JEE Main 2018 — 8 Apr Offline Set D (Previous Year Paper)

JEE Main · 2018 · with answers and explanations

questions
89
Total marks
356
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If the curves y2=6xy^2 = 6x, 9x2+by2=169x^2 + by^2 = 16 intersect each other at right angles, then the value of bb is :

  1. A.92\frac{9}{2}
  2. B.6
  3. C.72\frac{7}{2}
  4. D.4

Correct answer: A

Explanation: At the common point (x1,y1)(x_1,y_1): m1=3y1m_1=\frac{3}{y_1}, m2=−9x1by1m_2=-\frac{9x_1}{by_1}. m1m2=−1⇒27x1by12=1m_1m_2=-1 \Rightarrow \frac{27x_1}{by_1^2}=1 and y12=6x1y_1^2=6x_1 gives b=92b=\frac{9}{2}.

Question 2

Mathematics · 4 marks · −1

Let u⃗\vec{u} be a vector coplanar with the vectors a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=j^+k^\vec{b} = \hat{j} + \hat{k}. If u⃗\vec{u} is perpendicular to a⃗\vec{a} and u⃗⋅b⃗=24\vec{u}\cdot\vec{b} = 24, then ∣u⃗∣2|\vec{u}|^2 is equal to :

  1. A.84
  2. B.336
  3. C.315
  4. D.256

Correct answer: B

Explanation: Write u⃗=λa⃗+μb⃗\vec u=\lambda\vec a+\mu\vec b. u⃗⋅a⃗=0⇒14λ+2μ=0\vec u\cdot\vec a=0 \Rightarrow 14\lambda+2\mu=0; u⃗⋅b⃗=24⇒2λ+2μ=24\vec u\cdot\vec b=24 \Rightarrow 2\lambda+2\mu=24. So λ=−2,μ=14\lambda=-2,\mu=14, u⃗=−4i^+8j^+16k^\vec u=-4\hat i+8\hat j+16\hat k, ∣u⃗∣2=336|\vec u|^2=336.

Question 3

Mathematics · 4 marks · −1

For each t∈Rt \in \mathbb{R}, let [t][t] be the greatest integer less than or equal to tt. Then lim⁡x→0+x([1x]+[2x]+⋯+[15x])\lim_{x\to 0^+} x\left(\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right] + \dots + \left[\frac{15}{x}\right]\right)

  1. A.does not exist (in R\mathbb{R})
  2. B.is equal to 0
  3. C.is equal to 15
  4. D.is equal to 120

Correct answer: D

Explanation: [k/x]=k/x−{k/x}[k/x]=k/x-\{k/x\}, so the expression is (1+2+⋯+15)−x∑{k/x}→120(1+2+\dots+15) - x\sum\{k/x\} \to 120 since the fractional parts are bounded.

Question 4

Mathematics · 4 marks · −1

If L1L_1 is the line of intersection of the planes 2x−2y+3z−2=02x - 2y + 3z - 2 = 0, x−y+z+1=0x - y + z + 1 = 0 and L2L_2 is the line of intersection of the planes x+2y−z−3=0x + 2y - z - 3 = 0, 3x−y+2z−1=03x - y + 2z - 1 = 0, then the distance of the origin from the plane, containing the lines L1L_1 and L2L_2, is :

  1. A.12\frac{1}{\sqrt{2}}
  2. B.142\frac{1}{4\sqrt{2}}
  3. C.132\frac{1}{3\sqrt{2}}
  4. D.122\frac{1}{2\sqrt{2}}

Correct answer: C

Explanation: Directions: L1∥i^+j^L_1 \parallel \hat i+\hat j, L2∥3i^−5j^−7k^L_2 \parallel 3\hat i-5\hat j-7\hat k. A point of L1L_1 is (−5,0,4)(-5,0,4). Normal =(i^+j^)×(3i^−5j^−7k^)∝(7,−7,8)=(\hat i+\hat j)\times(3\hat i-5\hat j-7\hat k) \propto (7,-7,8); plane 7x−7y+8z+3=07x-7y+8z+3=0; distance =3/162=132=3/\sqrt{162}=\frac{1}{3\sqrt2}.

Question 5

Mathematics · 4 marks · −1

The value of ∫−π/2π/2sin⁡2x1+2x dx\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x}\,dx is :

  1. A.π4\frac{\pi}{4}
  2. B.π8\frac{\pi}{8}
  3. C.π2\frac{\pi}{2}
  4. D.4π4\pi

Correct answer: A

Explanation: Adding the integral with x→−xx\to -x gives 2I=∫−π/2π/2sin⁡2x dx=π22I=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\frac{\pi}{2}, so I=π4I=\frac{\pi}{4}.

Question 6

Mathematics · 4 marks · −1

Let g(x)=cos⁡x2g(x) = \cos x^2, f(x)=xf(x) = \sqrt{x}, and α,β\alpha, \beta (α<β)(\alpha < \beta) be the roots of the quadratic equation 18x2−9πx+π2=018x^2 - 9\pi x + \pi^2 = 0. Then the area (in sq. units) bounded by the curve y=(g∘f)(x)y = (g\circ f)(x) and the lines x=αx = \alpha, x=βx = \beta and y=0y = 0, is :

  1. A.12(2−1)\frac{1}{2}(\sqrt{2}-1)
  2. B.12(3−1)\frac{1}{2}(\sqrt{3}-1)
  3. C.12(3+1)\frac{1}{2}(\sqrt{3}+1)
  4. D.12(3−2)\frac{1}{2}(\sqrt{3}-\sqrt{2})

Correct answer: B

Explanation: (g∘f)(x)=cos⁡x(g\circ f)(x)=\cos x; roots are π/6,π/3\pi/6,\pi/3. Area =∫π/6π/3cos⁡x dx=3−12=\int_{\pi/6}^{\pi/3}\cos x\,dx=\frac{\sqrt3-1}{2}.

Question 7

Mathematics · 4 marks · −1

If sum of all the solutions of the equation 8cos⁡x(cos⁡(π6+x)cos⁡(π6−x)−12)=18\cos x\left(\cos\left(\frac{\pi}{6}+x\right)\cos\left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right)=1 in [0,π][0,\pi] is kπk\pi, then kk is equal to :

  1. A.209\frac{20}{9}
  2. B.23\frac{2}{3}
  3. C.139\frac{13}{9}
  4. D.89\frac{8}{9}

Correct answer: C

Explanation: cos⁡(π6+x)cos⁡(π6−x)=34−sin⁡2x\cos(\frac\pi6+x)\cos(\frac\pi6-x)=\frac34-\sin^2x, so the equation becomes 2cos⁡3x=12\cos3x=1. Solutions in [0,π][0,\pi]: π9,5π9,7π9\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}; sum 13π9\frac{13\pi}{9}.

Question 8

Mathematics · 4 marks · −1

Let f(x)=x2+1x2f(x) = x^2 + \frac{1}{x^2} and g(x)=x−1xg(x) = x - \frac{1}{x}, x∈R−{−1,0,1}x \in \mathbb{R} - \{-1, 0, 1\}. If h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}, then the local minimum value of h(x)h(x) is :

  1. A.222\sqrt{2}
  2. B.3
  3. C.−3-3
  4. D.−22-2\sqrt{2}

Correct answer: A

Explanation: With t=x−1xt=x-\frac1x, h=t+2th=t+\frac{2}{t}; for t>0t>0 the minimum is 222\sqrt2 at t=2t=\sqrt2.

