AIEEE 2006 — 30 Apr Offline (Previous Year Paper)

JEE Main · 2006 · with answers and explanations

questions
145
Total marks
580
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

ABC is a triangle, right angled at A. The resultant of the forces acting along AB⃗\vec{AB}, AC⃗\vec{AC} with magnitudes 1AB\frac{1}{AB} and 1AC\frac{1}{AC} respectively is the force along AD⃗\vec{AD}, where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is

  1. A.AB2+AC2(AB)2(AC)2\frac{AB^2+AC^2}{(AB)^2(AC)^2}
  2. B.(AB)(AC)AB+AC\frac{(AB)(AC)}{AB+AC}
  3. C.1AB+1AC\frac{1}{AB}+\frac{1}{AC}
  4. D.1AD\frac{1}{AD}

Correct answer: D

Explanation: R=1AB2+1AC2=BCAB⋅AC=1ADR=\sqrt{\frac1{AB^2}+\frac1{AC^2}}=\frac{BC}{AB\cdot AC}=\frac1{AD}.

Question 2

Mathematics · 4 marks · −1

Suppose a population A has 100 observations 101, 102, ..., 200, and another population B has 100 observations 151, 152, ..., 250. If VAV_A and VBV_B represent the variances of the two populations, respectively, then VAVB\frac{V_A}{V_B} is

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

Correct answer: A

Explanation: Shifted data have equal variance.

Question 3

Mathematics · 4 marks · −1

If the roots of the quadratic equation x2+px+q=0x^2 + px + q = 0 are tan⁡30∘\tan30^\circ and tan⁡15∘\tan15^\circ, respectively then the value of 2+q−p2 + q - p is

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

Correct answer: B

Explanation: tan⁡45∘=−p1−q=1⇒q−p=1\tan45^\circ=\frac{-p}{1-q}=1\Rightarrow q-p=1.

Question 4

Mathematics · 4 marks · −1

The value of the integral, ∫36x9−x+xdx\int_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx is

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

Correct answer: B

Explanation: King's rule: 2I=32I=3.

Question 5

Mathematics · 4 marks · −1

The number of values of x in the interval [0,3π][0, 3\pi] satisfying the equation 2sin⁡2x+5sin⁡x−3=02\sin^2x + 5\sin x - 3 = 0 is

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

Correct answer: A

Explanation: sin⁡x=12\sin x=\frac12: four solutions in [0,3π][0,3\pi].

Question 6

Mathematics · 4 marks · −1

If (a⃗×b⃗)×c⃗=a⃗×(b⃗×c⃗)(\vec{a}\times\vec{b})\times\vec{c} = \vec{a}\times(\vec{b}\times\vec{c}), where a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} are any three vectors such that a⃗⋅b⃗≠0\vec{a}\cdot\vec{b} \ne 0, b⃗⋅c⃗≠0\vec{b}\cdot\vec{c} \ne 0, then a⃗\vec{a} and c⃗\vec{c} are

  1. A.inclined at an angle of π/3\pi/3 between them
  2. B.inclined at an angle of π/6\pi/6 between them
  3. C.perpendicular
  4. D.parallel

Correct answer: D

Explanation: Expanding gives (a⃗⋅b⃗)c⃗=(b⃗⋅c⃗)a⃗(\vec a\cdot\vec b)\vec c=(\vec b\cdot\vec c)\vec a.

Question 7

Mathematics · 4 marks · −1

Let W denote the words in the English dictionary. Define the relation R by : R={(x,y)∈W×WR = \{(x, y) \in W\times W | the words x and y have at least one letter in common}\}. Then R is

  1. A.not reflexive, symmetric and transitive
  2. B.reflexive, symmetric and not transitive
  3. C.reflexive, symmetric and transitive
  4. D.reflexive, not symmetric and transitive

Correct answer: B

Explanation: Not transitive (DELHI, DWARKA, PARK).

Question 8

Mathematics · 4 marks · −1

If A and B are square matrices of size n×nn\times n such that A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B), then which of the following will be always true ?

  1. A.A = B
  2. B.AB = BA
  3. C.either of A or B is a zero matrix
  4. D.either of A or B is an identity matrix

Correct answer: B

Explanation: Expand.

Question 9

Mathematics · 4 marks · −1

The value of ∑k=110(sin⁡2kπ11+icos⁡2kπ11)\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11} + i\cos\frac{2k\pi}{11}\right) is

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

Correct answer: D

Explanation: ∑sin⁡=0\sum\sin=0, ∑cos⁡=−1\sum\cos=-1.

Question 10

Mathematics · 4 marks · −1

All the values of m for which both roots of the equations x2−2mx+m2−1=0x^2 - 2mx + m^2 - 1 = 0 are greater than -2 but less than 4, lie in the interval

  1. A.-2 < m < 0
  2. B.m > 3
  3. C.-1 < m < 3
  4. D.1 < m < 4

Correct answer: C

Explanation: Roots m±1m\pm1.

Question 11

Mathematics · 4 marks · −1

A particle has two velocities of equal magnitude inclined to each other at an angle θ\theta. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then θ\theta is

  1. A.90∘90^\circ
  2. B.120∘120^\circ
  3. C.45∘45^\circ
  4. D.60∘60^\circ

Correct answer: B

Explanation: tan⁡θ4=u2sin⁡θu+u2cos⁡θ\tan\frac\theta4=\frac{\frac u2\sin\theta}{u+\frac u2\cos\theta} gives sin⁡2θ4=14\sin^2\frac\theta4=\frac14, θ=120∘\theta=120^\circ.

Question 12

Mathematics · 4 marks · −1

At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

  1. A.65e\frac{6}{5^e}
  2. B.56\frac{5}{6}
  3. C.655\frac{6}{55}
  4. D.6e5\frac{6}{e^5}

Correct answer: D

Explanation: e−5(1+5)e^{-5}(1+5).

Question 13

Mathematics · 4 marks · −1

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2^2, then the height above the point P from where the body began to fall is

  1. A.720 m
  2. B.900 m
  3. C.320 m
  4. D.680 m

Correct answer: A

Explanation: 400=8g+4gt⇒t=8400=8g+4gt\Rightarrow t=8 s; height =320+400=720=320+400=720 m.

Question 14

Mathematics · 4 marks · −1

∫0πxf(sin⁡x) dx\int_0^\pi xf(\sin x)\,dx is equal to

  1. A.π∫0πf(cos⁡x)dx\pi\int_0^\pi f(\cos x)dx
  2. B.π∫0πf(sin⁡x)dx\pi\int_0^\pi f(\sin x)dx
  3. C.π2∫0π/2f(sin⁡x)dx\frac{\pi}{2}\int_0^{\pi/2}f(\sin x)dx
  4. D.π∫0π/2f(cos⁡x)dx\pi\int_0^{\pi/2}f(\cos x)dx

Correct answer: D

Explanation: I=π2∫0πf(sin⁡x)=π∫0π/2f(cos⁡x)I=\frac\pi2\int_0^\pi f(\sin x)=\pi\int_0^{\pi/2}f(\cos x).

Question 15

Mathematics · 4 marks · −1

A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is

  1. A.x + y = 7
  2. B.3x - 4y + 7 = 0
  3. C.4x + 3y = 24
  4. D.3x + 4y = 25

Correct answer: C

Explanation: Intercepts 6 and 8.

Question 16

Mathematics · 4 marks · −1

The two lines x = ay + b, z = cy + d; and x = a'y + b', z = c'y + d' are perpendicular to each other if

  1. A.aa' + cc' = -1
  2. B.aa' + cc' = 1
  3. C.aa′+cc′=−1\frac{a}{a'}+\frac{c}{c'}=-1
  4. D.aa′+cc′=1\frac{a}{a'}+\frac{c}{c'}=1

Correct answer: A

Explanation: Directions (a, 1, c) and (a', 1, c').

Question 17

Mathematics · 4 marks · −1

The locus of the vertices of the family of parabolas y=a3x23+a2x2−2ay = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a is

  1. A.xy=10564xy = \frac{105}{64}
  2. B.xy=34xy = \frac{3}{4}
  3. C.xy=3516xy = \frac{35}{16}
  4. D.xy=64105xy = \frac{64}{105}

Correct answer: A

Explanation: Vertex (−34a,−35a16)\left(-\frac{3}{4a},-\frac{35a}{16}\right).

