AIEEE 2011 — 1 May Offline Set Q (Previous Year Paper)

JEE Main · 2011 · with answers and explanations

questions
87
Total marks
348
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Mathematics · Chemistry

Question 1

Physics · 4 marks · −1

The transverse displacement y(x, t) of a wave on a string is given by y(x,t)=e−(ax2+bt2+2ab xt)y(x,t) = e^{-\left(ax^2 + bt^2 + 2\sqrt{ab}\,xt\right)}. This represents a

  1. A.wave moving in −x-x direction with speed ba\sqrt{\frac{b}{a}}
  2. B.standing wave of frequency b\sqrt{b}
  3. C.standing wave of frequency 1b\frac{1}{\sqrt{b}}
  4. D.wave moving in +x+x direction with ab\sqrt{\frac{a}{b}}

Correct answer: A

Explanation: y=e−(ax+bt)2=f(x+b/a t)y=e^{-(\sqrt a x+\sqrt b t)^2}=f(x+\sqrt{b/a}\,t): travels along −x-x at b/a\sqrt{b/a}.

Question 2

Physics · 4 marks · −1

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm. Circular scale reading : 52 divisions. Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is :

  1. A.0.052 cm
  2. B.0.026 cm
  3. C.0.005 cm
  4. D.0.52 cm

Correct answer: A

Explanation: LC 0.01 mm; 52×0.01=0.5252\times0.01=0.52 mm =0.052=0.052 cm.

Question 3

Physics · 4 marks · −1

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  1. A.g
  2. B.23g\frac{2}{3}g
  3. C.g3\frac{g}{3}
  4. D.32g\frac{3}{2}g

Correct answer: B

Explanation: a=mgm+I/R2=mgm+m/2=2g3a=\frac{mg}{m+I/R^2}=\frac{mg}{m+m/2}=\frac{2g}{3}.

Question 4

Physics · 4 marks · −1

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 Nm−1^{-1}):

  1. A.0.2π0.2\pi mJ
  2. B.2π2\pi mJ
  3. C.0.4π0.4\pi mJ
  4. D.4π4\pi mJ

Correct answer: C

Explanation: Two surfaces: W=2T⋅4π(r22−r12)=0.06⋅4π⋅16×10−4≈0.38πW=2T\cdot4\pi(r_2^2-r_1^2)=0.06\cdot4\pi\cdot16\times10^{-4}\approx0.38\pi mJ.

Question 5

Physics · 4 marks · −1

A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc:

  1. A.continuously decreases
  2. B.continuously increases
  3. C.first increases and then decreases
  4. D.remains unchanged

Correct answer: C

Explanation: Angular momentum is conserved; I first decreases (insect nears the axis) then increases.

Question 6

Physics · 4 marks · −1

Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

  1. A.−4Gmr-\frac{4Gm}{r}
  2. B.−6Gmr-\frac{6Gm}{r}
  3. C.−9Gmr-\frac{9Gm}{r}
  4. D.zero

Correct answer: C

Explanation: Null point at r3\frac r3 from m: V=−3Gmr−6Gmr=−9GmrV=-\frac{3Gm}{r}-\frac{6Gm}{r}=-\frac{9Gm}{r}.

Question 7

Physics · 4 marks · −1

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d (d≪1)(d \ll 1) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them,

  1. A.v∝x−1v \propto x^{-1}
  2. B.v∝x1/2v \propto x^{1/2}
  3. C.v∝xv \propto x
  4. D.v∝x−1/2v \propto x^{-1/2}

Correct answer: D

Explanation: Equilibrium: kq2x2=mgx2l⇒q2∝x3\frac{kq^2}{x^2}=\frac{mgx}{2l}\Rightarrow q^2\propto x^3. dqdt\frac{dq}{dt} constant gives x1/2dxdt∝x^{1/2}\frac{dx}{dt}\propto const, so v∝x−1/2v\propto x^{-1/2}.

Question 8

Physics · 4 marks · −1

A boat is moving due east in a region where the earth's magnetic field is 5.0×10−55.0\times10^{-5} NA−1^{-1}m−1^{-1} due north and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms−1^{-1}, the magnitude of the induced emf in the wire of aerial is :

  1. A.0.75 mV
  2. B.0.50 mV
  3. C.0.15 mV
  4. D.1 mV

Correct answer: C

Explanation: e=Bvl=5×10−5⋅1.5⋅2=0.15e=Bvl=5\times10^{-5}\cdot1.5\cdot2=0.15 mV.

Question 9

Physics · 4 marks · −1

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by : dvdt=−2.5v\frac{dv}{dt} = -2.5\sqrt{v} where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

  1. A.2 s
  2. B.4 s
  3. C.8 s
  4. D.1 s

Correct answer: A

Explanation: 2v∣6.250=−2.5t⇒t=2⋅2.52.5=22\sqrt v\big|_{6.25}^0=-2.5t\Rightarrow t=\frac{2\cdot2.5}{2.5}=2 s.

Question 10

Physics · 4 marks · −1

A fully charged capacitor C with initial charge q0q_0 is connected to a coil of self inductance L at t=0t = 0. The time at which the energy is stored equally between the electric and the magnetic field is :

  1. A.π4LC\frac{\pi}{4}\sqrt{LC}
  2. B.2πLC2\pi\sqrt{LC}
  3. C.LC\sqrt{LC}
  4. D.πLC\pi\sqrt{LC}

Correct answer: A

Explanation: q=q0cos⁡ωt=q02q=q_0\cos\omega t=\frac{q_0}{\sqrt2} first at ωt=π4\omega t=\frac\pi4.

Question 11

Physics · 4 marks · −1

Let the x - z plane be the boundary between two transparent media. Medium 1 in z≥0z \ge 0 has a refractive index of 2\sqrt{2} and medium 2 with z<0z < 0 has a refractive index of 3\sqrt{3}. A ray of light in medium 1 given by the vector A⃗=63i^+83j^−10k^\vec{A} = 6\sqrt{3}\hat{i} + 8\sqrt{3}\hat{j} - 10\hat{k} is incident on the plane of separation. The angle of refraction in medium 2 is

  1. A.45∘45^\circ
  2. B.60∘60^\circ
  3. C.75∘75^\circ
  4. D.30∘30^\circ

Correct answer: A

Explanation: The media are separated by z=0z=0 (normal along z): cos⁡i=1020\cos i=\frac{10}{20}, i=60∘i=60^\circ; 2sin⁡60∘=3sin⁡r⇒r=45∘\sqrt2\sin60^\circ=\sqrt3\sin r\Rightarrow r=45^\circ.

