JEE Main 2024 — 30 Jan Shift 2 (Previous Year Paper)

JEE Main · 2024 · with answers and explanations

questions
89
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If x2−y2+2hxy+2gx+2fy+c=0x^2-y^2+2hxy+2gx+2fy+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0x+2y+7=0 and 2x−y+8=02x-y+8=0, then the value of g+c+h−fg+c+h-f equals.

  1. A.14
  2. B.6
  3. C.8
  4. D.29

Correct answer: A

Explanation: (x+2y+7)2=(2x−y+8)2⇒x2−y2−83xy+6x−443y+5=0(x+2y+7)^2=(2x-y+8)^2\Rightarrow x^2-y^2-\frac83xy+6x-\frac{44}{3}y+5=0; g=3,h=−43,f=−223,c=5g=3,h=-\frac43,f=-\frac{22}{3},c=5.

Question 2

Mathematics · 4 marks · −1

Let f(x)=(x+3)2(x−2)3f(x)=(x+3)^2(x-2)^3, x∈[−4,4]x\in[-4,4]. If MM and mm are the maximum and minimum values of ff, respectively in [−4,4][-4,4], then the value of M−mM-m is:

  1. A.392
  2. B.600
  3. C.108
  4. D.608

Correct answer: D

Explanation: Candidates f(−4)=−216f(-4)=-216, f(−3)=0f(-3)=0, f(−1)=−108f(-1)=-108, f(2)=0f(2)=0, f(4)=392f(4)=392.

Question 3

Mathematics · 4 marks · −1

Let y=f(x)y=f(x) be a thrice differentiable function in (−5, 5). Let the tangents to the curve y=f(x)y=f(x) at (1,f(1))(1,f(1)) and (3,f(3))(3,f(3)) make angles π6\frac\pi6 and π4\frac\pi4, respectively with positive xx-axis. If 27∫13((f′(t))2+1)f′′(t)dt=α+β327\int_1^3\left((f'(t))^2+1\right)f''(t)dt=\alpha+\beta\sqrt3 where α,β\alpha,\beta are integers, then the value of α+β\alpha+\beta equals.

  1. A.36
  2. B.−16
  3. C.−14
  4. D.26

Correct answer: D

Explanation: 27[(f′)33+f′]1/31=36−10327\left[\frac{(f')^3}{3}+f'\right]_{1/\sqrt3}^{1}=36-10\sqrt3.

Question 4

Mathematics · 4 marks · −1

Let a⃗=i^+αj^+βk^\vec a=\hat i+\alpha\hat j+\beta\hat k, α,β∈R\alpha,\beta\in\mathbb{R}. Let a vector b⃗\vec b be such that the angle between a⃗\vec a and b⃗\vec b is π4\frac\pi4 and ∣b⃗∣2=6|\vec b|^2=6. If a⃗⋅b⃗=32\vec a\cdot\vec b=3\sqrt2, then the value of (α2+β2)∣a⃗×b⃗∣2(\alpha^2+\beta^2)|\vec a\times\vec b|^2 is equal to:

  1. A.85
  2. B.75
  3. C.90
  4. D.95

Correct answer: C

Explanation: ∣a⃗∣=6|\vec a|=\sqrt6 so α2+β2=5\alpha^2+\beta^2=5; ∣a⃗×b⃗∣2=36−18=18|\vec a\times\vec b|^2=36-18=18.

Question 5

Mathematics · 4 marks · −1

Let L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^)L_1:\vec r=(\hat i-\hat j+2\hat k)+\lambda(\hat i-\hat j+2\hat k), λ∈R\lambda\in\mathbb{R}, L2:r⃗=(j^−k^)+μ(3i^+j^+pk^)L_2:\vec r=(\hat j-\hat k)+\mu(3\hat i+\hat j+p\hat k), μ∈R\mu\in\mathbb{R}, and L3:r⃗=δ(li^+mj^+nk^)L_3:\vec r=\delta(l\hat i+m\hat j+n\hat k), δ∈R\delta\in\mathbb{R} be three lines such that L1L_1 is perpendicular to L2L_2 and L3L_3 is perpendicular to both L1L_1 and L2L_2. Then, the point which lies on L3L_3 is:

  1. A.(1, −7, 4)
  2. B.(−1, 7, 4)
  3. C.(−1, −7, 4)
  4. D.(1, 7, −4)

Correct answer: B

Explanation: p=−1p=-1; direction of L3L_3 is (1,−1,2)×(3,1,−1)=(−1,7,4)(1,-1,2)\times(3,1,-1)=(-1,7,4), through the origin.

Question 6

Mathematics · 4 marks · −1

Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is:

  1. A.19\frac19
  2. B.14\frac14
  3. C.13\frac13
  4. D.310\frac3{10}

Correct answer: C

Explanation: 0.30.3+0.6=13\frac{0.3}{0.3+0.6}=\frac13.

Question 7

Mathematics · 4 marks · −1

Let f:R−{0}→Rf:\mathbb{R}-\{0\}\to\mathbb{R} be a function satisfying f(xy)=f(x)f(y)f\left(\frac xy\right)=\frac{f(x)}{f(y)} for all x,yx,y, f(y)≠0f(y)\ne0. If f′(1)=2024f'(1)=2024, then:

  1. A.xf′(x)+2024f(x)=0xf'(x)+2024f(x)=0
  2. B.xf′(x)+f(x)=2024xf'(x)+f(x)=2024
  3. C.xf′(x)−2024f(x)=0xf'(x)-2024f(x)=0
  4. D.xf′(x)−2023f(x)=0xf'(x)-2023f(x)=0

Correct answer: C

Explanation: f(x)=xkf(x)=x^k with k=f′(1)=2024k=f'(1)=2024.

Question 8

Mathematics · 4 marks · −1

Let R=(x000y000z)R=\begin{pmatrix}x&0&0\\0&y&0\\0&0&z\end{pmatrix} be a non-zero 3×33\times3 matrix, where xsin⁡θ=ysin⁡(θ+2π3)=zsin⁡(θ+4π3)≠0x\sin\theta=y\sin\left(\theta+\frac{2\pi}{3}\right)=z\sin\left(\theta+\frac{4\pi}{3}\right)\ne0, θ∈(0,2π)\theta\in(0,2\pi). For a square matrix MM, Let trace (M)(M) denote the sum of all the diagonal entries of MM. Then, among the statements: (I) Trace (R)=0(R)=0 (II) If trace(adj(adj(R)))=0(\text{adj}(\text{adj}(R)))=0, then RR has exactly one non-zero entry.