Question 18

Mathematics · 4 marks · −1

The values of a, for which the points A, B, C with position vectors 2i^−j^+k^2\hat{i} - \hat{j} + \hat{k}, i^−3j^−5k^\hat{i} - 3\hat{j} - 5\hat{k} and ai^−3j^+k^a\hat{i} - 3\hat{j} + \hat{k} respectively are the vertices of a right-angled triangle with C=π2C = \frac{\pi}{2} are

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

Correct answer: A

Explanation: CA⃗⋅CB⃗=(2−a)(1−a)=0\vec{CA}\cdot\vec{CB}=(2-a)(1-a)=0.

Question 19

Mathematics · 4 marks · −1

∫−3π/2−π/2[(x+π)3+cos⁡2(x+3π)]dx\int_{-3\pi/2}^{-\pi/2}\left[(x+\pi)^3 + \cos^2(x+3\pi)\right]dx is equal to

  1. A.π432\frac{\pi^4}{32}
  2. B.π432+π2\frac{\pi^4}{32}+\frac{\pi}{2}
  3. C.π2\frac{\pi}{2}
  4. D.π4−1\frac{\pi}{4}-1

Correct answer: C

Explanation: With t=x+πt=x+\pi: odd part vanishes, ∫−π/2π/2cos⁡2t dt=π2\int_{-\pi/2}^{\pi/2}\cos^2t\,dt=\frac\pi2.

Question 20

Mathematics · 4 marks · −1

If x is real, the maximum value of 3x2+9x+173x2+9x+7\frac{3x^2+9x+17}{3x^2+9x+7} is

  1. A.1/4
  2. B.41
  3. C.1
  4. D.17/7

Correct answer: B

Explanation: Minimum of 3x2+9x+73x^2+9x+7 is 14\frac14.

Question 21

Mathematics · 4 marks · −1

In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is

  1. A.35\frac{3}{5}
  2. B.12\frac{1}{2}
  3. C.45\frac{4}{5}
  4. D.15\frac{1}{\sqrt{5}}

Correct answer: A

Explanation: ae = 3, b = 4, a = 5.

Question 22

Mathematics · 4 marks · −1

Let A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B=(a00b)B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, a,b∈Na, b \in N. Then

  1. A.there cannot exist any B such that AB = BA
  2. B.there exist more than one but finite number of B's such that AB = BA
  3. C.there exists exactly one B such that AB = BA
  4. D.there exist infinitely many B's such that AB = BA

Correct answer: D

Explanation: AB = BA iff a = b.

Question 23

Mathematics · 4 marks · −1

The function f(x)=x2+2xf(x) = \frac{x}{2} + \frac{2}{x} has a local minimum at

  1. A.x = 2
  2. B.x = -2
  3. C.x = 0
  4. D.x = 1

Correct answer: A

Explanation: f′=12−2x2=0f'=\frac12-\frac2{x^2}=0, f′′>0f''>0 at x = 2.

Question 24

Mathematics · 4 marks · −1

Angle between the tangents to the curve y=x2−5x+6y = x^2 - 5x + 6 at the points (2, 0) and (3, 0) is

  1. A.π2\frac{\pi}{2}
  2. B.π2\frac{\pi}{2} (duplicate as printed)
  3. C.π6\frac{\pi}{6}
  4. D.π4\frac{\pi}{4}

Correct answer: B

Explanation: Slopes -1 and 1: perpendicular.

Question 25

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \dots be terms of an A.P. If a1+a2+⋯+apa1+a2+⋯+aq=p2q2\frac{a_1+a_2+\dots+a_p}{a_1+a_2+\dots+a_q} = \frac{p^2}{q^2}, p≠qp \ne q, then a6a21\frac{a_6}{a_{21}} equals

  1. A.4111\frac{41}{11}
  2. B.72\frac{7}{2}
  3. C.27\frac{2}{7}
  4. D.1141\frac{11}{41}

Correct answer: D

Explanation: an∝2n−1a_n\propto2n-1.

Question 26

Mathematics · 4 marks · −1

The set of points where f(x)=x1+∣x∣f(x) = \frac{x}{1+|x|} is differentiable is

  1. A.(−∞,0)∪(0,∞)(-\infty, 0)\cup(0, \infty)
  2. B.(−∞,−1)∪(−1,∞)(-\infty, -1)\cup(-1, \infty)
  3. C.(−∞,∞)(-\infty, \infty)
  4. D.(0,∞)(0, \infty)

Correct answer: C

Explanation: One-sided derivatives at 0 are both 1.

Question 27

Mathematics · 4 marks · −1

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is

  1. A.32x2\frac{3}{2}x^2
  2. B.x38\sqrt{\frac{x^3}{8}}
  3. C.12x2\frac{1}{2}x^2
  4. D.πx2\pi x^2

Correct answer: C

Explanation: 12x2sin⁡θ\frac12x^2\sin\theta.

Question 28

Mathematics · 4 marks · −1

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is

  1. A.5040
  2. B.6210
  3. C.385
  4. D.1110

Correct answer: C

Explanation: ∑k=14(10k)=385\sum_{k=1}^4\binom{10}{k}=385.

Question 29

Mathematics · 4 marks · −1

If the expansion in powers of x of the function 1(1−ax)(1−bx)\frac{1}{(1-ax)(1-bx)} is a0+a1x+a2x2+a3x3+…a_0 + a_1x + a_2x^2 + a_3x^3 + \dots, then ana_n is

  1. A.bn−anb−a\frac{b^n-a^n}{b-a}
  2. B.an−bnb−a\frac{a^n-b^n}{b-a}
  3. C.an+1−bn+1b−a\frac{a^{n+1}-b^{n+1}}{b-a}
  4. D.bn+1−an+1b−a\frac{b^{n+1}-a^{n+1}}{b-a}

Correct answer: D

Explanation: ∑k=0nakbn−k\sum_{k=0}^na^kb^{n-k}.

Question 30

Mathematics · 4 marks · −1

For natural numbers m, n if (1−y)m(1+y)n=1+a1y+a2y2+…(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \dots, and a1=a2=10a_1 = a_2 = 10, then (m, n) is

  1. A.(20, 45)
  2. B.(35, 20)
  3. C.(45, 35)
  4. D.(35, 45)

Correct answer: D

Explanation: n−m=10n-m=10, (m−n)2−(m+n)=20(m-n)^2-(m+n)=20.

Question 31

Mathematics · 4 marks · −1

The value of ∫1a[x]f′(x)dx\int_1^a[x]f'(x)dx, a > 1, where [x] denotes the greatest integer not exceeding x is

  1. A.af(a)−{f(1)+f(2)+⋯+f([a])}af(a) - \{f(1) + f(2) + \dots + f([a])\}
  2. B.[a]f(a)−{f(1)+f(2)+⋯+f([a])}[a]f(a) - \{f(1) + f(2) + \dots + f([a])\}
  3. C.[a]f([a])−{f(1)+f(2)+⋯+f(a)}[a]f([a]) - \{f(1) + f(2) + \dots + f(a)\}
  4. D.af([a])−{f(1)+f(2)+⋯+f(a)}af([a]) - \{f(1) + f(2) + \dots + f(a)\}

Correct answer: B

Explanation: Split over unit intervals and telescope.

Question 32

Mathematics · 4 marks · −1

If the lines 3x−4y−7=03x - 4y - 7 = 0 and 2x−3y−5=02x - 3y - 5 = 0 are two diameters of a circle of area 49π49\pi square units, the equation of the circle is

  1. A.x2+y2+2x−2y−47=0x^2 + y^2 + 2x - 2y - 47 = 0
  2. B.x2+y2+2x−2y−62=0x^2 + y^2 + 2x - 2y - 62 = 0
  3. C.x2+y2−2x+2y−62=0x^2 + y^2 - 2x + 2y - 62 = 0
  4. D.x2+y2−2x+2y−47=0x^2 + y^2 - 2x + 2y - 47 = 0

Correct answer: D

Explanation: Centre (1, -1), r = 7.

Question 33

Mathematics · 4 marks · −1

The differential equation whose solution is Ax2+By2=1Ax^2 + By^2 = 1, where A and B are arbitrary constants is of

  1. A.second order and second degree
  2. B.first order and second degree
  3. C.first order and first degree
  4. D.second order and first degree

Correct answer: D

Explanation: xyy′′+xy′2−yy′=0xyy''+xy'^2-yy'=0.