Question 12

Physics · 4 marks · −1

A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is

  1. A.μ0I2π2R\frac{\mu_0I}{2\pi^2R}
  2. B.μ0I2πR\frac{\mu_0I}{2\pi R}
  3. C.μ0I4π2R\frac{\mu_0I}{4\pi^2R}
  4. D.μ0Iπ2R\frac{\mu_0I}{\pi^2R}

Correct answer: D

Explanation: Strips dI=IπdθdI=\frac{I}{\pi}d\theta, each giving μ0dI2πR\frac{\mu_0dI}{2\pi R}; integrating the non-cancelling component: B=μ0Iπ2RB=\frac{\mu_0I}{\pi^2R}.

Question 13

Physics · 4 marks · −1

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ\gamma. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

  1. A.(γ−1)2γRMv2\frac{(\gamma-1)}{2\gamma R}Mv^2 K
  2. B.γMv22R\frac{\gamma Mv^2}{2R} K
  3. C.(γ−1)2RMv2\frac{(\gamma-1)}{2R}Mv^2 K
  4. D.(γ−1)2(γ+1)RMv2\frac{(\gamma-1)}{2(\gamma+1)R}Mv^2 K

Correct answer: C

Explanation: 12Mv2=CvΔT=Rγ−1ΔT\frac12Mv^2=C_v\Delta T=\frac{R}{\gamma-1}\Delta T per mole.

Question 14

Physics · 4 marks · −1

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1A_1. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A2A_2. The ratio of (A1A2)\left(\frac{A_1}{A_2}\right) is :

  1. A.M+mM\frac{M+m}{M}
  2. B.(MM+m)1/2\left(\frac{M}{M+m}\right)^{1/2}
  3. C.(M+mM)1/2\left(\frac{M+m}{M}\right)^{1/2}
  4. D.MM+m\frac{M}{M+m}

Correct answer: C

Explanation: Momentum conserved: MA1ω1=(M+m)A2ω2MA_1\omega_1=(M+m)A_2\omega_2 with ω∝1mass\omega\propto\frac1{\sqrt{mass}}.

Question 15

Physics · 4 marks · −1

Water is flowing continuously from a tap having an internal diameter 8×10−38\times10^{-3} m. The water velocity as it leaves the tap is 0.4 ms−1^{-1}. The diameter of the water stream at a distance 2×10−12\times10^{-1} m below the tap is close to :

  1. A.7.5×10−37.5\times10^{-3} m
  2. B.9.6×10−39.6\times10^{-3} m
  3. C.3.6×10−33.6\times10^{-3} m
  4. D.5.0×10−35.0\times10^{-3} m

Correct answer: C

Explanation: v2=0.16+4≈2.04v_2=\sqrt{0.16+4}\approx2.04 m/s; d2=d1v1/v2≈3.6×10−3d_2=d_1\sqrt{v_1/v_2}\approx3.6\times10^{-3} m.

Question 16

Physics · 4 marks · −1

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals. Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is false.

Correct answer: A

Explanation: Sky waves depend on ionospheric reflection, which fluctuates, explaining their instability.

Question 17

Physics · 4 marks · −1

Three perfect gases at absolute temperatures T1T_1, T2T_2 and T3T_3 are mixed. The masses of molecules are m1m_1, m2m_2 and m3m_3 and the number of molecules are n1n_1, n2n_2 and n3n_3 respectively. Assuming no loss of energy, the final temperature of the mixture is :

  1. A.n1T1+n2T2+n3T3n1+n2+n3\frac{n_1T_1 + n_2T_2 + n_3T_3}{n_1 + n_2 + n_3}
  2. B.n1T1+n2T22+n3T32n1T1+n2T2+n3T3\frac{n_1T_1 + n_2T_2^2 + n_3T_3^2}{n_1T_1 + n_2T_2 + n_3T_3}
  3. C.n12T12+n22T22+n32T32n1T1+n2T2+n3T3\frac{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2}{n_1T_1 + n_2T_2 + n_3T_3}
  4. D.(T1+T2+T3)3\frac{(T_1 + T_2 + T_3)}{3}

Correct answer: A

Explanation: Total translational KE conserved: ∑niTi=T∑ni\sum n_iT_i=T\sum n_i (same degrees of freedom assumed).

Question 18

Physics · 4 marks · −1

A pulley of radius 2 m is rotated about its axis by a force F=(20t−5t2)F = (20t - 5t^2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2^2, the number of rotations made by the pulley before its direction of motion is reversed, is :

  1. A.more than 3 but less than 6
  2. B.more than 6 but less than 9
  3. C.more than 9
  4. D.less than 3

Correct answer: A

Explanation: α=2(20t−5t2)10=4t−t2\alpha=\frac{2(20t-5t^2)}{10}=4t-t^2; ω=2t2−t33=0\omega=2t^2-\frac{t^3}{3}=0 at t=6t=6 s; θ=2t33−t412=36\theta=\frac{2t^3}{3}-\frac{t^4}{12}=36 rad ≈5.7\approx5.7 rotations.

Question 19

Physics · 4 marks · −1

A resistor R and 2 μ2\,\muF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log⁡102.5=0.4\log_{10}2.5 = 0.4)

  1. A.1.7×105 Ω1.7\times10^5\,\Omega
  2. B.2.7×106 Ω2.7\times10^6\,\Omega
  3. C.3.3×107 Ω3.3\times10^7\,\Omega
  4. D.1.3×104 Ω1.3\times10^4\,\Omega

Correct answer: B

Explanation: 120=200(1−e−t/RC)⇒tRC=ln⁡2.5=0.92120=200(1-e^{-t/RC})\Rightarrow\frac{t}{RC}=\ln2.5=0.92; R=50.92⋅2×10−6≈2.7×106 ΩR=\frac{5}{0.92\cdot2\times10^{-6}}\approx2.7\times10^6\,\Omega.

