JEE Main 2014 — 6 Apr Offline Set H (Previous Year Paper)

JEE Main · 2014 · with answers and explanations

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

Subjects: Physics · Mathematics · Chemistry

Question 1

Physics · 4 marks · −1

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100∘100^\circC is : (For steel Young's modulus is 2×10112\times10^{11} N m−2^{-2} and coefficient of thermal expansion is 1.1×10−51.1\times10^{-5} K−1^{-1})

  1. A.2.2×1072.2\times10^7 Pa
  2. B.2.2×1062.2\times10^6 Pa
  3. C.2.2×1082.2\times10^8 Pa
  4. D.2.2×1092.2\times10^9 Pa

Correct answer: C

Explanation: Thermal stress =YαΔT=2×1011⋅1.1×10−5⋅100=2.2×108=Y\alpha\Delta T=2\times10^{11}\cdot1.1\times10^{-5}\cdot100=2.2\times10^8 Pa.

Question 2

Physics · 4 marks · −1

A conductor lies along the z-axis at −1.5≤z<1.5-1.5 \le z < 1.5 m and carries a fixed current of 10.0 A in −a^z-\hat{a}_z direction (see figure). For a field B⃗=3.0×10−4e−0.2x a^y\vec{B} = 3.0\times10^{-4}e^{-0.2x}\,\hat{a}_y T, find the power required to move the conductor at constant speed to x=2.0x = 2.0 m, y=0y = 0 m in 5×10−35\times10^{-3} s. Assume parallel motion along the x-axis

Figure for question 2
  1. A.14.85 W
  2. B.29.7 W
  3. C.1.57 W
  4. D.2.97 W

Correct answer: D

Explanation: W=∫02IℓB dx=∫0210⋅3⋅3×10−4e−0.2xdx=0.045(1−e−0.4)≈0.0148W=\int_0^2 I\ell B\,dx=\int_0^2 10\cdot3\cdot3\times10^{-4}e^{-0.2x}dx=0.045(1-e^{-0.4})\approx0.0148 J; P=W/t≈2.97P=W/t\approx2.97 W.

Question 3

Physics · 4 marks · −1

A bob of mass m attached to an inextensible string of length ℓ\ell is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω\omega rad/s about the vertical. About the point of suspension :

  1. A.angular momentum changes in direction but not in magnitude.
  2. B.angular momentum changes both in direction and magnitude.
  3. C.angular momentum is conserved.
  4. D.angular momentum changes in magnitude but not in direction.

Correct answer: A

Explanation: For a conical pendulum, L⃗\vec L about the suspension point has constant magnitude but precesses about the vertical.

Question 4

Physics · 4 marks · −1

The current voltage relation of diode is given by I=(e1000V/T−1)I = (e^{1000V/T} - 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ±0.01\pm0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?

  1. A.0.5 mA
  2. B.0.05 mA
  3. C.0.2 mA
  4. D.0.02 mA

Correct answer: C

Explanation: e1000V/T=6e^{1000V/T}=6; dI=1000Te1000V/TdV=1000300⋅6⋅0.01=0.2dI=\frac{1000}{T}e^{1000V/T}dV=\frac{1000}{300}\cdot6\cdot0.01=0.2 mA.

Question 5

Physics · 4 marks · −1

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

  1. A.38 cm
  2. B.6 cm
  3. C.16 cm
  4. D.22 cm

Correct answer: C

Explanation: Boyle: 76⋅8=(76−x)(54−x)76\cdot8=(76-x)(54-x) where x is the mercury rise; x=38x=38 cm, so the air column is 54−38=1654-38=16 cm.

Question 6

Physics · 4 marks · −1

Match List-I (Electromagnetic wave type) with List-II (Its association/application): (a) Infrared waves, (b) Radio waves, (c) X-rays, (d) Ultraviolet rays; (i) To treat muscular strain, (ii) For broadcasting, (iii) To detect fracture of bones, (iv) Absorbed by the ozone layer of the atmosphere

  1. A.(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
  2. B.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  3. C.(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
  4. D.(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

Correct answer: B

Explanation: IR: muscular strain; radio: broadcasting; X-rays: bone fractures; UV: absorbed by ozone.

Question 7

Physics · 4 marks · −1

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3×1043\times10^4 V/m, the charge density of the positive plate will be close to :

  1. A.3×1043\times10^4 C/m2^2
  2. B.6×1046\times10^4 C/m2^2
  3. C.6×10−76\times10^{-7} C/m2^2
  4. D.3×10−73\times10^{-7} C/m2^2

Correct answer: C

Explanation: σ=Kϵ0E=2.2⋅8.85×10−12⋅3×104≈5.8×10−7\sigma=K\epsilon_0E=2.2\cdot8.85\times10^{-12}\cdot3\times10^4\approx5.8\times10^{-7} C/m2^2.

Question 8

Physics · 4 marks · −1

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

  1. A.A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
  2. B.A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.
  3. C.A meter scale.
  4. D.A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

Correct answer: D

Explanation: 3.50 cm implies a least count of 0.01 cm = 0.1 mm, which is that vernier calliper's least count.

Question 9

Physics · 4 marks · −1

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :

  1. A.GMR(1+22)\sqrt{\frac{GM}{R}(1+2\sqrt{2})}
  2. B.12GMR(1+22)\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}
  3. C.GMR\sqrt{\frac{GM}{R}}
  4. D.22GMR\sqrt{2\sqrt{2}\frac{GM}{R}}

Correct answer: B

Explanation: Net inward force =GM2R2(14+12)=Mv2R⇒v=12GMR(1+22)=\frac{GM^2}{R^2}\left(\frac14+\frac1{\sqrt2}\right)=\frac{Mv^2}{R}\Rightarrow v=\frac12\sqrt{\frac{GM}{R}(1+2\sqrt2)}.

Question 10

Physics · 4 marks · −1

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

  1. A.12 A
  2. B.14 A
  3. C.8 A
  4. D.10 A

Correct answer: A

Explanation: Total 2500 W; I=2500220≈11.4I=\frac{2500}{220}\approx11.4 A, so a 12 A fuse.

Question 11

Physics · 4 marks · −1

A particle moves with simple harmonic motion in a straight line. In first τ\tau s, after starting from rest it travels a distance a, and in next τ\tau s it travels 2a, in same direction, then :

  1. A.amplitude of motion is 4a
  2. B.time period of oscillations is 6τ6\tau
  3. C.amplitude of motion is 3a
  4. D.time period of oscillations is 8τ8\tau

Correct answer: B

Explanation: A(1−cos⁡ωτ)=aA(1-\cos\omega\tau)=a, A(1−cos⁡2ωτ)=3aA(1-\cos2\omega\tau)=3a give A=2aA=2a, cos⁡ωτ=12\cos\omega\tau=\frac12, so T=6τT=6\tau.

Question 12

Physics · 4 marks · −1

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3×1033\times10^3 Am−1^{-1}. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is:

  1. A.3A
  2. B.6A
  3. C.30 mA
  4. D.60 mA

Correct answer: A

Explanation: H=ni=1000.1i=3×103⇒i=3H=ni=\frac{100}{0.1}i=3\times10^3\Rightarrow i=3 A.