Question 9

Mathematics · 4 marks · −1

The integral ∫sin⁡2xcos⁡2x(sin⁡5x+cos⁡3xsin⁡2x+sin⁡3xcos⁡2x+cos⁡5x)2 dx\int \frac{\sin^2 x\cos^2 x}{\left(\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x\right)^2}\,dx is equal to : (where CC is a constant of integration)

  1. A.−11+cot⁡3x+C\frac{-1}{1+\cot^3 x}+C
  2. B.13(1+tan⁡3x)+C\frac{1}{3(1+\tan^3 x)}+C
  3. C.−13(1+tan⁡3x)+C\frac{-1}{3(1+\tan^3 x)}+C
  4. D.11+cot⁡3x+C\frac{1}{1+\cot^3 x}+C

Correct answer: C

Explanation: Denominator base =(sin⁡3x+cos⁡3x)(sin⁡2x+cos⁡2x)=(\sin^3x+\cos^3x)(\sin^2x+\cos^2x). Dividing by cos⁡6x\cos^6x: ∫tan⁡2xsec⁡2x(1+tan⁡3x)2dx\int\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}dx; put t=1+tan⁡3xt=1+\tan^3x to get −13(1+tan⁡3x)+C-\frac{1}{3(1+\tan^3x)}+C.

Question 10

Mathematics · 4 marks · −1

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

  1. A.34\frac{3}{4}
  2. B.310\frac{3}{10}
  3. C.25\frac{2}{5}
  4. D.15\frac{1}{5}

Correct answer: C

Explanation: P=410⋅612+610⋅412=25P=\frac{4}{10}\cdot\frac{6}{12}+\frac{6}{10}\cdot\frac{4}{12}=\frac{2}{5}.

Question 11

Mathematics · 4 marks · −1

Let the orthocentre and centroid of a triangle be A(−3,5)A(-3, 5) and B(3,3)B(3, 3) respectively. If CC is the circumcentre of this triangle, then the radius of the circle having line segment ACAC as diameter, is :

  1. A.352\frac{3\sqrt{5}}{2}
  2. B.10\sqrt{10}
  3. C.2102\sqrt{10}
  4. D.3523\sqrt{\frac{5}{2}}

Correct answer: D

Explanation: Centroid divides orthocentre-circumcentre in 2:12:1, so C=(6,2)C=(6,2). AC=90=310AC=\sqrt{90}=3\sqrt{10}; radius =3102=35/2=\frac{3\sqrt{10}}{2}=3\sqrt{5/2}.

Question 12

Mathematics · 4 marks · −1

If the tangent at (1,7)(1, 7) to the curve x2=y−6x^2 = y - 6 touches the circle x2+y2+16x+12y+c=0x^2 + y^2 + 16x + 12y + c = 0 then the value of cc is :

  1. A.95
  2. B.195
  3. C.185
  4. D.85

Correct answer: A

Explanation: Tangent: 2x−y+5=02x-y+5=0. Distance from centre (−8,−6)(-8,-6) is 5\sqrt5, so 100−c=5100-c=5, c=95c=95.

Question 13

Mathematics · 4 marks · −1

If α,β∈C\alpha, \beta \in \mathbb{C} are the distinct roots of the equation x2−x+1=0x^2 - x + 1 = 0, then α101+β107\alpha^{101} + \beta^{107} is equal to :

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

Correct answer: D

Explanation: Roots are −ω,−ω2-\omega,-\omega^2. (−ω)101=−ω2(-\omega)^{101}=-\omega^2, (−ω2)107=−ω(-\omega^2)^{107}=-\omega; sum =−(ω+ω2)=1=-(\omega+\omega^2)=1.

Question 14

Mathematics · 4 marks · −1

PQR is a triangular park with PQ=PR=200PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45∘45^\circ, 30∘30^\circ and 30∘30^\circ, then the height of the tower (in m) is :

  1. A.50250\sqrt{2}
  2. B.100
  3. C.50
  4. D.1003100\sqrt{3}

Correct answer: B

Explanation: With tower height hh at midpoint T: PT=hPT=h, QT=h3QT=h\sqrt3, and PT⊥QTPT\perp QT, so h2+3h2=2002h^2+3h^2=200^2, h=100h=100.

Question 15

Mathematics · 4 marks · −1

If ∑i=19(xi−5)=9\sum_{i=1}^{9}(x_i - 5) = 9 and ∑i=19(xi−5)2=45\sum_{i=1}^{9}(x_i - 5)^2 = 45, then the standard deviation of the 9 items x1,x2,…,x9x_1, x_2, \dots, x_9 is :

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

Correct answer: D

Explanation: σ2=459−(99)2=4\sigma^2=\frac{45}{9}-\left(\frac{9}{9}\right)^2=4, σ=2\sigma=2.

Question 16

Mathematics · 4 marks · −1

The sum of the co-efficients of all odd degree terms in the expansion of (x+x3−1)5+(x−x3−1)5\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, (x>1)(x > 1) is :

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

Correct answer: A

Explanation: Sum =2[x5+10x3(x3−1)+5x(x3−1)2]=2[x5+10x6−10x3+5x7−10x4+5x]=2[x^5+10x^3(x^3-1)+5x(x^3-1)^2]=2[x^5+10x^6-10x^3+5x^7-10x^4+5x]. Odd-degree coefficients: 2(1−10+5+5)=22(1-10+5+5)=2.

Question 17

Mathematics · 4 marks · −1

Tangents are drawn to the hyperbola 4x2−y2=364x^2 - y^2 = 36 at the points P and Q. If these tangents intersect at the point T(0,3)T(0, 3) then the area (in sq. units) of △PTQ\triangle PTQ is :

  1. A.36536\sqrt{5}
  2. B.45545\sqrt{5}
  3. C.54354\sqrt{3}
  4. D.60360\sqrt{3}

Correct answer: B

Explanation: Chord of contact from (0,3)(0,3): y=−12y=-12, meeting the hyperbola at x=±35x=\pm3\sqrt5. Area =12⋅65⋅15=455=\frac12\cdot6\sqrt5\cdot15=45\sqrt5.

Question 18

Mathematics · 4 marks · −1

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

  1. A.at least 750 but less than 1000
  2. B.at least 1000
  3. C.less than 500
  4. D.at least 500 but less than 750

Correct answer: B

Explanation: (31)⋅(64)⋅4!=3⋅15⋅24=1080\binom{3}{1}\cdot\binom{6}{4}\cdot4!=3\cdot15\cdot24=1080.

Question 19

Mathematics · 4 marks · −1

If the system of linear equations x+ky+3z=0x + ky + 3z = 0, 3x+ky−2z=03x + ky - 2z = 0, 2x+4y−3z=02x + 4y - 3z = 0 has a non-zero solution (x,y,z)(x, y, z), then xzy2\frac{xz}{y^2} is equal to :

  1. A.30
  2. B.−10-10
  3. C.10
  4. D.−30-30

Correct answer: C

Explanation: Determinant zero gives k=11k=11. Solving with z=tz=t: x=52tx=\frac52t, y=−t2y=-\frac t2, so xzy2=10\frac{xz}{y^2}=10.

Question 20

Mathematics · 4 marks · −1

If ∣x−42x2x2xx−42x2x2xx−4∣=(A+Bx)(x−A)2\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = (A + Bx)(x - A)^2, then the ordered pair (A,B)(A, B) is equal to :

  1. A.(4,5)(4, 5)
  2. B.(−4,−5)(-4, -5)
  3. C.(−4,3)(-4, 3)
  4. D.(−4,5)(-4, 5)

Correct answer: D

Explanation: Adding columns: (5x−4)⋅(x+4)2(5x-4)\cdot(x+4)^2. Comparing, A=−4A=-4, B=5B=5.

