JEE Main 2020 — 7 Jan Shift 1 (Previous Year Paper)

JEE Main · 2020 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:

  1. A.L↔m, C↔k, R↔bL \leftrightarrow m,\ C \leftrightarrow k,\ R \leftrightarrow b
  2. B.L↔m, C↔1k, R↔bL \leftrightarrow m,\ C \leftrightarrow \frac{1}{k},\ R \leftrightarrow b
  3. C.L↔k, C↔b, R↔mL \leftrightarrow k,\ C \leftrightarrow b,\ R \leftrightarrow m
  4. D.L↔1b, C↔1m, R↔1kL \leftrightarrow \frac{1}{b},\ C \leftrightarrow \frac{1}{m},\ R \leftrightarrow \frac{1}{k}

Correct answer: B

Explanation: Lq¨+Rq˙+q/C=0L\ddot q + R\dot q + q/C = 0 vs mx¨+bx˙+kx=0m\ddot x + b\dot x + kx = 0: L→mL\to m, R→bR\to b, 1/C→k1/C\to k.

Question 2

Physics · 4 marks · −1

As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Figure for question 2
  1. A.1r4gh3\frac{1}{r}\sqrt{\frac{4gh}{3}}
  2. B.r34ghr\sqrt{\frac{3}{4gh}}
  3. C.1r2gh3\frac{1}{r}\sqrt{\frac{2gh}{3}}
  4. D.r32ghr\sqrt{\frac{3}{2gh}}

Correct answer: A

Explanation: mgh=12mv2+12⋅12mr2ω2=34mv2⇒v=4gh/3mgh = \frac12 mv^2 + \frac12\cdot\frac12 mr^2\omega^2 = \frac34 mv^2 \Rightarrow v=\sqrt{4gh/3}, ω=v/r\omega = v/r.

Question 3

Physics · 4 marks · −1

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g=10 m s−2g = 10\ \mathrm{m\,s^{-2}})

  1. A.2.0 m s−12.0\ \mathrm{m\,s^{-1}}
  2. B.1.9 m s−11.9\ \mathrm{m\,s^{-1}}
  3. C.1.7 m s−11.7\ \mathrm{m\,s^{-1}}
  4. D.1.5 m s−11.5\ \mathrm{m\,s^{-1}}

Correct answer: B

Explanation: P=(mg+f)v⇒44760=24000 v⇒v≈1.9P = (mg+f)v \Rightarrow 44760 = 24000\,v \Rightarrow v \approx 1.9 m/s.

Question 4

Physics · 4 marks · −1

Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The center of mass of the system is at a point:

Figure for question 4
  1. A.0.9 cm right and 2.0 cm above 1 kg mass
  2. B.2.0 cm right and 0.9 cm above 1 kg mass
  3. C.0.6 cm right and 2.0 cm above 1 kg mass
  4. D.1.5 cm right and 1.2 cm above 1 kg mass

Correct answer: A

Explanation: Origin at 1 kg: x=(1.5×3)/5=0.9x = (1.5\times3)/5 = 0.9 cm, y=(2.5×4)/5=2.0y = (2.5\times4)/5 = 2.0 cm.

Question 5

Physics · 4 marks · −1

The radius of gyration of a uniform rod of length ll, about an axis passing through a point l4\frac{l}{4} away from the centre of the rod, and perpendicular to it, is:

  1. A.14l\frac{1}{4}l
  2. B.38 l\sqrt{\frac{3}{8}}\,l
  3. C.748 l\sqrt{\frac{7}{48}}\,l
  4. D.18l\frac{1}{8}l

Correct answer: C

Explanation: k2=l212+l216=7l248k^2 = \frac{l^2}{12}+\frac{l^2}{16} = \frac{7l^2}{48}.