Question 13

Physics · 4 marks · −1

The forward biased diode connection is:

Figure for question 13
  1. A.Anode at 2V, cathode side (through resistor) at 4V
  2. B.Anode at −2-2V, cathode side at +2+2V
  3. C.Anode at +2+2V, cathode side at −2-2V
  4. D.Anode at −3-3V, cathode side at −3-3V

Correct answer: C

Explanation: Forward bias needs the p side (anode) at higher potential than the n side: +2+2 V to −2-2 V.

Question 14

Physics · 4 marks · −1

During the propagation of electromagnetic waves in a medium:

  1. A.Electric energy density is equal to the magnetic energy density.
  2. B.Both electric and magnetic energy densities are zero.
  3. C.Electric energy density is double of the magnetic energy density.
  4. D.Electric energy density is half of the magnetic energy density.

Correct answer: A

Explanation: In an EM wave the average electric and magnetic energy densities are equal.

Question 15

Physics · 4 marks · −1

In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t=0t = 0. Ratio of the voltage across resistance and the inductor at t=L/Rt = L/R will be equal to :

Figure for question 15
  1. A.−1-1
  2. B.1−ee\frac{1-e}{e}
  3. C.e1−e\frac{e}{1-e}
  4. D.1

Correct answer: A

Explanation: After switching, R and L form a closed loop with no source: VR+VL=0V_R+V_L=0 at every instant, so VR/VL=−1V_R/V_L=-1.

Question 16

Physics · 4 marks · −1

A mass 'm' is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Figure for question 16
  1. A.5g6\frac{5g}{6}
  2. B.g
  3. C.2g3\frac{2g}{3}
  4. D.g2\frac{g}{2}

Correct answer: D

Explanation: mg−T=mamg-T=ma, TR=mR2aR⇒T=maTR=mR^2\frac aR\Rightarrow T=ma; so a=g2a=\frac g2.

Question 17

Physics · 4 marks · −1

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:

Figure for question 17
  1. A.The change in internal energy in the process AB is −350-350R.
  2. B.The change in internal energy in the process BC is −500-500R.
  3. C.The change in internal energy in whole cyclic process is 250R.
  4. D.The change in internal energy in the process CA is 700R.

Correct answer: B

Explanation: ΔUBC=52R(600−800)=−500R\Delta U_{BC}=\frac52R(600-800)=-500R.

Question 18

Physics · 4 marks · −1

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

  1. A.2gH=nu2(n−2)2gH = nu^2(n-2)
  2. B.gH=(n−2)u2gH = (n-2)u^2
  3. C.2gH=n2u22gH = n^2u^2
  4. D.gH=(n−2)2u2gH = (n-2)^2u^2

Correct answer: A

Explanation: −H=unug−12gn2u2g2⇒2gH=nu2(n−2)-H=u\frac{nu}{g}-\frac12g\frac{n^2u^2}{g^2}\Rightarrow2gH=nu^2(n-2).

Question 19

Physics · 4 marks · −1

A thin convex lens made from crown glass (μ=32)\left(\mu = \frac{3}{2}\right) has focal length f. When it is measured in two different liquids having refractive indices 43\frac{4}{3} and 53\frac{5}{3}, it has the focal lengths f1f_1 and f2f_2 respectively. The correct relation between the focal lengths is:

  1. A.f2>ff_2 > f and f1f_1 becomes negative
  2. B.f1f_1 and f2f_2 both become negative
  3. C.f1=f2<ff_1 = f_2 < f
  4. D.f1>ff_1 > f and f2f_2 becomes negative

Correct answer: D

Explanation: fmf=μ−1μ/μm−1\frac{f_m}{f}=\frac{\mu-1}{\mu/\mu_m-1}: f1=4ff_1=4f, f2=−5ff_2=-5f.

Question 20

Physics · 4 marks · −1

Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm2^2. End of copper rod is maintained at 100∘100^\circC where as ends of brass and steel are kept at 0∘0^\circC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :

  1. A.4.8 cal/s
  2. B.6.0 cal/s
  3. C.1.2 cal/s
  4. D.2.4 cal/s

Correct answer: A

Explanation: Junction temperature T: 0.92(100−T)46=0.26T13+0.12T12⇒T=40∘\frac{0.92(100-T)}{46}=\frac{0.26T}{13}+\frac{0.12T}{12}\Rightarrow T=40^\circC; dQdt=0.92⋅4⋅6046=4.8\frac{dQ}{dt}=\frac{0.92\cdot4\cdot60}{46}=4.8 cal/s.

Question 21

Physics · 4 marks · −1

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

  1. A.6
  2. B.4
  3. C.12
  4. D.8

Correct answer: A

Explanation: Fundamental 3404⋅0.85=100\frac{340}{4\cdot0.85}=100 Hz; odd harmonics 100, 300, ..., 1100 Hz: 6 modes.

Question 22

Physics · 4 marks · −1

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1d_1 and d2d_2 are filled in the tube. Each liquid subtends 90∘90^\circ angle at centre. Radius joining their interface makes an angle α\alpha with vertical. Ratio d1/d2d_1/d_2 is

Figure for question 22
  1. A.1+tan⁡α1−tan⁡α\frac{1+\tan\alpha}{1-\tan\alpha}
  2. B.1+sin⁡α1−cos⁡α\frac{1+\sin\alpha}{1-\cos\alpha}
  3. C.1+sin⁡α1−sin⁡α\frac{1+\sin\alpha}{1-\sin\alpha}
  4. D.1+cos⁡α1−cos⁡α\frac{1+\cos\alpha}{1-\cos\alpha}

Correct answer: A

Explanation: Balancing pressures at the bottom of the tube through the two liquid columns gives d1(cos⁡α+sin⁡α)=d2(cos⁡α−sin⁡α)d_1(\cos\alpha+\sin\alpha)=d_2(\cos\alpha-\sin\alpha) for the geometry shown, i.e. d1d2=1+tan⁡α1−tan⁡α\frac{d_1}{d_2}=\frac{1+\tan\alpha}{1-\tan\alpha}.

Question 23

Physics · 4 marks · −1

A green light is incident from the water to the air - water interface at the critical angle (θ)(\theta). Select the correct statement

  1. A.The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
  2. B.The entire spectrum of visible light will come out of the water at various angles to the normal.
  3. C.The entire spectrum of visible light will come out of the water at an angle of 90∘90^\circ to the normal.
  4. D.The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

Correct answer: D

Explanation: Lower-frequency colours have smaller refractive index and larger critical angle, so they escape; higher frequencies are totally reflected.

Question 24

Physics · 4 marks · −1

Hydrogen (1H1)(_1H^1), Deuterium (1H2)(_1H^2), singly ionised Helium (2He4)+(_2He^4)^+ and doubly ionised lithium (3Li6)++(_3Li^6)^{++} all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are λ1\lambda_1, λ2\lambda_2, λ3\lambda_3 and λ4\lambda_4 respectively then approximately which one of the following is correct?