  1. A.Only (II) is true
  2. B.Only (I) is true
  3. C.Neither (I) nor (II) is true
  4. D.Both (I) and (II) are true

Correct answer: C

Explanation: ∑1sin⁡(θ+2kπ/3)=3sin⁡3θ≠0\sum\frac{1}{\sin(\theta+2k\pi/3)}=\frac{3}{\sin3\theta}\ne0, so trace(R)≠0(R)\ne0; and adj(adj RR) =∣R∣R=|R|R has non-zero trace, while all three diagonal entries are non-zero.

Question 9

Mathematics · 4 marks · −1

If zz is a complex number, then the number of common roots of the equations z1985+z100+1=0z^{1985}+z^{100}+1=0 and z3+2z2+2z+1=0z^3+2z^2+2z+1=0, is equal to:

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

Correct answer: C

Explanation: The cubic has roots −1,ω,ω2-1,\omega,\omega^2; ω\omega and ω2\omega^2 satisfy the first equation, −1-1 does not.

Question 10

Mathematics · 4 marks · −1

Let aa and bb be two distinct positive real numbers. Let 11th11^{th} term of a GP, whose first term is aa and third term is bb, is equal to pthp^{th} term of another GP, whose first term is a and fifth term is bb. Then pp is equal to:

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

Correct answer: D

Explanation: a(b/a)5=a(b/a)(p−1)/4⇒p=21a(b/a)^5=a(b/a)^{(p-1)/4}\Rightarrow p=21.

Question 11

Mathematics · 4 marks · −1

Let a⃗\vec a and b⃗\vec b be two vectors such that ∣b⃗∣=1|\vec b|=1 and ∣b⃗×a⃗∣=2|\vec b\times\vec a|=2. Then ∣(b⃗×a⃗)−b⃗∣2|(\vec b\times\vec a)-\vec b|^2 is equal to:

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

Correct answer: A

Explanation: 4+1−0=54+1-0=5.

Question 12

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be defined as f(x)=ae2x+bex+cxf(x)=ae^{2x}+be^x+cx. If f(0)=−1f(0)=-1, f′(log⁡e2)=21f'(\log_e2)=21 and ∫0log⁡e4(f(x)−cx)dx=392\int_0^{\log_e4}(f(x)-cx)dx=\frac{39}{2}, then the value of ∣a+b+c∣|a+b+c| equals.

  1. A.10
  2. B.8
  3. C.12
  4. D.16

Correct answer: B

Explanation: a+b=−1a+b=-1, 8a+2b+c=218a+2b+c=21, 15a2+3b=392\frac{15a}{2}+3b=\frac{39}{2} give a=5,b=−6,c=−7a=5,b=-6,c=-7.

Question 13

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a function defined by f(x)=x(1+x4)1/4f(x)=\frac{x}{(1+x^4)^{1/4}}, and g(x)=f(f(f(f(x))))g(x)=f(f(f(f(x)))). Then, 18∫025x2g(x)dx18\int_0^{\sqrt{2\sqrt5}}x^2g(x)dx is equal to:

  1. A.42
  2. B.39
  3. C.36
  4. D.33

Correct answer: B

Explanation: g(x)=x(1+4x4)1/4g(x)=\frac{x}{(1+4x^4)^{1/4}}; with u=1+4x4u=1+4x^4 the integral is 32[u3/4]181=39\frac32[u^{3/4}]_1^{81}=39.

Question 14

Mathematics · 4 marks · −1

Let PP be a point on the hyperbola H:x29−y24=1H:\frac{x^2}{9}-\frac{y^2}{4}=1, in the first quadrant such that the area of triangle formed by PP and the two foci of HH is 2132\sqrt{13}. Then, the square of the distance of PP from the origin is:

  1. A.20
  2. B.22
  3. C.18
  4. D.26

Correct answer: B

Explanation: Area =13 y=213⇒y=2=\sqrt{13}\,y=2\sqrt{13}\Rightarrow y=2, x2=18x^2=18; OP2=22OP^2=22.

Question 15

Mathematics · 4 marks · −1

Let aa and bb be real constants such that the function ff defined by f(x)={x2+3x+a,x≤1bx+2,x>1f(x)=\begin{cases}x^2+3x+a, & x\le1\\ bx+2, & x>1\end{cases} be differentiable on R\mathbb{R}. Then, the value of ∫−22f(x)dx\int_{-2}^{2}f(x)dx equals.

  1. A.17
  2. B.196\frac{19}{6}
  3. C.156\frac{15}{6}
  4. D.21

Correct answer: A

Explanation: b=5b=5, a=3a=3; ∫−21(x2+3x+3)dx+∫12(5x+2)dx=7.5+9.5\int_{-2}^1(x^2+3x+3)dx+\int_1^2(5x+2)dx=7.5+9.5.

Question 16

Mathematics · 4 marks · −1

If the domain of the function f(x)=log⁡e(2x+34x2+x−3)+cos⁡−1(2x−1x+2)f(x)=\log_e\left(\frac{2x+3}{4x^2+x-3}\right)+\cos^{-1}\left(\frac{2x-1}{x+2}\right) is (α,β](\alpha,\beta], then the value of 5β−4α5\beta-4\alpha is equal to:

  1. A.9
  2. B.11
  3. C.10
  4. D.12

Correct answer: D

Explanation: Log condition: (−32,−1)∪(34,∞)(-\frac32,-1)\cup(\frac34,\infty); cos⁡−1\cos^{-1} condition: [−13,3][-\frac13,3]; domain (34,3](\frac34,3].

Question 17

Mathematics · 4 marks · −1

Let A(α,0)A(\alpha,0) and B(0,β)B(0,\beta) be the points on the line 5x+7y=505x+7y=50. Let the point PP divide the line segment ABAB internally in the ratio 7 : 3. Let 3x−25=03x-25=0 be a directrix of the ellipse E:x2a2+y2b2=1E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and the corresponding focus be SS. If from SS, the perpendicular on the xx-axis passes through PP, then the length of the latus rectum of EE is equal to:

  1. A.329\frac{32}{9}
  2. B.259\frac{25}{9}
  3. C.253\frac{25}{3}
  4. D.325\frac{32}{5}

Correct answer: D

Explanation: P=(3,5)P=(3,5), so ae=3ae=3 and ae=253\frac ae=\frac{25}{3}: a=5a=5, b2=16b^2=16; latus rectum 2b2a=325\frac{2b^2}{a}=\frac{32}{5}.

