JEE Main 2024 — 29 Jan Shift 2 (Previous Year Paper)

JEE Main · 2024 · with answers and explanations

questions
90
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If RR is the smallest equivalence relation on the set {1,2,3,4}\{1,2,3,4\} such that {(1,2),(1,3)}⊂R\{(1,2),(1,3)\}\subset R, then the number of elements in RR is ______.

  1. A.10
  2. B.8
  3. C.12
  4. D.15

Correct answer: A

Explanation: Classes {1,2,3}\{1,2,3\} and {4}\{4\}: 9+1=109+1=10 pairs.

Question 2

Mathematics · 4 marks · −1

If the mean and variance of five observations are 245\frac{24}{5} and 19425\frac{194}{25} respectively and the mean of the first four observations is 72\frac72, then the variance of the first four observations in equal to:

  1. A.1054\frac{105}{4}
  2. B.54\frac54
  3. C.7712\frac{77}{12}
  4. D.45\frac45

Correct answer: B

Explanation: ∑x=24\sum x=24, ∑x2=154\sum x^2=154; the fifth value is 10, so the first four have ∑x=14\sum x=14, ∑x2=54\sum x^2=54; variance 544−494=54\frac{54}{4}-\frac{49}{4}=\frac54.

Question 3

Mathematics · 4 marks · −1

Let y=log⁡e(1−x21+x2)y=\log_e\left(\frac{1-x^2}{1+x^2}\right), −1<x<1-1<x<1. Then at x=12x=\frac12, the value of 225(y′−y′′)225(y'-y'') is equal to:

  1. A.746
  2. B.742
  3. C.732
  4. D.736

Correct answer: D

Explanation: y′=−3215y'=-\frac{32}{15}, y′′=−1216225y''=-\frac{1216}{225} at x=12x=\frac12; 225(y′−y′′)=−480+1216=736225(y'-y'')=-480+1216=736.

Question 4

Mathematics · 4 marks · −1

The function f(x)=xx2−6x−16f(x)=\frac{x}{x^2-6x-16}, x∈R−{−2,8}x\in\mathbb{R}-\{-2,8\}.

  1. A.decreases in (−∞,−2)∪(−2,8)∪(8,∞)(-\infty,-2)\cup(-2,8)\cup(8,\infty)
  2. B.increases in (−∞,−2)∪(−2,8)∪(8,∞)(-\infty,-2)\cup(-2,8)\cup(8,\infty)
  3. C.decreases in (−∞,−2)(-\infty,-2) and increases in (8,∞)(8,\infty)
  4. D.decreases in (−2,8)(-2,8) and increases in (−∞,−2)∪(8,∞)(-\infty,-2)\cup(8,\infty)

Correct answer: A

Explanation: f′(x)=−(x2+16)(x2−6x−16)2<0f'(x)=\frac{-(x^2+16)}{(x^2-6x-16)^2}<0.

Question 5

Mathematics · 4 marks · −1

If ∫sin⁡32x+cos⁡32xsin⁡3xcos⁡3xsin⁡(x−θ)dx=Acos⁡θtan⁡x−sin⁡θ+Bcos⁡θ−sin⁡θcot⁡x+C\int\frac{\sin^{\frac32}x+\cos^{\frac32}x}{\sqrt{\sin^3x\cos^3x\sin(x-\theta)}}dx=A\sqrt{\cos\theta\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta\cot x}+C, where CC is the integration constant, then AB is equal to:

  1. A.2sec⁡θ2\sec\theta
  2. B.4sec⁡θ4\sec\theta
  3. C.8cosec⁡(2θ)8\operatorname{cosec}(2\theta)
  4. D.4cosec⁡(2θ)4\operatorname{cosec}(2\theta)

Correct answer: C

Explanation: Split the integrand: ∫sec⁡2x dxcos⁡θtan⁡x−sin⁡θ\int\frac{\sec^2x\,dx}{\sqrt{\cos\theta\tan x-\sin\theta}} gives A=2sec⁡θA=2\sec\theta and ∫cosec⁡2x dxcos⁡θ−sin⁡θcot⁡x\int\frac{\operatorname{cosec}^2x\,dx}{\sqrt{\cos\theta-\sin\theta\cot x}} gives B=2cosec⁡θB=2\operatorname{cosec}\theta; AB=8cosec⁡2θAB=8\operatorname{cosec}2\theta.

Question 6

Mathematics · 4 marks · −1

The function f(x)=2x+3(x)23f(x)=2x+3(x)^{\frac23}, x∈Rx\in\mathbb{R} has:

  1. A.exactly one point of local minima and no point of local maxima
  2. B.exactly one point of local maxima and no point of local minima
  3. C.exactly two points of local maxima and exactly one point of local minima
  4. D.exactly one point of local maxima and exactly one point of local minima

Correct answer: D

Explanation: f′=2+2x−1/3f'=2+2x^{-1/3}: maximum at x=−1x=-1, minimum at x=0x=0 (cusp).

Question 7

Mathematics · 4 marks · −1

If log⁡ea\log_ea, log⁡eb\log_eb, log⁡ec\log_ec are in an A.P. and log⁡ea−log⁡e2b\log_ea-\log_e2b, log⁡e2b−log⁡e3c\log_e2b-\log_e3c, log⁡e3c−log⁡ea\log_e3c-\log_ea are also in an A.P., then a:b:ca:b:c is equal to:

  1. A.25 : 10 : 4
  2. B.6 : 3 : 2
  3. C.16 : 4 : 1
  4. D.9 : 6 : 4

Correct answer: D

Explanation: b2=acb^2=ac and (2b3c)2=3c2b\left(\frac{2b}{3c}\right)^2=\frac{3c}{2b}, so bc=32\frac bc=\frac32; a:b:c=9:6:4a:b:c=9:6:4.

