.008 V and 2.303RTF=0.06\frac{2.303RT}{F}=0.06 V)

Correct answer: 4

Explanation: 0.801=0.771−0.06n(−2)⇒n=40.801=0.771-\frac{0.06}{n}(-2)\Rightarrow n=4.

Question 83

Chemistry · 4 marks · −1

The half life for the decomposition of gaseous compound A is 240 s when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is ________. (Nearest integer)

Correct answer: 1

Explanation: Half-life independent of initial pressure: first order (1).

Question 84

Chemistry · 4 marks · −1

Consider the following metal complexes: [Co(NH3)6]3+[\mathrm{Co(NH_3)_6}]^{3+}, [CoCl(NH3)5]2+[\mathrm{CoCl(NH_3)_5}]^{2+}, [Co(CN)6]3−[\mathrm{Co(CN)_6}]^{3-}, [Co(NH3)5(H2O)]3+[\mathrm{Co(NH_3)_5(H_2O)}]^{3+}. The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is ________ B.M. (Nearest integer)

Correct answer: 0

Explanation: [Co(CN)6_6]3−^{3-} (largest Δo\Delta_o) is low-spin d6^6: 0 B.M.

Question 85

Chemistry · 4 marks · −1

Among Co3+^{3+}, Ti2+^{2+}, V2+^{2+} and Cr2+^{2+} ions, one if used as a reagent cannot liberate H2_2 from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is ________ B.M. (Nearest integer)

Correct answer: 5

Explanation: Co3+^{3+} (oxidising) cannot; gaseous d6^6: 4 unpaired, 24≈5\sqrt{24}\approx5 B.M.

Question 86

Chemistry · 4 marks · −1

While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.25 g of the compound neutralized 2.5 mL of 2 M H2_2SO4_4. The percentage of nitrogen present in organic compound is ________.

Correct answer: 56

Explanation: NH3_3 = 10 mmol, N = 0.14 g: 56%.

Question 87

Chemistry · 4 marks · −1

In the given reaction: a cyclopentane ring bearing, on C-1, a CH3_3 and a –C(=O)OEt group and, on the adjacent C-2, a –CH2_2–C(=O)–CH3_3 group, treated with −^-OEt gives A. (Where Et is –C2_2H5_5). The number of chiral carbon/s in product A is ________.

Figure for question 87

Correct answer: 2

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