Question 18

Mathematics · 4 marks · −1

For α,β∈(0,π2)\alpha,\beta\in\left(0,\frac\pi2\right), let 3sin⁡(α+β)=2sin⁡(α−β)3\sin(\alpha+\beta)=2\sin(\alpha-\beta) and a real number kk be such that tan⁡α=ktan⁡β\tan\alpha=k\tan\beta. Then, the value of kk is equal to:

  1. A.5
  2. B.23\frac23
  3. C.−5
  4. D.−23\frac{-2}{3}

Correct answer: C

Explanation: sin⁡αcos⁡β=−5cos⁡αsin⁡β\sin\alpha\cos\beta=-5\cos\alpha\sin\beta.

Question 19

Mathematics · 4 marks · −1

Consider the system of linear equations x+y+z=5x+y+z=5, x+2y+λ2z=9x+2y+\lambda^2z=9, x+3y+λz=μx+3y+\lambda z=\mu, where λ,μ∈R\lambda,\mu\in\mathbb{R}. Then, which of the following statement is NOT correct?

  1. A.System is consistent if λ≠1\lambda\ne1 and μ=13\mu=13
  2. B.System has unique solution if λ≠1\lambda\ne1 and μ≠13\mu\ne13
  3. C.System is inconsistent if λ=1\lambda=1 and μ≠13\mu\ne13
  4. D.System has infinite number of solutions if λ=1\lambda=1 and μ=13\mu=13

Correct answer: B

Explanation: Δ=−(2λ+1)(λ−1)\Delta=-(2\lambda+1)(\lambda-1) also vanishes at λ=−12\lambda=-\frac12, so "λ≠1\lambda\ne1" does not guarantee a unique solution.

Question 20

Mathematics · 4 marks · −1

Let SnS_n be the sum to nn-terms of an arithmetic progression 3, 7, 11, ....... If 40<(6n(n+1)∑k=1nSk)<4240<\left(\frac{6}{n(n+1)}\sum_{k=1}^nS_k\right)<42, then nn equals ______.

Correct answer: 9

Explanation: Sk=2k2+kS_k=2k^2+k; 6n(n+1)∑Sk=2(2n+1)+3=4n+5\frac{6}{n(n+1)}\sum S_k=2(2n+1)+3=4n+5; 40<4n+5<42⇒n=940<4n+5<42\Rightarrow n=9.

Question 21

Mathematics · 4 marks · −1

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A students is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C 6 questions, then the total number of ways a student can select 15 questions is ______.

Correct answer: 11376

Question 22

Mathematics · 4 marks · −1

Consider two circles C1:x2+y2=25C_1:x^2+y^2=25 and C2:(x−α)2+y2=16C_2:(x-\alpha)^2+y^2=16, where α∈(5,9)\alpha\in(5,9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C1C_1 and C2C_2 be sin⁡−1(638)\sin^{-1}\left(\frac{\sqrt{63}}{8}\right). If the length of common chord of C1C_1 and C2C_2 is β\beta, then the value of (αβ)2(\alpha\beta)^2 equals ______.

Correct answer: 1575

Question 23

Mathematics · 4 marks · −1

The variance σ2\sigma^2 of the data | xix_i | 0 | 1 | 5 | 6 | 10 | 12 | 17 | | fif_i | 3 | 2 | 3 | 2 | 6 | 3 | 3 | is ______.

Correct answer: 29

Question 24

Mathematics · 4 marks · −1

Let a line passing through the point (−1, 2, 3) intersects the lines L1:x−13=y−22=z+1−2L_1:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2} at M(α,β,γ)M(\alpha,\beta,\gamma) and L2:x+2−3=y−2−2=z−14L_2:\frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4} at N(a,b,c)N(a,b,c). Then, the value of (α+β+γ)2(a+b+c)2\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} equals ______.

Correct answer: 196

Question 25

Mathematics · 4 marks · −1

The number of symmetric relations defined on the set {1, 2, 3, 4} which are not reflexive is ______.

Correct answer: 960

Explanation: Symmetric relations 24+6=10242^{4+6}=1024; symmetric and reflexive 26=642^6=64; 1024−64=9601024-64=960.

Question 26

Mathematics · 4 marks · −1

Let Y=Y(X)Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x)Y-y=Y'(x)(X-x) and the co-ordinate axes, where (x,y)(x,y) is any point on the curve, is always −y22Y′(x)+1\frac{-y^2}{2Y'(x)}+1, Y′(x)≠0Y'(x)\ne0. If Y(1)=1Y(1)=1, then 12Y(2)12Y(2) equals ______.

Correct answer: 20

Question 27

Mathematics · 4 marks · −1

The area of the region enclosed by the parabola (y−2)2=x−1(y-2)^2=x-1, the line x−2y+4=0x-2y+4=0 and the positive coordinate axes is ______.

Correct answer: 5

Question 28

Mathematics · 4 marks · −1

The number of real solutions of the equation x(x2+3∣x∣+5∣x−1∣+6∣x−2∣)=0x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0 is ______.

Correct answer: 1

Explanation: The bracket is always positive, so only x=0x=0.

Question 29

Mathematics · 4 marks · −1

Let α=∑k=0n((nCk)2k+1)\alpha=\sum_{k=0}^{n}\left(\frac{\left({}^nC_k\right)^2}{k+1}\right) and β=∑k=0n−1(nCk nCk+1k+2)\beta=\sum_{k=0}^{n-1}\left(\frac{{}^nC_k\,{}^nC_{k+1}}{k+2}\right). If 5α=6β5\alpha=6\beta, then nn equals ______.

Correct answer: 10

Question 30

Physics · 4 marks · −1

A block of mass mm is placed on a surface having vertical cross section given by y=x24y=\frac{x^2}{4}. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

  1. A.13\frac13 m
  2. B.12\frac12 m
  3. C.16\frac16 m
  4. D.14\frac14 m

Correct answer: D

Explanation: tan⁡θ=x2≤0.5⇒x≤1\tan\theta=\frac x2\le0.5\Rightarrow x\le1, y=14y=\frac14 m.