Question 6

Physics · 4 marks · −1

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass m10\frac{m}{10} so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth):

  1. A.5m(u2−119200GMR)5m\left(u^2 - \frac{119}{200}\frac{GM}{R}\right)
  2. B.m20(u2+113200GMR)\frac{m}{20}\left(u^2 + \frac{113}{200}\frac{GM}{R}\right)
  3. C.3m8(u+5GM6R)2\frac{3m}{8}\left(u + \sqrt{\frac{5GM}{6R}}\right)^2
  4. D.m20(u−2GM3R)2\frac{m}{20}\left(u - \sqrt{\frac{2GM}{3R}}\right)^2

Correct answer: A

Explanation: At height R: v2=u2−GM/Rv^2 = u^2 - GM/R (vertical). Remaining 9m10\frac{9m}{10} moves horizontally with GM/2R\sqrt{GM/2R}; rocket gets vertical speed 10v10v and horizontal speed 9GM/2R9\sqrt{GM/2R}. KE=12m10(100v2+81GM2R)=5m(u2−119200GMR)KE = \frac12\frac{m}{10}\left(100v^2 + \frac{81GM}{2R}\right) = 5m\left(u^2 - \frac{119}{200}\frac{GM}{R}\right).

Question 7

Physics · 4 marks · −1

Two moles of an ideal gas with CPCV=53\frac{C_P}{C_V} = \frac{5}{3} are mixed with 3 moles of another ideal gas with CPCV=43\frac{C_P}{C_V} = \frac{4}{3}. The value of CPCV\frac{C_P}{C_V} for the mixture is:

  1. A.1.50
  2. B.1.47
  3. C.1.42
  4. D.1.45

Correct answer: C

Explanation: CV=2(1.5R)+3(3R)5=2.4RC_V = \frac{2(1.5R)+3(3R)}{5} = 2.4R, CP=3.4RC_P = 3.4R, γ=1.42\gamma = 1.42.

Question 8

Physics · 4 marks · −1

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ=1.40\gamma = 1.40, the work done by air is: (31.4=4.65553^{1.4} = 4.6555) [Take air to be an ideal gas]

  1. A.48 J
  2. B.60.7 J
  3. C.90.5 J
  4. D.100.8 J

Correct answer: C

Explanation: W=P1V1−P2V2γ−1=101.3(1−3/4.6555)0.4≈90W = \frac{P_1V_1 - P_2V_2}{\gamma-1} = \frac{101.3(1 - 3/4.6555)}{0.4} \approx 90 J.

Question 9

Physics · 4 marks · −1

Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2^2) is 90 m s−1^{-1}. If the Young's modulus of wire is 16×101116\times10^{11} N m−2^{-2}, the extension of wire over its natural length is:

  1. A.0.03 mm
  2. B.0.02 mm
  3. C.0.01 mm
  4. D.0.04 mm

Correct answer: A

Explanation: T=μv2=0.01×8100=81T = \mu v^2 = 0.01\times8100 = 81 N; Δl=TLAY=81×0.610−6×16×1011≈0.03\Delta l = \frac{TL}{AY} = \frac{81\times0.6}{10^{-6}\times16\times10^{11}} \approx 0.03 mm.

Question 10

Physics · 4 marks · −1

Two infinite planes each with uniform surface charge density +σ+\sigma are kept in such a way that the angle between them is 30°. The electric field in the region shown between them is given by:

Figure for question 10
  1. A.σ2ϵ0[(1−32)y^−x^2]\frac{\sigma}{2\epsilon_0}\left[\left(1-\frac{\sqrt3}{2}\right)\hat y - \frac{\hat x}{2}\right]
  2. B.σϵ0[(1+32)y^+x^2]\frac{\sigma}{\epsilon_0}\left[\left(1+\frac{\sqrt3}{2}\right)\hat y + \frac{\hat x}{2}\right]
  3. C.σ2ϵ0[(1+3)y^+x^2]\frac{\sigma}{2\epsilon_0}\left[\left(1+\sqrt3\right)\hat y + \frac{\hat x}{2}\right]
  4. D.σ2ϵ0[(1+3)y^−x^2]\frac{\sigma}{2\epsilon_0}\left[\left(1+\sqrt3\right)\hat y - \frac{\hat x}{2}\right]