Question 34

Mathematics · 4 marks · −1

Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of 2π3\frac{2\pi}{3} at its centre is

  1. A.x2+y2=32x^2 + y^2 = \frac{3}{2}
  2. B.x2+y2=1x^2 + y^2 = 1
  3. C.x2+y2=274x^2 + y^2 = \frac{27}{4}
  4. D.x2+y2=94x^2 + y^2 = \frac{9}{4}

Correct answer: D

Explanation: Distance 3cos⁡π3=323\cos\frac\pi3=\frac32.

Question 35

Mathematics · 4 marks · −1

If (a,a2)(a, a^2) falls inside the angle made by the lines y=x2y = \frac{x}{2}, x > 0 and y = 3x, x > 0, then a belongs to

  1. A.(0,12)\left(0, \frac{1}{2}\right)
  2. B.(3,∞)(3, \infty)
  3. C.(12,3)\left(\frac{1}{2}, 3\right)
  4. D.(−3,−12)\left(-3, -\frac{1}{2}\right)

Correct answer: C

Explanation: a2<3aa^2<3a and a2>a2a^2>\frac a2.

Question 36

Mathematics · 4 marks · −1

If z2+z+1=0z^2 + z + 1 = 0, where z is a complex number, then the value of (z+1z)2+(z2+1z2)2+(z3+1z3)2+⋯+(z6+1z6)2\left(z+\frac{1}{z}\right)^2 + \left(z^2+\frac{1}{z^2}\right)^2 + \left(z^3+\frac{1}{z^3}\right)^2 + \dots + \left(z^6+\frac{1}{z^6}\right)^2 is

  1. A.18
  2. B.54
  3. C.6
  4. D.12

Correct answer: D

Explanation: Terms 1, 1, 4, 1, 1, 4.

Question 37

Mathematics · 4 marks · −1

If 0<x<π0 < x < \pi and cos⁡x+sin⁡x=12\cos x + \sin x = \frac{1}{2}, then tan⁡x\tan x is

  1. A.1−74\frac{1-\sqrt{7}}{4}
  2. B.4−73\frac{4-\sqrt{7}}{3}
  3. C.−4+73-\frac{4+\sqrt{7}}{3}
  4. D.1+74\frac{1+\sqrt{7}}{4}

Correct answer: C

Explanation: sin⁡2x=−34\sin2x=-\frac34, x obtuse: 3t2+8t+3=03t^2+8t+3=0, t=−4−73t=\frac{-4-\sqrt7}{3}.

Question 38

Mathematics · 4 marks · −1

If a1,a2,…,ana_1, a_2, \dots, a_n are in H.P., then the expression a1a2+a2a3+⋯+an−1ana_1a_2 + a_2a_3 + \dots + a_{n-1}a_n is equal to

  1. A.n(a1−an)n(a_1 - a_n)
  2. B.(n−1)(a1−an)(n-1)(a_1 - a_n)
  3. C.na1anna_1a_n
  4. D.(n−1)a1an(n-1)a_1a_n

Correct answer: D

Explanation: Telescoping with the common difference of reciprocals.

Question 39

Mathematics · 4 marks · −1

If xm⋅yn=(x+y)m+nx^m\cdot y^n = (x + y)^{m+n}, then dydx\frac{dy}{dx} is

  1. A.yx\frac{y}{x}
  2. B.x+yxy\frac{x+y}{xy}
  3. C.xy
  4. D.xy\frac{x}{y}

Correct answer: A

Explanation: Logarithmic differentiation.

Question 40

Physics · 4 marks · −1

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as v=αxv = \alpha\sqrt{x}. The displacement of the particle varies with time as

  1. A.t3t^3
  2. B.t2t^2
  3. C.t
  4. D.t1/2t^{1/2}

Correct answer: B

Explanation: 2x=αt2\sqrt x=\alpha t.

Question 41

Physics · 4 marks · −1

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45∘45^\circ with the initial vertical direction is

  1. A.Mg(2−1)Mg(\sqrt{2}-1)
  2. B.Mg(2+1)Mg(\sqrt{2}+1)
  3. C.Mg2Mg\sqrt{2}
  4. D.Mg2\frac{Mg}{\sqrt{2}}

Correct answer: A

Explanation: Work-energy with constant horizontal F: Fℓsin⁡45∘=Mgℓ(1−cos⁡45∘)F\ell\sin45^\circ=Mg\ell(1-\cos45^\circ).

Question 42

Physics · 4 marks · −1

A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms−1^{-1}. The kinetic energy of the other mass is

  1. A.96 J
  2. B.144 J
  3. C.288 J
  4. D.192 J

Correct answer: C

Explanation: v = 12 m/s.

Question 43

Physics · 4 marks · −1

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is

  1. A.0.5 J
  2. B.-0.5 J
  3. C.-1.25 J
  4. D.1.25 J

Correct answer: C

Explanation: −12mv2-\frac12mv^2.

Question 44

Physics · 4 marks · −1

A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms−1^{-1}. The velocity of sound in air is 300 ms−1^{-1}. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the whistle is

  1. A.30 ms−1^{-1}
  2. B.15215\sqrt{2} ms−1^{-1}
  3. C.15/215/\sqrt{2} ms−1^{-1}
  4. D.15 ms−1^{-1}

Correct answer: D

Explanation: 10000=9500300300−v10000=9500\frac{300}{300-v}.

Question 45

Physics · 4 marks · −1

A electric dipole is placed at an angle of 30∘30^\circ to a non-uniform electric field. The dipole will experience

  1. A.a torque only
  2. B.a translational force only in the direction of the field
  3. C.a translational force only in a direction normal to the direction of the field
  4. D.a torque as well as a translational force

Correct answer: D

Explanation: Non-uniform field gives a net force as well.

Question 46

Physics · 4 marks · −1

The Kirchhoff's first law (∑i=0\sum i = 0) and second law (∑iR=∑E\sum iR = \sum E), where the symbols have their usual meanings, are respectively based on

  1. A.conservation of charge, conservation of energy
  2. B.conservation of charge, conservation of momentum
  3. C.conservation of energy, conservation of charge
  4. D.conservation of momentum, conservation of charge

Correct answer: A

Explanation: Standard.

Question 47

Physics · 4 marks · −1

In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

  1. A.circle
  2. B.helix
  3. C.straight line
  4. D.ellipse

Correct answer: C

Explanation: Velocity stays parallel to B.

Question 48

Physics · 4 marks · −1

Needles N1N_1, N2N_2 and N3N_3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

  1. A.attract all three of them
  2. B.attract N1N_1 and N2N_2 strongly but repel N3N_3
  3. C.attract N1N_1 strongly, N2N_2 weakly and repel N3N_3 weakly
  4. D.attract N1N_1 strongly, but repel N2N_2 and N3N_3 weakly

Correct answer: C

Explanation: Standard behaviour.

Question 49

Physics · 4 marks · −1

Which of the following units denotes the dimensions ML2/Q2ML^2/Q^2, where Q denotes the electric charge?

  1. A.Weber (Wb)
  2. B.Wb/m2^2
  3. C.Henry (H)
  4. D.H/m2^2

Correct answer: C

Explanation: L=ϕIL=\frac{\phi}{I} has ML2Q−2ML^2Q^{-2}.

Question 50

Physics · 4 marks · −1

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to

  1. A.300 N
  2. B.150 N
  3. C.3 N
  4. D.30 N

Correct answer: D

Explanation: 0.15⋅200.1\frac{0.15\cdot20}{0.1}.

Question 51

Physics · 4 marks · −1

Consider a two particle system with particles having masses m1m_1 and m2m_2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

  1. A.d
  2. B.m2m1d\frac{m_2}{m_1}d
  3. C.m1m1+m2d\frac{m_1}{m_1+m_2}d
  4. D.m1m2d\frac{m_1}{m_2}d

Correct answer: D

Explanation: m1d=m2xm_1d=m_2x.

Question 52

Physics · 4 marks · −1

Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?

  1. A.112\frac{1}{12} s
  2. B.16\frac{1}{6} s
  3. C.14\frac{1}{4} s
  4. D.13\frac{1}{3} s

Correct answer: B

Explanation: cos⁡2ωt=34\cos^2\omega t=\frac34, ω=π\omega=\pi.

Question 53

Physics · 4 marks · −1

The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is

  1. A.100 s
  2. B.0.01 s
  3. C.10 s
  4. D.0.1 s

Correct answer: B

Explanation: T=2πAvmaxT=\frac{2\pi A}{v_{max}}.

