JEE Main 2024 — 27 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

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is:

  1. A.3715\frac{3}{715}
  2. B.5256\frac{5}{256}
  3. C.5715\frac{5}{715}
  4. D.3256\frac{3}{256}

Correct answer: A

Explanation: 6C415C4⋅9C411C4=151365⋅126330=3715\frac{^6C_4}{^{15}C_4}\cdot\frac{^9C_4}{^{11}C_4}=\frac{15}{1365}\cdot\frac{126}{330}=\frac{3}{715}.

Question 2

Mathematics · 4 marks · −1

If y=y(x)y=y(x) is the solution curve of the differential equation (x2−4) dy−(y2−3y) dx=0(x^2-4)\,dy-(y^2-3y)\,dx=0, x>2x>2, y(4)=32y(4)=\frac32 and the slope of the curve is never zero, then the value of y(10)y(10) equals:

  1. A.31+(8)1/4\frac{3}{1+(8)^{1/4}}
  2. B.31−22\frac{3}{1-2\sqrt2}
  3. C.31+22\frac{3}{1+2\sqrt2}
  4. D.31−(8)1/4\frac{3}{1-(8)^{1/4}}

Correct answer: A

Explanation: Separating: 13ln⁡∣y−3y∣=14ln⁡x−2x+2+c\frac13\ln\left|\frac{y-3}{y}\right|=\frac14\ln\frac{x-2}{x+2}+c; y(4)=32y(4)=\frac32 gives c=14ln⁡3c=\frac14\ln3. At x=10x=10: ∣y−3y∣=23/4=81/4\left|\frac{y-3}{y}\right|=2^{3/4}=8^{1/4} and, since 0<y<30<y<3, y=31+81/4y=\frac{3}{1+8^{1/4}}.

Question 3

Mathematics · 4 marks · −1

The values of α\alpha, for which ∣132α+32113α+132α+33α+10∣=0\begin{vmatrix}1&\frac32&\alpha+\frac32\\1&\frac13&\alpha+\frac13\\2\alpha+3&3\alpha+1&0\end{vmatrix}=0, lie in the interval.

  1. A.(−3,0)(-3,0)
  2. B.(−2,1)(-2,1)
  3. C.(0,3)(0,3)
  4. D.(−32,32)\left(-\frac32,\frac32\right)

Correct answer: A

Explanation: The determinant equals 76(2α2+6α+1)\frac76(2\alpha^2+6\alpha+1); roots −3±72\frac{-3\pm\sqrt7}{2}, both in (−3,0)(-3,0).

Question 4

Mathematics · 4 marks · −1

The position vectors of the vertices AA, BB and CC of a triangle are 2i^−3j^+3k^2\hat i-3\hat j+3\hat k, 2i^+2j^+3k^2\hat i+2\hat j+3\hat k and −i^+j^+3k^-\hat i+\hat j+3\hat k respectively. Let ll denotes the length of the angle bisector ADAD of ∠BAC\angle BAC where DD is on the line segment BCBC, then 2l22l^2 equals:

  1. A.50
  2. B.49
  3. C.42
  4. D.45

Correct answer: D

Explanation: AB=AC=5AB=AC=5, so DD is the midpoint (0.5,1.5,3)(0.5,1.5,3); l2=2.25+20.25=22.5l^2=2.25+20.25=22.5, 2l2=452l^2=45.

Question 5

Mathematics · 4 marks · −1

Let g(x)=3f(x3)+f(3−x)g(x)=3f\left(\frac x3\right)+f(3-x) and f′′(x)>0f''(x)>0 for all x∈(0,3)x\in(0,3). If gg is decreasing in (0,α)(0,\alpha) and increasing in (α,3)(\alpha,3), then 8α8\alpha is:

  1. A.18
  2. B.24
  3. C.0
  4. D.20

Correct answer: A

Explanation: g′(x)=f′(x/3)−f′(3−x)g'(x)=f'(x/3)-f'(3-x), zero when x/3=3−xx/3=3-x, x=9/4x=9/4; 8α=188\alpha=18.

Question 6

Mathematics · 4 marks · −1

Let α=(4!)!(4!)3!\alpha=\frac{(4!)!}{(4!)^{3!}} and β=(5!)!(5!)4!\beta=\frac{(5!)!}{(5!)^{4!}}. Then:

  1. A.α∈N\alpha\in N and β∉N\beta\notin N
  2. B.α∉N\alpha\notin N and β∈N\beta\in N
  3. C.α∈N\alpha\in N and β∈N\beta\in N
  4. D.α∉N\alpha\notin N and β∉N\beta\notin N

Correct answer: C

Explanation: (mn)!(n!)m\frac{(mn)!}{(n!)^m} is the number of ways to split mnmn objects into mm labelled groups of nn; both are integers.

Question 7

Mathematics · 4 marks · −1

Consider the function f:(0,2)→Rf:(0,2)\to\mathbb{R} defined by f(x)=x2+2xf(x)=\frac x2+\frac2x and the function g(x)g(x) defined by g(x)={min⁡{f(t)},0<t≤x and 0<x≤132+x,1<x<2g(x)=\begin{cases}\min\{f(t)\}, & 0<t\le x \text{ and } 0<x\le1\\ \frac32+x, & 1<x<2\end{cases}. Then,

  1. A.gg is continuous but not differentiable at x=1x=1
  2. B.gg is not continuous for all x∈(0,2)x\in(0,2)
  3. C.gg is continuous and differentiable for all x∈(0,2)x\in(0,2)
  4. D.gg is neither continuous nor differentiable at x=1x=1

Correct answer: A

Explanation: ff decreases on (0,2)(0,2), so g=f(x)g=f(x) on (0,1](0,1]; g(1±)=2.5g(1^\pm)=2.5 but g′(1−)=−1.5≠g′(1+)=1g'(1^-)=-1.5\ne g'(1^+)=1.