Question 21

Mathematics · 4 marks · −1

Two sets A and B are as under : A={(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1}A = \{(a, b) \in \mathbb{R}\times\mathbb{R} : |a - 5| < 1 \text{ and } |b - 5| < 1\}; B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}B = \{(a, b) \in \mathbb{R}\times\mathbb{R} : 4(a - 6)^2 + 9(b - 5)^2 \le 36\}. Then :

  1. A.neither A⊂BA \subset B nor B⊂AB \subset A
  2. B.B⊂AB \subset A
  3. C.A⊂BA \subset B
  4. D.A∩B=ϕA \cap B = \phi (an empty set)

Correct answer: C

Explanation: A is the open square 4<a<64<a<6, 4<b<64<b<6; B is the ellipse (a−6)29+(b−5)24≤1\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le1. The square's farthest corners (4,4),(4,6)(4,4),(4,6) give 49+14<1\frac49+\frac14<1, so A⊂BA\subset B.

Question 22

Mathematics · 4 marks · −1

Tangent and normal are drawn at P(16,16)P(16, 16) on the parabola y2=16xy^2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠CPB=θ\angle CPB = \theta, then a value of tan⁡θ\tan\theta is :

  1. A.43\frac{4}{3}
  2. B.12\frac{1}{2}
  3. C.2
  4. D.3

Correct answer: C

Explanation: A=(−16,0)A=(-16,0), B=(24,0)B=(24,0); ∠APB=90∘\angle APB=90^\circ so C is the midpoint (4,0)(4,0) (the focus). mCP=43m_{CP}=\frac43, mPB=−2m_{PB}=-2; tan⁡θ=∣4/3+21−8/3∣=2\tan\theta=\left|\frac{4/3+2}{1-8/3}\right|=2.

Question 23

Mathematics · 4 marks · −1

Let S={t∈R:f(x)=∣x−π∣⋅(e∣x∣−1)sin⁡∣x∣ is not differentiable at t}S = \{t \in \mathbb{R} : f(x) = |x - \pi|\cdot(e^{|x|} - 1)\sin|x| \text{ is not differentiable at } t\}. Then the set S is equal to :

  1. A.{0,π}\{0, \pi\}
  2. B.ϕ\phi (an empty set)
  3. C.{0}\{0\}
  4. D.{π}\{\pi\}

Correct answer: B

Explanation: Near 0, (e∣x∣−1)sin⁡∣x∣∼x2(e^{|x|}-1)\sin|x|\sim x^2; near π\pi, ∣x−π∣sin⁡x∼−(x−π)∣x−π∣|x-\pi|\sin x\sim -(x-\pi)|x-\pi|. Both are differentiable, so S is empty.

Question 24

Mathematics · 4 marks · −1

The Boolean expression ∼(p∨q)∨(∼p∧q)\sim(p \vee q) \vee (\sim p \wedge q) is equivalent to :

  1. A.∼q\sim q
  2. B.∼p\sim p
  3. C.pp
  4. D.qq

Correct answer: B

Explanation: (∼p∧∼q)∨(∼p∧q)=∼p∧(∼q∨q)=∼p(\sim p\wedge\sim q)\vee(\sim p\wedge q)=\sim p\wedge(\sim q\vee q)=\sim p.

Question 25

Mathematics · 4 marks · −1

A straight line through a fixed point (2,3)(2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :

  1. A.3x+2y=6xy3x + 2y = 6xy
  2. B.3x+2y=63x + 2y = 6
  3. C.2x+3y=xy2x + 3y = xy
  4. D.3x+2y=xy3x + 2y = xy

Correct answer: D

Explanation: Line xa+yb=1\frac xa+\frac yb=1 through (2,3)(2,3): 2a+3b=1\frac2a+\frac3b=1, and R=(a,b)R=(a,b), giving 3x+2y=xy3x+2y=xy.

Question 26

Mathematics · 4 marks · −1

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12+2⋅22+32+2⋅42+52+2⋅62+…1^2 + 2\cdot2^2 + 3^2 + 2\cdot4^2 + 5^2 + 2\cdot6^2 + \dots. If B−2A=100λB - 2A = 100\lambda, then λ\lambda is equal to :

  1. A.496
  2. B.232
  3. C.248
  4. D.464

Correct answer: C

Explanation: For even nn, Sn=n(n+1)22S_n=\frac{n(n+1)^2}{2}: A=4410A=4410, B=33620B=33620, B−2A=24800B-2A=24800, λ=248\lambda=248.

Question 27

Mathematics · 4 marks · −1

Let y=y(x)y = y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x\sin x\frac{dy}{dx} + y\cos x = 4x, x∈(0,π)x \in (0, \pi). If y(π2)=0y\left(\frac{\pi}{2}\right) = 0, then y(π6)y\left(\frac{\pi}{6}\right) is equal to :

  1. A.−49π2-\frac{4}{9}\pi^2
  2. B.493π2\frac{4}{9\sqrt{3}}\pi^2
  3. C.−893π2\frac{-8}{9\sqrt{3}}\pi^2
  4. D.−89π2-\frac{8}{9}\pi^2

Correct answer: D

Explanation: ddx(ysin⁡x)=4x⇒ysin⁡x=2x2−π22\frac{d}{dx}(y\sin x)=4x \Rightarrow y\sin x=2x^2-\frac{\pi^2}{2}. At x=π/6x=\pi/6: y2=π218−π22\frac y2=\frac{\pi^2}{18}-\frac{\pi^2}{2}, y=−8π29y=-\frac{8\pi^2}{9}.

Question 28

Mathematics · 4 marks · −1

The length of the projection of the line segment joining the points (5,−1,4)(5, -1, 4) and (4,−1,3)(4, -1, 3) on the plane, x+y+z=7x + y + z = 7 is :

  1. A.23\sqrt{\frac{2}{3}}
  2. B.23\frac{2}{\sqrt{3}}
  3. C.23\frac{2}{3}
  4. D.13\frac{1}{3}

Correct answer: A

Explanation: AB⃗=(−1,0,−1)\vec{AB}=(-1,0,-1), ∣AB∣=2|AB|=\sqrt2; component along normal =23=\frac{2}{\sqrt3}. Projection =2−43=23=\sqrt{2-\frac43}=\sqrt{\frac23}.

Question 29

Mathematics · 4 marks · −1

Let S={x∈R:x≥0 and 2∣x−3∣+x(x−6)+6=0}S = \{x \in \mathbb{R} : x \ge 0 \text{ and } 2\left|\sqrt{x}-3\right| + \sqrt{x}\left(\sqrt{x}-6\right) + 6 = 0\}. Then S :

  1. A.contains exactly four elements
  2. B.is an empty set
  3. C.contains exactly one element
  4. D.contains exactly two elements

Correct answer: D

Explanation: With t=xt=\sqrt x: for t≤3t\le3, t2−8t+12=0⇒t=2t^2-8t+12=0\Rightarrow t=2; for t≥3t\ge3, t2−4t=0⇒t=4t^2-4t=0\Rightarrow t=4. So x=4,16x=4,16.

Question 30

Mathematics · 4 marks · −1

Let a1,a2,a3,…,a49a_1, a_2, a_3, \dots, a_{49} be in A.P. such that ∑k=012a4k+1=416\sum_{k=0}^{12} a_{4k+1} = 416 and a9+a43=66a_9 + a_{43} = 66. If a12+a22+⋯+a172=140ma_1^2 + a_2^2 + \dots + a_{17}^2 = 140m, then mm is equal to :

  1. A.33
  2. B.66
  3. C.68
  4. D.34

Correct answer: D

Explanation: 13a1+312d=416⇒a1+24d=3213a_1+312d=416\Rightarrow a_1+24d=32; 2a1+50d=662a_1+50d=66. So d=1d=1, a1=8a_1=8; ∑r=824r2=4760=140⋅34\sum_{r=8}^{24}r^2=4760=140\cdot34.