Question 8

Mathematics · 4 marks · −1

Let A=[2126211332]A=\begin{bmatrix}2&1&2\\6&2&11\\3&3&2\end{bmatrix} and P=[120502715]P=\begin{bmatrix}1&2&0\\5&0&2\\7&1&5\end{bmatrix}. The sum of the prime factors of ∣P−1AP−2I∣|P^{-1}AP-2I| is equal to:

  1. A.26
  2. B.27
  3. C.66
  4. D.23

Correct answer: A

Explanation: ∣P−1AP−2I∣=∣A−2I∣=69=3×23|P^{-1}AP-2I|=|A-2I|=69=3\times23.

Question 9

Mathematics · 4 marks · −1

The distance of the point (2, 3) from the line 2x−3y+28=02x-3y+28=0, measured parallel to the line 3x−y+1=0\sqrt3x-y+1=0, is equal to:

  1. A.424\sqrt2
  2. B.4+634+6\sqrt3
  3. C.3+423+4\sqrt2
  4. D.636\sqrt3

Correct answer: B

Explanation: Point (2+r2,3+3r2)(2+\frac r2,3+\frac{\sqrt3r}{2}) on the line gives r=4633−2=4+63r=\frac{46}{3\sqrt3-2}=4+6\sqrt3.

Question 10

Mathematics · 4 marks · −1

If each term of a geometric progression a1,a2,a3,…a_1,a_2,a_3,\ldots with a1=18a_1=\frac18 and a2≠a1a_2\ne a_1, is the arithmetic mean of the next two terms and Sn=a1+a2+…+anS_n=a_1+a_2+\ldots+a_n, then S20−S18S_{20}-S_{18} is equal to:

  1. A.−215-2^{15}
  2. B.2152^{15}
  3. C.2182^{18}
  4. D.−218-2^{18}

Correct answer: A

Explanation: 2=r+r2⇒r=−22=r+r^2\Rightarrow r=-2; a19+a20=18(−2)18(1−2)=−215a_{19}+a_{20}=\frac18(-2)^{18}(1-2)=-2^{15}.

Question 11

Mathematics · 4 marks · −1

An integer is chosen at random from the integers 1, 2, 3, ...., 50. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is:

  1. A.2150\frac{21}{50}
  2. B.1425\frac{14}{25}
  3. C.825\frac8{25}
  4. D.950\frac9{50}

Correct answer: A

Explanation: 12+8+7−4−1−1+0=2112+8+7-4-1-1+0=21.

Question 12

Mathematics · 4 marks · −1

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to:

  1. A.16
  2. B.12
  3. C.18
  4. D.15

Correct answer: D

Explanation: Partitions of 8 into at most 4 parts: 15.

Question 13

Mathematics · 4 marks · −1

Let P(3,2,3)P(3,2,3), Q(4,6,2)Q(4,6,2) and R(7,3,2)R(7,3,2) be the vertices of ΔPQR\Delta PQR. Then, the angle ∠QPR\angle QPR is:

  1. A.π3\frac\pi3
  2. B.cos⁡−1(118)\cos^{-1}\left(\frac1{18}\right)
  3. C.π6\frac\pi6
  4. D.cos⁡−1(718)\cos^{-1}\left(\frac7{18}\right)

Correct answer: A

Explanation: PQ⃗⋅PR⃗=9\vec{PQ}\cdot\vec{PR}=9, ∣PQ⃗∣=∣PR⃗∣=18|\vec{PQ}|=|\vec{PR}|=\sqrt{18}; cos⁡=12\cos=\frac12.

Question 14

Mathematics · 4 marks · −1

Let a unit vector u^=xi^+yj^+zk^\hat u=x\hat i+y\hat j+z\hat k make angles π2,π3\frac\pi2,\frac\pi3 and 2π3\frac{2\pi}{3} with the vectors 12i^+12k^\frac1{\sqrt2}\hat i+\frac1{\sqrt2}\hat k, 12j^+12k^\frac1{\sqrt2}\hat j+\frac1{\sqrt2}\hat k and 12i^+12j^\frac1{\sqrt2}\hat i+\frac1{\sqrt2}\hat j respectively. If v⃗=12i^+12k^+12j^\vec v=\frac1{\sqrt2}\hat i+\frac1{\sqrt2}\hat k+\frac1{\sqrt2}\hat j, then ∣u^−v⃗∣2|\hat u-\vec v|^2 is equal to:

  1. A.112\frac{11}{2}
  2. B.9
  3. C.52\frac52
  4. D.7

Correct answer: C

Explanation: x+z=0x+z=0, y+z=12y+z=\frac1{\sqrt2}, x+y=−12x+y=-\frac1{\sqrt2} give u^=(−12,0,12)\hat u=(-\frac1{\sqrt2},0,\frac1{\sqrt2}); ∣u^−v⃗∣2=2+12|\hat u-\vec v|^2=2+\frac12.

Question 15

Mathematics · 4 marks · −1

Let OA⃗=a⃗\vec{OA}=\vec a, OB⃗=12a⃗+4b⃗\vec{OB}=12\vec a+4\vec b and OC⃗=b⃗\vec{OC}=\vec b, where OO is the origin. If SS is the parallelogram with adjacent sides OA and OC, then area of the quadrilateral OABCarea of S\frac{\text{area of the quadrilateral }OABC}{\text{area of }S} is equal to ______.

  1. A.10
  2. B.7
  3. C.6
  4. D.8

Correct answer: D

Explanation: Area OABC=12∣OB⃗×AC⃗∣=12∣16 a⃗×b⃗∣=8∣a⃗×b⃗∣OABC=\frac12|\vec{OB}\times\vec{AC}|=\frac12|16\,\vec a\times\vec b|=8|\vec a\times\vec b|.