Question 8

Mathematics · 4 marks · −1

Let the position vectors of vertices AA, BB and CC of a triangle be 2i^+2j^+k^2\hat i+2\hat j+\hat k, i^+2j^+2k^\hat i+2\hat j+2\hat k and 2i^+j^+2k^2\hat i+\hat j+2\hat k respectively. Let l1,l2l_1,l_2 and l3l_3 be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides ABAB, BCBC and CACA respectively, then l12+l22+l32l_1^2+l_2^2+l_3^2 equals:

  1. A.13\frac13
  2. B.15\frac15
  3. C.14\frac14
  4. D.12\frac12

Correct answer: D

Explanation: Equilateral triangle of side 2\sqrt2; each distance is the inradius 223\frac{\sqrt2}{2\sqrt3}, so the sum is 3⋅16=123\cdot\frac16=\frac12.

Question 9

Mathematics · 4 marks · −1

Let the image of the point (1,0,7)(1,0,7) in the line x1=y−12=z−23\frac x1=\frac{y-1}{2}=\frac{z-2}{3} be the point (α,β,γ)(\alpha,\beta,\gamma). Then which one of the following points lies on the line passing through (α,β,γ)(\alpha,\beta,\gamma) and making angles 2π3\frac{2\pi}{3} and 3π4\frac{3\pi}{4} with yy-axis and zz-axis respectively and an acute angle with xx-axis ?

  1. A.(1,2,1−2)(1,2,1-\sqrt2)
  2. B.(3,−4,3+22)(3,-4,3+2\sqrt2)
  3. C.(3,4,3−22)(3,4,3-2\sqrt2)
  4. D.(1,−2,1+2)(1,-2,1+\sqrt2)

Correct answer: C

Explanation: Foot (1,3,5)(1,3,5), image (1,6,3)(1,6,3); direction (1,−1,−2)(1,-1,-\sqrt2); t=2t=2 gives (3,4,3−22)(3,4,3-2\sqrt2).

Question 10

Mathematics · 4 marks · −1

Let AA and BB be two finite sets with mm and nn elements respectively. The total number of subsets of the set AA is 56 more than the total number of subsets of BB. Then the distance of the point P(m,n)P(m,n) from the point Q(−2,−3)Q(-2,-3) is:

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

Correct answer: C

Explanation: 2m−2n=56⇒m=6,n=32^m-2^n=56\Rightarrow m=6,n=3; distance 64+36=10\sqrt{64+36}=10.

Question 11

Mathematics · 4 marks · −1

Let f:R−{−12}→Rf:\mathbb{R}-\left\{\frac{-1}{2}\right\}\to\mathbb{R} and g:R−{−52}→Rg:\mathbb{R}-\left\{\frac{-5}{2}\right\}\to\mathbb{R} be defined as f(x)=2x+32x+1f(x)=\frac{2x+3}{2x+1} and g(x)=∣x∣+12x+5g(x)=\frac{|x|+1}{2x+5}. Then, the domain of the function f∘gf\circ g is:

  1. A.R−{−74}\mathbb{R}-\left\{-\frac74\right\}
  2. B.R−{−52}\mathbb{R}-\left\{-\frac52\right\}
  3. C.R−{−52,−74}\mathbb{R}-\left\{-\frac52,-\frac74\right\}
  4. D.R\mathbb{R}

Correct answer: B

Explanation: g(x)=−12g(x)=-\frac12 has no solution, so only x=−52x=-\frac52 is excluded.

Question 12

Mathematics · 4 marks · −1

Let RR be the interior region between the lines 3x−y+1=03x-y+1=0 and x+2y−5=0x+2y-5=0 containing the origin. The set of all values of aa, for which the points (a2,a+1)(a^2,a+1) lie in RR, is:

  1. A.(−3,−1)∪(13,1)(-3,-1)\cup\left(\frac13,1\right)
  2. B.(−3,−1)∪(−13,1)(-3,-1)\cup\left(-\frac13,1\right)
  3. C.(−3,0)∪(23,1)(-3,0)\cup\left(\frac23,1\right)
  4. D.(−3,0)∪(13,1)(-3,0)\cup\left(\frac13,1\right)

Correct answer: D

Explanation: 3a2−a>03a^2-a>0 and a2+2a−3<0a^2+2a-3<0 give (−3,0)∪(13,1)(-3,0)\cup(\frac13,1).

Question 13

Mathematics · 4 marks · −1

The integral ∫(x8−x2) dx(x12+3x6+1)tan⁡−1(x3+1x3)\int\frac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\frac{1}{x^3}\right)} is equal to:

  1. A.log⁡e(∣tan⁡−1(x3+1x3)∣)+C\log_e\left(\left|\tan^{-1}\left(x^3+\frac1{x^3}\right)\right|\right)+C
  2. B.log⁡e(∣tan⁡−1(x3+1x3)∣)13+C\log_e\left(\left|\tan^{-1}\left(x^3+\frac1{x^3}\right)\right|\right)^{\frac13}+C
  3. C.log⁡e(∣tan⁡−1(x3+1x3)∣)12+C\log_e\left(\left|\tan^{-1}\left(x^3+\frac1{x^3}\right)\right|\right)^{\frac12}+C
  4. D.log⁡e(∣tan⁡−1(x3+1x3)∣)3+C\log_e\left(\left|\tan^{-1}\left(x^3+\frac1{x^3}\right)\right|\right)^{3}+C

Correct answer: B

Explanation: With t=x3+x−3t=x^3+x^{-3}, dt=3(x2−x−4)dxdt=3(x^2-x^{-4})dx and the integrand becomes dt3(1+t2)tan⁡−1t\frac{dt}{3(1+t^2)\tan^{-1}t}.