Question 31

Physics · 4 marks · −1

Three concentric metal shells A, B and C of respective radii a, b and c (a<b<c)(a < b < c) have surface charge densities +σ+\sigma, −σ-\sigma and +σ+\sigma respectively. The potential of shell B is :

  1. A.σϵ0[b2−c2c+a]\frac{\sigma}{\epsilon_0}\left[\frac{b^2-c^2}{c}+a\right]
  2. B.σϵ0[a2−b2a+c]\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{a}+c\right]
  3. C.σϵ0[a2−b2b+c]\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{b}+c\right]
  4. D.σϵ0[b2−c2b+a]\frac{\sigma}{\epsilon_0}\left[\frac{b^2-c^2}{b}+a\right]

Correct answer: C

Explanation: VB=14πϵ0[4πa2σb−4πb2σb+4πc2σc]=σϵ0[a2−b2b+c]V_B=\frac{1}{4\pi\epsilon_0}\left[\frac{4\pi a^2\sigma}{b}-\frac{4\pi b^2\sigma}{b}+\frac{4\pi c^2\sigma}{c}\right]=\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{b}+c\right].

Question 32

Physics · 4 marks · −1

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :

Figure for question 32
  1. A.1812MR2\frac{181}{2}MR^2
  2. B.192MR2\frac{19}{2}MR^2
  3. C.552MR2\frac{55}{2}MR^2
  4. D.732MR2\frac{73}{2}MR^2

Correct answer: A

Explanation: About O: MR22+6(MR22+M(2R)2)=552MR2\frac{MR^2}{2}+6\left(\frac{MR^2}{2}+M(2R)^2\right)=\frac{55}{2}MR^2. P is at 3R3R from O: add 7M(3R)27M(3R)^2, total 1812MR2\frac{181}{2}MR^2.

Question 33

Physics · 4 marks · −1

From a uniform circular disc of radius R and mass 9M, a small disc of radius R3\frac{R}{3} is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

Figure for question 33
  1. A.379MR2\frac{37}{9}MR^2
  2. B.4MR24MR^2
  3. C.409MR2\frac{40}{9}MR^2
  4. D.10MR210MR^2

Correct answer: B

Explanation: Removed disc has mass M (area ratio 1/9), centre at 2R/32R/3: I=M(R/3)22+M(2R3)2=MR22I=\frac{M(R/3)^2}{2}+M\left(\frac{2R}{3}\right)^2=\frac{MR^2}{2}. Remaining =9MR22−MR22=4MR2=\frac{9MR^2}{2}-\frac{MR^2}{2}=4MR^2.

Question 34

Physics · 4 marks · −1

The reading of the ammeter for a silicon diode in the given circuit is :

Figure for question 34
  1. A.13.5 mA
  2. B.0
  3. C.15 mA
  4. D.11.5 mA

Correct answer: D

Explanation: Diode is forward biased; silicon drop 0.7 V. I=3−0.7200=11.5I=\frac{3-0.7}{200}=11.5 mA.

Question 35

Physics · 4 marks · −1

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I2\frac{I}{2}. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I8\frac{I}{8}. The angle between polarizer A and C is :

  1. A.60∘60^\circ
  2. B.0∘0^\circ
  3. C.30∘30^\circ
  4. D.45∘45^\circ

Correct answer: D

Explanation: A and B are parallel. With C at angle θ\theta: I2cos⁡4θ=I8⇒cos⁡θ=12\frac I2\cos^4\theta=\frac I8\Rightarrow\cos\theta=\frac1{\sqrt2}, θ=45∘\theta=45^\circ.

Question 36

Physics · 4 marks · −1

For an RLC circuit driven with voltage of amplitude vmv_m and frequency ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}} the current exhibits resonance. The quality factor, Q is given by :

  1. A.CRω0\frac{CR}{\omega_0}
  2. B.ω0LR\frac{\omega_0 L}{R}
  3. C.ω0RL\frac{\omega_0 R}{L}
  4. D.Rω0C\frac{R}{\omega_0 C}

Correct answer: B

Explanation: Q=ω0LR=1RLCQ=\frac{\omega_0L}{R}=\frac1R\sqrt{\frac LC} (the only dimensionless option).

Question 37

Physics · 4 marks · −1

Two masses m1=5m_1 = 5 kg and m2=10m_2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2m_2 to stop the motion is :

Figure for question 37
  1. A.10.3 kg
  2. B.18.3 kg
  3. C.27.3 kg
  4. D.43.3 kg

Correct answer: C

Explanation: To stop: μ(10+m)g≥5g⇒m≥23.3\mu(10+m)g\ge 5g \Rightarrow m\ge 23.3 kg; smallest option satisfying it is 27.3 kg.

Question 38

Physics · 4 marks · −1

In a collinear collision, a particle with an initial speed v0v_0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

  1. A.v02\frac{v_0}{\sqrt{2}}
  2. B.v04\frac{v_0}{4}
  3. C.2 v0\sqrt{2}\,v_0
  4. D.v02\frac{v_0}{2}

Correct answer: C

Explanation: v1+v2=v0v_1+v_2=v_0, v12+v22=1.5v02⇒2v1v2=−0.5v02v_1^2+v_2^2=1.5v_0^2 \Rightarrow 2v_1v_2=-0.5v_0^2; (v2−v1)2=2v02(v_2-v_1)^2=2v_0^2.

Question 39

Physics · 4 marks · −1

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nthn^{th} power of R. If the period of rotation of the particle is T, then :

  1. A.T∝Rn/2T \propto R^{n/2}
  2. B.T∝R3/2T \propto R^{3/2} for any n.
  3. C.T∝Rn2+1T \propto R^{\frac{n}{2}+1}
  4. D.T∝R(n+1)/2T \propto R^{(n+1)/2}

Correct answer: D

Explanation: mv2R∝R−n⇒v∝R(1−n)/2\frac{mv^2}{R}\propto R^{-n}\Rightarrow v\propto R^{(1-n)/2}; T=2πRv∝R(n+1)/2T=\frac{2\pi R}{v}\propto R^{(n+1)/2}.

Question 40

Physics · 4 marks · −1

Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 Ω10\,\Omega. The internal resistances of the two batteries are 1 Ω1\,\Omega and 2 Ω2\,\Omega respectively. The voltage across the load lies between :

  1. A.11.7 V and 11.8 V
  2. B.11.6 V and 11.7 V
  3. C.11.5 V and 11.6 V
  4. D.11.4 V and 11.5 V

Correct answer: C

Explanation: Eeq=12/1+13/21+1/2=373E_{eq}=\frac{12/1+13/2}{1+1/2}=\frac{37}{3} V, req=23 Ωr_{eq}=\frac23\,\Omega. V=373⋅1032/3≈11.56V=\frac{37}{3}\cdot\frac{10}{32/3}\approx11.56 V.

Question 41

Physics · 4 marks · −1

In an a.c. circuit, the instantaneous e.m.f. and current are given by e=100sin⁡30te = 100\sin 30t, i=20sin⁡(30t−π4)i = 20\sin\left(30t - \frac{\pi}{4}\right). In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

  1. A.50, 0
  2. B.50, 10
  3. C.10002\frac{1000}{\sqrt{2}}, 10
  4. D.502\frac{50}{\sqrt{2}}, 0

Correct answer: C

Explanation: ⟨P⟩=100⋅202cos⁡45∘=10002\langle P\rangle=\frac{100\cdot20}{2}\cos45^\circ=\frac{1000}{\sqrt2} W; wattless current =Irmssin⁡ϕ=202⋅12=10=I_{rms}\sin\phi=\frac{20}{\sqrt2}\cdot\frac1{\sqrt2}=10 A.