Question 54

Physics · 4 marks · −1

A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

  1. A.10.5 Hz
  2. B.105 Hz
  3. C.1.05 Hz
  4. D.1050 Hz

Correct answer: B

Explanation: Successive harmonics differ by the fundamental.

Question 55

Physics · 4 marks · −1

Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth, at a distance r from the Sun. (where r0r_0 is the radius of the Earth and σ\sigma is Stefan's constant)

  1. A.R2σT4r2\frac{R^2\sigma T^4}{r^2}
  2. B.4πr02R2σT4r2\frac{4\pi r_0^2R^2\sigma T^4}{r^2}
  3. C.πr02R2σT4r2\frac{\pi r_0^2R^2\sigma T^4}{r^2}
  4. D.r02R2σT44πr2\frac{r_0^2R^2\sigma T^4}{4\pi r^2}

Correct answer: C

Explanation: σT44πR2⋅πr024πr2\sigma T^44\pi R^2\cdot\frac{\pi r_0^2}{4\pi r^2}.

Question 56

Physics · 4 marks · −1

The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1D_1 and D2D_2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then

  1. A.D1>D2D_1 > D_2
  2. B.D1<D2D_1 < D_2
  3. C.D1=D2D_1 = D_2
  4. D.D1D_1 can be less than or greater than depending upon the angle of prism

Correct answer: B

Explanation: Larger index, larger deviation.

Question 57

Physics · 4 marks · −1

In a Wheatstone's bridge, there resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1S_1 and S2S_2 connected in parallel. The condition for bridge to be balanced will be

  1. A.PQ=RS1+S2\frac{P}{Q} = \frac{R}{S_1+S_2}
  2. B.PQ=2RS1+S2\frac{P}{Q} = \frac{2R}{S_1+S_2}
  3. C.PQ=R(S1+S2)S1S2\frac{P}{Q} = \frac{R(S_1+S_2)}{S_1S_2}
  4. D.PQ=R(S1+S2)2S1S2\frac{P}{Q} = \frac{R(S_1+S_2)}{2S_1S_2}

Correct answer: C

Explanation: Fourth arm S1S2S1+S2\frac{S_1S_2}{S_1+S_2}.

Question 58

Physics · 4 marks · −1

The current I drawn from the 5 volt source will be

Figure for question 58
  1. A.0.17 A
  2. B.0.33 A
  3. C.0.5 A
  4. D.0.67 A

Correct answer: C

Explanation: The network is a balanced Wheatstone bridge; equivalent resistance 10 Ω10\,\Omega.

Question 59

Physics · 4 marks · −1

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ\Omega with C = 2μ\muF. The resonant frequency ω\omega is 200 rad/s. At resonance the voltage across L is

  1. A.4×10−34\times10^{-3} V
  2. B.2.5×10−22.5\times10^{-2} V
  3. C.40 V
  4. D.250 V

Correct answer: D

Explanation: I=0.1I=0.1 A; VL=VC=IωC=250V_L=V_C=\frac{I}{\omega C}=250 V.

Question 60

Physics · 4 marks · −1

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2−V1=20V_2 - V_1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6×10−19e = 1.6\times10^{-19} C, me=9.11×10−31m_e = 9.11\times10^{-31} kg)

Figure for question 60
  1. A.32×10−1932\times10^{-19} m/s
  2. B.2.65×1062.65\times10^6 m/s
  3. C.7.02×10127.02\times10^{12} m/s
  4. D.1.87×1061.87\times10^6 m/s

Correct answer: B

Explanation: v=2eV/mv=\sqrt{2eV/m}.

Question 61

Physics · 4 marks · −1

The resistance of a bulb filament is 100 Ω100\,\Omega at a temperature of 100∘100^\circC. If its temperature coefficient of resistance be 0.005 per ∘^\circC, its resistance will become 200 Ω200\,\Omega at a temperature of

  1. A.200∘200^\circC
  2. B.300∘300^\circC
  3. C.400∘400^\circC
  4. D.500∘500^\circC

Correct answer: B

Explanation: 200=100(1+0.005Δt)200=100(1+0.005\Delta t) taking 100 Ω100\,\Omega as reference.

Question 62

Physics · 4 marks · −1

In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω\omega in a magnetic field B. The maximum value of emf generated in the coil is

  1. A.N.A.B.ω\omega
  2. B.N.A.B.R.ω\omega
  3. C.N.A.B
  4. D.N.A.B.R

Correct answer: A

Explanation: e0=NABωe_0=NAB\omega.

Question 63

Physics · 4 marks · −1

The flux linked with a coil at any instant 't' is given by ϕ=10t2−50t+250\phi = 10t^2 - 50t + 250. The induced emf at t = 3 s is

  1. A.190 V
  2. B.-190 V
  3. C.-10 V
  4. D.10 V

Correct answer: C

Explanation: −(20t−50)-(20t-50) at t = 3.

Question 64

Physics · 4 marks · −1

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will

  1. A.flow from Antimony to Bismuth at the cold junction
  2. B.flow from Antimony to Bismuth at the hot junction
  3. C.flow from Bismuth to Antimony at the cold junction
  4. D.not flow through the thermocouple

Correct answer: A

Explanation: Seebeck: Sb to Bi at the cold junction.

Question 65

Physics · 4 marks · −1

The time by a photoelectron to come out after the photon strikes is approximately

  1. A.10−110^{-1} s
  2. B.10−410^{-4} s
  3. C.10−1010^{-10} s
  4. D.10−1610^{-16} s

Correct answer: C

Explanation: Photoemission is practically instantaneous (~10−1010^{-10} s).

Question 66

Physics · 4 marks · −1

An alpha nucleus of energy 12mv2\frac{1}{2}mv^2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

  1. A.1Ze\frac{1}{Ze}
  2. B.v2v^2
  3. C.1m\frac{1}{m}
  4. D.1v4\frac{1}{v^4}

Correct answer: C

Explanation: r0∝Zemv2r_0\propto\frac{Ze}{mv^2}.

Question 67

Physics · 4 marks · −1

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in

  1. A.X-ray region
  2. B.ultra-violet region
  3. C.infra-red region
  4. D.visible region

Correct answer: B

Explanation: Photon 11.2 eV, λ≈110\lambda\approx110 nm.

Question 68

Physics · 4 marks · −1

The energy spectrum of β\beta-particles [number N(E) as a function of β\beta-energy E] emitted from a radioactive source is

Figure for question 68
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: D

Explanation: Continuous spectrum from 0 to E0E_0 skewed to low energies.

Question 69

Physics · 4 marks · −1

When 3Li7_3Li^7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8_4Be^8, the emitted particles will be

  1. A.neutrons
  2. B.alpha particles
  3. C.beta particles
  4. D.gamma photons

Correct answer: D

Explanation: p+7Li→8Be+γp+{}^7Li\rightarrow{}^8Be+\gamma.

Question 70

Physics · 4 marks · −1

A solid which is transparent to visible light and whose conductivity increases with temperature is formed by

  1. A.Metallic binding
  2. B.Ionic binding
  3. C.Covalent binding
  4. D.Van der Waals binding

Correct answer: C

Explanation: A covalent semiconductor/insulator with a wide gap.

Question 71

Physics · 4 marks · −1

If the ratio of the concentration of electrons to that of holes in a semiconductor is 75\frac{7}{5} and the ratio of currents is 74\frac{7}{4}, then what is the ratio of their drift velocities?

  1. A.47\frac{4}{7}
  2. B.58\frac{5}{8}
  3. C.45\frac{4}{5}
  4. D.54\frac{5}{4}

Correct answer: D

Explanation: vevh=IeIhnhne\frac{v_e}{v_h}=\frac{I_e}{I_h}\frac{n_h}{n_e}.

Question 72

Physics · 4 marks · −1

In a common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β\beta) will be

  1. A.48
  2. B.49
  3. C.50
  4. D.51

Correct answer: B

Explanation: β=5.4880.112\beta=\frac{5.488}{0.112}.

Question 73

Physics · 4 marks · −1

The potential energy of a 1 kg particle free to move along the x-axis is given by V(x)=(x44−x22)V(x) = \left(\frac{x^4}{4} - \frac{x^2}{2}\right) J. The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

  1. A.2
  2. B.3/23/\sqrt{2}
  3. C.2\sqrt{2}
  4. D.1/21/\sqrt{2}

Correct answer: B

Explanation: Vmin=−14V_{min}=-\frac14 at x=±1x=\pm1; KEmax=94KE_{max}=\frac94.