Question 16

Mathematics · 4 marks · −1

The sum of the solutions x∈Rx\in\mathbb{R} of the equation 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\frac{3\cos2x+\cos^32x}{\cos^6x-\sin^6x}=x^3-x^2+6 is:

  1. A.1
  2. B.0
  3. C.−1
  4. D.3

Correct answer: C

Explanation: cos⁡6x−sin⁡6x=cos⁡2x(3+cos⁡22x)4\cos^6x-\sin^6x=\frac{\cos2x(3+\cos^22x)}{4}, so LHS =4=4; x3−x2+2=0x^3-x^2+2=0 has the single real root −1-1.

Question 17

Mathematics · 4 marks · −1

Let x=mnx=\frac mn (m,nm,n are co-prime natural numbers) be a solution of the equation cos⁡(2sin⁡−1x)=19\cos(2\sin^{-1}x)=\frac19 and let α,β (α>β)\alpha,\beta\,(\alpha>\beta) be the roots of the equation mx2−nx−m+n=0mx^2-nx-m+n=0. Then the point (α,β)(\alpha,\beta) lies on the line.

  1. A.5x+8y=95x+8y=9
  2. B.5x−8y=−95x-8y=-9
  3. C.3x+2y=23x+2y=2
  4. D.3x−2y=−23x-2y=-2

Correct answer: A

Explanation: 1−2x2=19⇒x=231-2x^2=\frac19\Rightarrow x=\frac23; 2x2−3x+1=02x^2-3x+1=0: α=1,β=12\alpha=1,\beta=\frac12.

Question 18

Mathematics · 4 marks · −1

Let AA be the point of intersection of the lines 3x+2y=143x+2y=14, 5x−y=65x-y=6 and BB be the point of intersection of the lines 4x+3y=84x+3y=8, 6x+y=56x+y=5. The distance of the point P(5,−2)P(5,-2) from the line ABAB is:

  1. A.52\frac52
  2. B.6
  3. C.8
  4. D.132\frac{13}{2}

Correct answer: B

Explanation: A(2,4)A(2,4), B(12,2)B(\frac12,2); AB:4x−3y+4=0AB:4x-3y+4=0; distance 305=6\frac{30}{5}=6.

Question 19

Mathematics · 4 marks · −1

Let rr and θ\theta respectively be the modulus and amplitude of the complex number z=2−i(2tan⁡5π8)z=2-i\left(2\tan\frac{5\pi}{8}\right), then (r,θ)(r,\theta) is equal to:

  1. A.(2sec⁡3π8,5π8)\left(2\sec\frac{3\pi}{8},\frac{5\pi}{8}\right)
  2. B.(2sec⁡5π8,3π8)\left(2\sec\frac{5\pi}{8},\frac{3\pi}{8}\right)
  3. C.(2sec⁡3π8,3π8)\left(2\sec\frac{3\pi}{8},\frac{3\pi}{8}\right)
  4. D.(2sec⁡11π8,11π8)\left(2\sec\frac{11\pi}{8},\frac{11\pi}{8}\right)

Correct answer: C

Explanation: z=2+2itan⁡3π8=2sec⁡3π8 ei3π/8z=2+2i\tan\frac{3\pi}{8}=2\sec\frac{3\pi}8\,e^{i3\pi/8}.

Question 20

Mathematics · 4 marks · −1

If sin⁡(yx)=log⁡e∣x∣+α2\sin\left(\frac yx\right)=\log_e|x|+\frac\alpha2 is the solution of the differential equation xcos⁡(yx)dydx=ycos⁡(yx)+xx\cos\left(\frac yx\right)\frac{dy}{dx}=y\cos\left(\frac yx\right)+x and y(1)=π3y(1)=\frac\pi3, then α2\alpha^2 is equal to:

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

Correct answer: D

Explanation: With y=vxy=vx: cos⁡v dv=dxx\cos v\,dv=\frac{dx}{x}; y(1)=π3y(1)=\frac\pi3 gives α2=32\frac\alpha2=\frac{\sqrt3}{2}.

Question 21

Mathematics · 4 marks · −1

Let the area of the region {(x,y):0≤x≤3, 0≤y≤min⁡{x2+2,2x+2}}\{(x,y):0\le x\le3,\ 0\le y\le\min\{x^2+2,2x+2\}\} be AA. Then 12A12A is equal to ______.

Correct answer: 164

Explanation: ∫02(x2+2)dx+∫23(2x+2)dx=83+4+7=413\int_0^2(x^2+2)dx+\int_2^3(2x+2)dx=\frac{8}{3}+4+7=\frac{41}{3}; 12A=16412A=164.

Question 22

Mathematics · 4 marks · −1

Remainder when 64323264^{32^{32}} is divided by 9 is equal to ______.

Correct answer: 1

Explanation: 64≡1(mod9)64\equiv1\pmod9.

Question 23

Mathematics · 4 marks · −1

Let α,β\alpha,\beta be the roots of the equation x2−6x+3=0x^2-\sqrt6x+3=0 such that Im(α)>Im(β)\text{Im}(\alpha)>\text{Im}(\beta). Let a,ba,b be integers not divisible by 3 and nn be a natural number such that α99β+α98=3n(a+ib)\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+ib), i=−1i=\sqrt{-1}. Then n+a+bn+a+b is equal to ______.

Correct answer: 49

Question 24

Mathematics · 4 marks · −1

Let the set C={(x,y)∣x2−2y=2023, x,y∈N}C=\{(x,y)\mid x^2-2^y=2023,\ x,y\in\mathbb{N}\}. Then ∑(x,y)∈C(x+y)\sum_{(x,y)\in C}(x+y) is equal to ______.

Correct answer: 46

Explanation: Only (x,y)=(45,1)(x,y)=(45,1) works (2025−2=20232025-2=2023), so the sum is 46.

Question 25

Mathematics · 4 marks · −1

If ∫π/6π/31−sin⁡2x dx=α+β2+γ3\int_{\pi/6}^{\pi/3}\sqrt{1-\sin2x}\,dx=\alpha+\beta\sqrt2+\gamma\sqrt3, where α,β\alpha,\beta and γ\gamma are rational numbers, then 3α+4β−γ3\alpha+4\beta-\gamma is equal to ______.