  1. A.λ1=λ2=4λ3=9λ4\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4
  2. B.λ1=2λ2=3λ3=4λ4\lambda_1 = 2\lambda_2 = 3\lambda_3 = 4\lambda_4
  3. C.4λ1=2λ2=2λ3=λ44\lambda_1 = 2\lambda_2 = 2\lambda_3 = \lambda_4
  4. D.λ1=2λ2=2λ3=λ4\lambda_1 = 2\lambda_2 = 2\lambda_3 = \lambda_4

Correct answer: A

Explanation: 1λ∝Z2\frac1\lambda\propto Z^2: Z = 1, 1, 2, 3.

Question 25

Physics · 4 marks · −1

The radiation corresponding to 3→23 \to 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3×10−43\times10^{-4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

  1. A.0.8 eV
  2. B.1.6 eV
  3. C.1.8 eV
  4. D.1.1 eV

Correct answer: D

Explanation: KEmax=(eBr)22m≈0.8KE_{max}=\frac{(eBr)^2}{2m}\approx0.8 eV; photon =13.6⋅536≈1.89=13.6\cdot\frac{5}{36}\approx1.89 eV; ϕ≈1.1\phi\approx1.1 eV.

Question 26

Physics · 4 marks · −1

A block of mass m is placed on a surface with a vertical cross section given by y=x36y = \frac{x^3}{6}. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is :

  1. A.13\frac{1}{3} m
  2. B.12\frac{1}{2} m
  3. C.16\frac{1}{6} m
  4. D.23\frac{2}{3} m

Correct answer: C

Explanation: Limiting: tan⁡θ=x22=μ=12⇒x=1\tan\theta=\frac{x^2}{2}=\mu=\frac12\Rightarrow x=1, y=16y=\frac16 m.

Question 27

Physics · 4 marks · −1

When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F=ax+bx2F = ax + bx^2 where a and b are constants. The work done in stretching the unstretched rubber band by L is :

  1. A.aL22+bL33\frac{aL^2}{2}+\frac{bL^3}{3}
  2. B.12(aL22+bL33)\frac{1}{2}\left(\frac{aL^2}{2}+\frac{bL^3}{3}\right)
  3. C.aL2+bL3aL^2 + bL^3
  4. D.12(aL2+bL3)\frac{1}{2}(aL^2 + bL^3)

Correct answer: A

Explanation: W=∫0L(ax+bx2)dxW=\int_0^L(ax+bx^2)dx.

Question 28

Physics · 4 marks · −1

On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r≪Rr \ll R, and the surface tension of water is T, value of r just before bubbles detach is : (density of water is ρw\rho_w)

Figure for question 28
  1. A.R2ρwgTR^2\sqrt{\frac{\rho_wg}{T}}
  2. B.R23ρwgTR^2\sqrt{\frac{3\rho_wg}{T}}
  3. C.R2ρwg3TR^2\sqrt{\frac{\rho_wg}{3T}}
  4. D.R2ρwg6TR^2\sqrt{\frac{\rho_wg}{6T}}

Correct answer: C

Explanation: Balancing the vertical surface-tension force on the contact circle (2πrTsin⁡θ2\pi rT\sin\theta with sin⁡θ≈r/R\sin\theta\approx r/R) against buoyancy 43πR3ρwg\frac43\pi R^3\rho_wg gives r=R22ρwg3Tr=R^2\sqrt{\frac{2\rho_wg}{3T}}; the official key accepts option 3.

Question 29

Physics · 4 marks · −1

Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of Polaroid through 30∘30^\circ makes the two beams appear equally bright. If the initial intensities of the two beams are IAI_A and IBI_B respectively, then IA/IBI_A/I_B equals :

  1. A.1
  2. B.13\frac{1}{3}
  3. C.3
  4. D.32\frac{3}{2}

Correct answer: B

Explanation: IAcos⁡230∘=IBcos⁡260∘⇒IAIB=13I_A\cos^230^\circ=I_B\cos^260^\circ\Rightarrow\frac{I_A}{I_B}=\frac13.

Question 30

Physics · 4 marks · −1

Assume that an electric field E⃗=30x2i^\vec{E} = 30x^2\hat{i} exists in space. Then the potential difference VA−VOV_A - V_O, where VOV_O is the potential at the origin and VAV_A the potential at x=2x = 2 m is :

  1. A.−80-80 J
  2. B.80 J
  3. C.120 J
  4. D.−120-120 J

Correct answer: A

Explanation: VA−VO=−∫0230x2dx=−80V_A-V_O=-\int_0^230x^2dx=-80 V (options are printed in J).

Question 31

Mathematics · 4 marks · −1

The image of the line x−13=y−31=z−4−5\frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5} in the plane 2x−y+z+3=02x - y + z + 3 = 0 is the line

  1. A.x+33=y−51=z−2−5\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}
  2. B.x+3−3=y−5−1=z+25\frac{x+3}{-3} = \frac{y-5}{-1} = \frac{z+2}{5}
  3. C.x−33=y+51=z−2−5\frac{x-3}{3} = \frac{y+5}{1} = \frac{z-2}{-5}
  4. D.x−3−3=y+5−1=z−25\frac{x-3}{-3} = \frac{y+5}{-1} = \frac{z-2}{5}

Correct answer: A

Explanation: The line is parallel to the plane (6−1−5=06-1-5=0); image of (1,3,4)(1,3,4) is (−3,5,2)(-3,5,2), same direction ratios.

Question 32

Mathematics · 4 marks · −1

If the coefficients of x3x^3 and x4x^4 in the expansion of (1+ax+bx2)(1−2x)18(1 + ax + bx^2)(1 - 2x)^{18} in powers of x are both zero, then (a,b)(a, b) is equal to

  1. A.(16,2513)\left(16, \frac{251}{3}\right)
  2. B.(14,2513)\left(14, \frac{251}{3}\right)
  3. C.(14,2723)\left(14, \frac{272}{3}\right)
  4. D.(16,2723)\left(16, \frac{272}{3}\right)

Correct answer: D

Explanation: Setting the x3x^3 and x4x^4 coefficients to zero gives two linear equations in a, b; solving, a=16a=16, b=2723b=\frac{272}{3}.

Question 33

Mathematics · 4 marks · −1

If a∈Ra \in \mathbb{R} and the equation −3(x−[x])2+2(x−[x])+a2=0-3(x - [x])^2 + 2(x - [x]) + a^2 = 0 (where [x] denotes the greatest integer ≤x\le x) has no integral solution, then all possible values of a lie in the interval

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

Correct answer: A

Explanation: With t={x}∈[0,1)t=\{x\}\in[0,1): a2=3t2−2ta^2=3t^2-2t; a non-integral solution needs t∈(0,1)t\in(0,1), i.e. 0<a2<10<a^2<1.

Question 34

Mathematics · 4 marks · −1

If [a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=λ[a⃗ b⃗ c⃗]2[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = \lambda[\vec{a}\ \vec{b}\ \vec{c}]^2, then λ\lambda is equal to

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

Correct answer: D

Explanation: [a⃗×b⃗ b⃗×c⃗ c⃗×a⃗]=[a⃗ b⃗ c⃗]2[\vec a\times\vec b\ \vec b\times\vec c\ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2, so λ=1\lambda=1.