Question 42

Physics · 4 marks · −1

An EM wave from air enters a medium. The electric fields are E⃗1=E01x^cos⁡[2πν(zc−t)]\vec{E}_1 = E_{01}\hat{x}\cos\left[2\pi\nu\left(\frac{z}{c} - t\right)\right] in air and E⃗2=E02x^cos⁡[k(2z−ct)]\vec{E}_2 = E_{02}\hat{x}\cos[k(2z - ct)] in medium, where the wave number k and frequency ν\nu refer to their values in air. The medium is non-magnetic. If ϵr1\epsilon_{r_1} and ϵr2\epsilon_{r_2} refer to relative permittivities of air and medium respectively, which of the following options is correct ?

  1. A.ϵr1ϵr2=12\frac{\epsilon_{r_1}}{\epsilon_{r_2}} = \frac{1}{2}
  2. B.ϵr1ϵr2=4\frac{\epsilon_{r_1}}{\epsilon_{r_2}} = 4
  3. C.ϵr1ϵr2=2\frac{\epsilon_{r_1}}{\epsilon_{r_2}} = 2
  4. D.ϵr1ϵr2=14\frac{\epsilon_{r_1}}{\epsilon_{r_2}} = \frac{1}{4}

Correct answer: D

Explanation: Speed in medium is c/2c/2, so refractive index 2 and ϵr1ϵr2=14\frac{\epsilon_{r_1}}{\epsilon_{r_2}}=\frac14.

Question 43

Physics · 4 marks · −1

A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

  1. A.2×1062\times10^6
  2. B.2×1032\times10^3
  3. C.2×1042\times10^4
  4. D.2×1052\times10^5

Correct answer: D

Explanation: 0.1×10×1095×103=2×105\frac{0.1\times10\times10^9}{5\times10^3}=2\times10^5.

Question 44

Physics · 4 marks · −1

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×1032.7\times10^3 kg/m3^3 and its Young's modulus is 9.27×10109.27\times10^{10} Pa. What will be the fundamental frequency of the longitudinal vibrations ?

  1. A.7.5 kHz
  2. B.5 kHz
  3. C.2.5 kHz
  4. D.10 kHz

Correct answer: B

Explanation: v=Y/ρ≈5.86×103v=\sqrt{Y/\rho}\approx5.86\times10^3 m/s; clamped at the middle, free ends: λ=2L=1.2\lambda=2L=1.2 m, f≈4.9f\approx4.9 kHz ≈5\approx5 kHz.

Question 45

Physics · 4 marks · −1

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pdp_d; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pcp_c. The values of pdp_d and pcp_c are respectively :

  1. A.(0, 1)
  2. B.(.89, .28)
  3. C.(.28, .89)
  4. D.(0, 0)

Correct answer: B

Explanation: Fractional loss =4A(1+A)2=\frac{4A}{(1+A)^2}: deuterium A=2A=2 gives 0.890.89, carbon A=12A=12 gives 0.280.28.

Question 46

Physics · 4 marks · −1

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:

  1. A.6%
  2. B.2.5%
  3. C.3.5%
  4. D.4.5%

Correct answer: D

Explanation: Δρρ=Δmm+3Δaa=1.5+3=4.5%\frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+3\frac{\Delta a}{a}=1.5+3=4.5\%.

Question 47

Physics · 4 marks · −1

Two moles of an ideal monoatomic gas occupies a volume V at 27∘27^\circC. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

  1. A.(a) 195 K (b) 2.7 kJ
  2. B.(a) 189 K (b) 2.7 kJ
  3. C.(a) 195 K (b) −2.7-2.7 kJ
  4. D.(a) 189 K (b) −2.7-2.7 kJ

Correct answer: D

Explanation: T2=300⋅2−2/3≈189T_2=300\cdot2^{-2/3}\approx189 K; ΔU=nCvΔT=2⋅32R(−111)≈−2.7\Delta U=nC_v\Delta T=2\cdot\frac32R(-111)\approx-2.7 kJ.

Question 48

Physics · 4 marks · −1

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (drr)\left(\frac{dr}{r}\right), is:

  1. A.mgKa\frac{mg}{Ka}
  2. B.Kamg\frac{Ka}{mg}
  3. C.Ka3mg\frac{Ka}{3mg}
  4. D.mg3Ka\frac{mg}{3Ka}

Correct answer: D

Explanation: Pressure mg/amg/a; ΔVV=mgKa=3Δrr\frac{\Delta V}{V}=\frac{mg}{Ka}=3\frac{\Delta r}{r}.

Question 49

Physics · 4 marks · −1

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K=53K = \frac{5}{3} is inserted between the plates, the magnitude of the induced charge will be :

  1. A.0.9 nC
  2. B.1.2 nC
  3. C.0.3 nC
  4. D.2.4 nC

Correct answer: B

Explanation: Q=KCV=3Q=KCV=3 nC; induced charge =Q(1−1K)=1.2=Q(1-\frac1K)=1.2 nC.

Question 50

Physics · 4 marks · −1

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1B_1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2B_2. The ratio B1B2\frac{B_1}{B_2} is:

  1. A.12\frac{1}{\sqrt{2}}
  2. B.2
  3. C.3\sqrt{3}
  4. D.2\sqrt{2}

Correct answer: D

Explanation: m∝r2m\propto r^2 so r′=2rr'=\sqrt2r; B∝1rB\propto\frac1r, so B1B2=2\frac{B_1}{B_2}=\sqrt2.

Question 51

Physics · 4 marks · −1

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn\lambda_n, λg\lambda_g be the de Broglie wavelength of the electron in the nthn^{th} state and the ground state respectively. Let Λn\Lambda_n be the wavelength of the emitted photon in the transition from the nthn^{th} state to the ground state. For large n, (A, B are constants)

  1. A.Λn2≈λ\Lambda_n^2 \approx \lambda
  2. B.Λn≈A+Bλn2\Lambda_n \approx A + \frac{B}{\lambda_n^2}
  3. C.Λn≈A+Bλn\Lambda_n \approx A + B\lambda_n
  4. D.Λn2≈A+Bλn2\Lambda_n^2 \approx A + B\lambda_n^2

Correct answer: B

Explanation: λn=2πr0n\lambda_n=2\pi r_0 n. 1Λn=R(1−1n2)⇒Λn≈1R(1+1n2)=A+Bλn2\frac1{\Lambda_n}=R(1-\frac1{n^2})\Rightarrow\Lambda_n\approx\frac1R(1+\frac1{n^2})=A+\frac{B}{\lambda_n^2}.

Question 52

Physics · 4 marks · −1

The mass of a hydrogen molecule is 3.32×10−273.32\times10^{-27} kg. If 102310^{23} hydrogen molecules strike, per second, a fixed wall of area 2 cm2^2 at an angle of 45∘45^\circ to the normal, and rebound elastically with a speed of 10310^3 m/s, then the pressure on the wall is nearly:

  1. A.4.70×1024.70\times10^2 N/m2^2
  2. B.2.35×1032.35\times10^3 N/m2^2
  3. C.4.70×1034.70\times10^3 N/m2^2
  4. D.2.35×1022.35\times10^2 N/m2^2

Correct answer: B

Explanation: P=2mvcos⁡45∘⋅nA=2⋅3.32×10−27⋅103⋅0.707⋅10232×10−4≈2.35×103P=\frac{2mv\cos45^\circ\cdot n}{A}=\frac{2\cdot3.32\times10^{-27}\cdot10^3\cdot0.707\cdot10^{23}}{2\times10^{-4}}\approx2.35\times10^3 Pa.

Question 53

Physics · 4 marks · −1

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Figure for question 53
  1. A.Graph (1): velocity vs time
  2. B.Graph (2): velocity vs position
  3. C.Graph (3): distance vs time
  4. D.Graph (4): position vs time

Correct answer: C

Explanation: The motion is uniformly decelerated motion with non-zero initial velocity (velocity reverses). Graph (3) starts with zero slope, i.e. zero initial speed, which contradicts the others.