Correct answer: 6

Question 26

Mathematics · 4 marks · −1

Let f(x)=lim⁡r→x{2r2[(f(r))2−f(x)f(r)]r2−x2−r3ef(r)r}f(x)=\sqrt{\lim_{r\to x}\left\{\frac{2r^2\left[(f(r))^2-f(x)f(r)\right]}{r^2-x^2}-r^3e^{\frac{f(r)}{r}}\right\}} be differentiable in (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty) and f(1)=1f(1)=1. Then the value of eaea, such that f(a)=0f(a)=0, is equal to ______.

Correct answer: 2

Question 27

Mathematics · 4 marks · −1

Let for any three distinct consecutive terms a,b,ca,b,c of an A.P. the lines ax+by+c=0ax+by+c=0 be concurrent at the point PP and Q(α,β)Q(\alpha,\beta) be a point such that the system of equations x+y+z=6x+y+z=6 2x+5y+αz=β2x+5y+\alpha z=\beta and x+2y+3z=4x+2y+3z=4, has infinitely many solutions. The (PQ)2(PQ)^2 is equal to ______.

Correct answer: 113

Explanation: a−2b+c=0a-2b+c=0 so every line passes through P(1,−2)P(1,-2). The system is dependent when 8−α=08-\alpha=0, and E2=3E3−E1E_2=3E_3-E_1 gives β=6\beta=6; (PQ)2=72+82=113(PQ)^2=7^2+8^2=113.

Question 28

Mathematics · 4 marks · −1

Let the slope of the line 45x+5y+3=045x+5y+3=0 be 27r1+9r2227r_1+\frac{9r_2}{2} for some r1,r2∈Rr_1,r_2\in\mathbb{R}. Then lim⁡x→3(∫3x8t23r2x2−r2x2−r1x3−3xdt)\lim_{x\to3}\left(\int_3^x\frac{8t^2}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}dt\right) is equal to ______.

Correct answer: 12

Question 29

Mathematics · 4 marks · −1

Let OO be the origin, and MM and NN be the points on the lines x−54=y−41=z−53\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3} and x+812=y+25=z−119\frac{x+8}{12}=\frac{y+2}{5}=\frac{z-11}{9} respectively such that MNMN is the shortest distance between the given lines. Then OM⃗⋅ON⃗\vec{OM}\cdot\vec{ON} is equal to ______.

Correct answer: 9

Question 30

Mathematics · 4 marks · −1

Let P(α,β)P(\alpha,\beta) be a point on the parabola y2=4xy^2=4x. If PP also lies on the chord of the parabola x2=8yx^2=8y whose mid-point is (1,54)\left(1,\frac54\right), then (α−28)(β−8)(\alpha-28)(\beta-8) is equal to ______.

Correct answer: 192

Question 31

Physics · 4 marks · −1

N moles of a polyatomic gas (f=6f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is:

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

Correct answer: A

Explanation: 6N+3⋅2N+2=5⇒N=4\frac{6N+3\cdot2}{N+2}=5\Rightarrow N=4.

Question 32

Physics · 4 marks · −1

The truth table for this given circuit is:

Figure for question 32
  1. A.A B Y: 0 0 0; 0 1 0; 1 0 0; 1 1 1
  2. B.A B Y: 0 0 1; 0 1 0; 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 0; 0 1 1; 1 0 0; 1 1 1

Correct answer: D

Explanation: Y=AB+AˉB=BY=AB+\bar AB=B.

Question 33

Physics · 4 marks · −1

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R1R_1 and R2R_2 respectively. The mass ratio of X and Y is:

  1. A.(R1R2)\left(\frac{R_1}{R_2}\right)
  2. B.(R1R2)2\left(\frac{R_1}{R_2}\right)^2
  3. C.(R2R1)2\left(\frac{R_2}{R_1}\right)^2
  4. D.(R2R1)\left(\frac{R_2}{R_1}\right)

Correct answer: B

Explanation: R=2mqVqB⇒m∝R2R=\frac{\sqrt{2mqV}}{qB}\Rightarrow m\propto R^2.

Question 34

Physics · 4 marks · −1

In the given circuit, the current in resistance R3R_3 is:

Figure for question 34
  1. A.1.5 A
  2. B.2.5 A
  3. C.2 A
  4. D.1 A

Correct answer: D

Explanation: Total 2+2+1=5 Ω2+2+1=5\ \Omega, I=2I=2 A, split equally: 1 A.

Question 35

Physics · 4 marks · −1

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7λ4\frac{7\lambda}{4}. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

  1. A.12\frac12
  2. B.34\frac34
  3. C.13\frac13
  4. D.14\frac14

Correct answer: A

Explanation: ϕ=7π2\phi=\frac{7\pi}{2}; cos⁡2ϕ2=cos⁡27π4=12\cos^2\frac\phi2=\cos^2\frac{7\pi}{4}=\frac12.

Question 36

Physics · 4 marks · −1

An electric field is given by (6i^+5j^+3k^)(6\hat i+5\hat j+3\hat k) N/C. The electric flux through a surface area 30i^30\hat i m² lying in YZ-plane (in SI unit) is:

  1. A.180
  2. B.150
  3. C.90
  4. D.60

Correct answer: A

Question 37

Physics · 4 marks · −1

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E⃗=9.6j^\vec E=9.6\hat j V/m. The value of magnetic field at this point is:

  1. A.9.6j^9.6\hat j T
  2. B.3.2×10−8i^3.2\times10^{-8}\hat i T
  3. C.9.6×10−8k^9.6\times10^{-8}\hat k T
  4. D.3.2×10−8k^3.2\times10^{-8}\hat k T

Correct answer: D

Explanation: B=E/cB=E/c and E^×B^=i^\hat E\times\hat B=\hat i.