Question 74

Physics · 4 marks · −1

A force of −Fk^-F\hat{k} acts on O, the origin of the coordinate system. The torque about the point (1, -1) is

  1. A.−F(i^−j^)-F(\hat{i}-\hat{j})
  2. B.F(i^−j^)F(\hat{i}-\hat{j})
  3. C.−F(i^+j^)-F(\hat{i}+\hat{j})
  4. D.F(i^+j^)F(\hat{i}+\hat{j})

Correct answer: C

Explanation: r⃗=(−1,1,0)\vec r=(-1,1,0); r⃗×(−Fk^)\vec r\times(-F\hat k).

Question 75

Physics · 4 marks · −1

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω\omega. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω′\omega' =

  1. A.ωm(m+2M)\frac{\omega m}{(m+2M)}
  2. B.ω(m+2M)m\frac{\omega(m+2M)}{m}
  3. C.ω(m−2M)(m+2M)\frac{\omega(m-2M)}{(m+2M)}
  4. D.ωm(m+M)\frac{\omega m}{(m+M)}

Correct answer: A

Explanation: Angular momentum conservation.

Question 76

Physics · 4 marks · −1

If the terminal speed of a sphere of gold (density = 19.5 kg/m3^3) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3^3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3^3) of the same size in the same liquid.

  1. A.0.2 m/s
  2. B.0.4 m/s
  3. C.0.133 m/s
  4. D.0.1 m/s

Correct answer: D

Explanation: v∝(ρ−σ)v\propto(\rho-\sigma): 0.2⋅9180.2\cdot\frac{9}{18}.

Question 77

Physics · 4 marks · −1

The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘7^\circC. The gas is (R = 8.3 J mol−1^{-1} K−1^{-1})

  1. A.monoatomic
  2. B.diatomic
  3. C.triatomic
  4. D.a mixture of monoatomic and diatomic

Correct answer: B

Explanation: Cv=1460001000⋅7≈20.9=52RC_v=\frac{146000}{1000\cdot7}\approx20.9=\frac52R.

Question 78

Physics · 4 marks · −1

The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is

  1. A.3.3×10−33.3\times10^{-3} J/m3^3
  2. B.4.58×10−64.58\times10^{-6} J/m3^3
  3. C.6.37×10−96.37\times10^{-9} J/m3^3
  4. D.81.35×10−1281.35\times10^{-12} J/m3^3

Correct answer: B

Explanation: u=ϵ0Erms2u=\epsilon_0E_{rms}^2.

Question 79

Physics · 4 marks · −1

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω\omega. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

  1. A.at the highest position of the platform
  2. B.at the mean position of the platform
  3. C.for an amplitude of gω2\frac{g}{\omega^2}
  4. D.for an amplitude of g2ω2\frac{g^2}{\omega^2}

Correct answer: C

Explanation: Contact lost when Aω2=gA\omega^2=g (at the top).

Question 80

Physics · 4 marks · −1

An electric bulb is rated 220 volt - 100 watt. The power consumed by it when operated on 110 volt will be

  1. A.50 watt
  2. B.75 watt
  3. C.40 watt
  4. D.25 watt

Correct answer: D

Explanation: P∝V2P\propto V^2.

Question 81

Physics · 4 marks · −1

The anode voltage of a photocell is kept fixed. The wavelength λ\lambda of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

Figure for question 81
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: C

Explanation: Current falls as λ\lambda increases, reaching zero at threshold.

Question 82

Physics · 4 marks · −1

If the binding energy per nucleon in 37Li^7_3Li and 24He^4_2He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+37Li→2 24Hep + ^7_3Li \rightarrow 2\,^4_2He energy of proton must be

  1. A.39.2 MeV
  2. B.28.24 MeV
  3. C.17.28 MeV
  4. D.1.46 MeV

Correct answer: C

Explanation: Q = 8(7.06) - 7(5.60) = 17.28 MeV (energy released).

Question 83

Physics · 4 marks · −1

If the lattice constant of this semiconductor is decreased, then which of the following is correct?

Figure for question 83
  1. A.All EcE_c, EgE_g, EvE_v decrease
  2. B.All EcE_c, EgE_g, EvE_v increase
  3. C.EcE_c, and EvE_v increase but EgE_g decreases
  4. D.EcE_c, and EvE_v decrease EgE_g increases

Correct answer: D

Explanation: Closer atoms broaden the bands and widen the gap (keyed).

Question 84

Physics · 4 marks · −1

In the following, which one of the diodes is reverse biased?

Figure for question 84
  1. A.Diagram (1)
  2. B.Diagram (2)
  3. C.Diagram (3)
  4. D.Diagram (4)

Correct answer: A

Explanation: In (1) the cathode side is at +5 V through R while the anode is grounded.

Question 85

Physics · 4 marks · −1

The circuit has two oppositely connected ideal diodes in parallel. What is the current following in the circuit?

Figure for question 85
  1. A.1.33 A
  2. B.1.71 A
  3. C.2.00 A
  4. D.2.31 A

Correct answer: C

Explanation: D1D_1 blocks; current through 4 + 2 ohm: 2 A.

Question 86

Physics · 4 marks · −1

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28×10−26.28\times10^{-2} Weber/m2^2. Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is

  1. A.1.05×10−41.05\times10^{-4} Weber/m2^2
  2. B.1.05×10−21.05\times10^{-2} Weber/m2^2
  3. C.1.05×10−51.05\times10^{-5} Weber/m2^2
  4. D.1.05×10−31.05\times10^{-3} Weber/m2^2

Correct answer: B

Explanation: B∝niB\propto ni: factor 16\frac16.

Question 87

Physics · 4 marks · −1

Four point masses, each of value m, are placed at the corners of a square ABCD of side ℓ\ell. The moment of inertia through A and parallel to BD is

  1. A.mℓ2m\ell^2
  2. B.2mℓ22m\ell^2
  3. C.3mℓ2\sqrt{3}m\ell^2
  4. D.3mℓ23m\ell^2

Correct answer: D

Explanation: B, D at ℓ2\frac{\ell}{\sqrt2}, C at 2ℓ\sqrt2\ell: mℓ22⋅2+2mℓ2m\frac{\ell^2}{2}\cdot2+2m\ell^2.

Question 88

Physics · 4 marks · −1

A wire elongates by ℓ\ell mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)

  1. A.ℓ/2\ell/2
  2. B.ℓ\ell
  3. C.2ℓ2\ell
  4. D.zero

Correct answer: B

Explanation: Tension is W in both cases.

Question 89

Physics · 4 marks · −1

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0T_0, while Box B contains one mole of helium at temperature (7/3)T0(7/3)T_0. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, TfT_f, in terms of T0T_0 is

  1. A.Tf=52T0T_f = \frac{5}{2}T_0
  2. B.Tf=37T0T_f = \frac{3}{7}T_0
  3. C.Tf=73T0T_f = \frac{7}{3}T_0
  4. D.Tf=32T0T_f = \frac{3}{2}T_0

Correct answer: D

Explanation: 52R(Tf−T0)+32R(Tf−73T0)=0\frac52R(T_f-T_0)+\frac32R(T_f-\frac73T_0)=0.

Question 90

Physics · 4 marks · −1

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is

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

Correct answer: D

Explanation: Equal potentials: E∝1rE\propto\frac1r.

Question 91

Physics · 4 marks · −1

An inductor (L = 100 mH), a resistor (R = 100 Ω\Omega) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the circuit is

Figure for question 91
  1. A.1 A
  2. B.1/e A
  3. C.e A
  4. D.0.1 A

Correct answer: B

Explanation: I=1⋅e−Rt/L=e−1I=1\cdot e^{-Rt/L}=e^{-1} A.

Question 92

Chemistry · 4 marks · −1

HBr reacts with CH2=CH−OCH3CH_2 = CH - OCH_3 under anhydrous conditions at room temperature to give

  1. A.CH3CHOCH_3CHO and CH3BrCH_3Br
  2. B.BrCH2CHOBrCH_2CHO and CH3OHCH_3OH
  3. C.BrCH2−CH2−OCH3BrCH_2 - CH_2 - OCH_3
  4. D.H3C−CHBr−OCH3H_3C - CHBr - OCH_3

Correct answer: D

Explanation: Markovnikov addition via the oxocarbenium ion.