Question 31

Physics · 4 marks · −1

If a nuclear fission reaction of an isotope of mass MM, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect ΔM\Delta M will be:

  1. A.c2ΔMMc\sqrt{\frac{2\Delta M}{M}}
  2. B.ΔMc23\frac{\Delta Mc^2}{3}
  3. C.c3ΔMMc\sqrt{\frac{3\Delta M}{M}}
  4. D.2cΔMM\sqrt{\frac{2c\Delta M}{M}}

Correct answer: A

Explanation: ΔMc2=3⋅12M3v2\Delta Mc^2=3\cdot\frac12\frac M3v^2.

Question 32

Physics · 4 marks · −1

A beam of unpolarized light of intensity I0I_0 passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of emergent light is:

  1. A.I0I_0
  2. B.I08\frac{I_0}{8}
  3. C.I02\frac{I_0}{2}
  4. D.I04\frac{I_0}{4}

Correct answer: D

Question 33

Physics · 4 marks · −1

If the total energy transferred to a surface in time tt is 6.48×1056.48\times10^5 J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

  1. A.2.16×10−32.16\times10^{-3} kg m/s
  2. B.4.32×10−34.32\times10^{-3} kg m/s
  3. C.1.58×10−31.58\times10^{-3} kg m/s
  4. D.2.46×10−32.46\times10^{-3} kg m/s

Correct answer: A

Explanation: p=U/cp=U/c.

Question 34

Physics · 4 marks · −1

A particle of charge '−q' and mass 'm' moves in a circle of radius 'r' around an infinitely long line charge of linear charge density '+λ'. Then time period will be given as: (Consider k as Coulomb's constant)

  1. A.T=12π2kλqmT=\frac{1}{2\pi}\sqrt{\frac{2k\lambda q}{m}}
  2. B.T=2πrm2kλqT=2\pi r\sqrt{\frac{m}{2k\lambda q}}
  3. C.T=12πrm2kλqT=\frac{1}{2\pi r}\sqrt{\frac{m}{2k\lambda q}}
  4. D.T2=4π2m2kλqr3T^2=\frac{4\pi^2m}{2k\lambda q}r^3

Correct answer: B

Explanation: mv2r=2kλqr\frac{mv^2}{r}=\frac{2k\lambda q}{r} gives a radius-independent speed.

Question 35

Physics · 4 marks · −1

If mass is written as m=kcPG−1/2h1/2m=kc^PG^{-1/2}h^{1/2} then the value of P will be: (Constants have their usual meaning with k a dimensionless constant).

  1. A.12\frac12
  2. B.−13\frac{-1}{3}
  3. C.13\frac13
  4. D.2

Correct answer: A

Explanation: Planck mass hc/G\sqrt{hc/G}.

Question 36

Physics · 4 marks · −1

If 50 Vernier divisions are equal to 49 main scale division of a traveling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of traveling microscope is:

  1. A.0.1 mm
  2. B.0.1 cm
  3. C.0.01 cm
  4. D.0.01 mm

Correct answer: D

Explanation: 0.5×150=0.010.5\times\frac{1}{50}=0.01 mm.

Question 37

Physics · 4 marks · −1

In the given circuit, the voltage across load resistance (RL)(R_L) is:

Figure for question 37
  1. A.14.00 V
  2. B.8.75 V
  3. C.9.00 V
  4. D.8.50 V

Correct answer: B

Explanation: I=15−0.3−0.74 kΩ=3.5I=\frac{15-0.3-0.7}{4\text{ k}\Omega}=3.5 mA; VL=8.75V_L=8.75 V.

Question 38

Physics · 4 marks · −1

If three moles of monoatomic gas (γ=53)\left(\gamma=\frac53\right) is mixed with two moles of a diatomic gas (γ=75)\left(\gamma=\frac75\right), the value of adiabatic exponent γ for the mixture is:

  1. A.1.35
  2. B.1.40
  3. C.1.52
  4. D.1.75

Correct answer: C

Explanation: CV=3(1.5)+2(2.5)5R=1.9RC_V=\frac{3(1.5)+2(2.5)}{5}R=1.9R; γ=2.91.9≈1.52\gamma=\frac{2.9}{1.9}\approx1.52.

Question 39

Physics · 4 marks · −1

For the photoelectric effect, the maximum kinetic energy (Ek)(E_k) of the photoelectrons is plotted against the frequency (ν) of the incident photons as shown in figure. The slope of the graph gives

Figure for question 39
  1. A.Ratio of Planck's constant to electric charge
  2. B.Planck's constant
  3. C.Work function of the metal
  4. D.Charge of electron

Correct answer: B

Question 40

Physics · 4 marks · −1

Choose the correct statement for processes A & B shown in figure.

Figure for question 40
  1. A.TγPγ−1=k\frac{T^\gamma}{P^{\gamma-1}}=k for process A and PV=kPV=k for process B.
  2. B.PV=kPV=k for process B and A.
  3. C.Pγ−1Tγ=k\frac{P^{\gamma-1}}{T^\gamma}=k for process B and T=kT=k for process A.
  4. D.PVγ=kPV^\gamma=k for process B and PV=kPV=k for process A.

Correct answer: D

Question 41

Physics · 4 marks · −1

Projectiles A and B are thrown at angles of 45° and 60° with vertical respectively from top of a 400 m high tower. If their range and times of flight are same, the ratio of their speeds of projection vA:vBv_A:v_B is: [Take g=10g=10 m s−2^{-2}]

  1. A.1 : 2
  2. B.1:31:\sqrt3
  3. C.1:21:\sqrt2
  4. D.2:1\sqrt2:1

Correct answer: C

Explanation: Equal times of flight need equal vertical components: vAcos⁡45∘=vBcos⁡60∘v_A\cos45^\circ=v_B\cos60^\circ.

Question 42

Physics · 4 marks · −1

A block of ice at −10°C is slowly heated and converted to steam at 100°C. Which of the following curves represent the phenomenon qualitatively:

Figure for question 42
  1. A.Temperature rises, stays constant, rises, then stays constant (two plateaus, second longer)
  2. B.Temperature rises, a plateau, then rises steeply (one plateau)
  3. C.Plateau at the start, rise, plateau, rise
  4. D.Rise, plateau, rise (single plateau)

Correct answer: A

Explanation: Two phase changes (melting at 0°C, boiling at 100°C) give two plateaus, the second much longer.