Correct answer: A

Question 11

Physics · 4 marks · −1

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x)=K(1+αx)k(x) = K(1+\alpha x) where 'x' is the distance measured from one of the plates. If (αd)≪1(\alpha d) \ll 1, the total capacitance of the system is best given by the expression:

Figure for question 11
  1. A.Aϵ0Kd(1+α2d22)\frac{A\epsilon_0 K}{d}\left(1+\frac{\alpha^2d^2}{2}\right)
  2. B.AKϵ0d(1+αd)\frac{AK\epsilon_0}{d}(1+\alpha d)
  3. C.Aϵ0Kd(1+(αd2)2)\frac{A\epsilon_0 K}{d}\left(1+\left(\frac{\alpha d}{2}\right)^2\right)
  4. D.AKϵ0d(1+αd2)\frac{AK\epsilon_0}{d}\left(1+\frac{\alpha d}{2}\right)

Correct answer: D

Explanation: 1C=∫0ddxKϵ0A(1+αx)=ln⁡(1+αd)Kϵ0Aα≈d(1−αd/2)Kϵ0A\frac1C = \int_0^d \frac{dx}{K\epsilon_0A(1+\alpha x)} = \frac{\ln(1+\alpha d)}{K\epsilon_0 A\alpha} \approx \frac{d(1-\alpha d/2)}{K\epsilon_0A}.

Question 12

Physics · 4 marks · −1

The current I1I_1 (in A) flowing through 1 Ω resistor in the following circuit is:

Figure for question 12
  1. A.0.4
  2. B.0.5
  3. C.0.2
  4. D.0.25

Correct answer: C

Explanation: Branch (1 Ω ∥ 1 Ω) + 2 Ω = 2.5 Ω across 1 V carries 0.4 A, split equally: I1=0.2I_1 = 0.2 A.

Question 13

Physics · 4 marks · −1

Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by ϕi\phi_i. The magnetic flux through the area of the circular coil area is given by ϕ0\phi_0. Which of the following option is correct?

  1. A.ϕi=ϕ0\phi_i = \phi_0
  2. B.ϕi<ϕ0\phi_i < \phi_0
  3. C.ϕi>ϕ0\phi_i > \phi_0
  4. D.ϕi=−ϕ0\phi_i = -\phi_0

Correct answer: D

Explanation: Field lines are closed, so the total flux through the infinite plane is zero.

Question 14

Physics · 4 marks · −1

A long solenoid of radius R carries a time (t)-dependent current I(t)=I0t(1−t)I(t) = I_0t(1-t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0≤t≤10 \le t \le 1, the induced current (IRI_R) and the induced EMF (VRV_R) in the ring change as:

  1. A.Direction of IRI_R remains unchanged and VRV_R is maximum at t=0.5t = 0.5
  2. B.At t=0.5t = 0.5 direction of IRI_R reverses and VRV_R is zero
  3. C.Direction of IRI_R remains unchanged and VRV_R is zero at t=0.25t = 0.25
  4. D.At t=0.25t = 0.25 direction of IRI_R reverses and VRV_R is maximum

Correct answer: B

Explanation: VR∝dI/dt=I0(1−2t)V_R \propto dI/dt = I_0(1-2t): zero and changes sign at t=0.5t = 0.5.

Question 15

Physics · 4 marks · −1

If the magnetic field in a plane electromagnetic wave is given by B⃗=3×10−8sin⁡(1.6×103x+48×1010t) j^\vec B = 3\times10^{-8}\sin(1.6\times10^3x + 48\times10^{10}t)\,\hat j T, then what will be expression for electric field?