Question 20

Physics · 4 marks · −1

A Carnot engine operating between temperatures T1T_1 and T2T_2 has efficiency 16\frac{1}{6}. When T2T_2 is lowered by 62 K, its efficiency increases to 13\frac{1}{3}. Then T1T_1 and T2T_2 are, respectively :

  1. A.372 K and 330 K
  2. B.330 K and 268 K
  3. C.310 K and 248 K
  4. D.372 K and 310 K

Correct answer: D

Explanation: 62T1=13−16⇒T1=372\frac{62}{T_1}=\frac13-\frac16\Rightarrow T_1=372 K, T2=56T1=310T_2=\frac56T_1=310 K.

Question 21

Physics · 4 marks · −1

If a wire is stretched to make it 0.1% longer, its resistance will :

  1. A.increase by 0.2%
  2. B.decrease by 0.2%
  3. C.decrease by 0.05%
  4. D.increases by 0.05%

Correct answer: A

Explanation: Volume constant: R∝l2R\propto l^2, so +0.2%+0.2\%.

Question 22

Physics · 4 marks · −1

A car is fitted with a convex side-view mirror of focal length 20cm. A second car 2.8 m behind the first car is overtaking the first car at relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is :

  1. A.115\frac{1}{15} m/s
  2. B.10 m/s
  3. C.15 m/s
  4. D.110\frac{1}{10} m/s

Correct answer: A

Explanation: v=ufu−f=28015v=\frac{uf}{u-f}=\frac{280}{15} cm; image speed =v2u2⋅15=(115)2⋅15=115=\frac{v^2}{u^2}\cdot15=\left(\frac1{15}\right)^2\cdot15=\frac1{15} m/s.

Question 23

Physics · 4 marks · −1

Energy required for the electron excitation in Li++Li^{++} from the first to the third Bohr orbit is :

  1. A.36.3 eV
  2. B.108.8 eV
  3. C.122.4 eV
  4. D.12.1 eV

Correct answer: B

Explanation: 13.6⋅9(1−19)=108.813.6\cdot9\left(1-\frac19\right)=108.8 eV.

Question 24

Physics · 4 marks · −1

The electrostatic potential inside a charged spherical ball is given by ϕ=aρ2+b\phi = a\rho^2 + b where ρ\rho is the distance from the centre; a, b are constants. Then the charge density inside ball is :

  1. A.−6aϵ0ρ-6a\epsilon_0\rho
  2. B.−24πaϵ0ρ-24\pi a\epsilon_0\rho
  3. C.−6aϵ0-6a\epsilon_0
  4. D.−24πaϵ0ρ-24\pi a\epsilon_0\rho

Correct answer: C

Explanation: E=−2aρE=-2a\rho; Gauss: ρch=ϵ0∇⋅E⃗=−6aϵ0\rho_{ch}=\epsilon_0\nabla\cdot\vec E=-6a\epsilon_0.

Question 25

Physics · 4 marks · −1

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :

  1. A.πv4g2\pi\frac{v^4}{g^2}
  2. B.πv42g2\frac{\pi v^4}{2g^2}
  3. C.πv2g2\pi\frac{v^2}{g^2}
  4. D.πv4g\pi\frac{v^4}{g}

Correct answer: A

Explanation: Maximum range v2g\frac{v^2}{g}; area πv4g2\pi\frac{v^4}{g^2}.

Question 26

Physics · 4 marks · −1

100g of water is heated from 30∘30^\circC to 50∘50^\circC. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4148 J/kg/K):

  1. A.8.4 kJ
  2. B.84 kJ
  3. C.2.1 kJ
  4. D.4.2 kJ

Correct answer: A

Explanation: ΔU=Q=0.1⋅4148⋅20≈8.3\Delta U=Q=0.1\cdot4148\cdot20\approx8.3 kJ.

Question 27

Physics · 4 marks · −1

The half life of a radioactive substance is 20 minutes. The approximate time interval (t2−t1)(t_2 - t_1) between the time t2t_2 when 23\frac{2}{3} of it has decayed and time t1t_1 when 13\frac{1}{3} of it had decayed is :

  1. A.14 min
  2. B.20 min
  3. C.28 min
  4. D.7 min

Correct answer: B

Explanation: Remaining 23\frac23 at t1t_1 and 13\frac13 at t2t_2: ratio 2, i.e. one half-life, 20 min.

Question 28

Physics · 4 marks · −1

Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency ν>ν0\nu > \nu_0 (the threshold frequency). The maximum kinetic energy and the stopping potential are KmaxK_{max} and V0V_0 respectively. If the frequency incident on the surface doubled, both the KmaxK_{max} and V0V_0 are also doubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is false.

Correct answer: C

Explanation: Kmax=hν−ϕK_{max}=h\nu-\phi is linear but not proportional, so doubling ν\nu more than doubles KmaxK_{max}.

Question 29

Mathematics · 4 marks · −1

The lines L1:y−x=0L_1 : y - x = 0 and L2:2x+y=0L_2 : 2x + y = 0 intersect the line L3:y+2=0L_3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1L_1 and L2L_2 intersect L3L_3 at R. Statement-1 : The ratio PR : RQ equals 22:52\sqrt{2} : \sqrt{5}. Statement-2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Correct answer: B

Explanation: P=(−2,−2)P=(-2,-2), Q=(1,−2)Q=(1,-2); bisector theorem: PR:RQ=OP:OQ=22:5PR:RQ=OP:OQ=2\sqrt2:\sqrt5. Statement-2 is false in general.

Question 30

Mathematics · 4 marks · −1

If A=sin⁡2x+cos⁡4xA = \sin^2x + \cos^4x, then for all real x

  1. A.1316≤A≤1\frac{13}{16} \le A \le 1
  2. B.1≤A≤21 \le A \le 2
  3. C.34≤A≤1316\frac{3}{4} \le A \le \frac{13}{16}
  4. D.34≤A≤1\frac{3}{4} \le A \le 1

Correct answer: D

Explanation: A=1−sin⁡2xcos⁡2x=1−sin⁡22x4∈[34,1]A=1-\sin^2x\cos^2x=1-\frac{\sin^22x}{4}\in[\frac34,1].