Question 43

Physics · 4 marks · −1

Three block A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure. The tension T1T_1 and T2T_2 in the string are respectively:

Figure for question 43
  1. A.60 N, 80 N
  2. B.40 N, 64 N
  3. C.80 N, 100 N
  4. D.88 N, 96 N

Correct answer: B

Explanation: a=8a=8 m/s²; T1=5a=40T_1=5a=40 N, T2=8a=64T_2=8a=64 N.

Question 44

Physics · 4 marks · −1

An electron revolving in nthn^{th} Bohr orbit has magnetic moment μn\mu_n. If μn∝nx\mu_n\propto n^x, the value of xx is:

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

Correct answer: C

Explanation: μ=nhe4πm∝n\mu=\frac{nh e}{4\pi m}\propto n.

Question 45

Physics · 4 marks · −1

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is: [g=10g=10 m/s²]

Figure for question 45
  1. A.10 J
  2. B.5×1035\times10^3 J
  3. C.5 J
  4. D.535\sqrt3 J

Correct answer: C

Explanation: μmgcos⁡60∘×10=0.1×10×0.5×10=5\mu mg\cos60^\circ\times10=0.1\times10\times0.5\times10=5 J.

Question 46

Physics · 4 marks · −1

When a potential difference V is applied across a wire of resistance R, it dissipate energy at a rate W. If the wire is cut into two halve and these halve are connected mutually parallel across the same supply, the energy dissipation rate will become:

  1. A.4W
  2. B.14W\frac14W
  3. C.12W\frac12W
  4. D.2W

Correct answer: A

Question 47

Physics · 4 marks · −1

Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:

  1. A.4.2 km/s
  2. B.8.4 km/s
  3. C.11.2 km/s
  4. D.7.9 km/s

Correct answer: D

Explanation: v∝M/R=12v\propto\sqrt{M/R}=\frac1{\sqrt2}.

Question 48

Physics · 4 marks · −1

An alternating voltage V(t)=220sin⁡100πtV(t)=220\sin100\pi t volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:

  1. A.5 ms
  2. B.3.3 ms
  3. C.7.2 ms
  4. D.2.2 ms

Correct answer: B

Explanation: Phase from π/6\pi/6 to π/2\pi/2: t=π/3100π=3.3t=\frac{\pi/3}{100\pi}=3.3 ms.

Question 49

Physics · 4 marks · −1

Match List I with List II. List-I: A. Gauss's law of magnetostatics; B. Faraday's law of electro magnetic induction; C. Ampere's Law; D. Gauss's law of electrostatics List-II: I. ∮E⃗⋅da⃗=1ϵ0∫ρdV\oint\vec E\cdot d\vec a=\frac1{\epsilon_0}\int\rho dV; II. ∮B⃗⋅da⃗=0\oint\vec B\cdot d\vec a=0; III. ∮E⃗⋅dl⃗=−ddt∫B⃗⋅da⃗\oint\vec E\cdot d\vec l=\frac{-d}{dt}\int\vec B\cdot d\vec a; IV. ∮B⃗⋅dl⃗=μ0I\oint\vec B\cdot d\vec l=\mu_0I Choose the correct answer from the options given below:

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

Correct answer: A

Question 50

Physics · 4 marks · −1

In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure. The focal length of the lens is ______ cm.

Figure for question 50

Correct answer: 20

Explanation: Newton's formula xy=f2xy=f^2; the symmetric point (20, 20) gives f=20f=20 cm.

Question 51

Physics · 4 marks · −1

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillation will be ______ s. [take g=π2g=\pi^2 m s−2^{-2}]

Correct answer: 8

Explanation: g′=g/4g'=g/4; T=2π4π2/4=8T=2\pi\sqrt{\frac{4}{\pi^2/4}}=8 s.

Question 52

Physics · 4 marks · −1

A big drop is formed by coalescing 1000 small identical drops of water. If E1E_1 be the total surface energy of 1000 small drops of water and E2E_2 be the surface energy of single big drop of water, then E1:E2E_1:E_2 is x:1x:1 where x=x= ______.

Correct answer: 10

Explanation: R=10rR=10r; 1000r2100r2=10\frac{1000r^2}{100r^2}=10.

Question 53

Physics · 4 marks · −1

Two resistance of 100 Ω and 200 Ω are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100 Ω resistance, which give reading as 1 V. The resistance of voltmeter must be ______ Ω.

Correct answer: 200

Explanation: 200 Ω carries 3 V → 15 mA; 100 Ω carries 10 mA, so the voltmeter draws 5 mA at 1 V: 200 Ω.

Question 54

Physics · 4 marks · −1

A vector has magnitude same as that of A⃗=3i^+4j^\vec A=3\hat i+4\hat j and is parallel to B⃗=4i^+3j^\vec B=4\hat i+3\hat j. The xx and yy components of this vector in first quadrant are xx and 3 respectively were x=x= ______.

Correct answer: 4

Explanation: 5⋅(4,3)5=(4,3)5\cdot\frac{(4,3)}{5}=(4,3).

Question 55

Physics · 4 marks · −1

A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5A and it efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is ______ A.

Correct answer: 45

Explanation: 0.9×2300×5=230I⇒I=450.9\times2300\times5=230I\Rightarrow I=45 A.

Question 56

Physics · 4 marks · −1

A point source is emitting sound waves of intensity 16×10−816\times10^{-8} W m−2^{-2} at the origin. The difference in intensity (magnitude only) at two points located at a distance of 2 m and 4 m from the origin respectively will be ______ ×10−8\times10^{-8} W m−2^{-2}.

Correct answer: 3

Explanation: Taking the given intensity at 1 m: 164−1616=3\frac{16}{4}-\frac{16}{16}=3.

Question 57

Physics · 4 marks · −1

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 37° with each other. When suspended in a liquid of density 0.7 g/cm³, the angle remains same. If density of material of the sphere is 1.4 g/cm³, the dielectric constant of the liquid is ______ (tan⁡37∘=34)\left(\tan37^\circ=\frac34\right).