Question 38

Physics · 4 marks · −1

A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K.E.)A(K.E.)B\frac{(K.E.)_A}{(K.E.)_B} is:

Figure for question 38
  1. A.5 : 1
  2. B.3 : 2
  3. C.1 : 5
  4. D.2 : 5

Correct answer: A

Explanation: vB2=gLv_B^2=gL, vA2=5gLv_A^2=5gL.

Question 39

Physics · 4 marks · −1

A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

  1. A.2 times
  2. B.32\frac32 times
  3. C.4 times
  4. D.3 times

Correct answer: A

Explanation: Δl∝F/r2\Delta l\propto F/r^2: 1/21/4=2\frac{1/2}{1/4}=2.

Question 40

Physics · 4 marks · −1

If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm. the focal length of the mirror must be:

  1. A.−12 cm
  2. B.15 cm
  3. C.10/3 cm
  4. D.−10 cm

Correct answer: D

Explanation: v=−2uv=-2u, ∣v−u∣=3∣u∣=15⇒u=−5|v-u|=3|u|=15\Rightarrow u=-5, v=10v=10; f=−10f=-10 cm.

Question 41

Physics · 4 marks · −1

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, g : 10 m s−2^{-2}]

  1. A.656\sqrt5 m s−1^{-1}
  2. B.252\sqrt5 m s−1^{-1}
  3. C.565\sqrt6 m s−1^{-1}
  4. D.555\sqrt5 m s−1^{-1}

Correct answer: A

Explanation: v2=0.9×2gL=180v^2=0.9\times2gL=180.

Question 42

Physics · 4 marks · −1

A physical quantity QQ is found to depend on quantities a, b, c by the relation Q=a4b3c2Q=\frac{a^4b^3}{c^2}. The percentage error in a, b and c are 3%, 4% and 5% respectively. Then, the percentage error in Q is:

  1. A.43%
  2. B.66%
  3. C.34%
  4. D.14%

Correct answer: C

Explanation: 12+12+10=34%12+12+10=34\%.

Question 43

Physics · 4 marks · −1

Given below are two statements: Statement I : Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. 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 true
  4. D.Both statement I and statement II are false

Correct answer: B

Explanation: Statement II describes Thomson's model.

Question 44

Physics · 4 marks · −1

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point is (if π2=9.8\pi^2=9.8 and g=9.8g=9.8 m/s²):

  1. A.9.8 N
  2. B.8.82 N
  3. C.97 N
  4. D.17.8 N

Correct answer: A

Explanation: ω=π3\omega=\frac\pi3; T=m(g+ω2r)=0.9(9.8+9.89)=9.8T=m(g+\omega^2r)=0.9(9.8+\frac{9.8}{9})=9.8 N.

Question 45

Physics · 4 marks · −1

A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be:

  1. A.18πR2T\frac18\pi R^2T
  2. B.4πR2T4\pi R^2T
  3. C.8πR2T8\pi R^2T
  4. D.3πR2T3\pi R^2T

Correct answer: C

Explanation: Area grows from 4πR24\pi R^2 to 12πR212\pi R^2.

Question 46

Physics · 4 marks · −1

Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be:

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

Correct answer: D

Explanation: n∝λn\propto\lambda at fixed power.

Question 47

Physics · 4 marks · −1

The temperature of a gas having 2.0×10252.0\times10^{25} molecules per cubic meter at 1.38 atm (Given, k=1.38×10−23k=1.38\times10^{-23} J K−1^{-1})

  1. A.100 K
  2. B.300 K
  3. C.200 K
  4. D.500 K

Correct answer: D

Explanation: T=Pnk=1.38×1.013×1052×1025×1.38×10−23≈500T=\frac{P}{nk}=\frac{1.38\times1.013\times10^5}{2\times10^{25}\times1.38\times10^{-23}}\approx500 K.

Question 48

Physics · 4 marks · −1

A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution:

  1. A.25
  2. B.50
  3. C.20
  4. D.100

Correct answer: A

Explanation: T∝r3/2T\propto r^{3/2}: 200/8=25200/8=25.

Question 49

Physics · 4 marks · −1

A particle is moving in a straight line. The variation of position 'x' as a function of time 't' is given as x=(t3−6t2+20t+15)x=(t^3-6t^2+20t+15) m. The velocity of the body when its acceleration becomes zero is:

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

Correct answer: C

Explanation: a=6t−12=0⇒t=2a=6t-12=0\Rightarrow t=2; v=12−24+20=8v=12-24+20=8 m/s.

Question 50

Physics · 4 marks · −1

In an a.c. circuit, voltage and current are given by: V=100sin⁡(100t)V=100\sin(100t) V and I=100sin⁡(100t+π3)I=100\sin\left(100t+\frac\pi3\right) mA respectively. The average power dissipated in one cycle is:

  1. A.2.5 W
  2. B.10 W
  3. C.5 W
  4. D.25 W

Correct answer: A

Explanation: P=100×0.12cos⁡60∘=2.5P=\frac{100\times0.1}{2}\cos60^\circ=2.5 W.

Question 51

Physics · 4 marks · −1

In the given figure, the charge stored in 6 μF capacitor. when points A and B are joined by a connecting wire is ______ μC.

Figure for question 51

Correct answer: 36

Explanation: Steady state: current flows 9 V → 6 Ω → A–B → 3 Ω → ground, I=1I=1 A, VA=VB=3V_A=V_B=3 V; the 6 μF capacitor has 6 V across it: Q=36Q=36 μC.

Question 52

Physics · 4 marks · −1

Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4: 1 and force F1F_1 is applied to P, an extension of Δl\Delta l is produced. The force which is required to produce same extension in Q is F2F_2. The value of F1F2\frac{F_1}{F_2} is ______.

Correct answer: 16

Explanation: F∝AL∝A2F\propto\frac{A}{L}\propto A^2 at fixed volume: ratio 16.