Question 31

Mathematics · 4 marks · −1

The coefficient of x7x^7 in the expansion of (1−x−x2+x3)6\left(1 - x - x^2 + x^3\right)^6 is

  1. A.−132-132
  2. B.−144-144
  3. C.132
  4. D.144

Correct answer: B

Explanation: (1−x)6(1−x2)6(1-x)^6(1-x^2)^6; x7x^7 from (i,j)=(1,3),(3,2),(5,1)(i,j)=(1,3),(3,2),(5,1): 120−300+36=−144120-300+36=-144.

Question 32

Mathematics · 4 marks · −1

lim⁡x→2(1−cos⁡{2(x−2)}x−2)\lim_{x\to2}\left(\frac{\sqrt{1-\cos\{2(x-2)\}}}{x-2}\right)

  1. A.equals 2\sqrt{2}
  2. B.equals −2-\sqrt{2}
  3. C.equals 12\frac{1}{\sqrt{2}}
  4. D.does not exist

Correct answer: D

Explanation: =2∣sin⁡(x−2)∣x−2=\frac{\sqrt2|\sin(x-2)|}{x-2}: limits ±2\pm\sqrt2 from the two sides.

Question 33

Mathematics · 4 marks · −1

Statement-1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3{}^9C_3. Statement-2 : The number of ways of choosing any 3 places from 9 different places is 9C3{}^9C_3.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Correct answer: D

Explanation: Place separators in 3 of the 9 gaps between the 10 balls: exactly the choice described in Statement-2.

Question 34

Mathematics · 4 marks · −1

d2xdy2\frac{d^2x}{dy^2} equals

  1. A.−(d2ydx2)−1(dydx)−3-\left(\frac{d^2y}{dx^2}\right)^{-1}\left(\frac{dy}{dx}\right)^{-3}
  2. B.(d2ydx2)(dydx)−2\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-2}
  3. C.−(d2ydx2)(dydx)−3-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}
  4. D.(d2ydx2)−1\left(\frac{d^2y}{dx^2}\right)^{-1}

Correct answer: C

Explanation: ddy(1y′)=−y′′y′2⋅1y′\frac{d}{dy}\left(\frac{1}{y'}\right)=-\frac{y''}{y'^2}\cdot\frac1{y'}.

Question 35

Mathematics · 4 marks · −1

If dydx=y+3>0\frac{dy}{dx} = y + 3 > 0 and y(0)=2y(0) = 2, then y(ln⁡2)y(\ln2) is equal to

  1. A.5
  2. B.13
  3. C.−2-2
  4. D.7

Correct answer: D

Explanation: y+3=5exy+3=5e^x; y(ln⁡2)=10−3=7y(\ln2)=10-3=7.

Question 36

Mathematics · 4 marks · −1

Let R be the set of real numbers. Statement-1 : A={(x,y)∈R×R:y−x is an integer}A = \{(x, y) \in R\times R : y - x \text{ is an integer}\} is an equivalence relation on R. Statement-2 : B={(x,y)∈R×R:x=αy for some rational number α}B = \{(x, y) \in R\times R : x = \alpha y \text{ for some rational number } \alpha\} is an equivalence relation on R.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Correct answer: B

Explanation: A is reflexive, symmetric, transitive. B is not symmetric: (0,1)∈B(0,1)\in B (α=0\alpha=0) but (1,0)∉B(1,0)\notin B.

Question 37

Mathematics · 4 marks · −1

The value of ∫018log⁡(1+x)1+x2 dx\int_0^1\frac{8\log(1+x)}{1+x^2}\,dx is

  1. A.π8log⁡2\frac{\pi}{8}\log2
  2. B.π2log⁡2\frac{\pi}{2}\log2
  3. C.log⁡2\log2
  4. D.πlog⁡2\pi\log2

Correct answer: D

Explanation: With x=tan⁡θx=\tan\theta: 8∫0π/4log⁡(1+tan⁡θ)dθ=8⋅π8log⁡28\int_0^{\pi/4}\log(1+\tan\theta)d\theta=8\cdot\frac\pi8\log2.

Question 38

Mathematics · 4 marks · −1

Let α,β\alpha, \beta be real and z be a complex number. If z2+αz+β=0z^2 + \alpha z + \beta = 0 has two distinct roots on the line Re z=1\text{Re}\,z = 1, then it is necessary that

  1. A.β∈(−1,0)\beta \in (-1, 0)
  2. B.∣β∣=1|\beta| = 1
  3. C.β∈(1,∞)\beta \in (1, \infty)
  4. D.β∈(0,1)\beta \in (0, 1)

Correct answer: C

Explanation: Roots 1±iq1\pm iq (q≠0q\ne0) give β=1+q2>1\beta=1+q^2>1.

Question 39

Mathematics · 4 marks · −1

Consider 5 independent Bernoulli's trials each with probability of success p. If the probability of at least one failure is greater than or equal to 3132\frac{31}{32}, then p lies in the interval

  1. A.(34,1112]\left(\frac{3}{4}, \frac{11}{12}\right]
  2. B.[0,12]\left[0, \frac{1}{2}\right]
  3. C.(1112,1]\left(\frac{11}{12}, 1\right]
  4. D.(12,34]\left(\frac{1}{2}, \frac{3}{4}\right]

Correct answer: B

Explanation: 1−p5≥3132⇒p≤121-p^5\ge\frac{31}{32}\Rightarrow p\le\frac12.

Question 40

Mathematics · 4 marks · −1

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

  1. A.19 months
  2. B.20 months
  3. C.21 months
  4. D.18 months

Correct answer: C

Explanation: 600+∑k=1n−3(200+40k)=11040⇒n=21600+\sum_{k=1}^{n-3}(200+40k)=11040\Rightarrow n=21.

Question 41

Mathematics · 4 marks · −1

The domain of the function f(x)=1∣x∣−xf(x) = \frac{1}{\sqrt{|x| - x}} is

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

Correct answer: B

Explanation: ∣x∣−x>0  ⟺  x<0|x|-x>0\iff x<0.