Correct answer: 2

Explanation: K=ρρ−σ=1.40.7=2K=\frac{\rho}{\rho-\sigma}=\frac{1.4}{0.7}=2.

Question 58

Physics · 4 marks · −1

The current of 5A flows in a square loop of side 1 m is placed in air. The magnetic field at the centre of the loop is X2×10−7X\sqrt2\times10^{-7} T. The value of X is ______.

Correct answer: 40

Explanation: B=22μ0Iπa=22×4π×10−7×5π=402×10−7B=\frac{2\sqrt2\mu_0I}{\pi a}=\frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi}=40\sqrt2\times10^{-7}.

Question 59

Physics · 4 marks · −1

Two discs of moment of inertia I1=4I_1=4 kg m² and I2=2I_2=2 kg m², about their central axes & normal to their planes, rotating with angular speeds 10 rad/s & 4 rad/s respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is ______ J.

Correct answer: 24

Explanation: ΔK=I1I22(I1+I2)(ω1−ω2)2=812×36=24\Delta K=\frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2=\frac{8}{12}\times36=24 J.

Question 60

Chemistry · 4 marks · −1

Salicylaldehyde is synthesized from phenol, when reacted with.

  1. A.HCCl3\text{HCCl}_3, NaOH
  2. B.Formyl chloride (HCOCl), NaOH
  3. C.CCl4\text{CCl}_4, NaOH
  4. D.CO2\text{CO}_2, NaOH

Correct answer: A

Question 61

Chemistry · 4 marks · −1

Given below are two statements: Statement – I: High concentration of strong nucleophilic reagent with secondary alkyl halide which do not have bulky substituents will follow SN2S_N2 mechanism. Statement – II: A secondary alkyl halide when treated with a large excess of ethanol follows SN1S_N1 mechanism. In the light of the above statements, choose the most appropriate from the options given below:

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

Correct answer: C

Question 62

Chemistry · 4 marks · −1

Reduction potential of ion are given below: ClO4−\text{ClO}_4^- E∘=1.19E^\circ=1.19 V; IO4−\text{IO}_4^- E∘=1.65E^\circ=1.65 V; BrO4−\text{BrO}_4^- E∘=1.74E^\circ=1.74 V. The correct order of their oxidizing power is:

  1. A.BrO4−>ClO4−>IO4−\text{BrO}_4^->\text{ClO}_4^->\text{IO}_4^-
  2. B.BrO4−>IO4−>ClO4−\text{BrO}_4^->\text{IO}_4^->\text{ClO}_4^-
  3. C.ClO4−>IO4−>BrO4−\text{ClO}_4^->\text{IO}_4^->\text{BrO}_4^-
  4. D.IO4−>BrO4−>ClO4−\text{IO}_4^->\text{BrO}_4^->\text{ClO}_4^-

Correct answer: B

Question 63

Chemistry · 4 marks · −1

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: H2Te\text{H}_2\text{Te} is more acidic than H2S\text{H}_2\text{S}. Reason R: Bond dissociation enthalpy of H2Te\text{H}_2\text{Te} is lower than H2S\text{H}_2\text{S}. In the light of the above statements, choose the most appropriate from the options given below:

  1. A.A is true but R is false
  2. B.Both A and R are true but R is NOT the correct explanation of A.
  3. C.A is false but R is true.
  4. D.Both A and R are true and R is the correct explanation of A.

Correct answer: D

Question 64

Chemistry · 4 marks · −1

The solution from the following with highest depression in freezing point/lowest freezing point is:

  1. A.180 g of acetic acid dissolved in water
  2. B.180 g of glucose dissolved in water
  3. C.180 g of acetic acid dissolved in benzene
  4. D.180 g of benzoic acid dissolved in benzene

Correct answer: A

Explanation: 3 mol of acetic acid (no dimerisation in water, slight ionisation) gives the largest number of solute particles.

Question 65

Chemistry · 4 marks · −1

If a substance 'A' dissolves in solution of a mixture of 'B' and 'C' with their respective number of moles as nAn_A, nBn_B and nCn_C, Mole fraction of C in the solution is:

  1. A.nCnA×nB×nC\frac{n_C}{n_A\times n_B\times n_C}
  2. B.nCnA−nB−nC\frac{n_C}{n_A-n_B-n_C}
  3. C.nBnA+nB\frac{n_B}{n_A+n_B}
  4. D.nCnA+nB+nC\frac{n_C}{n_A+n_B+n_C}

Correct answer: D

Question 66

Chemistry · 4 marks · −1

Given below are two statements: Statement-I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18. Statement-II: The nature of oxides formed by group 1 elements is basic while that of group 17 elements is acidic. In the light of the above statements, choose the most appropriate 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: B

Question 67

Chemistry · 4 marks · −1

Which among the following purification methods is based on the principle of "Solubility" in two different solvents?

  1. A.Distillation
  2. B.Differential Extraction
  3. C.Column Chromatography
  4. D.Sublimation

Correct answer: B

Question 68

Chemistry · 4 marks · −1

Given below are two statement: Statement-I: Since Fluorine is more electronegative than nitrogen, the net dipole moment of NF3\text{NF}_3 is greater than NH3\text{NH}_3. Statement-II: In NH3\text{NH}_3, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite direction, but in NF3\text{NF}_3 the orbital dipole due to lone pair and dipole moment of N – F bonds are in same direction. In the light of the above statements, choose the most appropriate 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: C

Explanation: NH3_3 (1.47 D) > NF3_3 (0.23 D); the directions in Statement II are reversed.