  1. A.E⃗=(3×10−8sin⁡(1.6×103x+48×1010t) j^ V/m)\vec E = (3\times10^{-8}\sin(1.6\times10^3x + 48\times10^{10}t)\,\hat j\ \mathrm{V/m})
  2. B.E⃗=(9sin⁡(1.6×103x+48×1010t) k^ V/m)\vec E = (9\sin(1.6\times10^3x + 48\times10^{10}t)\,\hat k\ \mathrm{V/m})
  3. C.E⃗=(3×10−8sin⁡(1.6×103x+48×1010t) i^ V/m)\vec E = (3\times10^{-8}\sin(1.6\times10^3x + 48\times10^{10}t)\,\hat i\ \mathrm{V/m})
  4. D.E⃗=(60sin⁡(1.6×103x+48×1010t) k^ V/m)\vec E = (60\sin(1.6\times10^3x + 48\times10^{10}t)\,\hat k\ \mathrm{V/m})

Correct answer: B

Explanation: E0=cB0=9E_0 = cB_0 = 9 V/m; wave travels along −x-x, so E^×j^=−i^⇒E^=k^\hat E\times\hat j = -\hat i \Rightarrow \hat E = \hat k.

Question 16

Physics · 4 marks · −1

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to:

  1. A.12 mm
  2. B.22 mm
  3. C.33 mm
  4. D.2 mm

Correct answer: B

Explanation: m=Lfo(1+Dfe)⇒12.5=1+250/fe⇒fe≈22m = \frac{L}{f_o}\left(1+\frac{D}{f_e}\right) \Rightarrow 12.5 = 1 + 250/f_e \Rightarrow f_e \approx 22 mm.

Question 17

Physics · 4 marks · −1

A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is:

  1. A.90°
  2. B.18.4°
  3. C.71.6°
  4. D.45°

Correct answer: B

Explanation: cos⁡2θ=0.1⇒θ=71.6°\cos^2\theta = 0.1 \Rightarrow \theta = 71.6°; further rotation 90°−71.6°=18.4°90° - 71.6° = 18.4°.

Question 18

Physics · 4 marks · −1

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6×10−161.6\times10^{-16} s. The frequency of revolution of the electron in its first excited state (in s−1^{-1}) is:

  1. A.6.2×10156.2\times10^{15}
  2. B.1.6×10141.6\times10^{14}
  3. C.7.8×10147.8\times10^{14}
  4. D.5.6×10125.6\times10^{12}

Correct answer: C

Explanation: T∝n3T \propto n^3: f2=18×1.6×10−16=7.8×1014f_2 = \frac{1}{8\times1.6\times10^{-16}} = 7.8\times10^{14} s−1^{-1}.

Question 19

Physics · 4 marks · −1

Which of the following gives a reversible operation?

Figure for question 19
  1. A.Option 1 (drawn): a single-input gate with an output bubble — a NOT gate
  2. B.Option 2 (drawn): a two-input AND-shaped gate with an output bubble — a NAND gate
  3. C.Option 3 (drawn): a two-input OR gate
  4. D.Option 4 (drawn): a two-input OR-shaped gate with an output bubble — a NOR gate

Correct answer: A

Explanation: Only the NOT gate maps inputs to outputs one-to-one, so its input can be recovered from its output.

Question 20

Physics · 4 marks · −1

Visible light of wavelength 6000×10−86000\times10^{-8} cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at θ1\theta_1, then θ1\theta_1 is close to:

  1. A.45°
  2. B.30°
  3. C.25°
  4. D.20°

Correct answer: C

Explanation: asin⁡60°=2λa\sin60° = 2\lambda, sin⁡θ1=λ/a=0.433⇒θ1≈25.7°\sin\theta_1 = \lambda/a = 0.433 \Rightarrow \theta_1 \approx 25.7°.

Question 21

Physics · 4 marks

A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaching its highest point P (height 1 m), the kinetic energy of the particle (in J) is: (Figure drawn is schematic and not to scale; take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}) ________.

Figure for question 21

Correct answer: 10

Explanation: KEP=mg(hA−hP)=1×10×(2−1)=10KE_P = mg(h_A - h_P) = 1\times10\times(2-1) = 10 J.

Question 22

Physics · 4 marks

A non-isotropic solid metal cube has coefficients of linear expansion as: 5×10−55\times10^{-5}/°C along the x-axis and 5×10−65\times10^{-6}/°C along the y and the z-axis. If the coefficient of volume expansion of the solid is C×10−6C\times10^{-6}/°C then the value of C is ________.