Question 42

Mathematics · 4 marks · −1

If the angle between the line x=y−12=z−3λx = \frac{y-1}{2} = \frac{z-3}{\lambda} and the plane x+2y+3z=4x + 2y + 3z = 4 is cos⁡−1(514)\cos^{-1}\left(\sqrt{\frac{5}{14}}\right), then λ\lambda equals

  1. A.32\frac{3}{2}
  2. B.25\frac{2}{5}
  3. C.53\frac{5}{3}
  4. D.23\frac{2}{3}

Correct answer: D

Explanation: sin⁡θ=314=5+3λ5+λ214⇒λ=23\sin\theta=\frac{3}{\sqrt{14}}=\frac{5+3\lambda}{\sqrt{5+\lambda^2}\sqrt{14}}\Rightarrow\lambda=\frac23.

Question 43

Mathematics · 4 marks · −1

If a⃗=110(3i^+k^)\vec{a} = \frac{1}{\sqrt{10}}\left(3\hat{i} + \hat{k}\right) and b⃗=17(2i^+3j^−6k^)\vec{b} = \frac{1}{7}\left(2\hat{i} + 3\hat{j} - 6\hat{k}\right), then the value of (2a⃗−b⃗)⋅[(a⃗×b⃗)×(a⃗+2b⃗)](2\vec{a} - \vec{b})\cdot\left[(\vec{a}\times\vec{b})\times(\vec{a} + 2\vec{b})\right] is

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

Correct answer: D

Explanation: a⃗,b⃗\vec a,\vec b are orthogonal unit vectors; (a⃗×b⃗)×(a⃗+2b⃗)=b⃗−2a⃗(\vec a\times\vec b)\times(\vec a+2\vec b)=\vec b-2\vec a, so the value is −∣2a⃗−b⃗∣2=−5-|2\vec a-\vec b|^2=-5.

Question 44

Mathematics · 4 marks · −1

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (−3,1)(-3, 1) and has eccentricity 25\sqrt{\frac{2}{5}} is

  1. A.5x2+3y2−48=05x^2 + 3y^2 - 48 = 0
  2. B.3x2+5y2−15=03x^2 + 5y^2 - 15 = 0
  3. C.5x2+3y2−32=05x^2 + 3y^2 - 32 = 0
  4. D.3x2+5y2−32=03x^2 + 5y^2 - 32 = 0

Correct answer: D

Explanation: b2=35a2b^2=\frac35a^2; 9a2+53a2=1⇒a2=323\frac9{a^2}+\frac{5}{3a^2}=1\Rightarrow a^2=\frac{32}3: 3x2+5y2=323x^2+5y^2=32.

Question 45

Mathematics · 4 marks · −1

Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation dV(t)dt=−k(T−t)\frac{dV(t)}{dt} = -k(T - t), where k>0k > 0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is

  1. A.I−kT22I - \frac{kT^2}{2}
  2. B.I−k(T−t)22I - \frac{k(T-t)^2}{2}
  3. C.e−kTe^{-kT}
  4. D.T2−IkT^2 - \frac{I}{k}

Correct answer: A

Explanation: V=k(T−t)22+I−kT22V=\frac{k(T-t)^2}{2}+I-\frac{kT^2}{2}; at t=Tt=T: I−kT22I-\frac{kT^2}{2}.

Question 46

Mathematics · 4 marks · −1

The vector a⃗\vec{a} and b⃗\vec{b} are not perpendicular and c⃗\vec{c} and d⃗\vec{d} are two vectors satisfying: b⃗×c⃗=b⃗×d⃗\vec{b}\times\vec{c} = \vec{b}\times\vec{d} and a⃗⋅d⃗=0\vec{a}\cdot\vec{d} = 0. Then the vector d⃗\vec{d} is equal to

  1. A.c⃗+(a⃗⋅c⃗a⃗⋅b⃗)b⃗\vec{c} + \left(\frac{\vec{a}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\right)\vec{b}
  2. B.b⃗+(b⃗⋅c⃗a⃗⋅b⃗)c⃗\vec{b} + \left(\frac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\right)\vec{c}
  3. C.c⃗−(a⃗⋅c⃗a⃗⋅b⃗)b⃗\vec{c} - \left(\frac{\vec{a}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\right)\vec{b}
  4. D.b⃗−(b⃗⋅c⃗a⃗⋅b⃗)c⃗\vec{b} - \left(\frac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\right)\vec{c}

Correct answer: C

Explanation: d⃗=c⃗+λb⃗\vec d=\vec c+\lambda\vec b; a⃗⋅d⃗=0⇒λ=−a⃗⋅c⃗a⃗⋅b⃗\vec a\cdot\vec d=0\Rightarrow\lambda=-\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}.

Question 47

Mathematics · 4 marks · −1

The two circles x2+y2=axx^2 + y^2 = ax and x2+y2=c2x^2 + y^2 = c^2 (c>0)(c > 0) touch each other if

  1. A.∣a∣=c|a| = c
  2. B.a=2ca = 2c
  3. C.∣a∣=2c|a| = 2c
  4. D.2∣a∣=c2|a| = c

Correct answer: A

Explanation: First circle: centre (a2,0)(\frac a2,0), radius ∣a∣2\frac{|a|}2, passing through the origin; internal tangency needs c−∣a∣2=∣a∣2c-\frac{|a|}2=\frac{|a|}2.

Question 48

Mathematics · 4 marks · −1

If C and D are two events such that C⊂DC \subset D and P(D)≠0P(D) \ne 0, then the correct statement among the following is

  1. A.P(C∣D)≥P(C)P(C|D) \ge P(C)
  2. B.P(C∣D)<P(C)P(C|D) < P(C)
  3. C.P(C∣D)=P(D)P(C)P(C|D) = \frac{P(D)}{P(C)}
  4. D.P(C∣D)=P(C)P(C|D) = P(C)

Correct answer: A

Explanation: P(C∣D)=P(C)P(D)≥P(C)P(C|D)=\frac{P(C)}{P(D)}\ge P(C).

Question 49

Mathematics · 4 marks · −1

The number of values of k for which the linear equations 4x+ky+2z=04x + ky + 2z = 0; kx+4y+z=0kx + 4y + z = 0; 2x+2y+z=02x + 2y + z = 0 possess a non-zero solution is

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

Correct answer: A

Explanation: Determinant =−k2+6k−8=0⇒k=2,4=-k^2+6k-8=0\Rightarrow k=2,4.