Question 93

Chemistry · 4 marks · −1

The IUPAC name of the compound shown below is (a cyclohexene ring with Cl on the alkene carbon C1 and Br on C3)

Figure for question 93
  1. A.2-bromo-6-chlorocyclohex-1-ene
  2. B.6-bromo-2-chlorocyclohexene
  3. C.3-bromo-1-chlorocyclohexene
  4. D.1-bromo-3-chlorocyclohexene

Correct answer: C

Explanation: Number the double bond first, then lowest locants.

Question 94

Chemistry · 4 marks · −1

The increasing order of the rate of HCN addition to compounds A - D is (A) HCHO (B) CH3COCH3CH_3COCH_3 (C) PhCOCH3PhCOCH_3 (D) PhCOPh

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

Correct answer: C

Explanation: Steric and electronic hindrance.

Question 95

Chemistry · 4 marks · −1

How many moles of magnesium phosphate, Mg3(PO4)2Mg_3(PO_4)_2 will contain 0.25 mole of oxygen atoms?

  1. A.0.02
  2. B.3.125×10−23.125\times10^{-2}
  3. C.1.25×10−21.25\times10^{-2}
  4. D.2.5×10−22.5\times10^{-2}

Correct answer: B

Explanation: 8 O per formula.

Question 96

Chemistry · 4 marks · −1

According to Bohr's theory, the angular momentum of an electron in 5th5^{th} orbit is

  1. A.25hπ25\frac{h}{\pi}
  2. B.1.0hπ1.0\frac{h}{\pi}
  3. C.10hπ10\frac{h}{\pi}
  4. D.2.5hπ2.5\frac{h}{\pi}

Correct answer: D

Explanation: 5h2π\frac{5h}{2\pi}.

Question 97

Chemistry · 4 marks · −1

Which of the following molecules/ions does not contain unpaired electrons?

  1. A.O22−O_2^{2-}
  2. B.B2B_2
  3. C.N2+N_2^+
  4. D.O2O_2

Correct answer: A

Explanation: Peroxide ion is diamagnetic.

Question 98

Chemistry · 4 marks · −1

Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius)

  1. A.203πr3\frac{20}{3}\pi r^3
  2. B.243πr3\frac{24}{3}\pi r^3
  3. C.123πr3\frac{12}{3}\pi r^3
  4. D.163πr3\frac{16}{3}\pi r^3

Correct answer: D

Explanation: 4 atoms.

Question 99

Chemistry · 4 marks · −1

A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will

  1. A.remain unchanged
  2. B.triple
  3. C.increase by a factor of 4
  4. D.double

Correct answer: C

Explanation: 222^2.

Question 100

Chemistry · 4 marks · −1

Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4H_2SO_4?

  1. A.2HI+H2SO4→I2+SO2+2H2O2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O
  2. B.Ca(OH)2+H2SO4→CaSO4+2H2OCa(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O
  3. C.NaCl+H2SO4→NaHSO4+HClNaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl
  4. D.2PCl5+H2SO4→2POCl3+2HCl+SO2Cl22PCl_5 + H_2SO_4 \rightarrow 2POCl_3 + 2HCl + SO_2Cl_2

Correct answer: A

Explanation: S reduced from +6 to +4.

Question 101

Chemistry · 4 marks · −1

The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2[Co(NO_2)(NH_3)_5]Cl_2 is

  1. A.nitrito-N-pentaamminecobalt (III) chloride
  2. B.nitrito-N-pentaamminecobalt (II) chloride
  3. C.pentaammine nitrito-N-cobalt (II) chloride
  4. D.pentaammine nitrito-N-cobalt (III) chloride

Correct answer: D

Explanation: Alphabetical ligand order; Co(III).

Question 102

Chemistry · 4 marks · −1

The term anomers of glucose refers to

  1. A.isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)
  2. B.a mixture of (D)-glucose and (L)-glucose
  3. C.enantiomers of glucose
  4. D.isomers of glucose that differ in configuration at carbon one (C-1)

Correct answer: D

Explanation: Anomers differ at C-1.

Question 103

Chemistry · 4 marks · −1

In the transformation of 92238U^{238}_{92}U to 92234U^{234}_{92}U, if one emission is an α\alpha-particle, what should be the other emission(s)?

  1. A.Two β−\beta^-
  2. B.Two β−\beta^- and one β+\beta^+
  3. C.One β−\beta^- and one γ\gamma
  4. D.One β+\beta^+ and one β−\beta^-

Correct answer: A

Explanation: Restore Z by two β−\beta^-.

Question 104

Chemistry · 4 marks · −1

Phenyl magnesium bromide reacts with methanol to give

  1. A.a mixture of anisole and Mg(OH)Br
  2. B.a mixture of benzene and Mg(OMe)Br
  3. C.a mixture of toluene and Mg(OH)Br
  4. D.a mixture of phenol and Mg(Me)Br

Correct answer: B

Explanation: Acid-base: PhMgBr+MeOH→PhH+Mg(OMe)BrPhMgBr+MeOH\rightarrow PhH+Mg(OMe)Br.

Question 105

Chemistry · 4 marks · −1

CH3Br+Nu−→CH3−Nu+Br−CH_3Br + Nu^- \rightarrow CH_3 - Nu + Br^-. The decreasing order of the rate of the above reaction with nucleophiles (Nu−Nu^-) A to D is [Nu−Nu^- = (A) PhO−PhO^-, (B) AcO−AcO^-, (C) HO−HO^-, (D) CH3O−CH_3O^-]

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

Correct answer: A

Explanation: Nucleophilicity follows basicity here.

Question 106

Chemistry · 4 marks · −1

The pyrimidine bases present in DNA are

  1. A.cytosine and adenine
  2. B.cytosine and guanine
  3. C.cytosine and thymine
  4. D.cytosine and uracil

Correct answer: C

Explanation: Pyrimidines in DNA: C and T.

Question 107

Chemistry · 4 marks · −1

Among the following the one that gives positive iodoform test upon reaction with I2I_2 and NaOH is

Figure for question 107
  1. A.CH3CH2CH(OH)CH2CH3CH_3CH_2CH(OH)CH_2CH_3
  2. B.C6H5CH2CH2OHC_6H_5CH_2CH_2OH
  3. C.2-methylpropan-1-ol ((CH3)2CHCH2OH(CH_3)_2CHCH_2OH)
  4. D.PhCH(OH)CH3PhCH(OH)CH_3

Correct answer: D

Explanation: Contains CH3CH(OH)−CH_3CH(OH)-.

Question 108

Chemistry · 4 marks · −1

The increasing order of stability of the following free radicals is

  1. A.(CH3)2C˙H<(CH3)3C˙<(C6H5)2C˙H<(C6H5)3C˙(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}
  2. B.(C6H5)3C˙<(C6H5)2C˙H<(CH3)3C˙<(CH3)2C˙H(C_6H_5)_3\dot{C} < (C_6H_5)_2\dot{C}H < (CH_3)_3\dot{C} < (CH_3)_2\dot{C}H
  3. C.(C6H5)2C˙H<(C6H5)3C˙<(CH3)3C˙<(CH3)2C˙H(C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C} < (CH_3)_3\dot{C} < (CH_3)_2\dot{C}H
  4. D.(CH3)2C˙H<(CH3)3C˙<(C6H5)3C˙<(C6H5)2C˙H(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_3\dot{C} < (C_6H_5)_2\dot{C}H

Correct answer: A

Explanation: Resonance > hyperconjugation.

Question 109

Chemistry · 4 marks · −1

Uncertainty in the position of an electron (mass = 9.1×10−319.1\times10^{-31} kg) moving with a velocity 300 ms−1^{-1}, accurate upto 0.001%, will be (h=6.63×10−34h = 6.63\times10^{-34} Js)

  1. A.19.2×10−219.2\times10^{-2} m
  2. B.5.76×10−25.76\times10^{-2} m
  3. C.1.92×10−21.92\times10^{-2} m
  4. D.3.84×10−23.84\times10^{-2} m

Correct answer: C

Explanation: Δx=h4πmΔv\Delta x=\frac{h}{4\pi m\Delta v} with Δv=3×10−3\Delta v=3\times10^{-3}.

Question 110

Chemistry · 4 marks · −1

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g). If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5PCl_5 is x, the partial pressure of PCl3PCl_3 will be

  1. A.(xx+1)P\left(\frac{x}{x+1}\right)P
  2. B.(2x1−x)P\left(\frac{2x}{1-x}\right)P
  3. C.(xx−1)P\left(\frac{x}{x-1}\right)P
  4. D.(x1−x)P\left(\frac{x}{1-x}\right)P

Correct answer: A

Explanation: Total moles 1+x1+x.