Question 53

Physics · 4 marks · −1

In the given circuit, the current flowing through the resistance 20 Ω is 0.3 A, while the ammeter reads 0.9 A. The value of R1R_1 is ______ Ω.

Figure for question 53

Correct answer: 30

Explanation: V=6V=6 V; 15 Ω carries 0.4 A, so R1R_1 carries 0.2 A: R1=30 ΩR_1=30\ \Omega.

Question 54

Physics · 4 marks · −1

A simple harmonic oscillator has an amplitude A and time period 6π6\pi second. Assuming the oscillation starts from its mean position. the time required by it to travel from x=Ax=A to x=32Ax=\frac{\sqrt3}{2}A will be πx\frac\pi x s, where x=x= ______.

Correct answer: 2

Explanation: ω=13\omega=\frac13; cos⁡ωt=32⇒t=π2\cos\omega t=\frac{\sqrt3}{2}\Rightarrow t=\frac\pi2.

Question 55

Physics · 4 marks · −1

In a single slit diffraction pattern, a light of wavelength 6000 Å is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen in placed 50 cm away from slits. The width of the slit is ______ ×10−4\times10^{-4} m.

Correct answer: 2

Explanation: 2λDa=3×10−3⇒a=2×10−4\frac{2\lambda D}{a}=3\times10^{-3}\Rightarrow a=2\times10^{-4} m.

Question 56

Physics · 4 marks · −1

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of it's acceleration are equal. If its speed at t=0t=0 is 4 m/s, the time taken to complete the first revolution will be 1α[1−e−2π]\frac1\alpha[1-e^{-2\pi}] s, where α=\alpha= ______.

Correct answer: 8

Explanation: dvdt=v2R\frac{dv}{dt}=\frac{v^2}{R} gives v=11/4−2tv=\frac{1}{1/4-2t} and s=−12ln⁡(1−8t)s=-\frac12\ln(1-8t); s=πs=\pi gives t=18(1−e−2π)t=\frac18(1-e^{-2\pi}).

Question 57

Physics · 4 marks · −1

A charge of 4.0 μC is moving with a velocity of 4.0×1064.0\times10^6 m s−1^{-1} along the positive y-axis under a magnetic field B⃗\vec B of strength (2k^)(2\hat k) T. The force acting on the charge is xi^x\hat i N. The value of xx is ______.

Correct answer: 32

Explanation: qv⃗×B⃗=4×10−6×4×106×2(j^×k^)=32i^q\vec v\times\vec B=4\times10^{-6}\times4\times10^6\times2(\hat j\times\hat k)=32\hat i N.

Question 58

Physics · 4 marks · −1

A horizontal straight wire 5 m long extending from east to west falling freely at right angle to horizontal component of earths magnetic field 0.60×10−40.60\times10^{-4} Wb m−2^{-2}. The instantaneous value of emf induced in the wire when its velocity is 10 m s−1^{-1} is ______ ×10−3\times10^{-3} V.

Correct answer: 3

Explanation: Blv=0.6×10−4×5×10=3×10−3Blv=0.6\times10^{-4}\times5\times10=3\times10^{-3} V.

Question 59

Physics · 4 marks · −1

A body of mass 5 kg moving with a uniform speed 323\sqrt2 m s−1^{-1} in X-Y plane along the line y=x+4y=x+4. The angular momentum of the particle about the origin will be ______ kg m² s−1^{-1}.

Correct answer: 60

Explanation: Perpendicular distance of the line from O is 222\sqrt2; L=5×32×22=60L=5\times3\sqrt2\times2\sqrt2=60.

Question 60

Physics · 4 marks · −1

Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being performed at T = 0 K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α10\frac\alpha{10} V, where α=\alpha= ______.

Correct answer: 121

Explanation: Excitation to n=3n=3 needs 12.09 eV ≈ 12.1 V.

Question 61

Chemistry · 4 marks · −1

According to IUPAC system, the compound (cyclohexene ring bearing OH on the carbon adjacent to the C=C, as shown) is named as:

  1. A.Cyclohex-1-en-2-ol
  2. B.Cyclohex-1-en-3-ol
  3. C.Cyclohex-2-en-1-ol
  4. D.1-Hydroxyhex-2-ene

Correct answer: C

Question 62

Chemistry · 4 marks · −1

Match List I with List II. List I (Spectral Series for Hydrogen): A. Lyman; B. Balmer; C. Paschen; D. Pfund List II (Spectral Region/Higher Energy State): I. Infrared region; II. UV region; III. Infrared region; IV. Visible region Choose the correct answer from the options given below:

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

Correct answer: C

Question 63

Chemistry · 4 marks · −1

Alkyl halide is converted into alkyl isocyanide by reaction with:

  1. A.NH4CN\text{NH}_4\text{CN}
  2. B.AgCN
  3. C.KCN
  4. D.NaCN

Correct answer: B

Question 64

Chemistry · 4 marks · −1

Match List I with List II. List I (Compound): A. Ethanol; B. Phenol; C. m-Nitrophenol; D. p-Nitrophenol List II (pKapK_a Value): I. 10.0; II. 15.9; III. 7.1; IV. 8.3 Choose the correct answer from the options given below:

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

Correct answer: A

Question 65

Chemistry · 4 marks · −1

A reagent which gives brilliant red precipitate with Nickel ions in basic medium is:

  1. A.dimethyl glyoxime
  2. B.sodium nitroprusside
  3. C.meta-dinitrobenzene
  4. D.neutral FeCl3\text{FeCl}_3

Correct answer: A

Question 66

Chemistry · 4 marks · −1

Given below are two statements: Statement I : Fluorine has most negative electron gain enthalpy in its group. Statement II : Oxygen has least negative electron gain enthalpy in its group. 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 true
  4. D.Both Statement I and Statement II are false

Correct answer: A

Explanation: Chlorine, not fluorine, has the most negative electron gain enthalpy in group 17.