Question 69

Chemistry · 4 marks · −1

The coordination geometry around the manganese in decacarbonyldimanganese (0) is:

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

Correct answer: B

Question 70

Chemistry · 4 marks · −1

Alkaline oxidative fusion of MnO2\text{MnO}_2 gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are

  1. A.Mn2O7\text{Mn}_2\text{O}_7 and MnO4−\text{MnO}_4^-
  2. B.MnO42−\text{MnO}_4^{2-} and Mn2O7\text{Mn}_2\text{O}_7
  3. C.MnO42−\text{MnO}_4^{2-} and MnO4−\text{MnO}_4^-
  4. D.Mn2O3\text{Mn}_2\text{O}_3 and MnO42−\text{MnO}_4^{2-}

Correct answer: C

Question 71

Chemistry · 4 marks · −1

The correct stability order of carbocations is:

  1. A.(CH3)3C+>(CH3)2C+H>CH3C+H2>C+H3(\text{CH}_3)_3\overset{+}{\text{C}}>(\text{CH}_3)_2\overset{+}{\text{C}}\text{H}>\text{CH}_3\overset{+}{\text{C}}\text{H}_2>\overset{+}{\text{C}}\text{H}_3
  2. B.(CH3)3C+>CH3C+H2>(CH3)2C+H>C+H3(\text{CH}_3)_3\overset{+}{\text{C}}>\text{CH}_3\overset{+}{\text{C}}\text{H}_2>(\text{CH}_3)_2\overset{+}{\text{C}}\text{H}>\overset{+}{\text{C}}\text{H}_3
  3. C.C+H3>CH3C+H2>(CH3)2C+H>(CH3)3C+\overset{+}{\text{C}}\text{H}_3>\text{CH}_3\overset{+}{\text{C}}\text{H}_2>(\text{CH}_3)_2\overset{+}{\text{C}}\text{H}>(\text{CH}_3)_3\text{C}^+
  4. D.C+H3>(CH3)2C+H>CH3C+H2>(CH3)3C+\overset{+}{\text{C}}\text{H}_3>(\text{CH}_3)_2\overset{+}{\text{C}}\text{H}>\text{CH}_3\overset{+}{\text{C}}\text{H}_2>(\text{CH}_3)_3\overset{+}{\text{C}}

Correct answer: A

Question 72

Chemistry · 4 marks · −1

The products A and B formed in the following reaction scheme are respectively Benzene →(i) conc. HNO3/conc. H2SO4, 323−333 K; (ii) Sn/HCl\xrightarrow{\text{(i) conc. HNO}_3/\text{conc. H}_2\text{SO}_4,\ 323-333\text{ K; (ii) Sn/HCl}} A →(i) NaNO2, HCl, 273–278 K; (ii) Phenol\xrightarrow{\text{(i) NaNO}_2\text{, HCl, 273–278 K; (ii) Phenol}} B

Figure for question 72
  1. A.1-Chloro-3-nitrobenzene, 3'-nitrobiphenyl-4-ol
  2. B.Aniline, 2-amino-4'-hydroxybiphenyl
  3. C.Aniline, 4-hydroxyazobenzene (HO-C6H4-N=N-C6H5\text{HO-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5, para)
  4. D.Aniline, 2-hydroxyazobenzene (ortho)

Correct answer: C

Explanation: Nitration, reduction to aniline, diazotisation and para coupling with phenol.

Question 73

Chemistry · 4 marks · −1

A and B formed in the following reactions are: CrO2Cl2+4NaOH→A+2NaCl+2H2O\text{CrO}_2\text{Cl}_2+4\text{NaOH}\to A+2\text{NaCl}+2\text{H}_2\text{O} A+2HCl+2H2O2→B+3H2OA+2\text{HCl}+2\text{H}_2\text{O}_2\to B+3\text{H}_2\text{O}

  1. A.A = Na2Cr2O7\text{Na}_2\text{Cr}_2\text{O}_7, B = CrO3\text{CrO}_3
  2. B.A = Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4, B = CrO4\text{CrO}_4
  3. C.A = Na2Cr2O7\text{Na}_2\text{Cr}_2\text{O}_7, B = CrO5\text{CrO}_5
  4. D.A = Na2CrO4\text{Na}_2\text{CrO}_4, B = CrO5\text{CrO}_5

Correct answer: D

Question 74

Chemistry · 4 marks · −1

Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order NH3>PH3>AsH3>SbH3>BiH3\text{NH}_3>\text{PH}_3>\text{AsH}_3>\text{SbH}_3>\text{BiH}_3 B. The reducing ability of the hydride increases in the order NH3<PH3<AsH3<SbH3<BiH3\text{NH}_3<\text{PH}_3<\text{AsH}_3<\text{SbH}_3<\text{BiH}_3 C. Among the hydrides, NH3\text{NH}_3 is strong reducing agent while BiH3\text{BiH}_3 is mild reducing agent. D. The basicity of the hydrides increases in the order NH3<PH3<AsH3<SbH3<BiH3\text{NH}_3<\text{PH}_3<\text{AsH}_3<\text{SbH}_3<\text{BiH}_3.

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

Correct answer: D

Question 75

Chemistry · 4 marks · −1

The molecule / ion with square pyramidal shape is:

  1. A.PF5\text{PF}_5
  2. B.[Ni(CN)4]2−[\text{Ni(CN)}_4]^{2-}
  3. C.BrF5\text{BrF}_5
  4. D.PCl5\text{PCl}_5

Correct answer: C

Question 76

Chemistry · 4 marks · −1

IUPAC name of following compound is: CH3-CH(NH2)-CH2-CN\text{CH}_3\text{-CH(NH}_2\text{)-CH}_2\text{-CN}

  1. A.2-Aminobutanenitrile
  2. B.2-Aminopentanenitrile
  3. C.3-Aminopropanenitrile
  4. D.3-Aminobutanenitrile

Correct answer: D

Question 77

Chemistry · 4 marks · −1

m-chlorobenzaldehyde on treatment with 50% KOH solution yields

Figure for question 77
  1. A.2-hydroxy-1,2-bis(3-chlorophenyl)ethanone (benzoin-type product)
  2. B.3-hydroxybenzaldehyde (two molecules)
  3. C.3-hydroxybenzoate + 3-hydroxybenzyl alcohol
  4. D.3-chlorobenzoate + 3-chlorobenzyl alcohol

Correct answer: D

Explanation: Cannizzaro disproportionation of a non-enolisable aldehyde.

Question 78

Chemistry · 4 marks · −1

Products A and B formed in the following set of reactions are 1-Methylcyclohexene: →H+/H2O\xrightarrow{\text{H}^+/\text{H}_2\text{O}} A; →B2H6; H2O2, NaOH (aq)\xrightarrow{\text{B}_2\text{H}_6;\ \text{H}_2\text{O}_2\text{, NaOH (aq)}} B

Figure for question 78
  1. A.A = 1-methylcyclohexan-1-ol; B = 2-methylcyclohexan-1-ol
  2. B.A = cyclohexylmethanol; B = 2-(hydroxymethyl)cyclohexan-1-ol
  3. C.A = 2-methylcyclohexan-1-ol; B = 1-methylcyclohexan-1-ol
  4. D.A = cyclohexylmethanol; B = 2-methylcyclohexan-1-ol

Correct answer: A

Explanation: Acid-catalysed hydration is Markovnikov; hydroboration–oxidation is anti-Markovnikov.