Question 54

Physics · 4 marks · −1

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii rer_e, rpr_p, rαr_\alpha respectively in a uniform magnetic field B. The relation between rer_e, rpr_p, rαr_\alpha is:

  1. A.re<rα<rpr_e < r_\alpha < r_p
  2. B.re>rp=rαr_e > r_p = r_\alpha
  3. C.re<rp=rαr_e < r_p = r_\alpha
  4. D.re<rp<rαr_e < r_p < r_\alpha

Correct answer: C

Explanation: r=2KmqB∝mqr=\frac{\sqrt{2Km}}{qB}\propto\frac{\sqrt m}{q}: proton 11\frac{\sqrt1}{1}, alpha 42=1\frac{\sqrt4}{2}=1, electron much smaller.

Question 55

Physics · 4 marks · −1

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ1\,k\Omega. How much was the resistance on the left slot before interchanging the resistances?

  1. A.910 Ω910\,\Omega
  2. B.990 Ω990\,\Omega
  3. C.505 Ω505\,\Omega
  4. D.550 Ω550\,\Omega

Correct answer: D

Explanation: R1R2=l100−l\frac{R_1}{R_2}=\frac{l}{100-l} and after swapping R2R1=l−10110−l\frac{R_2}{R_1}=\frac{l-10}{110-l} give l=55l=55 cm; R1=55100⋅1000=550 ΩR_1=\frac{55}{100}\cdot1000=550\,\Omega.

Question 56

Physics · 4 marks · −1

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω5\,\Omega, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

  1. A.2.5 Ω2.5\,\Omega
  2. B.1 Ω1\,\Omega
  3. C.1.5 Ω1.5\,\Omega
  4. D.2 Ω2\,\Omega

Correct answer: C

Explanation: r=R(l1l2−1)=5(5240−1)=1.5 Ωr=R\left(\frac{l_1}{l_2}-1\right)=5\left(\frac{52}{40}-1\right)=1.5\,\Omega.

Question 57

Physics · 4 marks · −1

If the series limit frequency of the Lyman series is νL\nu_L, then the series limit frequency of the Pfund series is:

  1. A.νL/25\nu_L/25
  2. B.25νL25\nu_L
  3. C.16νL16\nu_L
  4. D.νL/16\nu_L/16

Correct answer: A

Explanation: Series limit ∝1n12\propto\frac1{n_1^2}; Pfund has n1=5n_1=5, so νL/25\nu_L/25.

Question 58

Physics · 4 marks · −1

The angular width of the central maximum in a single slit diffraction pattern is 60∘60^\circ. The width of the slit is 1 μ1\,\mum. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

  1. A.100 μ100\,\mum
  2. B.25 μ25\,\mum
  3. C.50 μ50\,\mum
  4. D.75 μ75\,\mum

Correct answer: B

Explanation: sin⁡30∘=λb⇒λ=0.5 μ\sin30^\circ=\frac\lambda b\Rightarrow\lambda=0.5\,\mum. d=λDβ=5×10−7⋅0.510−2=25 μd=\frac{\lambda D}{\beta}=\frac{5\times10^{-7}\cdot0.5}{10^{-2}}=25\,\mum.

Question 59

Physics · 4 marks · −1

A particle is moving in a circular path of radius a under the action of an attractive potential U=−k2r2U = -\frac{k}{2r^2}. Its total energy is:

  1. A.−32ka2-\frac{3}{2}\frac{k}{a^2}
  2. B.−k4a2-\frac{k}{4a^2}
  3. C.k2a2\frac{k}{2a^2}
  4. D.Zero

Correct answer: D

Explanation: F=kr3F=\frac{k}{r^3} (attractive), mv2=ka2mv^2=\frac{k}{a^2}, KE =k2a2=\frac{k}{2a^2}, PE =−k2a2=-\frac{k}{2a^2}; total zero.

Question 60

Physics · 4 marks · −1

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 101210^{12}/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02×10236.02\times10^{23} gm mole−1^{-1})

  1. A.5.5 N/m
  2. B.6.4 N/m
  3. C.7.1 N/m
  4. D.2.2 N/m

Correct answer: C

Explanation: k=m(2πf)2=0.1086.02×1023⋅4π2⋅1024≈7.1k=m(2\pi f)^2=\frac{0.108}{6.02\times10^{23}}\cdot4\pi^2\cdot10^{24}\approx7.1 N/m.

Question 61

Chemistry · 4 marks · −1

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

  1. A.[Co(H2O)3Cl3]⋅3H2O[Co(H_2O)_3Cl_3]\cdot3H_2O
  2. B.[Co(H2O)6]Cl3[Co(H_2O)_6]Cl_3
  3. C.[Co(H2O)5Cl]Cl2⋅H2O[Co(H_2O)_5Cl]Cl_2\cdot H_2O
  4. D.[Co(H2O)4Cl2]Cl⋅2H2O[Co(H_2O)_4Cl_2]Cl\cdot2H_2O

Correct answer: A

Explanation: The non-ionising complex (i=1i=1) gives the smallest depression, hence the highest freezing point.

Question 62

Chemistry · 4 marks · −1

Hydrogen peroxide oxidises [Fe(CN)6]4−[Fe(CN)_6]^{4-} to [Fe(CN)6]3−[Fe(CN)_6]^{3-} in acidic medium but reduces [Fe(CN)6]3−[Fe(CN)_6]^{3-} to [Fe(CN)6]4−[Fe(CN)_6]^{4-} in alkaline medium. The other products formed are, respectively

  1. A.H2OH_2O and (H2O+OH−)(H_2O + OH^-)
  2. B.(H2O+O2)(H_2O + O_2) and H2OH_2O
  3. C.(H2O+O2)(H_2O + O_2) and (H2O+OH−)(H_2O + OH^-)
  4. D.H2OH_2O and (H2O+O2)(H_2O + O_2)

Correct answer: D

Explanation: As oxidant, H2O2H_2O_2 is reduced to H2OH_2O; as reductant in base it is oxidised to O2O_2 (with water).

Question 63

Chemistry · 4 marks · −1

Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

Figure for question 63
  1. A.Benzenediazonium chloride (C6H5N2+Cl−C_6H_5N_2^+Cl^-)
  2. B.Pyridine
  3. C.Aniline (C6H5NH2C_6H_5NH_2)
  4. D.Nitrobenzene (C6H5NO2C_6H_5NO_2)

Correct answer: C

Explanation: Kjeldahl's method fails for nitro/azo/diazo compounds and ring nitrogen (pyridine); aniline is suitable.

Question 64

Chemistry · 4 marks · −1

Glucose on prolonged heating with HI gives

  1. A.6-iodohexanal
  2. B.n-Hexane
  3. C.1-Hexene
  4. D.Hexanoic acid

Correct answer: B

Explanation: HI/heat reduces glucose completely to n-hexane, showing a straight six-carbon chain.

Question 65

Chemistry · 4 marks · −1

An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination? (Base, Acid, End point)

  1. A.Strong, Strong, Pink to colourless
  2. B.Weak, Strong, Colourless to pink
  3. C.Strong, Strong, Pinkish red to yellow
  4. D.Weak, Strong, Yellow to pinkish red

Correct answer: D

Explanation: Methyl orange suits weak base vs strong acid; the alkaline (yellow) solution turns pinkish red at the end point as acid is added.