Question 50

Mathematics · 4 marks · −1

Consider the following statements P : Suman is brilliant, Q : Suman is rich, R : Suman is honest. The negation of the statement 'Suman is brilliant and dishonest if and only if Suman is rich' can be expressed as

  1. A.∼(Q↔(P∧∼R))\sim(Q \leftrightarrow (P\wedge\sim R))
  2. B.∼Q↔∼P∧R\sim Q \leftrightarrow \sim P \wedge R
  3. C.∼(P∧∼R)↔Q\sim(P\wedge\sim R) \leftrightarrow Q
  4. D.∼P∧(Q↔∼R)\sim P \wedge (Q \leftrightarrow \sim R)

Correct answer: A

Explanation: The statement is (P∧∼R)↔Q(P\wedge\sim R)\leftrightarrow Q; its negation is option 1.

Question 51

Mathematics · 4 marks · −1

The shortest distance between line y−x=1y - x = 1 and curve x=y2x = y^2 is

  1. A.328\frac{3\sqrt{2}}{8}
  2. B.832\frac{8}{3\sqrt{2}}
  3. C.43\frac{4}{\sqrt{3}}
  4. D.34\frac{\sqrt{3}}{4}

Correct answer: A

Explanation: Distance from (y2,y)(y^2,y): y2−y+12\frac{y^2-y+1}{\sqrt2}, minimum 3/42=328\frac{3/4}{\sqrt2}=\frac{3\sqrt2}{8}.

Question 52

Mathematics · 4 marks · −1

If the mean deviation about the median of the numbers a, 2a, ..., 50a is 50, then |a| equals

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

Correct answer: B

Explanation: Median 25.5a25.5a; mean deviation =∣a∣50⋅2(0.5+1.5+⋯+24.5)=12.5∣a∣=50=\frac{|a|}{50}\cdot2(0.5+1.5+\dots+24.5)=12.5|a|=50.

Question 53

Mathematics · 4 marks · −1

Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line x1=y−12=z−23\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}. Statement-2 : The line x1=y−12=z−23\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Correct answer: A

Explanation: Midpoint (1,3,5)(1,3,5) lies on the line and AB⃗⊥\vec{AB}\perp the line, so S1 is true; S2 is true but bisection alone does not establish the mirror image.

Question 54

Mathematics · 4 marks · −1

Let A and B be two symmetric matrices of order 3. Statement-1 : A(BA) and (AB)A are symmetric matrices. Statement-2 : AB is symmetric matrix if matrix multiplication of A and B is commutative.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Correct answer: A

Explanation: (ABA)T=ATBTAT=ABA(ABA)^T=A^TB^TA^T=ABA; S2 also true but not the reason.

Question 55

Mathematics · 4 marks · −1

If ω(≠1)\omega(\ne1) is a cube root of unity, and (1+ω)7=A+Bω(1 + \omega)^7 = A + B\omega. Then (A, B) equals

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

Correct answer: A

Explanation: (1+ω)7=(−ω2)7=−ω2=1+ω(1+\omega)^7=(-\omega^2)^7=-\omega^2=1+\omega.

Question 56

Mathematics · 4 marks · −1

The value of p and q for which the function f(x)={sin⁡(p+1)x+sin⁡xx,x<0q,x=0x+x2−xx3/2,x>0f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases} is continuous for all x in R, is

  1. A.p=52,q=12p = \frac{5}{2}, q = \frac{1}{2}
  2. B.p=−32,q=12p = -\frac{3}{2}, q = \frac{1}{2}
  3. C.p=12,q=32p = \frac{1}{2}, q = \frac{3}{2}
  4. D.p=12,q=−32p = \frac{1}{2}, q = -\frac{3}{2}

Correct answer: B

Explanation: Right limit 1+x−1x→12\frac{\sqrt{1+x}-1}{x}\to\frac12; left limit p+2=12⇒p=−32p+2=\frac12\Rightarrow p=-\frac32.

Question 57

Mathematics · 4 marks · −1

The area of the region enclosed by the curves y=xy = x, x=ex = e, y=1xy = \frac{1}{x} and the positive x-axis is

  1. A.1 square units
  2. B.32\frac{3}{2} square units
  3. C.52\frac{5}{2} square units
  4. D.12\frac{1}{2} square units

Correct answer: B

Explanation: ∫01x dx+∫1edxx=12+1\int_0^1x\,dx+\int_1^e\frac{dx}{x}=\frac12+1.

Question 58

Mathematics · 4 marks · −1

For x∈(0,5π2)x \in \left(0, \frac{5\pi}{2}\right), define f(x)=∫0xtsin⁡t dtf(x) = \int_0^x\sqrt{t}\sin t\,dt. Then f has

  1. A.local minimum at π\pi and 2π2\pi
  2. B.local minimum at π\pi and local maximum at 2π2\pi
  3. C.local maximum at π\pi and local minimum at 2π2\pi
  4. D.local maximum at π\pi and 2π2\pi

Correct answer: C

Explanation: f′=xsin⁡xf'=\sqrt x\sin x changes + to - at π\pi and - to + at 2π2\pi.

Question 59

Chemistry · 4 marks · −1

Among the following the maximum covalent character is shown by the compound :

  1. A.SnCl2SnCl_2
  2. B.AlCl3AlCl_3
  3. C.MgCl2MgCl_2
  4. D.FeCl2FeCl_2

Correct answer: B

Explanation: Small, highly charged Al3+Al^{3+} polarises Cl−Cl^- most (Fajans' rules).

Question 60

Chemistry · 4 marks · −1

The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA ?

  1. A.2nd2^{nd}
  2. B.3rd3^{rd}
  3. C.4th4^{th}
  4. D.1st1^{st}

Correct answer: A

Explanation: DNA contains 2-deoxyribose.

Question 61

Chemistry · 4 marks · −1

Trichloroacetaldehyde was subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :

  1. A.Trichloromethanol
  2. B.2, 2, 2-Trichloropropanol
  3. C.Chloroform
  4. D.2, 2, 2-Trichloroethanol

Correct answer: D

Explanation: Disproportionation: one aldehyde is reduced to CCl3CH2OHCCl_3CH_2OH.