Question 14

Mathematics · 4 marks · −1

If 2tan⁡2θ−5sec⁡θ=12\tan^2\theta-5\sec\theta=1 has exactly 7 solutions in the interval [0,nπ2]\left[0,\frac{n\pi}{2}\right], for the least value of n∈Nn\in N, then ∑k=1nk2k\sum_{k=1}^{n}\frac{k}{2^k} is equal to:

  1. A.1213(214−15)\frac{1}{2^{13}}\left(2^{14}-15\right)
  2. B.1215(214−14)\frac{1}{2^{15}}\left(2^{14}-14\right)
  3. C.1−152131-\frac{15}{2^{13}}
  4. D.1214(215−15)\frac{1}{2^{14}}\left(2^{15}-15\right)

Correct answer: A

Explanation: sec⁡θ=3\sec\theta=3; the 7th solution lies in (6π,6π+π2)(6\pi,6\pi+\frac\pi2), so n=13n=13 and ∑=2−15213\sum=2-\frac{15}{2^{13}}.

Question 15

Mathematics · 4 marks · −1

If α,β\alpha,\beta are the roots of the equation, x2−x−1=0x^2-x-1=0 and Sn=2023αn+2024βnS_n=2023\alpha^n+2024\beta^n, then:

  1. A.2S11=S12+S102S_{11}=S_{12}+S_{10}
  2. B.S11=S10+S12S_{11}=S_{10}+S_{12}
  3. C.2S12=S11+S102S_{12}=S_{11}+S_{10}
  4. D.S12=S11+S10S_{12}=S_{11}+S_{10}

Correct answer: D

Explanation: α2=α+1\alpha^2=\alpha+1 gives Sn+2=Sn+1+SnS_{n+2}=S_{n+1}+S_n.

Question 16

Mathematics · 4 marks · −1

Let e1e_1 be the eccentricity of the hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 and e2e_2 be the eccentricity of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>ba>b, which passes through the foci of the hyperbola. If e1e2=1e_1e_2=1, then the length of the chord of the ellipse parallel to the xx-axis and passing through (0,2)(0,2) is:

  1. A.454\sqrt5
  2. B.1053\frac{10\sqrt5}{3}
  3. C.853\frac{8\sqrt5}{3}
  4. D.353\sqrt5

Correct answer: B

Explanation: e1=54e_1=\frac54, a=5a=5, e2=45e_2=\frac45, b=3b=3; at y=2y=2, x=±553x=\pm\frac{5\sqrt5}{3}.

Question 17

Mathematics · 4 marks · −1

For 0<a<10<a<1, the value of the integral ∫0πdx1−2acos⁡x+a2\int_0^\pi\frac{dx}{1-2a\cos x+a^2} is:

  1. A.π2π+a2\frac{\pi^2}{\pi+a^2}
  2. B.π1+a2\frac{\pi}{1+a^2}
  3. C.π2π−a2\frac{\pi^2}{\pi-a^2}
  4. D.π1−a2\frac{\pi}{1-a^2}

Correct answer: D

Question 18

Mathematics · 4 marks · −1

If lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13\lim_{x\to0}\frac{3+\alpha\sin x+\beta\cos x+\log_e(1-x)}{3\tan^2x}=\frac13, then 2α−β2\alpha-\beta is equal to:

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

Correct answer: D

Explanation: β=−3\beta=-3, α=1\alpha=1 (the xx term must vanish); 2α−β=52\alpha-\beta=5.

Question 19

Mathematics · 4 marks · −1

The 20th20^{th} term from the end of the progression 20,1914,1812,1734,…,−1291420,19\frac14,18\frac12,17\frac34,\ldots,-129\frac14 is:

  1. A.−100
  2. B.−110
  3. C.−118
  4. D.−115

Correct answer: D

Explanation: −129.25+19(0.75)=−115-129.25+19(0.75)=-115.

Question 20

Mathematics · 4 marks · −1

Considering only the principal values of inverse trigonometric functions, the number of positive real values of xx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan^{-1}(x)+\tan^{-1}(2x)=\frac\pi4 is:

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

Correct answer: C

Explanation: 3x1−2x2=1⇒2x2+3x−1=0\frac{3x}{1-2x^2}=1\Rightarrow2x^2+3x-1=0; one positive root.

Question 21

Mathematics · 4 marks · −1

If the area of the region {(x,y):0≤y≤min⁡{2x,6x−x2}}\{(x,y):0\le y\le\min\{2x,6x-x^2\}\} is AA, then 12A12A is equal to ______.

Correct answer: 304

Question 22

Mathematics · 4 marks · −1

If the solution curve, of the differential equation dydx=x+y−2x−y\frac{dy}{dx}=\frac{x+y-2}{x-y} passing through the point (2,1)(2,1) is tan⁡−1(y−1x−1)−1βlog⁡e(α+(y−1x−1)2)=log⁡e∣x−1∣\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac1\beta\log_e\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log_e|x-1|, then 5β+α5\beta+\alpha is equal to:

Correct answer: 11

Question 23

Mathematics · 4 marks · −1

The coefficient of x2012x^{2012} in the expansion of (1−x)2008(1+x+x2)2007(1-x)^{2008}(1+x+x^2)^{2007} is equal to ______.

Correct answer: 0

Explanation: (1−x)2008(1+x+x2)2007=(1−x)(1−x3)2007(1-x)^{2008}(1+x+x^2)^{2007}=(1-x)(1-x^3)^{2007}; 2012 and 2011 are not multiples of 3, so the coefficient is 0.

Question 24

Mathematics · 4 marks · −1

The lines x−22=y−2=z−716\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16} and x+34=y+23=z+21\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1} intersect at the point PP. If the distance of PP from the line x+12=y−13=z−11\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1} is ll, then 14l214l^2 is equal to ______.

Correct answer: 108

Question 25

Mathematics · 4 marks · −1

Let the complex numbers α\alpha and 1αˉ\frac1{\bar\alpha} lie on the circles ∣z−z0∣2=4|z-z_0|^2=4 and ∣z−z0∣2=16|z-z_0|^2=16 respectively, where z0=1+iz_0=1+i. Then, the value of 100∣α∣2100|\alpha|^2 is ______.