Correct answer: 60

Explanation: γ=αx+αy+αz=50+5+5=60×10−6\gamma = \alpha_x+\alpha_y+\alpha_z = 50+5+5 = 60\times10^{-6}/°C.

Question 23

Physics · 4 marks

A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle, is ________.

Correct answer: 600

Explanation: η=1−300/900=2/3\eta = 1 - 300/900 = 2/3, Q1=1800Q_1 = 1800 J, Q2=Q1−W=600Q_2 = Q_1 - W = 600 J.

Question 24

Physics · 4 marks

A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is B⃗=(3i^+4k^)\vec B = (3\hat i + 4\hat k) T. The quantity of flux through the loop ABCDEFA (in Wb) is ________.

Correct answer: 175

Explanation: Loop = ABCDA (A⃗=25k^\vec A = 25\hat k) + ADEFA (A⃗=25i^\vec A = 25\hat i). ϕ=3(25)+4(25)=175\phi = 3(25)+4(25) = 175 Wb.

Question 25

Physics · 4 marks

A beam of electromagnetic radiation of intensity 6.4×10−56.4\times10^{-5} W/cm2^2 is comprised of wavelength, λ=310\lambda = 310 nm. It falls normally on a metal (work function φ=2\varphi = 2 eV) of surface area of 1 cm2^2. If one in 10310^3 photons ejects an electron, total number of electrons ejected in 1 s is 10x10^x. (hc = 1240 eV nm, 1 eV = 1.6×10−191.6\times10^{-19} J), then x is ________.

Correct answer: 11

Explanation: Eph=1240/310=4E_{ph} = 1240/310 = 4 eV =6.4×10−19= 6.4\times10^{-19} J; photons/s =6.4×10−5/6.4×10−19=1014= 6.4\times10^{-5}/6.4\times10^{-19} = 10^{14}; electrons =1011= 10^{11}, x = 11.

Question 26

Chemistry · 4 marks · −1

At 35°C, the vapour pressure of CS2_2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2_2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is:

  1. A.heat must be absorbed in order to produce the solution at 35°C
  2. B.a mixture of 100 mL CS2_2 and 100 mL acetone has a volume < 200 mL
  3. C.CS2_2 and acetone are less attracted to each other than to themselves
  4. D.Raoult's law is not obeyed by this system

Correct answer: B

Explanation: Total VP exceeds both pure values: positive deviation, so ΔV_mix > 0; statement 2 is false.

Question 27

Chemistry · 4 marks · −1

Given that the standard potentials (E°) of Cu2+^{2+}/Cu and Cu+^+/Cu are 0.34 V and 0.522 V respectively, the E° of Cu2+^{2+}/Cu+^+ is:

  1. A.0.182 V
  2. B.−0.182 V
  3. C.+0.158 V
  4. D.−0.158 V

Correct answer: C

Explanation: E°=2(0.34)−1(0.522)=0.158E° = 2(0.34) - 1(0.522) = 0.158 V.

Question 28

Chemistry · 4 marks · −1

The dipole moments of CCl4_4, CHCl3_3 and CH4_4 are in the order:

  1. A.CH4_4 < CCl4_4 < CHCl3_3
  2. B.CCl4_4 < CH4_4 < CHCl3_3
  3. C.CH4_4 = CCl4_4 < CHCl3_3
  4. D.CHCl3_3 < CH4_4 = CCl4_4

Correct answer: C

Question 29

Chemistry · 4 marks · −1

The number of orbitals associated with quantum numbers n=5n = 5, ms=+12m_s = +\frac{1}{2} is:

  1. A.50
  2. B.15
  3. C.25
  4. D.11

Correct answer: C

Explanation: n = 5 has n2=25n^2 = 25 orbitals; each holds one electron with ms=+1/2m_s = +1/2.

Question 30

Chemistry · 4 marks · −1

The relative strength of interionic/intermolecular forces in decreasing order is:

  1. A.ion-dipole > ion-ion > dipole-dipole
  2. B.ion-ion > ion-dipole > dipole-dipole
  3. C.ion-dipole > dipole-dipole > ion-ion
  4. D.dipole-dipole > ion-dipole > ion-ion

Correct answer: B

Question 31

Chemistry · 4 marks · −1

Amongst the following statements, that which was not proposed by Dalton was:

  1. A.matter consists of indivisible atoms.
  2. B.all the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
  3. C.when gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.
  4. D.chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.