Question 62

Chemistry · 4 marks · −1

Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is :

  1. A.2-Butanone
  2. B.Ethyl chloride
  3. C.Ethyl ethanoate
  4. D.Diethyl ether

Correct answer: C

Explanation: C2H5ONa+CH3COCl→CH3COOC2H5C_2H_5ONa+CH_3COCl\rightarrow CH_3COOC_2H_5.

Question 63

Chemistry · 4 marks · −1

The reduction potential of hydrogen half cell will be negative if :

  1. A.p(H2)=1p(H_2) = 1 atm and [H+]=1.0[H^+] = 1.0 M
  2. B.p(H2)=2p(H_2) = 2 atm and [H+]=1.0[H^+] = 1.0 M
  3. C.p(H2)=2p(H_2) = 2 atm and [H+]=2.0[H^+] = 2.0 M
  4. D.p(H2)=1p(H_2) = 1 atm and [H+]=2.0[H^+] = 2.0 M

Correct answer: B

Explanation: E=−0.0592log⁡pH2[H+]2<0E=-\frac{0.059}{2}\log\frac{p_{H_2}}{[H^+]^2}<0 requires pH2>[H+]2p_{H_2}>[H^+]^2: only option 2.

Question 64

Chemistry · 4 marks · −1

The strongest acid amongst the following compounds is :

  1. A.HCOOH
  2. B.CH3CH2CH(Cl)CO2HCH_3CH_2CH(Cl)CO_2H
  3. C.ClCH2CH2CH2COOHClCH_2CH_2CH_2COOH
  4. D.CH3COOHCH_3COOH

Correct answer: B

Explanation: The α\alpha-chloro (-I close to COOH) acid is strongest.

Question 65

Chemistry · 4 marks · −1

The degree of dissociation (α)(\alpha) of a weak electrolyte, AxByA_xB_y is related to van't Hoff factor (i) by the expression :

  1. A.α=i−1x+y+1\alpha = \frac{i-1}{x+y+1}
  2. B.α=x+y−1i−1\alpha = \frac{x+y-1}{i-1}
  3. C.α=x+y+1i−1\alpha = \frac{x+y+1}{i-1}
  4. D.α=i−1(x+y−1)\alpha = \frac{i-1}{(x+y-1)}

Correct answer: D

Explanation: i=1+α(x+y−1)i=1+\alpha(x+y-1).

Question 66

Chemistry · 4 marks · −1

'a' and 'b' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because

  1. A.a and b for Cl2Cl_2 < a and b for C2H6C_2H_6
  2. B.a for Cl2Cl_2 < a for C2H6C_2H_6 but b for Cl2Cl_2 > b for C2H6C_2H_6
  3. C.a for Cl2Cl_2 > a for C2H6C_2H_6 but b for Cl2Cl_2 < b for C2H6C_2H_6
  4. D.a and b for Cl2Cl_2 > a and b for C2H6C_2H_6

Correct answer: C

Explanation: Larger a (attraction) and smaller b favour liquefaction (Cl2Cl_2: a 6.49, b 0.0562; C2H6C_2H_6: 5.49, 0.0638).

Question 67

Chemistry · 4 marks · −1

A vessel at 1000 K contains CO2CO_2 with a pressure of 0.5 atm. Some of the CO2CO_2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is

  1. A.3 atm
  2. B.0.3 atm
  3. C.0.18 atm
  4. D.1.8 atm

Correct answer: D

Explanation: CO2+C⇌2COCO_2+C\rightleftharpoons2CO: 0.5−x+2x=0.8⇒x=0.30.5-x+2x=0.8\Rightarrow x=0.3; Kp=0.620.2=1.8K_p=\frac{0.6^2}{0.2}=1.8 atm.

Question 68

Chemistry · 4 marks · −1

Boron cannot form which one of the following anions ?

  1. A.BH4−BH_4^-
  2. B.B(OH)4−B(OH)_4^-
  3. C.BO2−BO_2^-
  4. D.BF63−BF_6^{3-}

Correct answer: D

Explanation: Boron has no d orbitals; maximum covalency 4.

Question 69

Chemistry · 4 marks · −1

Which of the following facts about the complex [Cr(NH3)6]Cl3[Cr(NH_3)_6]Cl_3 is wrong ?

  1. A.The complex is paramagnetic
  2. B.The complex is an outer orbital complex
  3. C.The complex gives white precipitate with silver nitrate solution
  4. D.The complex involves d2sp3d^2sp^3 hybridization and is octahedral in shape.

Correct answer: B

Explanation: Cr3+Cr^{3+} (d3d^3) uses inner 3d orbitals: d2sp3d^2sp^3, inner-orbital.

Question 70

Chemistry · 4 marks · −1

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6∘-6^\circC will be : (KfK_f for water = 1.86 K kg mol−1^{-1}, and molar mass of ethylene glycol = 62g mol−1^{-1})

  1. A.204.30g
  2. B.400.00 g
  3. C.304.60 g
  4. D.804.32g

Correct answer: D

Explanation: w=6⋅4⋅621.86=800w=\frac{6\cdot4\cdot62}{1.86}=800 g ≈804.32\approx804.32 g (option 4).

Question 71

Chemistry · 4 marks · −1

Which one of the following order represents the correct sequence of the increasing basic nature of the given oxides ?

  1. A.MgO<K2O<Al2O3<Na2OMgO < K_2O < Al_2O_3 < Na_2O
  2. B.Na2O<K2O<MgO<Al2O3Na_2O < K_2O < MgO < Al_2O_3
  3. C.K2O<Na2O<Al2O3<MgOK_2O < Na_2O < Al_2O_3 < MgO
  4. D.Al2O3<MgO<Na2O<K2OAl_2O_3 < MgO < Na_2O < K_2O

Correct answer: D

Explanation: Basicity rises with metallic character.

Question 72

Chemistry · 4 marks · −1

The rate of a chemical reaction doubles for every 10∘10^\circC rise of temperature. If the temperature is raised by 50∘50^\circC, the rate of the reaction increases by about :

  1. A.24 times
  2. B.32 times
  3. C.64 times
  4. D.10 times

Correct answer: B

Explanation: 25=322^5=32.

Question 73

Chemistry · 4 marks · −1

The magnetic moment (spin only) of [NiCl4]2−[NiCl_4]^{2-} is

  1. A.5.46 BM
  2. B.2.83 BM
  3. C.1.41 BM
  4. D.1.82 BM

Correct answer: B

Explanation: Tetrahedral Ni2+Ni^{2+} (d8d^8), 2 unpaired: 8=2.83\sqrt8=2.83 BM.