Correct answer: 20

Question 26

Mathematics · 4 marks · −1

Let f(x)=∫0xg(t)log⁡e(1−t1+t)dtf(x)=\int_0^x g(t)\log_e\left(\frac{1-t}{1+t}\right)dt, where gg is a continuous odd function. If ∫−π/2π/2(f(x)+x2cos⁡x1+ex)dx=(πα)2−α\int_{-\pi/2}^{\pi/2}\left(f(x)+\frac{x^2\cos x}{1+e^x}\right)dx=\left(\frac\pi\alpha\right)^2-\alpha, then α\alpha is equal to ______.

Correct answer: 2

Explanation: ff is odd so it integrates to 0; ∫x2cos⁡x1+ex=∫0π/2x2cos⁡x dx=π24−2\int\frac{x^2\cos x}{1+e^x}=\int_0^{\pi/2}x^2\cos x\,dx=\frac{\pi^2}{4}-2; α=2\alpha=2.

Question 27

Mathematics · 4 marks · −1

Let AA be a 2×22\times2 real matrix and II be the identity matrix of order 2. If the roots of the equation ∣A−xI∣=0|A-xI|=0 be −1-1 and 3, then the sum of the diagonal elements of the matrix A2A^2 is ______.

Correct answer: 10

Explanation: Eigenvalues of A2A^2 are 1 and 9; trace 10.

Question 28

Mathematics · 4 marks · −1

If the sum of squares of all real values of α\alpha, for which the lines 2x−y+3=02x-y+3=0, 6x+3y+1=06x+3y+1=0 and αx+2y−2=0\alpha x+2y-2=0 do not form a triangle is pp, then the greatest integer less than or equal to pp is ______.

Correct answer: 32

Question 29

Mathematics · 4 marks · −1

Consider a circle (x−α)2+(y−β)2=50(x-\alpha)^2+(y-\beta)^2=50, where α,β>0\alpha,\beta>0. If the circle touches the line y+x=0y+x=0 at the point PP, whose distance from the origin is 424\sqrt2, then (α+β)2(\alpha+\beta)^2 is equal to ______.

Correct answer: 100

Explanation: Distance of centre from x+y=0x+y=0 is 50\sqrt{50}, so α+β=10\alpha+\beta=10.

Question 30

Mathematics · 4 marks · −1

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If μ\mu and σ2\sigma^2 denote the mean and variance of the correct observations respectively, then 15(μ+μ2+σ2)15(\mu+\mu^2+\sigma^2) is equal to ______.

Correct answer: 2521

Explanation: Corrected ∑x=182\sum x=182, ∑x2=2295−100+144=2339\sum x^2=2295-100+144=2339; 15(μ+μ2+σ2)=182+2339=252115(\mu+\mu^2+\sigma^2)=182+2339=2521.

Question 31

Physics · 4 marks · −1

An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity 6×1086\times10^8 W/m² falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space 3×1083\times10^8 m/s):

  1. A.2 N m−2^{-2}
  2. B.6 N m−2^{-2}
  3. C.36 N m−2^{-2}
  4. D.18 N m−2^{-2}

Correct answer: B

Explanation: P=Iv=Inc=6×108×33×108=6P=\frac{I}{v}=\frac{In}{c}=\frac{6\times10^8\times3}{3\times10^8}=6 N m−2^{-2}.

Question 32

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R): Electric lines of forces are always perpendicular to equipotential surfaces. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A.(A) is not correct but (R) is correct
  2. B.(A) is correct but (R) is not correct
  3. C.Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. D.Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Correct answer: C

Question 33

Physics · 4 marks · −1

Primary side of a transformer is connected to 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10 : 1. Load resistance connected to secondary side is 46 Ω. The power consumed in it is:

  1. A.10.0 W
  2. B.11.5 W
  3. C.12.0 W
  4. D.12.5 W

Correct answer: B

Explanation: Vs=23V_s=23 V; P=232/46=11.5P=23^2/46=11.5 W.

Question 34

Physics · 4 marks · −1

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of CpCv\frac{C_p}{C_v} for the gas is:

  1. A.75\frac75
  2. B.32\frac32
  3. C.97\frac97
  4. D.53\frac53

Correct answer: B

Explanation: Adiabatic: P∝Tγ/(γ−1)P\propto T^{\gamma/(\gamma-1)}; γγ−1=3⇒γ=32\frac{\gamma}{\gamma-1}=3\Rightarrow\gamma=\frac32.

Question 35

Physics · 4 marks · −1

The threshold frequency of a metal with work function 6.63 eV is:

  1. A.1.6×10151.6\times10^{15} Hz
  2. B.16×101216\times10^{12} Hz
  3. C.1.6×10121.6\times10^{12} Hz
  4. D.16×101516\times10^{15} Hz

Correct answer: A

Explanation: ν0=6.63×1.6×10−196.63×10−34=1.6×1015\nu_0=\frac{6.63\times1.6\times10^{-19}}{6.63\times10^{-34}}=1.6\times10^{15} Hz.

Question 36

Physics · 4 marks · −1

Given below are two statements: Statement (I): The limiting force of static friction depends on the area of contact and independent of materials. Statement (II): The limiting forces of kinetic friction is independent of the area of contact and depends on materials. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: D

Question 37

Physics · 4 marks · −1

The atomic mass of 6C12{}_6C^{12} is 12.000000 u and that of 6C13{}_6C^{13} is 13.003354 u. The required energy to remove a neutron from 6C13{}_6C^{13}, if mass of neutron is 1.008665 u, will be:

  1. A.4.95 MeV
  2. B.62.5 MeV
  3. C.49.5 MeV
  4. D.6.25 MeV

Correct answer: A

Explanation: Δm=12+1.008665−13.003354=0.005311\Delta m=12+1.008665-13.003354=0.005311 u ≈4.95\approx4.95 MeV.