Question 111

Chemistry · 4 marks · −1

The standard enthalpy of formation (ΔfH∘\Delta_fH^\circ) at 298 K for methane, CH4CH_4(g), is -74.8 kJ mol−1^{-1}. The additional information required to determine the average energy for C - H bond formation would be

  1. A.the dissociation energy of H2H_2 and enthalpy of sublimation of carbon
  2. B.latent heat of vapourization of methane
  3. C.the first four ionization energies of carbon and electron gain enthalpy of hydrogen
  4. D.the dissociation energy of hydrogen molecule, H2H_2

Correct answer: A

Explanation: Born-Haber style cycle needs atomisation enthalpies.

Question 112

Chemistry · 4 marks · −1

Among the following mixtures, dipole-dipole as the major interaction, is present in

  1. A.benzene and ethanol
  2. B.acetonitrile and acetone
  3. C.KCl and water
  4. D.benzene and carbon tetrachloride

Correct answer: B

Explanation: Both polar aprotic molecules.

Question 113

Chemistry · 4 marks · −1

Fluorobenzene (C6H5FC_6H_5F) can be synthesized in the laboratory

  1. A.by heating phenol with HF and KF
  2. B.from aniline by diazotisation followed by heating the diazonium salt with HBF4HBF_4
  3. C.by direct fluorination of benzene with F2F_2 gas
  4. D.by reacting bromobenzene with NaF solution

Correct answer: B

Explanation: Balz-Schiemann reaction.

Question 114

Chemistry · 4 marks · −1

A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

  1. A.MCl2MCl_2 is more volatile than MCl4MCl_4
  2. B.MCl2MCl_2 is more soluble in anhydrous ethanol than MCl4MCl_4
  3. C.MCl2MCl_2 is more ionic than MCl4MCl_4
  4. D.MCl2MCl_2 is more easily hydrolysed than MCl4MCl_4

Correct answer: C

Explanation: Lower oxidation state is more ionic (Fajans).

Question 115

Chemistry · 4 marks · −1

Which of the following statements is true?

  1. A.H3PO3H_3PO_3 is a stronger acid than H2SO3H_2SO_3
  2. B.In aqueous medium HF is a stronger acid than HCl
  3. C.HClO4HClO_4 is a weaker acid than HClO3HClO_3
  4. D.HNO3HNO_3 is a stronger acid than HNO2HNO_2

Correct answer: D

Explanation: More oxygen, stronger oxoacid.

Question 116

Chemistry · 4 marks · −1

The molar conductivities ΛNaOAc∘\Lambda^\circ_{NaOAc} and ΛHCl∘\Lambda^\circ_{HCl} at infinite dilution in water at 25∘25^\circC are 91.0 and 426.2 S cm2^2/mol respectively. To calculate ΛHOAc∘\Lambda^\circ_{HOAc}, the additional value required is

  1. A.ΛH2O∘\Lambda^\circ_{H_2O}
  2. B.ΛKCl∘\Lambda^\circ_{KCl}
  3. C.ΛNaOH∘\Lambda^\circ_{NaOH}
  4. D.ΛNaCl∘\Lambda^\circ_{NaCl}

Correct answer: D

Explanation: Kohlrausch's law.

Question 117

Chemistry · 4 marks · −1

Which one of the following sets of ions represents a collection of isoelectronic species?

  1. A.K+K^+, Cl−Cl^-, Ca2+Ca^{2+}, Sc3+Sc^{3+}
  2. B.Ba2+Ba^{2+}, Sr2+Sr^{2+}, K+K^+, S2−S^{2-}
  3. C.N3−N^{3-}, O2−O^{2-}, F−F^-, S2−S^{2-}
  4. D.Li+Li^+, Na+Na^+, Mg2+Mg^{2+}, Ca2+Ca^{2+}

Correct answer: A

Explanation: All 18 electrons.

Question 118

Chemistry · 4 marks · −1

The correct order of increasing acid strength of the compounds (a) CH3CO2HCH_3CO_2H (b) MeOCH2CO2HMeOCH_2CO_2H (c) CF3CO2HCF_3CO_2H (d) (CH3)2CHCO2H(CH_3)_2CHCO_2H is

Figure for question 118
  1. A.b < d < a < c
  2. B.d < a < c < b
  3. C.d < a < b < c
  4. D.a < d < c < b

Correct answer: C

Explanation: +I lowers, -I raises acidity.

Question 119

Chemistry · 4 marks · −1

In which of the following molecules/ions are all the bonds not equal?

  1. A.SF4SF_4
  2. B.SiF4SiF_4
  3. C.XeF4XeF_4
  4. D.BF4−BF_4^-

Correct answer: A

Explanation: See-saw SF4SF_4 has axial and equatorial bonds.

Question 120

Chemistry · 4 marks · −1

What products are expected from the disproportionation reaction of hypochlorous acid?

  1. A.HClO3HClO_3 and Cl2OCl_2O
  2. B.HClO2HClO_2 and HClO4HClO_4
  3. C.HCl and Cl2OCl_2O
  4. D.HCl and HClO3HClO_3

Correct answer: D

Explanation: 3HOCl→2HCl+HClO33HOCl\rightarrow2HCl+HClO_3.

Question 121

Chemistry · 4 marks · −1

Nickel (Z = 28) combines with a uninegative monodentate ligand X−X^- to form a paramagnetic complex [NiX4]2−[NiX_4]^{2-}. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

  1. A.one, tetrahedral
  2. B.two, tetrahedral
  3. C.one, square planar
  4. D.two, square planar

Correct answer: B

Explanation: d8d^8, sp3sp^3.

Question 122

Chemistry · 4 marks · −1

In Fe(CO)5Fe(CO)_5, the Fe - C bond possesses

  1. A.π\pi-character only
  2. B.both σ\sigma and π\pi characters
  3. C.ionic character
  4. D.σ\sigma-character only

Correct answer: B

Explanation: Synergic bonding.

Question 123

Chemistry · 4 marks · −1

The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is

  1. A.F < S < P < B
  2. B.P < S < B < F
  3. C.B < P < S < F
  4. D.B < S < P < F

Correct answer: D

Explanation: P > S (half-filled p).

Question 124

Chemistry · 4 marks · −1

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If TiT_i is the initial temperature and TfT_f is the final temperature, which of the following statements is correct?

  1. A.(Tf)irrev>(Tf)rev(T_f)_{irrev} > (T_f)_{rev}
  2. B.Tf>TiT_f > T_i for reversible process but Tf=TiT_f = T_i for irreversible process
  3. C.(Tf)rev=(Tf)irrev(T_f)_{rev} = (T_f)_{irrev}
  4. D.Tf=TiT_f = T_i for both reversible and irreversible processes

Correct answer: A

Explanation: Adiabatic reversible expansion does more work, cooling more.

Question 125

Chemistry · 4 marks · −1

In Langmuir's model of adsorption of a gas on a solid surface

  1. A.the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered
  2. B.the adsorption at a single site on the surface may involve multiple molecules at the same time
  3. C.the mass of gas striking a given area of surface is proportional to the pressure of the gas
  4. D.the mass of gas striking a given area of surface is independent of the pressure of the gas

Correct answer: C

Explanation: Kinetic theory: collision rate proportional to pressure.

Question 126

Chemistry · 4 marks · −1

Rate of a reaction can be expressed by Arrhenius equation as: k=Ae−E/RTk = Ae^{-E/RT}. In this equation, E represents

  1. A.the energy above which all the colliding molecules will react
  2. B.the energy below which colliding molecules will not react
  3. C.the total energy of the reacting molecules at a temperature, T
  4. D.the fraction of molecules with energy greater than the activation energy of the reaction

Correct answer: B

Explanation: Activation energy threshold.

Question 127

Chemistry · 4 marks · −1

The structure of the major product formed in the following reaction: 3-iodobenzyl chloride →NaCN, DMF\xrightarrow{NaCN,\ DMF} ?

Figure for question 127
  1. A.3-cyanobenzyl cyanide (both I and Cl replaced by CN)
  2. B.benzyl chloride with CN para to CH2Cl and I retained
  3. C.3-cyanobenzyl chloride (aryl I replaced)
  4. D.3-iodophenylacetonitrile (only the benzylic Cl replaced)

Correct answer: D

Explanation: SN2S_N2 at the benzylic C; the aryl iodide is inert.

Question 128

Chemistry · 4 marks · −1

Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

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

Correct answer: D

Explanation: Anti-periplanar H is only available away from the phenyl carbon.