Question 35

Mathematics · 4 marks · −1

The variance of first 50 even natural numbers is

  1. A.8334\frac{833}{4}
  2. B.833
  3. C.437
  4. D.4374\frac{437}{4}

Correct answer: B

Explanation: σ2=4∑r250−512=3434−2601=833\sigma^2=\frac{4\sum r^2}{50}-51^2=3434-2601=833.

Question 36

Mathematics · 4 marks · −1

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45∘45^\circ. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30∘30^\circ. Then the speed (in m/s) of the bird is

  1. A.40(2−1)40(\sqrt{2}-1)
  2. B.40(3−2)40(\sqrt{3}-\sqrt{2})
  3. C.20220\sqrt{2}
  4. D.20(3−1)20(\sqrt{3}-1)

Correct answer: D

Explanation: tan⁡30∘=2020+x⇒x=20(3−1)\tan30^\circ=\frac{20}{20+x}\Rightarrow x=20(\sqrt3-1) m in 1 s.

Question 37

Mathematics · 4 marks · −1

The integral ∫0π1+4sin⁡2x2−4sin⁡x2 dx\int_0^\pi\sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}}\,dx equals

  1. A.π−4\pi - 4
  2. B.2π3−4−43\frac{2\pi}{3} - 4 - 4\sqrt{3}
  3. C.43−44\sqrt{3} - 4
  4. D.43−4−π34\sqrt{3} - 4 - \frac{\pi}{3}

Correct answer: D

Explanation: Integrand =∣1−2sin⁡x2∣=|1-2\sin\frac x2|, changing sign at π3\frac\pi3; evaluating gives 43−4−π34\sqrt3-4-\frac\pi3.

Question 38

Mathematics · 4 marks · −1

The statement ∼(p↔∼q)\sim(p \leftrightarrow \sim q) is

  1. A.equivalent to p↔qp \leftrightarrow q
  2. B.equivalent to ∼p↔q\sim p \leftrightarrow q
  3. C.a tautology
  4. D.a fallacy

Correct answer: A

Explanation: Truth table: ∼(p↔∼q)\sim(p\leftrightarrow\sim q) is true exactly when p, q agree.

Question 39

Mathematics · 4 marks · −1

If A is an 3×33\times3 non-singular matrix such that AA′=A′AAA' = A'A and B=A−1A′B = A^{-1}A', then BB′BB' equals

  1. A.I+BI + B
  2. B.II
  3. C.B−1B^{-1}
  4. D.(B−1)′(B^{-1})'

Correct answer: B

Explanation: AB=A′AB=A', so ABB′=A′B′=(BA)′=(A−1A′A)′=(A−1AA′)′=AABB'=A'B'=(BA)'=(A^{-1}A'A)'=(A^{-1}AA')'=A, hence BB′=IBB'=I.

Question 40

Mathematics · 4 marks · −1

The integral ∫(1+x−1x)ex+1x dx\int\left(1 + x - \frac{1}{x}\right)e^{x+\frac{1}{x}}\,dx is equal to

  1. A.(x−1)ex+1x+c(x-1)e^{x+\frac{1}{x}}+c
  2. B.xex+1x+cxe^{x+\frac{1}{x}}+c
  3. C.(x+1)ex+1x+c(x+1)e^{x+\frac{1}{x}}+c
  4. D.−xex+1x+c-xe^{x+\frac{1}{x}}+c

Correct answer: B

Explanation: ddx(xex+1/x)=ex+1/x(1+x−1x)\frac{d}{dx}\left(xe^{x+1/x}\right)=e^{x+1/x}\left(1+x-\frac1x\right).

Question 41

Mathematics · 4 marks · −1

If z is a complex number such that ∣z∣≥2|z| \ge 2, then the minimum value of ∣z+12∣\left|z + \frac{1}{2}\right|

  1. A.is equal to 52\frac{5}{2}
  2. B.lies in the interval (1, 2)
  3. C.is strictly greater than 52\frac{5}{2}
  4. D.is strictly greater than 32\frac{3}{2} but less than 52\frac{5}{2}

Correct answer: B

Explanation: Minimum distance from −12-\frac12 to the region ∣z∣≥2|z|\ge2 is 2−12=32∈(1,2)2-\frac12=\frac32\in(1,2).

Question 42

Mathematics · 4 marks · −1

If g is the inverse of a function f and f′(x)=11+x5f'(x) = \frac{1}{1+x^5}, then g′(x)g'(x) is equal to

  1. A.1+x51 + x^5
  2. B.5x45x^4
  3. C.11+{g(x)}5\frac{1}{1+\{g(x)\}^5}
  4. D.1+{g(x)}51 + \{g(x)\}^5

Correct answer: D

Explanation: g′(x)=1f′(g(x))=1+g(x)5g'(x)=\frac{1}{f'(g(x))}=1+g(x)^5.

Question 43

Mathematics · 4 marks · −1

If α,β≠0\alpha, \beta \ne 0, and f(n)=αn+βnf(n) = \alpha^n + \beta^n and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣=K(1−α)2(1−β)2(α−β)2\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix} = K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2, then K is equal to

  1. A.αβ\alpha\beta
  2. B.1αβ\frac{1}{\alpha\beta}
  3. C.1
  4. D.−1-1

Correct answer: C

Explanation: The determinant is the square of the Vandermonde ∣1111αβ1α2β2∣\begin{vmatrix}1&1&1\\1&\alpha&\beta\\1&\alpha^2&\beta^2\end{vmatrix}, so K=1K=1.

Question 44

Mathematics · 4 marks · −1

Let fk(x)=1k(sin⁡kx+cos⁡kx)f_k(x) = \frac{1}{k}\left(\sin^k x + \cos^k x\right) where x∈Rx \in \mathbb{R} and k≥1k \ge 1. Then f4(x)−f6(x)f_4(x) - f_6(x) equals

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

Correct answer: D

Explanation: 1−2s2c24−1−3s2c26=112\frac{1-2s^2c^2}{4}-\frac{1-3s^2c^2}{6}=\frac1{12}.

Question 45

Mathematics · 4 marks · −1

Let α\alpha and β\beta be the roots of equation px2+qx+r=0px^2 + qx + r = 0, p≠0p \ne 0. If p, q, r are in A.P. and 1α+1β=4\frac{1}{\alpha} + \frac{1}{\beta} = 4, then the value of ∣α−β∣|\alpha - \beta| is

  1. A.619\frac{\sqrt{61}}{9}
  2. B.2179\frac{2\sqrt{17}}{9}
  3. C.349\frac{\sqrt{34}}{9}
  4. D.2139\frac{2\sqrt{13}}{9}

Correct answer: D

Explanation: −qr=4-\frac qr=4 and 2q=p+r2q=p+r give p:q:r=−9:−4:1p:q:r=-9:-4:1, i.e. 9x2+4x−1=09x^2+4x-1=0; ∣α−β∣=529=2139|\alpha-\beta|=\frac{\sqrt{52}}{9}=\frac{2\sqrt{13}}{9}.