Question 66

Chemistry · 4 marks · −1

The predominant form of histamine present in human blood is (pKapK_a, Histidine = 6.0)

Figure for question 66
  1. A.Imidazole ring neutral, side-chain amino group protonated (−NH3+-NH_3^+)
  2. B.Imidazole ring neutral, side-chain amino group neutral (−NH2-NH_2)
  3. C.Imidazole ring protonated (imidazolium) and side-chain −NH3+-NH_3^+
  4. D.Imidazole ring protonated (imidazolium) and side-chain −NH2-NH_2

Correct answer: A

Explanation: Blood pH 7.4 exceeds the ring pKapK_a (6.0), so the ring stays unprotonated; the aliphatic amine (pKa>7.4pK_a>7.4) is protonated.

Question 67

Chemistry · 4 marks · −1

The increasing order of basicity of the following compounds is: (a) CH2=CH−CH2−NH2CH_2=CH-CH_2-NH_2 (b) CH3CH2CH=NHCH_3CH_2CH=NH (c) CH3C(=NH)NH2CH_3C(=NH)NH_2 (d) CH3CH2NHCH3CH_3CH_2NHCH_3

Figure for question 67
  1. A.(d) < (b) < (a) < (c)
  2. B.(a) < (b) < (c) < (d)
  3. C.(b) < (a) < (c) < (d)
  4. D.(b) < (a) < (d) < (c)

Correct answer: D

Explanation: The sp2sp^2 imine N (b) is least basic; the secondary amine (d) beats the allylic primary amine (a); the amidine (c) is most basic since its conjugate acid is resonance-stabilised.

Question 68

Chemistry · 4 marks · −1

Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

Figure for question 68
  1. A.A and D
  2. B.A and B
  3. C.B and C
  4. D.C and D

Correct answer: B

Explanation: ln⁡K=−ΔH∘RT+ΔS∘R\ln K=-\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R}; for ΔH∘<0\Delta H^\circ<0 the plot of ln⁡K\ln K vs 1/T1/T has positive slope: lines A and B.

Question 69

Chemistry · 4 marks · −1

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

  1. A.1.6 hours
  2. B.6.4 hours
  3. C.0.8 hours
  4. D.3.2 hours

Correct answer: D

Explanation: 27.66 g B2H6B_2H_6 = 1 mol, needs 3 mol O2O_2 = 12 F. t=12⋅96500100t=\frac{12\cdot96500}{100} s ≈3.2\approx3.2 h.

Question 70

Chemistry · 4 marks · −1

Consider the following reaction and statements: [Co(NH3)4Br2]++Br−→[Co(NH3)3Br3]+NH3[Co(NH_3)_4Br_2]^+ + Br^- \rightarrow [Co(NH_3)_3Br_3] + NH_3. (I) Two isomers are produced if the reactant complex ion is a cis-isomer (II) Two isomers are produced if the reactant complex ion is a trans-isomer (III) Only one isomer is produced if the reactant complex ion is a trans-isomer (IV) Only one isomer is produced if the reactant complex ion is a cis-isomer. The correct statements are

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

Correct answer: C

Explanation: Replacing an NH3NH_3 in the cis isomer can give fac or mer; in the trans isomer every replacement gives the mer isomer.

Question 71

Chemistry · 4 marks · −1

Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2Br_2 to form product B. A and B are respectively:

Figure for question 71
  1. A.Methyl salicylate and methyl 2-bromo-6-hydroxybenzoate
  2. B.Methyl salicylate and methyl 4-bromo-2-hydroxybenzoate
  3. C.Methyl phenyl carbonate and 3-bromophenyl methyl carbonate
  4. D.Methyl phenyl carbonate and 4-bromophenyl methyl carbonate

Correct answer: D

Explanation: Phenoxide attacks the acyl chloride to give PhOCOOCH3PhOCOOCH_3 (A); the −OCOOMe-OCOOMe group is o,p-directing, bromination gives the para product (B).

Question 72

Chemistry · 4 marks · −1

An aqueous solution contains an unknown concentration of Ba2+Ba^{2+}. When 50 mL of a 1 M solution of Na2SO4Na_2SO_4 is added, BaSO4BaSO_4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4BaSO_4 is 1×10−101\times10^{-10}. What is the original concentration of Ba2+Ba^{2+}?

  1. A.1.0×10−101.0\times10^{-10} M
  2. B.5×10−95\times10^{-9} M
  3. C.2×10−92\times10^{-9} M
  4. D.1.1×10−91.1\times10^{-9} M

Correct answer: D

Explanation: [SO42−]=0.1[SO_4^{2-}]=0.1 M in 500 mL, so [Ba2+]=10−9[Ba^{2+}]=10^{-9} M there; original (450 mL) =10−9⋅500450≈1.1×10−9=10^{-9}\cdot\frac{500}{450}\approx1.1\times10^{-9} M.

Question 73

Chemistry · 4 marks · −1

At 518∘518^\circC the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s−1^{-1} when 5% had reacted and 0.5 Torr s−1^{-1} when 33% had reacted. The order of the reaction is

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

Correct answer: B

Explanation: 10.5=(344.85243.21)n=1.418n⇒n≈2\frac{1}{0.5}=\left(\frac{344.85}{243.21}\right)^n=1.418^n\Rightarrow n\approx2.

Question 74

Chemistry · 4 marks · −1

The combustion of benzene (l) gives CO2CO_2(g) and H2OH_2O(l). Given that heat of combustion of benzene at constant volume is −3263.9-3263.9 kJ mol−1^{-1} at 25∘25^\circC; heat of combustion (in kJ mol−1^{-1}) of benzene at constant pressure will be (R=8.314R = 8.314 JK−1^{-1} mol−1^{-1})

  1. A.−3267.6-3267.6
  2. B.4152.6
  3. C.−452.46-452.46
  4. D.3260

Correct answer: A

Explanation: Δng=6−7.5=−1.5\Delta n_g=6-7.5=-1.5; ΔH=ΔU+ΔngRT=−3263.9−3.7=−3267.6\Delta H=\Delta U+\Delta n_gRT=-3263.9-3.7=-3267.6 kJ/mol.

Question 75

Chemistry · 4 marks · −1

The ratio of mass percent of C and H of an organic compound (CXHYOZ)(C_XH_YO_Z) is 6 : 1. If one molecule of the above compound (CXHYOZ)(C_XH_YO_Z) contains half as much oxygen as required to burn one molecule of compound CXHYC_XH_Y completely to CO2CO_2 and H2OH_2O. The empirical formula of compound CXHYOZC_XH_YO_Z is

  1. A.C2H4O3C_2H_4O_3
  2. B.C3H6O3C_3H_6O_3
  3. C.C2H4OC_2H_4O
  4. D.C3H4O2C_3H_4O_2

Correct answer: A

Explanation: C:H mass 6:1 gives mole ratio 1:2. O atoms needed to burn CxHyC_xH_y =2x+y2=2x+\frac y2; zz is half of that =x+y4=x+\frac y4. For x=2,y=4x=2,y=4: z=3z=3, i.e. C2H4O3C_2H_4O_3.

Question 76

Chemistry · 4 marks · −1

The trans-alkenes are formed by the reduction of alkynes with

  1. A.Sn - HCl
  2. B.H2H_2 - Pd/C, BaSO4BaSO_4
  3. C.NaBH4NaBH_4
  4. D.Na/liq. NH3NH_3

Correct answer: D

Explanation: Birch-type reduction with Na in liquid NH3NH_3 gives trans-alkenes (anti addition).

Question 77

Chemistry · 4 marks · −1

When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

  1. A.Fe
  2. B.Zn
  3. C.Ca
  4. D.Al

Correct answer: D

Explanation: Al(OH)3Al(OH)_3 is white, gelatinous and amphoteric (dissolves as [Al(OH)4]−[Al(OH)_4]^-); ignition gives Al2O3Al_2O_3, a chromatography adsorbent.

Question 78

Chemistry · 4 marks · −1

According to molecular orbital theory, which of the following will not be a viable molecule?