Question 67

Chemistry · 4 marks · −1

Which of the following acts as a strong reducing agent? (Atomic number: Ce = 58, Eu = 63, Gd = 64, Lu = 71)

  1. A.Lu3+\text{Lu}^{3+}
  2. B.Ce4+\text{Ce}^{4+}
  3. C.Gd3+\text{Gd}^{3+}
  4. D.Eu2+\text{Eu}^{2+}

Correct answer: D

Question 68

Chemistry · 4 marks · −1

Which of the following reaction is correct?

  1. A.C2H5CONH2+Br2+NaOH→C2H5CH2NH2+Na2CO3+NaBr+H2O\text{C}_2\text{H}_5\text{CONH}_2+\text{Br}_2+\text{NaOH}\to\text{C}_2\text{H}_5\text{CH}_2\text{NH}_2+\text{Na}_2\text{CO}_3+\text{NaBr}+\text{H}_2\text{O}
  2. B.Cyclohexene + Br2→Δ, UV light+\ \text{Br}_2\xrightarrow{\Delta,\ \text{UV light}} 1,2-dibromocyclohexane
  3. C.CH3CH2CH2NH2→HNO2, 0∘C, H2OCH3CH2OH+N2+HCl\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\xrightarrow{\text{HNO}_2,\ 0^\circ\text{C},\ \text{H}_2\text{O}}\text{CH}_3\text{CH}_2\text{OH}+\text{N}_2+\text{HCl}
  4. D.1-Methylcyclohexene ++ HI →\to 1-iodo-1-methylcyclohexane

Correct answer: D

Explanation: Markovnikov addition of HI; the others give wrong products (Hofmann gives C2_2H5_5NH2_2; UV/Δ gives allylic bromide; propylamine gives propanol).

Question 69

Chemistry · 4 marks · −1

On passing a gas, 'X', through Nessler's reagent, a brown precipitate is obtained. The gas 'X' is:

  1. A.Cl2\text{Cl}_2
  2. B.CO2\text{CO}_2
  3. C.NH3\text{NH}_3
  4. D.H2S\text{H}_2\text{S}

Correct answer: C

Question 70

Chemistry · 4 marks · −1

Match List I with List II. List I (Bio polymer): A. Starch; B. Cellulose; C. Nucleic acid; D. Protein List II (Monomer): I. nucleotide; II. α-glucose; III. β-glucose; IV. α-amino acid Choose the correct answer from the options given below:

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

Correct answer: D

Question 71

Chemistry · 4 marks · −1

Which one of the following will show geometrical isomerism?

Figure for question 71
  1. A.4-Bromo-1-methylidenecyclohexane (=CH2_2 exocyclic, Br para)
  2. B.3-Bromo-1-(bromomethylidene)cyclohexane (=CHBr exocyclic, Br meta)
  3. C.4-Bromo-1-(bromomethylidene)cyclohexane (=CHBr exocyclic, Br para)
  4. D.3-Bromo-1-methylidenecyclohexane (=CH2_2 exocyclic, Br meta)

Correct answer: B

Explanation: Only with =CHBr and an unsymmetrically substituted ring (Br at C-3) are the two ends of the double bond each differently substituted.

Question 72

Chemistry · 4 marks · −1

The correct IUPAC name of K2MnO4\text{K}_2\text{MnO}_4 is:

  1. A.Potassium tetraoxidomanganate (VI)
  2. B.Dipotassium tetraoxidomanganate (VII)
  3. C.Potassium tetraoxopermanganate (VI)
  4. D.Potassium tetraoxidomanganese (VI)

Correct answer: A

Question 73

Chemistry · 4 marks · −1

Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolyzed in presence of an acid results.

  1. A.Benzene-1,2-diol
  2. B.2-Hydroxybenzaldehyde
  3. C.Benzene-1,3-diol
  4. D.Salicylic acid

Correct answer: B

Explanation: Reimer–Tiemann reaction.

Question 74

Chemistry · 4 marks · −1

Chromatographic technique/s based on the principle of differential adsorption is/are: A. Column chromatography B. Thin layer chromatography C. Paper chromatography Choose the most appropriate answer from the options given below:

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

Correct answer: B

Explanation: Paper chromatography is partition chromatography.

Question 75

Chemistry · 4 marks · −1

The product A formed in the following reaction is: Aniline →NaNO2, HCl, 0∘C; then Cu2Cl2\xrightarrow{\text{NaNO}_2\text{, HCl, }0^\circ\text{C; then Cu}_2\text{Cl}_2} A

  1. A.Chlorobenzene
  2. B.Anilinium chloride (C6H5N+H3Cl−\text{C}_6\text{H}_5\overset{+}{\text{N}}\text{H}_3\text{Cl}^-)
  3. C.1,2-Dichlorobenzene
  4. D.2-Chloroaniline

Correct answer: A

Explanation: Diazotisation followed by Sandmeyer reaction.

Question 76

Chemistry · 4 marks · −1

The element having the highest first ionization enthalpy is:

  1. A.N
  2. B.Al
  3. C.Si
  4. D.C

Correct answer: A

Question 77

Chemistry · 4 marks · −1

Identify the reagents used for the following conversion. A keto-ester-aldehyde bicyclic precursor (a cyclohexane ring bearing OH, a −CH2COOCH3-\text{CH}_2\text{COOCH}_3 arm and a −CH2CHO-\text{CH}_2\text{CHO} arm) →(A)\xrightarrow{(A)} the dialdehyde →(B)\xrightarrow{(B)} a bicyclic α,β-unsaturated aldehyde (with OH) →(C)\xrightarrow{(C)} a bicyclic diene bearing a methyl group (OH removed)

Figure for question 77
  1. A.A = LiAlH4\text{LiAlH}_4, B = NaOH(alc)\text{NaOH}_{(alc)}, C = Zn/HCl
  2. B.A = DIBAL – H, B = NaOH(alc)\text{NaOH}_{(alc)}, C = Zn/HCl
  3. C.A = DIBAL – H, B = NaOH(aq)\text{NaOH}_{(aq)}, C = NH2NH2\text{NH}_2\text{NH}_2/KOH, ethylene glycol
  4. D.A = LiAlH4\text{LiAlH}_4, B = NaOH(aq)\text{NaOH}_{(aq)}, C = NH2-NH2\text{NH}_2\text{-NH}_2/KOH, ethylene glycol

Correct answer: B

Explanation: DIBAL-H reduces the ester only to the aldehyde; alcoholic NaOH effects aldol condensation; Clemmensen (Zn/HCl) reduces CHO to CH3_3 while the acid dehydrates the alcohol.