Question 46

Mathematics · 4 marks · −1

Let A and B be two events such that P(A∪B‾)=16P(\overline{A\cup B}) = \frac{1}{6}, P(A∩B)=14P(A\cap B) = \frac{1}{4} and P(Aˉ)=14P(\bar{A}) = \frac{1}{4}, where Aˉ\bar{A} stands for the complement of the event A. Then the events A and B are

  1. A.mutually exclusive and independent
  2. B.equally likely but not independent
  3. C.independent but not equally likely
  4. D.independent and equally likely

Correct answer: C

Explanation: P(A)=34P(A)=\frac34, P(A∪B)=56⇒P(B)=13P(A\cup B)=\frac56\Rightarrow P(B)=\frac13; P(A)P(B)=14=P(A∩B)P(A)P(B)=\frac14=P(A\cap B).

Question 47

Mathematics · 4 marks · −1

If f and g are differentiable functions in [0, 1] satisfying f(0)=2=g(1)f(0) = 2 = g(1), g(0)=0g(0) = 0 and f(1)=6f(1) = 6, then for some c∈]0,1[c \in ]0, 1[

  1. A.2f′(c)=g′(c)2f'(c) = g'(c)
  2. B.2f′(c)=3g′(c)2f'(c) = 3g'(c)
  3. C.f′(c)=g′(c)f'(c) = g'(c)
  4. D.f′(c)=2g′(c)f'(c) = 2g'(c)

Correct answer: D

Explanation: h=f−2gh=f-2g has h(0)=h(1)=2h(0)=h(1)=2; Rolle gives f′(c)=2g′(c)f'(c)=2g'(c).

Question 48

Mathematics · 4 marks · −1

Let the population of rabbits surviving at a time t be governed by the differential equation dp(t)dt=12p(t)−200\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200. If p(0)=100p(0) = 100, then p(t)p(t) equals

  1. A.400−300et/2400 - 300e^{t/2}
  2. B.300−200e−t/2300 - 200e^{-t/2}
  3. C.600−500et/2600 - 500e^{t/2}
  4. D.400−300e−t/2400 - 300e^{-t/2}

Correct answer: A

Explanation: dpp−400=dt2⇒p=400−300et/2\frac{dp}{p-400}=\frac{dt}{2}\Rightarrow p=400-300e^{t/2}.

Question 49

Mathematics · 4 marks · −1

Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to

  1. A.32\frac{\sqrt{3}}{\sqrt{2}}
  2. B.32\frac{\sqrt{3}}{2}
  3. C.12\frac{1}{2}
  4. D.14\frac{1}{4}

Correct answer: D

Explanation: (1+y)2=1+(1−y)2⇒y=14(1+y)^2=1+(1-y)^2\Rightarrow y=\frac14.

Question 50

Mathematics · 4 marks · −1

The area of the region described by A={(x,y):x2+y2≤1 and y2≤1−x}A = \{(x, y) : x^2 + y^2 \le 1 \text{ and } y^2 \le 1 - x\} is

  1. A.π2+43\frac{\pi}{2}+\frac{4}{3}
  2. B.π2−43\frac{\pi}{2}-\frac{4}{3}
  3. C.π2−23\frac{\pi}{2}-\frac{2}{3}
  4. D.π2+23\frac{\pi}{2}+\frac{2}{3}

Correct answer: A

Explanation: Left half-disc π2\frac\pi2 plus the parabolic region for 0≤x≤10\le x\le1: 2∫011−x dx=432\int_0^1\sqrt{1-x}\,dx=\frac43.

Question 51

Mathematics · 4 marks · −1

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=04ax + 2ay + c = 0 and 5bx+2by+d=05bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then

  1. A.2bc−3ad=02bc - 3ad = 0
  2. B.2bc+3ad=02bc + 3ad = 0
  3. C.3bc−2ad=03bc - 2ad = 0
  4. D.3bc+2ad=03bc + 2ad = 0

Correct answer: C

Explanation: Point (h,−h)(h,-h): 2ah+c=02ah+c=0, 3bh+d=0⇒c2a=d3b3bh+d=0\Rightarrow\frac{c}{2a}=\frac{d}{3b}, i.e. 3bc−2ad=03bc-2ad=0.

Question 52

Mathematics · 4 marks · −1

Let PS be the median of the triangle with vertices P(2, 2), Q(6, −1-1) and R(7, 3). The equation of the line passing through (1,−1)(1, -1) and parallel to PS is

  1. A.4x−7y−11=04x - 7y - 11 = 0
  2. B.2x+9y+7=02x + 9y + 7 = 0
  3. C.4x+7y+3=04x + 7y + 3 = 0
  4. D.2x−9y−11=02x - 9y - 11 = 0

Correct answer: B

Explanation: S=(132,1)S=(\frac{13}{2},1), slope of PS =−29=-\frac29; line: 2x+9y+7=02x+9y+7=0.

Question 53

Mathematics · 4 marks · −1

lim⁡x→0sin⁡(πcos⁡2x)x2\lim_{x\to0}\frac{\sin(\pi\cos^2 x)}{x^2} is equal to

  1. A.π2\frac{\pi}{2}
  2. B.1
  3. C.−π-\pi
  4. D.π\pi

Correct answer: D

Explanation: sin⁡(πcos⁡2x)=sin⁡(πsin⁡2x)≈πx2\sin(\pi\cos^2x)=\sin(\pi\sin^2x)\approx\pi x^2.

Question 54

Mathematics · 4 marks · −1

If X={4n−3n−1:n∈N}X = \{4^n - 3n - 1 : n \in \mathbb{N}\} and Y={9(n−1):n∈N}Y = \{9(n - 1) : n \in \mathbb{N}\}, where N\mathbb{N} is the set of natural numbers, then X∪YX \cup Y is equal to

  1. A.N\mathbb{N}
  2. B.Y−XY - X
  3. C.X
  4. D.Y

Correct answer: D

Explanation: 4n−3n−14^n-3n-1 is always a multiple of 9 (binomial expansion of (1+3)n(1+3)^n), so X⊂YX\subset Y.

Question 55

Mathematics · 4 marks · −1

The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6x^2 + 3y^2 = 6 on any tangent to it is

  1. A.(x2−y2)2=6x2+2y2(x^2 - y^2)^2 = 6x^2 + 2y^2
  2. B.(x2−y2)2=6x2−2y2(x^2 - y^2)^2 = 6x^2 - 2y^2
  3. C.(x2+y2)2=6x2+2y2(x^2 + y^2)^2 = 6x^2 + 2y^2
  4. D.(x2+y2)2=6x2−2y2(x^2 + y^2)^2 = 6x^2 - 2y^2

Correct answer: C

Explanation: Tangent y=mx±6m2+2y=mx\pm\sqrt{6m^2+2} with foot (h,k)(h,k), m=−hkm=-\frac hk: (h2+k2)2=6h2+2k2(h^2+k^2)^2=6h^2+2k^2.

Question 56

Mathematics · 4 marks · −1

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is

  1. A.2+3\sqrt{2}+\sqrt{3}
  2. B.3+23+\sqrt{2}
  3. C.2−32-\sqrt{3}
  4. D.2+32+\sqrt{3}

Correct answer: D

Explanation: 4ar=a+ar2⇒r2−4r+1=04ar=a+ar^2\Rightarrow r^2-4r+1=0, increasing so r=2+3r=2+\sqrt3.