Question 38

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is Elasticity. Reason (R): The restoring force depends upon the bonded inter atomic and inter molecular force of solid. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: A

Question 39

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun. Reason (R): The moon takes less time to move around the earth than the time taken by the earth to move around the sun. In the light of the above statements, choose the correct answer from the options given below:

  1. A.(A) is not correct but (R) is correct
  2. B.Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. C.(A) is correct but (R) is not correct
  4. D.Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct answer: D

Explanation: ω=2π/T\omega=2\pi/T and the moon's period is shorter.

Question 40

Physics · 4 marks · −1

The truth table of the given circuit diagram is:

Figure for question 40
  1. A.A B Y: 0 0 0; 0 1 0; 1 0 0; 1 1 1
  2. B.A B Y: 0 0 0; 0 1 1; 1 0 1; 1 1 0
  3. C.A B Y: 0 0 1; 0 1 1; 1 0 1; 1 1 0
  4. D.A B Y: 0 0 1; 0 1 0; 1 0 0; 1 1 1

Correct answer: B

Explanation: Y=ABˉ+AˉBY=A\bar B+\bar AB is XOR.

Question 41

Physics · 4 marks · −1

A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the weight experienced by the man is:

  1. A.3 kg
  2. B.636\sqrt3 kg
  3. C.6 kg
  4. D.12 kg

Correct answer: A

Explanation: Taking the shoulder force perpendicular to the rod: N⋅L=WL2cos⁡60∘⇒N=W4=3N\cdot L=W\frac L2\cos60^\circ\Rightarrow N=\frac W4=3 kg-wt.

Question 42

Physics · 4 marks · −1

When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of:

  1. A.45°
  2. B.30°
  3. C.60°
  4. D.90°

Correct answer: A

Explanation: I∝sin⁡22θI\propto\sin^22\theta, maximum at 45°.

Question 43

Physics · 4 marks · −1

A current of 200 μA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π10\frac{\pi}{10} radian is:

  1. A.60 μA
  2. B.120 μA
  3. C.30 μA
  4. D.180 μA

Correct answer: A

Explanation: I∝θI\propto\theta: 200×π/10π/3=60200\times\frac{\pi/10}{\pi/3}=60 μA.

Question 44

Physics · 4 marks · −1

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1V_1, V2V_2 and V3V_3. Choose the correct option.

Figure for question 44
  1. A.V1=V2V_1=V_2
  2. B.V1+V2=V3V_1+V_2=V_3
  3. C.V1+V2>V3V_1+V_2>V_3
  4. D.V1≠V3−V2V_1\ne V_3-V_2

Correct answer: B

Question 45

Physics · 4 marks · −1

The equation of state of a real gas is given by (P+aV2)(V−b)=RT\left(P+\frac{a}{V^2}\right)(V-b)=RT, where P, V and T are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of ab2\frac{a}{b^2} is similar to that of:

  1. A.PVPV
  2. B.RR
  3. C.RTRT
  4. D.PP

Correct answer: D

Explanation: a∼PV2a\sim PV^2, b∼Vb\sim V, so a/b2∼Pa/b^2\sim P.

Question 46

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): In Vernier Calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. Reason (R): The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. In the light of the above statements, choose the correct answer from the options give below:

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

Correct answer: D

Question 47

Physics · 4 marks · −1

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (θ)(\theta) of thread deflection in the extreme position will be:

  1. A.2tan⁡−1(15)2\tan^{-1}\left(\frac1{\sqrt5}\right)
  2. B.2tan⁡−1(12)2\tan^{-1}\left(\frac12\right)
  3. C.tan⁡−1(12)\tan^{-1}\left(\frac12\right)
  4. D.tan⁡−1(2)\tan^{-1}(\sqrt2)

Correct answer: B

Explanation: gsin⁡θ=2g(1−cos⁡θ)⇒tan⁡θ2=12g\sin\theta=2g(1-\cos\theta)\Rightarrow\tan\frac\theta2=\frac12.

Question 48

Physics · 4 marks · −1

The total kinetic energy of 1 mole of oxygen at 27°C is: [Use universal gas constant (R) = 8.31 J/mole K]

  1. A.5670.5 J
  2. B.5942.0 J
  3. C.6845.5 J
  4. D.6232.5 J

Correct answer: D

Explanation: 52RT=2.5×8.31×300=6232.5\frac52RT=2.5\times8.31\times300=6232.5 J.

Question 49

Physics · 4 marks · −1

Wheatstone bridge principle is used to measure the specific resistance (S1)(S_1) of given wire, having length LL, radius rr. If XX is the resistance of wire, then specific resistance is; S1=X(πr2L)S_1=X\left(\frac{\pi r^2}{L}\right). If the length of the wire gets doubled then the value of specific resistance will be:

  1. A.S14\frac{S_1}{4}
  2. B.S1S_1
  3. C.S12\frac{S_1}{2}
  4. D.2S12S_1

Correct answer: B

Explanation: Resistivity is a material property.

Question 50

Physics · 4 marks · −1

A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ______ (degree).

Correct answer: 30

Explanation: sin⁡θ=λ/a=0.5\sin\theta=\lambda/a=0.5, θ=30∘\theta=30^\circ.

Question 51

Physics · 4 marks · −1

Two charges of −4 μC-4\ \mu C and +4 μC+4\ \mu C are placed at the points A(1,0,4)A(1,0,4) m and B(2,−1,5)B(2,-1,5) m located in an electric field E⃗=0.20 i^\vec E=0.20\,\hat i V/cm. The magnitude of the torque acting on the dipole is 8α×10−58\sqrt\alpha\times10^{-5} N m, where α=\alpha= ______.

Correct answer: 2

Explanation: p⃗=4×10−6(1,−1,1)\vec p=4\times10^{-6}(1,-1,1), E⃗=20i^\vec E=20\hat i V/m; ∣p⃗×E⃗∣=82×10−5|\vec p\times\vec E|=8\sqrt2\times10^{-5}.

Question 52

Physics · 4 marks · −1

A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is ______ m/s.

Correct answer: 294

Explanation: v6−v7=7⇒v=294\frac{v}{6}-\frac{v}{7}=7\Rightarrow v=294 m/s.