  1. A.H22−H_2^{2-}
  2. B.He22+He_2^{2+}
  3. C.He2+He_2^{+}
  4. D.H2−H_2^{-}

Correct answer: A

Explanation: Bond orders: H22−H_2^{2-} 0, He22+He_2^{2+} 1, He2+He_2^+ 0.5, H2−H_2^- 0.5. Zero bond order is not viable.

Question 79

Chemistry · 4 marks · −1

The major product formed in the following reaction is: 1-(1-methoxyethyl)-2-ethoxybenzene (a benzene ring bearing −CH(CH3)OCH3-CH(CH_3)OCH_3 and, ortho to it, −OCH2CH3-OCH_2CH_3) →HI, heat\xrightarrow{HI,\ \text{heat}} ?

Figure for question 79
  1. A.2-(1-iodoethyl)phenol (ring with −CH(CH3)I-CH(CH_3)I and ortho −OH-OH)
  2. B.2-(1-hydroxyethyl)phenol (ring with −CH(CH3)OH-CH(CH_3)OH and ortho −OH-OH)
  3. C.1-iodo-2-(1-iodoethyl)benzene (ring with −CH(CH3)I-CH(CH_3)I and ortho −I-I)
  4. D.1-(2-iodophenyl)ethanol (ring with −CH(CH3)OH-CH(CH_3)OH and ortho −I-I)

Correct answer: A

Explanation: The aryl ether cleaves to phenol + C2H5IC_2H_5I (aryl C-O is not broken). The benzylic ether cleaves via a stable benzylic carbocation, which takes I−I^- to give the benzylic iodide.

Question 80

Chemistry · 4 marks · −1

Phenol on treatment with CO2CO_2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O(CH_3CO)_2O in the presence of catalytic amount of H2SO4H_2SO_4 produces:

Figure for question 80
  1. A.4-acetoxybenzene-1,3-dicarboxylic acid (acetylated 4-hydroxyisophthalic acid)
  2. B.2-acetoxybenzoic acid (aspirin)
  3. C.4-acetoxybenzoic acid
  4. D.Salicylic acid mixed anhydride with acetic acid (2-hydroxybenzoyl acetate)

Correct answer: B

Explanation: Kolbe reaction gives salicylic acid (X); acetylation of its OH gives acetylsalicylic acid (aspirin).

Question 81

Chemistry · 4 marks · −1

Which of the following compounds contain(s) no covalent bond(s)? KClKCl, PH3PH_3, O2O_2, B2H6B_2H_6, H2SO4H_2SO_4

  1. A.KClKCl, B2H6B_2H_6
  2. B.KClKCl, B2H6B_2H_6, PH3PH_3
  3. C.KClKCl, H2SO4H_2SO_4
  4. D.KClKCl

Correct answer: D

Explanation: Only KCl is purely ionic; the others contain covalent bonds.

Question 82

Chemistry · 4 marks · −1

Which type of 'defect' has the presence of cations in the interstitial sites?

  1. A.Metal deficiency defect
  2. B.Schottky defect
  3. C.Vacancy defect
  4. D.Frenkel defect

Correct answer: D

Explanation: In a Frenkel defect the (smaller) cation leaves its site and occupies an interstitial site.

Question 83

Chemistry · 4 marks · −1

The major product of the following reaction is: 2-bromo-1,1-dimethylcyclohexane (Br on the carbon adjacent to the gem-dimethyl carbon, drawn with a wedge) →NaOMe/MeOH\xrightarrow{NaOMe/MeOH} ?

Figure for question 83
  1. A.2-methoxy-1,1-dimethylcyclohexane (OMe on a hashed bond)
  2. B.2-methoxy-1,1-dimethylcyclohexane (OMe on a wedged bond)
  3. C.3,3-dimethylcyclohexene
  4. D.1,2-dimethylcyclohexene

Correct answer: C

Explanation: Strong base/nucleophile on a hindered secondary halide: E2. The only β\beta-H is on C3 (C1 is quaternary), giving 3,3-dimethylcyclohexene.

Question 84

Chemistry · 4 marks · −1

The compound that does not produce nitrogen gas by the thermal decomposition is

  1. A.(NH4)2SO4(NH_4)_2SO_4
  2. B.Ba(N3)2Ba(N_3)_2
  3. C.(NH4)2Cr2O7(NH_4)_2Cr_2O_7
  4. D.NH4NO2NH_4NO_2

Correct answer: A

Explanation: (NH4)2SO4(NH_4)_2SO_4 gives NH3NH_3 (and H2SO4H_2SO_4); the others release N2N_2.

Question 85

Chemistry · 4 marks · −1

An aqueous solution contains 0.10 M H2SH_2S and 0.20 M HCl. If the equilibrium constants for the formation of HS−HS^- from H2SH_2S is 1.0×10−71.0\times10^{-7} and that of S2−S^{2-} from HS−HS^- ions is 1.2×10−131.2\times10^{-13} then the concentration of S2−S^{2-} ions in aqueous solution is

  1. A.5×10−195\times10^{-19}
  2. B.5×10−85\times10^{-8}
  3. C.3×10−203\times10^{-20}
  4. D.6×10−216\times10^{-21}

Correct answer: C

Explanation: K1K2=1.2×10−20=(0.2)2[S2−]0.1⇒[S2−]=3×10−20K_1K_2=1.2\times10^{-20}=\frac{(0.2)^2[S^{2-}]}{0.1}\Rightarrow[S^{2-}]=3\times10^{-20} M.

Question 86

Chemistry · 4 marks · −1

The oxidation states of Cr in [Cr(H2O)6]Cl3[Cr(H_2O)_6]Cl_3, [Cr(C6H6)2][Cr(C_6H_6)_2] and K2[Cr(CN)2(O)2(O2)(NH3)]K_2[Cr(CN)_2(O)_2(O_2)(NH_3)] respectively are:

  1. A.+3, 0 and +4
  2. B.+3, +4 and +6
  3. C.+3, +2 and +4
  4. D.+3, 0 and +6

Correct answer: D

Explanation: +3; 0 (neutral benzene); in the last, −2=x−2−4−2-2=x-2-4-2 (2 CN, 2 oxide, 1 peroxide) gives x=+6x=+6.

Question 87

Chemistry · 4 marks · −1

The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2⋅Ca(OH)2][3Ca_3(PO_4)_2\cdot Ca(OH)_2] to

  1. A.[3{Ca(OH)2}⋅CaF2][3\{Ca(OH)_2\}\cdot CaF_2]
  2. B.[CaF2][CaF_2]
  3. C.[3(CaF2)⋅Ca(OH)2][3(CaF_2)\cdot Ca(OH)_2]
  4. D.[3Ca3(PO4)2⋅CaF2][3Ca_3(PO_4)_2\cdot CaF_2]

Correct answer: D

Explanation: Fluoride replaces hydroxide in hydroxyapatite to form fluorapatite [3Ca3(PO4)2⋅CaF2][3Ca_3(PO_4)_2\cdot CaF_2].

Question 88

Chemistry · 4 marks · −1

Which of the following salts is the most basic in aqueous solution?

  1. A.Pb(CH3COO)2Pb(CH_3COO)_2
  2. B.Al(CN)3Al(CN)_3
  3. C.CH3COOKCH_3COOK
  4. D.FeCl3FeCl_3

Correct answer: C

Explanation: CH3COOKCH_3COOK is a salt of a strong base and weak acid, so its solution is basic; the others involve weak-base cations.

Question 89

Chemistry · 4 marks · −1

Total number of lone pair of electrons in I3−I_3^- ion is

  1. A.12
  2. B.3
  3. C.6
  4. D.9

Correct answer: D

Explanation: Central I has 3 lone pairs, each terminal I has 3: total 9.

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