Correct answer: C

Explanation: The law of combining volumes is Gay-Lussac's, not Dalton's.

Question 32

Chemistry · 4 marks · −1

The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are:

  1. A.−333, −349, −325 and −296
  2. B.−349, −333, −325 and −296
  3. C.−333, −325, −349 and −296
  4. D.−296, −325, −333 and −349

Correct answer: A

Question 33

Chemistry · 4 marks · −1

The purest form of commercial iron is:

  1. A.wrought iron
  2. B.pig iron
  3. C.cast iron
  4. D.scrap iron and pig iron

Correct answer: A

Question 34

Chemistry · 4 marks · −1

In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is:

  1. A.less efficient as it exchanges only anions
  2. B.more efficient as it can exchange both cations as well as anions
  3. C.more efficient as it can exchange only cations
  4. D.less efficient as the resins cannot be regenerated

Correct answer: B

Question 35

Chemistry · 4 marks · −1

Oxidation number of potassium in K2_2O, K2_2O2_2 and KO2_2, respectively, is:

  1. A.+1, +2 and +4
  2. B.+1, +4 and +2
  3. C.+2, +1 and +12+\frac{1}{2}
  4. D.+1, +1 and +1

Correct answer: D

Question 36

Chemistry · 4 marks · −1

The atomic radius of Ag is closest to:

  1. A.Ni
  2. B.Cu
  3. C.Hg
  4. D.Au

Correct answer: D

Question 37

Chemistry · 4 marks · −1

The theory that can completely/properly explain the nature of bonding in [Ni(CO)4_4] is:

  1. A.Werner's theory
  2. B.Valence bond theory
  3. C.Crystal field theory
  4. D.Molecular orbital theory

Correct answer: D

Question 38

Chemistry · 4 marks · −1

The IUPAC name of the complex [Pt(NH3_3)2_2Cl(NH2_2CH3_3)]Cl is:

  1. A.Bisammine(methanamine)chlorido platinum(II)chloride
  2. B.Diammine(methanamine)chlorido platinum(II)chloride
  3. C.Diamminechlorido(methanamine) platinum(II)chloride
  4. D.Diamminechlorido(aminomethane) platinum(II)chloride

Correct answer: C

Question 39

Chemistry · 4 marks · −1

Consider the following reaction: N,N-dimethylaniline + Na+ −O3S−C6H4−N2+Cl−\mathrm{Na^{+}\,{}^{-}O_3S{-}C_6H_4{-}N_2^{+}Cl^{-}} (p-sulphonatobenzenediazonium chloride) →OH−\xrightarrow{\mathrm{OH^-}} 'X'. The product 'X' is used:

Figure for question 39
  1. A.as food grade colourant
  2. B.in protein estimation as an alternative to ninhydrin
  3. C.in acid base titration as an indicator
  4. D.in laboratory test for phenols

Correct answer: C

Explanation: Coupling gives methyl orange, an acid–base indicator.

Question 40

Chemistry · 4 marks · −1

The increasing order of pKb_b for the following compounds will be: (A) NH2_2–CH=NH, (B) the bicyclic amidine shown in the figure, (C) CH3_3NHCH3_3

Figure for question 40
  1. A.(A) < (B) < (C)
  2. B.(B) < (C) < (A)
  3. C.(C) < (A) < (B)
  4. D.(B) < (A) < (C)

Correct answer: D

Explanation: Amidines are stronger bases than secondary amines (resonance-stabilised conjugate acid); the cyclic, alkyl-substituted amidine (B) is the strongest base: pKb B < A < C.