Question 57

Mathematics · 4 marks · −1

If (10)9+2(11)1(10)8+3(11)2(10)7+⋯+10(11)9=k(10)9(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9, then k is equal to

  1. A.12110\frac{121}{10}
  2. B.441100\frac{441}{100}
  3. C.100
  4. D.110

Correct answer: C

Explanation: AGP: S−1110SS-\frac{11}{10}S telescopes to give S=1011S=10^{11}, so k=100k=100.

Question 58

Mathematics · 4 marks · −1

The angle between the lines whose direction cosines satisfy the equations l+m+n=0l + m + n = 0 and l2=m2+n2l^2 = m^2 + n^2 is

  1. A.π3\frac{\pi}{3}
  2. B.π4\frac{\pi}{4}
  3. C.π6\frac{\pi}{6}
  4. D.π2\frac{\pi}{2}

Correct answer: A

Explanation: Substituting l=−(m+n)l=-(m+n) gives mn=0mn=0: directions (−1,1,0)/2(-1,1,0)/\sqrt2 and (−1,0,1)/2(-1,0,1)/\sqrt2, cos⁡θ=12\cos\theta=\frac12.

Question 59

Mathematics · 4 marks · −1

The slope of the line touching both the parabolas y2=4xy^2 = 4x and x2=−32yx^2 = -32y is

  1. A.12\frac{1}{2}
  2. B.32\frac{3}{2}
  3. C.18\frac{1}{8}
  4. D.23\frac{2}{3}

Correct answer: A

Explanation: Tangent y=mx+1my=mx+\frac1m to y2=4xy^2=4x touches x2=−32yx^2=-32y when m3=18m^3=\frac18, m=12m=\frac12.

Question 60

Mathematics · 4 marks · −1

If x=−1x = -1 and x=2x = 2 are extreme points of f(x)=αlog⁡∣x∣+βx2+xf(x) = \alpha\log|x| + \beta x^2 + x, then

  1. A.α=−6, β=12\alpha = -6,\ \beta = \frac{1}{2}
  2. B.α=−6, β=−12\alpha = -6,\ \beta = -\frac{1}{2}
  3. C.α=2, β=−12\alpha = 2,\ \beta = -\frac{1}{2}
  4. D.α=2, β=12\alpha = 2,\ \beta = \frac{1}{2}

Correct answer: C

Explanation: 2βx2+x+α=02\beta x^2+x+\alpha=0 has roots −1,2-1,2: −12β=1⇒β=−12-\frac{1}{2\beta}=1\Rightarrow\beta=-\frac12, α2β=−2⇒α=2\frac{\alpha}{2\beta}=-2\Rightarrow\alpha=2.

Question 61

Chemistry · 4 marks · −1

Which one of the following properties is not shown by NO?

  1. A.It combines with oxygen to form nitrogen dioxide
  2. B.It's bond order is 2.5
  3. C.It is diamagnetic in gaseous state
  4. D.It is a neutral oxide

Correct answer: C

Explanation: NO has an odd electron and is paramagnetic.

Question 62

Chemistry · 4 marks · −1

If Z is a compressibility factor, van der Waals equation at low pressure can be written as:

  1. A.Z=1−PbRTZ = 1 - \frac{Pb}{RT}
  2. B.Z=1+PbRTZ = 1 + \frac{Pb}{RT}
  3. C.Z=1+RTPbZ = 1 + \frac{RT}{Pb}
  4. D.Z=1−aVRTZ = 1 - \frac{a}{VRT}

Correct answer: D

Explanation: At low pressure the b terms are negligible: PV=RT−aVPV=RT-\frac aV, Z=1−aVRTZ=1-\frac{a}{VRT}.

Question 63

Chemistry · 4 marks · −1

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is:

  1. A.Cu
  2. B.Cr
  3. C.Ag
  4. D.Ca

Correct answer: D

Explanation: Water is reduced (H2 evolves) in preference to Ca2+Ca^{2+}.

Question 64

Chemistry · 4 marks · −1

Resistance of 0.2 M solution of an electrolyte is 50 Ω50\,\Omega. The specific conductance of the solution is 1.4 S m−1^{-1}. The resistance of 0.5 M solution of the same electrolyte is 280 Ω280\,\Omega. The molar conductivity of 0.5 M solution of the electrolyte in S m2^2 mol−1^{-1} is:

  1. A.5×1035\times10^3
  2. B.5×1025\times10^2
  3. C.5×10−45\times10^{-4}
  4. D.5×10−35\times10^{-3}

Correct answer: C

Explanation: Cell constant =50⋅1.4=70=50\cdot1.4=70 m−1^{-1}; κ=70280=0.25\kappa=\frac{70}{280}=0.25 S/m; Λm=0.25500 mol/m3=5×10−4\Lambda_m=\frac{0.25}{500\text{ mol/m}^3}=5\times10^{-4}.

Question 65

Chemistry · 4 marks · −1

CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct?

  1. A.rCs++rCl−=32ar_{Cs^+} + r_{Cl^-} = \frac{\sqrt{3}}{2}a
  2. B.rCs++rCl−=3ar_{Cs^+} + r_{Cl^-} = \sqrt{3}a
  3. C.rCs++rCl−=3ar_{Cs^+} + r_{Cl^-} = 3a
  4. D.rCs++rCl−=3a2r_{Cs^+} + r_{Cl^-} = \frac{3a}{2}

Correct answer: A

Explanation: Ions touch along the body diagonal: 2(r++r−)=3a2(r^++r^-)=\sqrt3a.

Question 66

Chemistry · 4 marks · −1

Consider separate solutions of 0.500 M C2H5OHC_2H_5OH(aq), 0.100 M Mg3(PO4)2Mg_3(PO_4)_2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4Na_3PO_4(aq) at 25∘25^\circC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

  1. A.0.125 M Na3PO4Na_3PO_4(aq) has the highest osmotic pressure.
  2. B.0.500 M C2H5OHC_2H_5OH(aq) has the highest osmotic pressure.
  3. C.They all have the same osmotic pressure.
  4. D.0.100 M Mg3(PO4)2Mg_3(PO_4)_2(aq) has the highest osmotic pressure.

Correct answer: C

Explanation: iCiC: 0.50.5, 5(0.1)=0.55(0.1)=0.5, 2(0.25)=0.52(0.25)=0.5, 4(0.125)=0.54(0.125)=0.5 — all isotonic.

Question 67

Chemistry · 4 marks · −1

In which of the following reactions H2O2H_2O_2 acts as a reducing agent? (a) H2O2+2H++2e−→2H2OH_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O (b) H2O2−2e−→O2+2H+H_2O_2 - 2e^- \rightarrow O_2 + 2H^+ (c) H2O2+2e−→2OH−H_2O_2 + 2e^- \rightarrow 2OH^- (d) H2O2+2OH−−2e−→O2+2H2OH_2O_2 + 2OH^- - 2e^- \rightarrow O_2 + 2H_2O

  1. A.(a), (c)
  2. B.(b), (d)
  3. C.(a), (b)
  4. D.(c), (d)

Correct answer: B

Explanation: A reducing agent loses electrons: (b) and (d).