Question 79

Chemistry · 4 marks · −1

The orange colour of K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 and purple colour of KMnO4\text{KMnO}_4 is due to:

  1. A.d → d transitions in KMnO4\text{KMnO}_4 and charge transfer transitions in K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7.
  2. B.d → d transitions in K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 and charge transfer transitions in KMnO4\text{KMnO}_4.
  3. C.Charge transfer transition in both.
  4. D.d → d transitions in both.

Correct answer: C

Explanation: Both are d0^0: colour from ligand-to-metal charge transfer.

Question 80

Chemistry · 4 marks · −1

Number of metal ions characterized by flame test among the following is ______. Sr2+\text{Sr}^{2+}, Ba2+\text{Ba}^{2+}, Ca2+\text{Ca}^{2+}, Cu2+\text{Cu}^{2+}, Zn2+\text{Zn}^{2+}, Co2+\text{Co}^{2+}, Fe2+\text{Fe}^{2+}

Correct answer: 4

Explanation: Sr, Ba, Ca, Cu.

Question 81

Chemistry · 4 marks · −1

The total number of correct statements, regarding the nucleic acids is ______. A. RNA is regarded as the reserve of genetic information. B. DNA molecule self-duplicates during cell division. C. DNA synthesizes proteins in the cell D. The message for the syntheses of particular proteins is present in DNA E. Identical DNA strands are transferred to daughter cells.

Correct answer: 3

Explanation: B, D, E are correct.

Question 82

Chemistry · 4 marks · −1

2 – chlorobutane + Cl2\text{Cl}_2 → C4H8Cl2\text{C}_4\text{H}_8\text{Cl}_2 (isomers) Total number of optically active isomers shown by C4H8Cl2\text{C}_4\text{H}_8\text{Cl}_2, obtained in the above reaction is ______.

Correct answer: 3

Question 83

Chemistry · 4 marks · −1

NO2\text{NO}_2 required for a reaction is produced by decomposition of N2O5\text{N}_2\text{O}_5 in CCl4\text{CCl}_4 as by equation 2N2O5(g)→4NO2(g)+O2(g)2\text{N}_2\text{O}_{5(g)}\to4\text{NO}_{2(g)}+\text{O}_{2(g)}. The initial concentration of N2O5\text{N}_2\text{O}_5 is 3 mol L−1^{-1} and it is 2.75 mol L−1^{-1} after 30 minutes. The rate of formation of NO2\text{NO}_2 is x×10−3x\times10^{-3} mol L−1^{-1} min−1^{-1}, value of xx is ______. (nearest integer).

Correct answer: 17

Explanation: −Δ[N2O5]Δt=0.2530-\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}=\frac{0.25}{30}; NO2_2 forms twice as fast: 16.7×10−316.7\times10^{-3}.

Question 84

Chemistry · 4 marks · −1

Number of geometrical isomers possible for the given structure is/are ______.

Figure for question 84

Correct answer: 4

Question 85

Chemistry · 4 marks · −1

Total number of species from the following which can undergo disproportionation reaction is ______. H2O2\text{H}_2\text{O}_2, ClO3−\text{ClO}_3^-, P4\text{P}_4, Cl2\text{Cl}_2, Ag, Cu+1\text{Cu}^{+1}, F2\text{F}_2, NO2\text{NO}_2, K+\text{K}^+

Correct answer: 6

Explanation: H2_2O2_2, ClO3−_3^-, P4_4, Cl2_2, Cu+^+, NO2_2.

Question 86

Chemistry · 4 marks · −1

Number of complexes which show optical isomerism among the following is ______. cis-[Cr(ox)2Cl2]3−[\text{Cr(ox)}_2\text{Cl}_2]^{3-}, [Co(en)3]3+[\text{Co(en)}_3]^{3+}, cis-[Pt(en)2Cl2]2+[\text{Pt(en)}_2\text{Cl}_2]^{2+}, cis-[Co(en)2Cl2]+[\text{Co(en)}_2\text{Cl}_2]^+, trans-[Pt(en)2Cl2]2+[\text{Pt(en)}_2\text{Cl}_2]^{2+}, trans-[Cr(ox)2Cl2]3−[\text{Cr(ox)}_2\text{Cl}_2]^{3-}

Correct answer: 4

Explanation: All four cis/tris-chelate complexes; the trans ones have a mirror plane.

Question 87

Chemistry · 4 marks · −1

The pH of an aqueous solution containing 1M benzoic acid (pKa=4.20pK_a=4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL. (given : log2 = 0.3)

Correct answer: 100

Explanation: [salt][acid]=100.3=2\frac{[\text{salt}]}{[\text{acid}]}=10^{0.3}=2: acid 100 mL, salt 200 mL.

Question 88

Chemistry · 4 marks · −1

Number of spectral lines obtained in He+\text{He}^+ spectra, when an electron makes transition from fifth excited state to first excited state will be:

Correct answer: 10

Explanation: n=6→n=2n=6\to n=2: (6−2)(6−2+1)2=10\frac{(6-2)(6-2+1)}{2}=10.

Question 89

Chemistry · 4 marks · −1

Two reactions are given below: 2Fe(s)+32O2(g)→Fe2O3(s)2\text{Fe(s)}+\frac32\text{O}_2\text{(g)}\to\text{Fe}_2\text{O}_3\text{(s)}, ΔH∘=−822\Delta H^\circ=-822 kJ/mol C(s)+12O2(g)→CO(g)\text{C(s)}+\frac12\text{O}_2\text{(g)}\to\text{CO(g)}, ΔH∘=−110\Delta H^\circ=-110 kJ/mol Then enthalpy change for following reaction 3C(s)+Fe2O3(s)→2Fe(s)+3CO(g)3\text{C(s)}+\text{Fe}_2\text{O}_3\text{(s)}\to2\text{Fe(s)}+3\text{CO(g)} is ______ kJ/mol.

Correct answer: 492

Explanation: 3(−110)+822=4923(-110)+822=492.

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