Question 53

Physics · 4 marks · −1

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1=2πR_1=2\pi m and R2=4πR_2=4\pi m, carrying current I=4I=4 A as per figure given below is α×10−7\alpha\times10^{-7} T. The value of α\alpha is ______. (Centre O is common for all segments)

Figure for question 53

Correct answer: 3

Explanation: B=μ0I4(1R1+1R2)=3×10−7B=\frac{\mu_0I}{4}\left(\frac1{R_1}+\frac1{R_2}\right)=3\times10^{-7} T.

Question 54

Physics · 4 marks · −1

A series LCR circuit with L=100πL=\frac{100}{\pi} mH, C=10−3πC=\frac{10^{-3}}{\pi} F and R=10 ΩR=10\ \Omega, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ______.

Correct answer: 1

Explanation: XL=XC=10 ΩX_L=X_C=10\ \Omega: resonance, power factor 1.

Question 55

Physics · 4 marks · −1

A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is ______ m (use g=10g=10 m s−2^{-2}).

Correct answer: 45

Explanation: 80=2u+20⇒u=3080=2u+20\Rightarrow u=30; h=u2/2g=45h=u^2/2g=45 m.

Question 56

Physics · 4 marks · −1

The reading of pressure metre attached with a closed pipe is 4.5×1044.5\times10^4 N/m². On opening the valve, water starts flowing and the reading of pressure metre falls to 2.0×1042.0\times10^4 N/m². The velocity of water is found to be V\sqrt V m/s. The value of VV is ______.

Correct answer: 50

Explanation: 12ρv2=2.5×104⇒v2=50\frac12\rho v^2=2.5\times10^4\Rightarrow v^2=50.

Question 57

Physics · 4 marks · −1

The electric potential at the surface of an atomic nucleus (z=50)(z=50) of radius 9×10−139\times10^{-13} cm is ______ ×106\times10^6 V.

Correct answer: 8

Explanation: V=kZeR=9×109×50×1.6×10−199×10−15=8×106V=\frac{kZe}{R}=\frac{9\times10^9\times50\times1.6\times10^{-19}}{9\times10^{-15}}=8\times10^6 V.

Question 58

Physics · 4 marks · −1

If Rydberg's constant is RR, the longest wavelength of radiation in Paschen series will be α7R\frac{\alpha}{7R}, where α=\alpha= ______.

Correct answer: 144

Explanation: 1λ=R(19−116)=7R144\frac1\lambda=R\left(\frac19-\frac1{16}\right)=\frac{7R}{144}.

Question 59

Physics · 4 marks · −1

A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7x\frac7x, where xx is ______.

Correct answer: 7

Explanation: Both lose the same potential energy on reaching the bottom (same mass implied), so the ratio is 1.

Question 60

Chemistry · 4 marks · −1

The final product A, formed in the following reaction sequence is: Ph-CH=CH2→(i) BH3 (ii) H2O2, O⊖H (iii) HBr (iv) Mg, ether, then HCHO/H3O+A\text{Ph-CH=CH}_2\xrightarrow{\text{(i) BH}_3\ \text{(ii) H}_2\text{O}_2,\ \overset{\ominus}{\text{O}}\text{H}\ \text{(iii) HBr}\ \text{(iv) Mg, ether, then HCHO/H}_3\text{O}^+}A

  1. A.Ph-CH(CH3)-CH3\text{Ph-CH(CH}_3\text{)-CH}_3
  2. B.Ph-CH2-CH2-CH2-OH\text{Ph-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}
  3. C.Ph-CH(CH3)-CH2OH\text{Ph-CH(CH}_3\text{)-CH}_2\text{OH}
  4. D.Ph-CH2-CH2-CH3\text{Ph-CH}_2\text{-CH}_2\text{-CH}_3

Correct answer: B

Explanation: Anti-Markovnikov alcohol → PhCH2_2CH2_2Br → Grignard + HCHO adds one carbon: PhCH2_2CH2_2CH2_2OH.

Question 61

Chemistry · 4 marks · −1

Which of the following cannot function as an oxidizing agent?

  1. A.SO42−\text{SO}_4^{2-}
  2. B.MnO4−\text{MnO}_4^-
  3. C.N3−\text{N}^{3-}
  4. D.BrO3−\text{BrO}_3^-

Correct answer: C

Explanation: N is in its lowest oxidation state (−3).

Question 62

Chemistry · 4 marks · −1

Which among the following halide/s will not show SN1S_N1 reaction: (A) H2C=CH-CH2Cl\text{H}_2\text{C=CH-CH}_2\text{Cl} (B) CH3-CH=CH-Cl\text{CH}_3\text{-CH=CH-Cl} (C) benzyl chloride (C6H5CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{Cl}) (D) (CH3)2CH-Cl(\text{CH}_3)_2\text{CH-Cl} Choose the most appropriate answer from the options given below:

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

Correct answer: D

Explanation: Only the vinylic halide cannot form a stable carbocation.

Question 63

Chemistry · 4 marks · −1

Identify B formed in the reaction. Cl-(CH2)4-Cl→excess NH3A→NaOHB+H2O+NaCl\text{Cl-(CH}_2)_4\text{-Cl}\xrightarrow{\text{excess NH}_3}A\xrightarrow{\text{NaOH}}B+\text{H}_2\text{O}+\text{NaCl}

  1. A.Cl− N+H3-(CH2)4-N+H3 Cl−\text{Cl}^-\,\overset{+}{\text{N}}\text{H}_3\text{-(CH}_2)_4\text{-}\overset{+}{\text{N}}\text{H}_3\,\text{Cl}^-
  2. B.1,1'-bipiperidine (two piperidine rings joined N–N)
  3. C.H2N-(CH2)4-NH2\text{H}_2\text{N-(CH}_2)_4\text{-NH}_2
  4. D.Piperidine

Correct answer: C

Question 64

Chemistry · 4 marks · −1

The molecular formula of second homologue in the homologous series of mono carboxylic acid is:

  1. A.C2H2O2\text{C}_2\text{H}_2\text{O}_2
  2. B.C3H6O2\text{C}_3\text{H}_6\text{O}_2
  3. C.CH2O\text{CH}_2\text{O}
  4. D.C2H4O2\text{C}_2\text{H}_4\text{O}_2

Correct answer: D

Question 65

Chemistry · 4 marks · −1

Match List-I with List-II. List-I (Reaction): (A) phenol → salicylic acid (o-hydroxybenzoic acid); (B) phenol → salicylaldehyde (o-hydroxybenzaldehyde); (C) phenol → p-benzoquinone; (D) phenol → anisole (methoxybenzene) List-II (Reagent(s)): (I) Na2Cr2O7\text{Na}_2\text{Cr}_2\text{O}_7, H2SO4\text{H}_2\text{SO}_4 (II) (i) NaOH (ii) CH3Cl\text{CH}_3\text{Cl} (III) (i) NaOH, CHCl3\text{CHCl}_3 (ii) NaOH (iii) HCl (IV) (i) NaOH (ii) CO2\text{CO}_2 (iii) HCl

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

Correct answer: B

Explanation: Kolbe (A), Reimer–Tiemann (B), oxidation (C), Williamson (D).

Question 66

Chemistry · 4 marks · −1

The quantity which changes with temperature is:

  1. A.Molarity
  2. B.Molality
  3. C.Mole fraction
  4. D.Mass percentage

Correct answer: A

Question 67

Chemistry · 4 marks · −1

The order of relative stability of the contributing structure is: I: CH2=CH-CH=O\text{CH}_2\text{=CH-CH=O} (acrolein, neutral); II: C⊕H2-CH=CH-O⊖\overset{\oplus}{\text{C}}\text{H}_2\text{-CH=CH-}\overset{\ominus}{\text{O}}; III: C⊖H2-CH=CH-O⊕\overset{\ominus}{\text{C}}\text{H}_2\text{-CH=CH-}\overset{\oplus}{\text{O}} (as drawn in the figure, with H on the carbonyl carbon) Choose the correct answer from the options given below:

Figure for question 67
  1. A.III > II > I
  2. B.I = II = III
  3. C.II > I > III
  4. D.I > II > III

Correct answer: D

Question 68

Chemistry · 4 marks · −1

Identify the incorrect pair from the following:

  1. A.Photography – AgBr
  2. B.Haber process – Iron
  3. C.Polythene preparation – TiCl4\text{TiCl}_4, Al(CH3)3\text{Al(CH}_3)_3
  4. D.Wacker process – PtCl2\text{PtCl}_2

Correct answer: D

Explanation: The Wacker process uses PdCl2_2.

Question 69

Chemistry · 4 marks · −1

The incorrect statement regarding conformations of ethane is:

  1. A.Ethane has infinite number of conformations
  2. B.The conformations of ethane are inter-convertible to one-another
  3. C.The dihedral angle in staggered conformation is 60°.
  4. D.Eclipsed conformation is the most stable conformation.

Correct answer: D

Question 70

Chemistry · 4 marks · −1

The technique used for purification of steam volatile water immiscible substances is:

  1. A.distillation
  2. B.steam distillation
  3. C.fractional distillation
  4. D.fractional distillation under reduced pressure

Correct answer: B

Question 71

Chemistry · 4 marks · −1

Which structure of protein remains intact after coagulation of egg white on boiling?

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

Correct answer: A

Question 72

Chemistry · 4 marks · −1

Bond line formula of HOCH(CN)2\text{HOCH(CN)}_2 is:

Figure for question 72
  1. A.Structure with explicit H: central C bearing H, OH and two CN (drawn with bonds)
  2. B.HO-CH(C≡N)2\text{HO-CH(C}\equiv\text{N)}_2 (condensed, triple bonds shown)
  3. C.Bond-line: a vertex carbon carrying OH above and two CN groups on the lower branches
  4. D.HO-CH(CN)2\text{HO-CH(CN)}_2 (condensed)

Correct answer: C

Question 73

Chemistry · 4 marks · −1

Identify from the following species in which d2sp3d^2sp^3 hybridization is shown by central atom:

  1. A.[Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+}
  2. B.BrF5\text{BrF}_5
  3. C.[Pt(Cl4)]2−[\text{Pt(Cl}_4)]^{2-}
  4. D.SF6\text{SF}_6

Correct answer: A

Explanation: Low-spin Co(III) with NH3_3 uses inner 3d orbitals.

Question 74

Chemistry · 4 marks · −1

Phenolic group can be identified by a positive:

  1. A.Tollen's test
  2. B.Lucas test
  3. C.Phthalein dye test
  4. D.Carbylamine test

Correct answer: C

Question 75

Chemistry · 4 marks · −1

Given below are two statements: Statement (I): In the lanthanoids, the formation of Ce+4\text{Ce}^{+4} is favoured by its noble gas configuration. Statement (II): Ce+4\text{Ce}^{+4} is a strong oxidant reverting to the common +3 state. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: A

Question 76

Chemistry · 4 marks · −1

Which of the following statements is not correct about rusting of iron?

  1. A.When pH lies above 9 or 10, rusting of iron does not take place.
  2. B.Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.
  3. C.Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.
  4. D.Dissolved acidic oxides SO2\text{SO}_2, NO2\text{NO}_2 in water act as catalyst in the process of rusting.

Correct answer: C

Question 77

Chemistry · 4 marks · −1

Given below are two statements: Statement (I): Oxygen being the first member of group 16 exhibits only −2 oxidation state. Statement (II): Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A.Both Statement I and Statement II are correct
  2. B.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: B

Explanation: Oxygen also shows +2, −1 etc.; down the group +4 becomes more stable (inert pair effect).