Question 129

Chemistry · 4 marks · −1

Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is

  1. A.Eclipse, Gauche, Anti
  2. B.Gauche, Eclipse, Anti
  3. C.Eclipse, Anti, Gauche
  4. D.Anti, Gauche, Eclipse

Correct answer: C

Explanation: Gauche stabilised by intramolecular H-bonding/gauche effect.

Question 130

Chemistry · 4 marks · −1

The structure of the compound that gives a tribromo derivative on treatment with bromine water is

Figure for question 130
  1. A.m-cresol (3-methylphenol)
  2. B.benzyl alcohol
  3. C.o-cresol (2-methylphenol)
  4. D.p-cresol (4-methylphenol)

Correct answer: A

Explanation: m-Cresol has three free o/p positions to OH.

Question 131

Chemistry · 4 marks · −1

The decreasing values of bond angles from NH3NH_3 (106∘^\circ) to SbH3SbH_3 (101∘^\circ) down group-15 of the periodic table is due to

  1. A.increasing bp-bp repulsion
  2. B.increasing p-orbital character in sp3sp^3
  3. C.decreasing lp-bp repulsion
  4. D.decreasing electronegativity

Correct answer: D

Explanation: Decreasing electronegativity of the central atom.

Question 132

Chemistry · 4 marks · −1

The alkene formed as a major product in the elimination reaction on heating (2-methylcyclohexyl)ethylmethyl-n-butylammonium hydroxide (Δ\Delta) is

Figure for question 132
  1. A.1-butene
  2. B.CH2=CH2CH_2 = CH_2
  3. C.1-methylcyclohexene
  4. D.3-methylcyclohexene

Correct answer: B

Explanation: Hofmann elimination favours the least substituted alkene: ethene from the N-ethyl group.

Question 133

Chemistry · 4 marks · −1

The 'spin-only' magnetic moment [in units of Bohr magneton, (μB\mu_B)] of Ni2+Ni^{2+} in aqueous solution would be (Atomic number of Ni = 28)

  1. A.2.84
  2. B.4.90
  3. C.0
  4. D.1.73

Correct answer: A

Explanation: Two unpaired: 8\sqrt8.

Question 134

Chemistry · 4 marks · −1

The equilibrium constant for the reaction SO3(g)⇌SO2(g)+12O2(g)SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g) is Kc=4.9×10−2K_c = 4.9\times10^{-2}. The value of KcK_c for the reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) will be

  1. A.416
  2. B.2.40×10−32.40\times10^{-3}
  3. C.9.8×10−29.8\times10^{-2}
  4. D.4.9×10−24.9\times10^{-2}

Correct answer: A

Explanation: (1/0.049)2(1/0.049)^2.

Question 135

Chemistry · 4 marks · −1

Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?

  1. A.The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group
  2. B.In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group
  3. C.Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens
  4. D.In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group

Correct answer: D

Explanation: Standard trends.

Question 136

Chemistry · 4 marks · −1

Given the data at 25∘25^\circC, Ag+I−→AgI+e−Ag + I^- \rightarrow AgI + e^-; E∘=0.152E^\circ = 0.152 V; Ag→Ag++e−Ag \rightarrow Ag^+ + e^-; E∘=−0.800E^\circ = -0.800 V. What is the value of log KspK_{sp} for AgI? (2.303RTF=0.059 V)\left(2.303\frac{RT}{F} = 0.059\ V\right)

  1. A.-8.12
  2. B.+8.612
  3. C.-37.83
  4. D.-16.13

Correct answer: D

Explanation: E∘=−0.952E^\circ=-0.952 V; log⁡Ksp=−0.9520.059\log K_{sp}=-\frac{0.952}{0.059}.

Question 137

Chemistry · 4 marks · −1

The following mechanism has been proposed for the reaction of NO with Br2Br_2 to form NOBr: NO(g)+Br2(g)⇌NOBr2(g)NO(g) + Br_2(g) \rightleftharpoons NOBr_2(g); NOBr2(g)+NO(g)→2NOBr(g)NOBr_2(g) + NO(g) \rightarrow 2NOBr(g). If the second step is the rate determining step, the order of the reaction with respect to NO(g) is

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

Correct answer: D

Explanation: Rate =kK[NO]2[Br2]=kK[NO]^2[Br_2].

Question 138

Chemistry · 4 marks · −1

Lanthanoid contraction is caused due to

  1. A.the appreciable shielding on outer electrons by 4f electrons from the nuclear charge
  2. B.the appreciable shielding on outer electrons by 5d electrons from the nuclear charge
  3. C.the same effective nuclear charge from Ce to Lu
  4. D.the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Correct answer: D

Explanation: Poor 4f shielding.

Question 139

Chemistry · 4 marks · −1

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω\Omega. The conductivity of this solution is 1.29 S m−1^{-1}. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω\Omega. The molar conductivity of 0.02 M solution of the electrolyte will be

  1. A.124×10−4124\times10^{-4} S m2^2 mol−1^{-1}
  2. B.1240×10−41240\times10^{-4} S m2^2 mol−1^{-1}
  3. C.1.24×10−41.24\times10^{-4} S m2^2 mol−1^{-1}
  4. D.12.4×10−412.4\times10^{-4} S m2^2 mol−1^{-1}

Correct answer: D

Explanation: Cell constant 129 m−1^{-1}; for the 0.2 M solution κ=129520\kappa=\frac{129}{520}, Λm=κ200=12.4×10−4\Lambda_m=\frac{\kappa}{200}=12.4\times10^{-4}.

Question 140

Chemistry · 4 marks · −1

The ionic mobility of alkali metal ions in aqueous solution is maximum for

  1. A.K+K^+
  2. B.Rb+Rb^+
  3. C.Li+Li^+
  4. D.Na+Na^+

Correct answer: B

Explanation: Least hydrated ion moves fastest.

Question 141

Chemistry · 4 marks · −1

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is

  1. A.1.14 mol kg−1^{-1}
  2. B.3.28 mol kg−1^{-1}
  3. C.2.28 mol kg−1^{-1}
  4. D.0.44 mol kg−1^{-1}

Correct answer: C

Explanation: Solvent 1020−123=8971020-123=897 g.

Question 142

Chemistry · 4 marks · −1

The enthalpy changes for the following processes are listed below: Cl2(g)=2Cl(g)Cl_2(g) = 2Cl(g), 242.3 kJ mol−1^{-1}; I2(g)=2I(g)I_2(g) = 2I(g), 151.0 kJ mol−1^{-1}; ICl(g) = I(g) + Cl(g), 211.3 kJ mol−1^{-1}; I2(s)=I2(g)I_2(s) = I_2(g), 62.76 kJ mol−1^{-1}. Given that the standard states for iodine and chlorine are I2I_2(s) and Cl2Cl_2(g), the standard enthalpy of formation for ICl(g) is

  1. A.-14.6 kJ mol−1^{-1}
  2. B.-16.8 kJ mol−1^{-1}
  3. C.+16.8 kJ mol−1^{-1}
  4. D.+244.8 kJ mol−1^{-1}

Correct answer: C

Explanation: 12(62.76+151.0+242.3)−211.3=16.73\frac12(62.76+151.0+242.3)-211.3=16.73.

Question 143

Chemistry · 4 marks · −1

How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+Ca^{2+} ion?

  1. A.Six
  2. B.Three
  3. C.One
  4. D.Two

Correct answer: C

Explanation: EDTA is hexadentate.

Question 144

Chemistry · 4 marks · −1

In the Reimer-Tiemann reaction, phenol + CHCl3CHCl_3 + NaOH →\rightarrow sodium salicylaldehyde phenoxide. The electrophile involved in the above reaction is

Figure for question 144
  1. A.dichloromethyl cation (C+HCl2\overset{+}{C}HCl_2)
  2. B.dichlorocarbene (:CCl2:CCl_2)
  3. C.trichloromethyl anion (C−Cl3\overset{-}{C}Cl_3)
  4. D.formyl cation (C+HO\overset{+}{C}HO)

Correct answer: B

Explanation: Dichlorocarbene.

Question 145

Chemistry · 4 marks · −1

18 g of glucose (C6H12O6C_6H_{12}O_6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100∘100^\circC is

  1. A.759.00 Torr
  2. B.7.60 Torr
  3. C.76.00 Torr
  4. D.752.40 Torr

Correct answer: D

Explanation: xwater=0.99x_{water}=0.99.

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