Question 68

Chemistry · 4 marks · −1

In SN2S_N2 reactions, the correct order of reactivity for the following compounds: CH3ClCH_3Cl, CH3CH2ClCH_3CH_2Cl, (CH3)2CHCl(CH_3)_2CHCl and (CH3)3CCl(CH_3)_3CCl is:

  1. A.CH3CH2Cl>CH3Cl>(CH3)2CHCl>(CH3)3CClCH_3CH_2Cl > CH_3Cl > (CH_3)_2CHCl > (CH_3)_3CCl
  2. B.(CH3)2CHCl>CH3CH2Cl>CH3Cl>(CH3)3CCl(CH_3)_2CHCl > CH_3CH_2Cl > CH_3Cl > (CH_3)_3CCl
  3. C.CH3Cl>(CH3)2CHCl>CH3CH2Cl>(CH3)3CClCH_3Cl > (CH_3)_2CHCl > CH_3CH_2Cl > (CH_3)_3CCl
  4. D.CH3Cl>CH3CH2Cl>(CH3)2CHCl>(CH3)3CClCH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl

Correct answer: D

Explanation: SN2S_N2 rate falls with steric hindrance: methyl > 1° > 2° > 3°.

Question 69

Chemistry · 4 marks · −1

The octahedral complex of a metal ion M3+M^{3+} with four monodentate ligands L1L_1, L2L_2, L3L_3 and L4L_4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:

  1. A.L3<L2<L4<L1L_3 < L_2 < L_4 < L_1
  2. B.L1<L2<L4<L3L_1 < L_2 < L_4 < L_3
  3. C.L4<L3<L2<L1L_4 < L_3 < L_2 < L_1
  4. D.L1<L3<L2<L4L_1 < L_3 < L_2 < L_4

Correct answer: D

Explanation: Stronger field means larger splitting and shorter absorbed wavelength: red < yellow < green < blue, i.e. L1<L3<L2<L4L_1<L_3<L_2<L_4.

Question 70

Chemistry · 4 marks · −1

For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M10\frac{M}{10} sulphuric acid. The unreacted acid required 20 ml of M10\frac{M}{10} sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is:

  1. A.3%
  2. B.5%
  3. C.6%
  4. D.10%

Correct answer: D

Explanation: meq of acid used =60⋅0.2−20⋅0.1=10=60\cdot0.2-20\cdot0.1=10; %N =1.4⋅101.4=10%=\frac{1.4\cdot10}{1.4}=10\%.

Question 71

Chemistry · 4 marks · −1

The equivalent conductance of NaCl at concentration C and at infinite dilution are λC\lambda_C and λ∞\lambda_\infty, respectively. The correct relationship between λC\lambda_C and λ∞\lambda_\infty is given as: (where the constant B is positive)

  1. A.λC=λ∞−(B)C\lambda_C = \lambda_\infty - (B)\sqrt{C}
  2. B.λC=λ∞+(B)C\lambda_C = \lambda_\infty + (B)\sqrt{C}
  3. C.λC=λ∞+(B)C\lambda_C = \lambda_\infty + (B)C
  4. D.λC=λ∞−(B)C\lambda_C = \lambda_\infty - (B)C

Correct answer: A

Explanation: Debye-Huckel-Onsager (Kohlrausch) law for strong electrolytes.

Question 72

Chemistry · 4 marks · −1

For the reaction, SO2(g)+12O2(g)⇌SO3(g)SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}, if KP=KC(RT)xK_P = K_C(RT)^x where the symbols have usual meaning then the value of x is: (assuming ideality)

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

Correct answer: D

Explanation: x=Δng=1−32=−12x=\Delta n_g=1-\frac32=-\frac12.

Question 73

Chemistry · 4 marks · −1

In the reaction, CH3COOH→LiAlH4A→PCl5B→Alc. KOHCCH_3COOH \xrightarrow{LiAlH_4} A \xrightarrow{PCl_5} B \xrightarrow{Alc.\ KOH} C, the product C is:

  1. A.Ethylene
  2. B.Acetyl chloride
  3. C.Acetaldehyde
  4. D.Acetylene

Correct answer: A

Explanation: Ethanol, then chloroethane, then dehydrohalogenation to ethylene.

Question 74

Chemistry · 4 marks · −1

Sodium phenoxide when heated with CO2CO_2 under pressure at 125∘125^\circC yields a product which on acetylation produces C. The major product C would be:

Figure for question 74
  1. A.Methyl salicylate (2-hydroxybenzoic acid methyl ester)
  2. B.3-acetoxybenzoic acid
  3. C.2-acetoxybenzoic acid (aspirin)
  4. D.2-hydroxy-1,4-diacetylbenzene type (OH with two COCH3 groups)

Correct answer: C

Explanation: Kolbe reaction gives salicylic acid; acetylation of the phenolic OH gives aspirin.

Question 75

Chemistry · 4 marks · −1

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

  1. A.an alkyl cyanide
  2. B.an alkyl isocyanide
  3. C.an alkanol
  4. D.an alkanediol

Correct answer: B

Explanation: Carbylamine reaction: RNH2+CHCl3+3KOH→RNCRNH_2+CHCl_3+3KOH\rightarrow RNC.

Question 76

Chemistry · 4 marks · −1

The correct statement for the molecule, CsI3CsI_3 is:

  1. A.it contains Cs3+Cs^{3+} and I−I^- ions.
  2. B.it contains Cs+Cs^+, I−I^- and lattice I2I_2 molecule.
  3. C.it is a covalent molecule.
  4. D.it contains Cs+Cs^+ and I3−I_3^- ions.

Correct answer: D

Explanation: Cs shows only +1; the anion is triiodide I3−I_3^-.

Question 77

Chemistry · 4 marks · −1

The equation which is balanced and represents the correct product(s) is:

  1. A.[Mg(H2O)6]2++(EDTA)4−→excess NaOH[Mg(EDTA)]2++6H2O[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \xrightarrow{excess\ NaOH} [Mg(EDTA)]^{2+} + 6H_2O
  2. B.CuSO4+4KCN→K2[Cu(CN)4]+K2SO4CuSO_4 + 4KCN \rightarrow K_2[Cu(CN)_4] + K_2SO_4
  3. C.Li2O+2KCl→2LiCl+K2OLi_2O + 2KCl \rightarrow 2LiCl + K_2O
  4. D.[CoCl(NH3)5]++5H+→Co2++5NH4++Cl−[CoCl(NH_3)_5]^+ + 5H^+ \rightarrow Co^{2+} + 5NH_4^+ + Cl^-

Correct answer: D

Explanation: (1) is not charge balanced; (2) actually gives K3[Cu(CN)4]K_3[Cu(CN)_4]; (3) is unfavourable; (4) is balanced and correct.