Question 74

Chemistry · 4 marks · −1

The hybridization of orbitals of N atom in NO3−NO_3^-, NO2+NO_2^+ and NH4+NH_4^+ are respectively :

  1. A.sp2,sp,sp3sp^2, sp, sp^3
  2. B.sp,sp3,sp2sp, sp^3, sp^2
  3. C.sp2,sp3,spsp^2, sp^3, sp
  4. D.sp,sp2,sp3sp, sp^2, sp^3

Correct answer: A

Explanation: Trigonal planar, linear, tetrahedral.

Question 75

Chemistry · 4 marks · −1

In context of the lanthanoids, which of the following statements is not correct ?

  1. A.All the members exhibit +3 oxidation state
  2. B.Because of similar properties the separation of lanthanoids is not easy.
  3. C.Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series.
  4. D.There is a gradual decrease in the radii of the members with increasing atomic number in the series.

Correct answer: C

Explanation: Only a few lanthanoids (e.g. Ce) show +4.

Question 76

Chemistry · 4 marks · −1

A 5.2 molal aqueous solution of methyl alcohol, CH3OHCH_3OH, is supplied. What is the mole fraction of methyl alcohol in the solution ?

  1. A.0.190
  2. B.0.086
  3. C.0.050
  4. D.0.100

Correct answer: B

Explanation: x=5.25.2+55.5≈0.086x=\frac{5.2}{5.2+55.5}\approx0.086.

Question 77

Chemistry · 4 marks · −1

Which of the following statement is wrong ?

  1. A.Nitrogen cannot form dπ−pπd\pi - p\pi bond.
  2. B.Single N - N bond is weaker than the single P - P bond.
  3. C.N2O4N_2O_4 has two resonance structures
  4. D.The stability of hydrides increases from NH3NH_3 to BiH3BiH_3 in group 15 of the periodic table

Correct answer: D

Explanation: Hydride stability decreases down group 15.

Question 78

Chemistry · 4 marks · −1

The outer electron configuration of Gd (Atomic No : 64) is :

  1. A.4f85d06s24f^8 5d^0 6s^2
  2. B.4f45d46s24f^4 5d^4 6s^2
  3. C.4f75d16s24f^7 5d^1 6s^2
  4. D.4f34d56s24f^3 4d^5 6s^2

Correct answer: C

Explanation: Half-filled 4f stability: [Xe]4f75d16s2[Xe]4f^75d^16s^2.

Question 79

Chemistry · 4 marks · −1

Which of the following statements regarding sulphur is incorrect ?

  1. A.The vapour at 200∘200^\circC consists mostly of S8S_8 rings
  2. B.At 600∘600^\circC the gas mainly consists of S2S_2 molecules
  3. C.The oxidation state of sulphur is never less than +4 in its compounds
  4. D.S2S_2 molecule is paramagnetic.

Correct answer: C

Explanation: Sulphur shows −2-2 (e.g. H2SH_2S).

Question 80

Chemistry · 4 marks · −1

The structure of IF7IF_7 is :

  1. A.trigonal bipyramid
  2. B.octahedral
  3. C.pentagonal bipyramid
  4. D.square pyramid

Correct answer: C

Explanation: sp3d3sp^3d^3: pentagonal bipyramid.

Question 81

Chemistry · 4 marks · −1

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :

  1. A.a vinyl group
  2. B.an isopropyl group
  3. C.an acetylenic triple bond
  4. D.two ethylenic double bonds

Correct answer: A

Explanation: A terminal CH2=CH_2= (vinyl) gives HCHO.

Question 82

Chemistry · 4 marks · −1

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

  1. A.325 nm
  2. B.743 nm
  3. C.518 nm
  4. D.1035 nm

Correct answer: B

Explanation: 1355=1680+1λ⇒λ≈743\frac1{355}=\frac1{680}+\frac1{\lambda}\Rightarrow\lambda\approx743 nm.

Question 83

Chemistry · 4 marks · −1

Which of the following reagents may be used to distinguish between phenol and benzoic acid ?

  1. A.Tollen's reagent
  2. B.Molisch reagent
  3. C.Neutral FeCl3FeCl_3
  4. D.Aqueous NaOH

Correct answer: C

Explanation: Phenol gives a violet colour with neutral FeCl3FeCl_3; benzoic acid a buff precipitate.

Question 84

Chemistry · 4 marks · −1

Phenol is heated with a solution of mixture of KBr and KBrO3KBrO_3. The major product obtained in the above reaction is

  1. A.3-Bromophenol
  2. B.4-Bromophenol
  3. C.2, 4, 6- Tribromophenol
  4. D.2-Bromophenol

Correct answer: C

Explanation: KBr/KBrO3KBr/KBrO_3 in acid liberates Br2Br_2 (aq), giving 2,4,6-tribromophenol.

Question 85

Chemistry · 4 marks · −1

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is :

  1. A.AB2AB_2
  2. B.A2B3A_2B_3
  3. C.A2B5A_2B_5
  4. D.A2BA_2B

Correct answer: C

Explanation: A: 1; B: 52\frac52; formula A2B5A_2B_5.

Question 86

Chemistry · 4 marks · −1

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3^3 to a volume of 100 dm3^3 at 27∘27^\circC is :

  1. A.35.8J mol−1^{-1}K−1^{-1}
  2. B.32.3J mol−1^{-1}K−1^{-1}
  3. C.42.3J mol−1^{-1}K−1^{-1}
  4. D.38.3J mol−1^{-1}K−1^{-1}

Correct answer: D

Explanation: ΔS=nRln⁡10=2⋅8.314⋅2.303≈38.3\Delta S=nR\ln10=2\cdot8.314\cdot2.303\approx38.3 J/K.

Question 87

Chemistry · 4 marks · −1

Identify the compound that exhibits tautomerism.

  1. A.Lactic acid
  2. B.2-Pentanone
  3. C.Phenol
  4. D.2- Butene

Correct answer: B

Explanation: 2-Pentanone has α\alpha-hydrogens and shows keto-enol tautomerism.

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