Question 78

Chemistry · 4 marks · −1

The ascending acidity order of the following H atoms is: A: HC≡C-H‾\text{HC}\equiv\text{C-}\underline{\text{H}}; B: H2C=CH-H‾\text{H}_2\text{C=CH-}\underline{\text{H}}; C: (CH3)3C-H‾(\text{CH}_3)_3\text{C-}\underline{\text{H}}; D: H3C-CH2-H‾\text{H}_3\text{C-CH}_2\text{-}\underline{\text{H}}

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

Correct answer: A

Explanation: sp C–H > sp2^2 C–H > 1° sp3^3 > 3° sp3^3.

Question 79

Chemistry · 4 marks · −1

Which of the following statements are correct about Zn, Cd and Hg? A. They exhibit high enthalpy of atomization as the d-subshell is full. B. Zn and Cd do not show variable oxidation state while Hg shows +I and +II. C. Compounds of Zn, Cd and Hg are paramagnetic in nature. D. Zn, Cd and Hg are called soft metals. Choose the most appropriate from the options given below:

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

Correct answer: B

Question 80

Chemistry · 4 marks · −1

Anomalous behavior of oxygen is due to its:

  1. A.small size and high electronegativity
  2. B.small size and low electronegativity
  3. C.large size and high electronegativity
  4. D.large size and low electronegativity

Correct answer: A

Question 81

Chemistry · 4 marks · −1

If 50 mL of 0.5M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is ______ g.

Correct answer: 4

Explanation: 0.025 mol oxalic acid needs 0.05 mol NaOH in 25 mL (2 M); 50 mL holds 0.1 mol = 4 g.

Question 82

Chemistry · 4 marks · −1

A constant current was passed through a solution of AuCl4−\text{AuCl}_4^- ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is ______ ×10−2\times10^{-2} F. (Given atomic mass of Au = 197)

Correct answer: 2

Explanation: 1.314197×3=0.02\frac{1.314}{197}\times3=0.02 F.

Question 83

Chemistry · 4 marks · −1

The following concentrations were observed at 500 K for the formation of NH3\text{NH}_3 from N2\text{N}_2 and H2\text{H}_2. At equilibrium: [N2]=2×10−2[\text{N}_2]=2\times10^{-2} M, [H2]=3×10−2[\text{H}_2]=3\times10^{-2} M and [NH3]=1.5×10−2[\text{NH}_3]=1.5\times10^{-2} M. Equilibrium constant for the reaction is ______.

Correct answer: 417

Explanation: Kc=(1.5×10−2)2(2×10−2)(3×10−2)3≈417K_c=\frac{(1.5\times10^{-2})^2}{(2\times10^{-2})(3\times10^{-2})^3}\approx417.

Question 84

Chemistry · 4 marks · −1

The total number of anti bonding molecular orbitals, formed from 2s and 2p atomic orbitals in a diatomic molecule is ______.

Correct answer: 4

Explanation: σ∗2s\sigma^*2s, σ∗2pz\sigma^*2p_z, π∗2px\pi^*2p_x, π∗2py\pi^*2p_y.

Question 85

Chemistry · 4 marks · −1

Standard enthalpy of vapourisation for CCl4\text{CCl}_4 is 30.5 kJ mol−1^{-1}. Heat required for vapourisation of 284 g of CCl4\text{CCl}_4 at constant temperature is ______ kJ. (Given molar mass in g mol−1^{-1}: C = 12, Cl = 35.5)

Correct answer: 56

Explanation: 284154×30.5≈56\frac{284}{154}\times30.5\approx56 kJ.

Question 86

Chemistry · 4 marks · −1

The total number of 'Sigma' and 'Pi' bonds in 2-formylhex-4-enoic acid is ______.

Correct answer: 22

Question 87

Chemistry · 4 marks · −1

The oxidation number of iron in the compound formed during brown ring test for NO3−\text{NO}_3^- ion is ______.

Correct answer: 1

Explanation: [Fe(H2O)5(NO)]2+[\text{Fe(H}_2\text{O)}_5(\text{NO})]^{2+} with NO+^+: Fe is +1.

Question 88

Chemistry · 4 marks · −1

The total number of molecules with zero dipole moment among CH4\text{CH}_4, BF3\text{BF}_3, H2O\text{H}_2\text{O}, HF, NH3\text{NH}_3, CO2\text{CO}_2 and SO2\text{SO}_2 is ______.

Correct answer: 3

Explanation: CH4_4, BF3_3, CO2_2.

Question 89

Chemistry · 4 marks · −1

The half-life of radioisotope bromine – 82 is 36 hours. The fraction which remains after one day is ______ ×10−2\times10^{-2}. (Given antilog 0.2006 = 1.587)

Correct answer: 63

Explanation: (1/2)24/36=2−2/3=11.587=0.63(1/2)^{24/36}=2^{-2/3}=\frac1{1.587}=0.63.

Question 90

Chemistry · 4 marks · −1

Molality of 0.8 M H2SO4\text{H}_2\text{SO}_4 solution (density 1.06 g cm−3^{-3}) is ______ ×10−3\times10^{-3} m.

Correct answer: 815

Explanation: 1 L = 1060 g, solute 78.4 g, solvent 981.6 g; m=0.80.9816=0.815m=\frac{0.8}{0.9816}=0.815.

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