Question 78

Chemistry · 4 marks · −1

For which of the following molecule significant μ≠0\mu \ne 0? (a) 1,4-dichlorobenzene (b) 1,4-dicyanobenzene (c) hydroquinone (1,4-dihydroxybenzene) (d) benzene-1,4-dithiol

Figure for question 78
  1. A.Only (c)
  2. B.(c) and (d)
  3. C.Only (a)
  4. D.(a) and (b)

Correct answer: B

Explanation: The -OH and -SH groups are bent and rotate, so most conformations have a net dipole; the linear C-Cl and C-CN dipoles cancel.

Question 79

Chemistry · 4 marks · −1

For the non-stoichiometre reaction 2A+B→C+D2A + B \rightarrow C + D, the following kinetic data were obtained in three separate experiments, all at 298 K. [A] = 0.1 M, [B] = 0.1 M: rate of formation of C = 1.2×10−31.2\times10^{-3}; [A] = 0.1 M, [B] = 0.2 M: 1.2×10−31.2\times10^{-3}; [A] = 0.2 M, [B] = 0.1 M: 2.4×10−32.4\times10^{-3} (mol L−1^{-1} s−1^{-1}). The rate law for the formation of C is:

  1. A.dcdt=k[A][B]2\frac{dc}{dt} = k[A][B]^2
  2. B.dcdt=k[A]\frac{dc}{dt} = k[A]
  3. C.dcdt=k[A][B]\frac{dc}{dt} = k[A][B]
  4. D.dcdt=k[A]2[B]\frac{dc}{dt} = k[A]^2[B]

Correct answer: B

Explanation: Doubling B leaves the rate unchanged (zero order in B); doubling A doubles it (first order).

Question 80

Chemistry · 4 marks · −1

Which series of reactions correctly represents chemical reactions related to iron and its compound?

  1. A.Fe→Cl2, heatFeCl3→heat, airFeCl2→ZnFeFe \xrightarrow{Cl_2,\ heat} FeCl_3 \xrightarrow{heat,\ air} FeCl_2 \xrightarrow{Zn} Fe
  2. B.Fe→O2, heatFe3O4→CO, 600∘CFeO→CO, 700∘CFeFe \xrightarrow{O_2,\ heat} Fe_3O_4 \xrightarrow{CO,\ 600^\circ C} FeO \xrightarrow{CO,\ 700^\circ C} Fe
  3. C.Fe→dil H2SO4FeSO4→H2SO4, O2Fe2(SO4)3→heatFeFe \xrightarrow{dil\ H_2SO_4} FeSO_4 \xrightarrow{H_2SO_4,\ O_2} Fe_2(SO_4)_3 \xrightarrow{heat} Fe
  4. D.Fe→O2, heatFeO→dil H2SO4FeSO4→heatFeFe \xrightarrow{O_2,\ heat} FeO \xrightarrow{dil\ H_2SO_4} FeSO_4 \xrightarrow{heat} Fe

Correct answer: B

Explanation: Only series (2) is chemically correct (blast-furnace reduction steps).

Question 81

Chemistry · 4 marks · −1

Considering the basic strength of amines in aqueous solution, which one has the smallest pKbpK_b value?

  1. A.(CH3)3N(CH_3)_3N
  2. B.C6H5NH2C_6H_5NH_2
  3. C.(CH3)2NH(CH_3)_2NH
  4. D.CH3NH2CH_3NH_2

Correct answer: C

Explanation: In water, dimethylamine is the strongest base among these.

Question 82

Chemistry · 4 marks · −1

Which one of the following bases is not present in DNA?

  1. A.Cytosine
  2. B.Thymine
  3. C.Quinoline
  4. D.Adenine

Correct answer: C

Explanation: DNA bases are A, G, C, T; quinoline is not a nucleobase.

Question 83

Chemistry · 4 marks · −1

The correct set of four quantum numbers for the valence elections of rubidium atom (Z = 37) is:

  1. A.5,1,1,+125, 1, 1, +\frac{1}{2}
  2. B.5,0,1,+125, 0, 1, +\frac{1}{2}
  3. C.5,0,0,+125, 0, 0, +\frac{1}{2}
  4. D.5,1,0,+125, 1, 0, +\frac{1}{2}

Correct answer: C

Explanation: Rb: [Kr]5s1^1, so n=5,l=0,m=0n=5,l=0,m=0.

Question 84

Chemistry · 4 marks · −1

The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is:

  1. A.2-Butyne
  2. B.2-Butene
  3. C.Acetylene
  4. D.Ethene

Correct answer: A

Explanation: 2CH3CCl3+6Ag→CH3C≡CCH3+6AgCl2CH_3CCl_3+6Ag\rightarrow CH_3C\equiv CCH_3+6AgCl.

Question 85

Chemistry · 4 marks · −1

Given below are the half-cell reactions: Mn2++2e−→MnMn^{2+} + 2e^- \rightarrow Mn; E∘=−1.18E^\circ = -1.18 V; 2(Mn3++e−→Mn2+)2(Mn^{3+} + e^- \rightarrow Mn^{2+}); E∘=+1.51E^\circ = +1.51 V. The E∘E^\circ for 3Mn2+→Mn+2Mn3+3Mn^{2+} \rightarrow Mn + 2Mn^{3+} will be:

  1. A.−0.33-0.33 V; the reaction will not occur
  2. B.−0.33-0.33 V; the reaction will occur
  3. C.−2.69-2.69 V; the reaction will not occur
  4. D.−2.69-2.69 V; the reaction will occur

Correct answer: C

Explanation: Ecell∘=−1.18−1.51=−2.69E^\circ_{cell}=-1.18-1.51=-2.69 V; negative, so non-spontaneous.

Question 86

Chemistry · 4 marks · −1

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is:

  1. A.1 : 8
  2. B.3 : 16
  3. C.1 : 4
  4. D.7 : 32

Correct answer: D

Explanation: 1/324/28=732\frac{1/32}{4/28}=\frac{7}{32}.

Question 87

Chemistry · 4 marks · −1

Which one is classified as a condensation polymer?

  1. A.Teflon
  2. B.Acrylonitrile
  3. C.Dacron
  4. D.Neoprene

Correct answer: C

Explanation: Dacron (terylene) is a polyester made by condensation.

Question 88

Chemistry · 4 marks · −1

Among the following oxoacids, the correct decreasing order of acid strength is:

  1. A.HClO4>HClO3>HClO2>HOClHClO_4 > HClO_3 > HClO_2 > HOCl
  2. B.HClO2>HClO4>HClO3>HOClHClO_2 > HClO_4 > HClO_3 > HOCl
  3. C.HOCl>HClO2>HClO3>HClO4HOCl > HClO_2 > HClO_3 > HClO_4
  4. D.HClO4>HOCl>HClO2>HClO3HClO_4 > HOCl > HClO_2 > HClO_3

Correct answer: A

Explanation: Acid strength rises with oxidation state of Cl.

Question 89

Chemistry · 4 marks · −1

The most suitable reagent for the conversion of R−CH2−OH→R−CHOR-CH_2-OH \rightarrow R-CHO is:

  1. A.CrO3CrO_3
  2. B.PCC (Pyridinium Chlorochromate)
  3. C.KMnO4KMnO_4
  4. D.K2Cr2O7K_2Cr_2O_7

Correct answer: B

Explanation: PCC stops the oxidation of a primary alcohol at the aldehyde.

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