Question 41

Chemistry · 4 marks · −1

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3_3 to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C, contain respectively:

  1. A.m-chloroaniline, m-chlorobenzoic acid and m-chlorophenol
  2. B.m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol
  3. C.m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline
  4. D.m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline

Correct answer: C

Question 42

Chemistry · 4 marks · −1

Match the following: (i) Riboflavin, (ii) Thiamine, (iii) Pyridoxine, (iv) Ascorbic acid with (a) Beriberi, (b) Scurvy, (c) Cheilosis, (d) Convulsions

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

Correct answer: A

Question 43

Chemistry · 4 marks · −1

What is the product of following reaction? Hex-3-ynal →(i) NaBH4 (ii) PBr3 (iii) Mg/ether (iv) CO2/H3O+\xrightarrow{\text{(i) NaBH}_4\ \text{(ii) PBr}_3\ \text{(iii) Mg/ether}\ \text{(iv) CO}_2/\text{H}_3\text{O}^+} ?

Figure for question 43
  1. A.Hept-4-enoic acid drawn as a zig-zag chain with a C=C double bond, ending in COOH
  2. B.Heptanoic acid (saturated zig-zag chain ending in COOH)
  3. C.Oct-4-enoic-type chain: longer zig-zag chain with a C=C double bond ending in COOH
  4. D.Hept-4-ynoic acid: CH3CH2–C≡C–CH2CH2–COOH (zig-zag with a triple bond, ending in COOH)

Correct answer: D

Explanation: CHO → CH2OH → CH2Br → CH2MgBr → CH2COOH; the triple bond survives: CH3CH2C≡CCH2CH2COOH.

Question 44

Chemistry · 4 marks · −1

1-methyl ethylene oxide when treated with an excess of HBr produces:

Figure for question 44
  1. A.Br–CH=CH–CH3 (1-bromopropene, drawn as skeletal structure)
  2. B.CH2=C(Br)–CH3 (2-bromopropene)
  3. C.BrCH2–CH(Br)–CH3 (1,2-dibromopropane)
  4. D.(CH3)2CBr2 (2,2-dibromopropane)

Correct answer: C

Explanation: Excess HBr opens the epoxide and then substitutes the OH: 1,2-dibromopropane.

Question 45

Chemistry · 4 marks · −1

Consider the following reactions: (a) (CH3_3)3_3CCH(OH)CH3_3 →conc. H2SO4\xrightarrow{\text{conc. H}_2\text{SO}_4} (b) (CH3_3)2_2CHCH(Br)CH3_3 →alc. KOH\xrightarrow{\text{alc. KOH}} (c) (CH3_3)2_2CHCH(Br)CH3_3 →(CH3)3O−K+\xrightarrow{(\text{CH}_3)_3\text{O}^-\text{K}^+} (d) (CH3_3)2_2C(OH)–CH2_2–CHO →Δ\xrightarrow{\Delta}. Which of these reaction(s) will not produce Saytzeff product?

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

Correct answer: D

Explanation: Bulky tert-butoxide gives the Hofmann (less substituted) alkene; the others give the more substituted/conjugated product.

Question 46

Chemistry · 4 marks

During the nuclear explosion, one of the products is 90^{90}Sr with half life of 6.93 years. If 1 μg of 90^{90}Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically ________.

Correct answer: 23.03

Explanation: t=t1/2ln⁡2ln⁡N0N=6.930.693×2.303=23.03t = \frac{t_{1/2}}{\ln2}\ln\frac{N_0}{N} = \frac{6.93}{0.693}\times2.303 = 23.03 years.

Question 47

Chemistry · 4 marks

Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2_2SO4_4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is ________.

Correct answer: 10.6

Explanation: OH−^-: 0.04 mol; H+^+: 2×0.01 = 0.02 mol; excess OH−^- = 0.02 mol in 50 L = 4×10−44\times10^{-4} M; pOH = 3.4, pH = 10.6.

Question 48

Chemistry · 4 marks

For the reaction; A(l) → 2B(g), ΔU = 2.1 kcal, ΔS = 20 cal K−1^{-1} at 300 K. Hence ΔG in kcal is ________.

Correct answer: -2.7

Explanation: ΔH=ΔU+ΔngRT=2.1+2(0.002)(300)=3.3\Delta H = \Delta U + \Delta n_gRT = 2.1 + 2(0.002)(300) = 3.3 kcal; ΔG=3.3−300(0.020)=−2.7\Delta G = 3.3 - 300(0.020) = -2.7 kcal.