Question 78

Chemistry · 4 marks · −1

Major product formed in the following reaction is a mixture of: Cyclohexyl tert-butyl ether C6H11-O-C(CH3)3→HI\text{C}_6\text{H}_{11}\text{-O-C(CH}_3)_3\xrightarrow{\text{HI}} Major Product

  1. A.Iodocyclohexane and (CH3)3COH(\text{CH}_3)_3\text{COH}
  2. B.Iodocyclohexane and (CH3)3Cl(\text{CH}_3)_3\text{Cl}
  3. C.Cyclohexanol and (CH3)3COH(\text{CH}_3)_3\text{COH}
  4. D.Cyclohexanol and (CH3)3C-I(\text{CH}_3)_3\text{C-I}

Correct answer: D

Explanation: Cleavage proceeds via the stable tert-butyl cation (S$_N$1): tert-butyl iodide + cyclohexanol.

Question 79

Chemistry · 4 marks · −1

Choose the correct having all the elements with d10d^{10} electronic configuration from the following:

  1. A.27{}^{27}Co, 28{}^{28}Ni, 26{}^{26}Fe, 24{}^{24}Cr
  2. B.29{}^{29}Cu, 30{}^{30}Zn, 48{}^{48}Cd, 47{}^{47}Ag
  3. C.28{}^{28}Ni, 24{}^{24}Cr, 26{}^{26}Fe, 29{}^{29}Cu
  4. D.46{}^{46}Pd, 28{}^{28}Ni, 26{}^{26}Fe, 24{}^{24}Cr

Correct answer: B

Question 80

Chemistry · 4 marks · −1

Time required for completion of 99.9% of a First order reaction is ______ times of half life (t1/2)(t_{1/2}) of the reaction.

Correct answer: 10

Explanation: t=ln⁡1000k≈10 t1/2t=\frac{\ln1000}{k}\approx10\,t_{1/2}.

Question 81

Chemistry · 4 marks · −1

The spin only magnetic moment value of square planar complex [Pt(NH3)2Cl(NH2CH3)]Cl[\text{Pt(NH}_3)_2\text{Cl(NH}_2\text{CH}_3)]\text{Cl} is ______ B.M. (Nearest integer) (Given atomic number for Pt = 78)

Correct answer: 0

Explanation: Pt(II), d8^8 square planar: no unpaired electrons.

Question 82

Chemistry · 4 marks · −1

Total number of compounds with Chiral carbon atoms from following is ______. (i) 2-methylcyclohexanone-type structure (cyclohexanone with a CH3_3 on the carbon next to C=O, as drawn); (ii) cyclopentane-1,3-dione (as drawn); (iii) CH3-CH2-CH(NO2)-COOH\text{CH}_3\text{-CH}_2\text{-CH(NO}_2\text{)-COOH}; (iv) CH3-CH2-CHBr-CH2-CH3\text{CH}_3\text{-CH}_2\text{-CHBr-CH}_2\text{-CH}_3; (v) CH3-CH(I)-CH2-NO2\text{CH}_3\text{-CH(I)-CH}_2\text{-NO}_2; (vi) CH3-CH2-CH(OH)-CH2OH\text{CH}_3\text{-CH}_2\text{-CH(OH)-CH}_2\text{OH}; (vii) CH3-CH(I)-CH(I)-C2H5\text{CH}_3\text{-CH(I)-CH(I)-C}_2\text{H}_5

Figure for question 82

Correct answer: 5

Question 83

Chemistry · 4 marks · −1

The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The potential of the electrode will be − ______ ×10−2\times10^{-2} V. (2.303RTF=0.059 V)\left(\frac{2.303RT}{F}=0.059\text{ V}\right)

Correct answer: 18

Explanation: E=−0.059×3=−0.177E=-0.059\times3=-0.177 V ≈ −18×10−2-18\times10^{-2} V.

Question 84

Chemistry · 4 marks · −1

The number of non-polar molecules from the following is ______. HF, H₂O, SO₂, H₂, CO₂, CH₄, NH₃, HCl, CHCl₃, BF₃

Correct answer: 4

Explanation: H2_2, CO2_2, CH4_4, BF3_3.

Question 85

Chemistry · 4 marks · −1

For a certain thermochemical reaction M → N at T = 400 K, ΔH⊖=77.2\Delta H^\ominus=77.2 kJ mol−1^{-1}, ΔS=122\Delta S=122 J K−1^{-1}, log equilibrium constant (log K) is − ______ ×10−1\times10^{-1}.

Correct answer: 37

Explanation: ΔG∘=77200−48800=28400\Delta G^\circ=77200-48800=28400 J; log⁡K=−284002.303×8.314×400=−3.7\log K=-\frac{28400}{2.303\times8.314\times400}=-3.7.

Question 86

Chemistry · 4 marks · −1

Volume of 3 M NaOH (formula weight 40 g mol⁻¹) which can be prepared from 84 g of NaOH is ______ ×10−1\times10^{-1} dm³.

Correct answer: 7

Explanation: 2.1 mol / 3 M = 0.7 dm3^3.

Question 87

Chemistry · 4 marks · −1

Total number of ions from the following with noble gas configuration is ______. Sr2+\text{Sr}^{2+} (z = 38), Cs+\text{Cs}^+ (z = 55), La2+\text{La}^{2+} (z = 57), Pb2+\text{Pb}^{2+} (z = 82), Yb2+\text{Yb}^{2+} (z = 70) and Fe2+\text{Fe}^{2+} (z = 26)

Correct answer: 2

Explanation: Only Sr2+^{2+} and Cs+^+.

Question 88

Chemistry · 4 marks · −1

9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ ×10−1\times10^{-1} g. (Given molar mass in g mol⁻¹ N : 14, O : 16, C : 12, H : 1)

Correct answer: 135

Explanation: 0.1 mol aniline → 0.1 mol acetanilide (135 g/mol) = 13.5 g.

Question 89

Chemistry · 4 marks · −1

1 mole of PbS is oxidized by "X" moles of O₃ to get "Y" moles of O₂. X + Y = ______.

Correct answer: 8

Explanation: PbS+4O3→PbSO4+4O2\text{PbS}+4\text{O}_3\to\text{PbSO}_4+4\text{O}_2.

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