Question 49

Chemistry · 4 marks

Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is ________.

Correct answer: 1.67

Explanation: X = NaCl, Y = NaClO3_3; ClO3−_3^- has 2 Cl=O + 1 Cl–O−^- resonating over 3 bonds: bond order = 5/3 ≈ 1.67.

Question 50

Chemistry · 4 marks

The number of chiral carbons in chloramphenicol is ________.

Correct answer: 2

Explanation: Chloramphenicol, O2N–C6H4–CH(OH)–CH(NHCOCHCl2)–CH2OH, has two stereocentres (the CH(OH) and CH(NH) carbons).

Question 51

Mathematics · 4 marks · −1

If g(x)=x2+x−1g(x) = x^2 + x - 1 and (g∘f)(x)=4x2−10x+5(g\circ f)(x) = 4x^2 - 10x + 5, then f(54)f\left(\frac{5}{4}\right) is equal to:

  1. A.−12-\frac{1}{2}
  2. B.12\frac{1}{2}
  3. C.−32-\frac{3}{2}
  4. D.32\frac{3}{2}

Correct answer: A

Explanation: At x=5/4x = 5/4: f2+f−1=−54f^2+f-1 = -\frac54 ⇒ (f+12)2=0(f+\frac12)^2 = 0 ⇒ f=−12f = -\frac12.

Question 52

Mathematics · 4 marks · −1

If Re(z−12z+i)=1\mathrm{Re}\left(\frac{z-1}{2z+i}\right) = 1, where z=x+iyz = x + iy, then the point (x,y)(x, y) lies on a:

  1. A.straight line whose slope is −23-\frac{2}{3}.
  2. B.straight line whose slope is 32\frac{3}{2}.
  3. C.circle whose centre is at (−12,−32)\left(-\frac{1}{2}, -\frac{3}{2}\right).
  4. D.circle whose diameter is 52\frac{\sqrt5}{2}.

Correct answer: D

Explanation: Simplifies to x2+y2+x+32y+12=0x^2+y^2+x+\frac32y+\frac12 = 0: radius 54\frac{\sqrt5}{4}, diameter 52\frac{\sqrt5}{2}.

Question 53

Mathematics · 4 marks · −1

Let α\alpha be a root of the equation x2+x+1=0x^2 + x + 1 = 0 and the matrix A=13[1111αα21α2α4]A = \frac{1}{\sqrt3}\begin{bmatrix}1 & 1 & 1\\ 1 & \alpha & \alpha^2\\ 1 & \alpha^2 & \alpha^4\end{bmatrix}, then the matrix A31A^{31} is equal to:

  1. A.AA
  2. B.A2A^2
  3. C.A3A^3
  4. D.I3I_3

Correct answer: C

Explanation: A4=IA^4 = I, so A31=A28A3=A3A^{31} = A^{28}A^3 = A^3.

Question 54

Mathematics · 4 marks · −1

If the system of linear equations 2x+2ay+az=02x + 2ay + az = 0, 2x+3by+bz=02x + 3by + bz = 0, 2x+4cy+cz=02x + 4cy + cz = 0, where a,b,c∈Ra, b, c \in \mathbb{R} are non-zero and distinct; has a non-zero solution, then:

  1. A.a+b+c=0a + b + c = 0
  2. B.1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A.P.
  3. C.a, b, c are in G.P.
  4. D.a, b, c are in A.P.

Correct answer: B

Explanation: Determinant =0= 0 gives 2ac=ab+bc2ac = ab + bc, i.e. 2b=1a+1c\frac2b = \frac1a+\frac1c.

Question 55

Mathematics · 4 marks · −1

Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is:

  1. A.6!6!
  2. B.12(6!)\frac{1}{2}(6!)
  3. C.52(6!)\frac{5}{2}(6!)
  4. D.565^6

Correct answer: C

Explanation: Choose the repeated digit (5 ways), arrange: 5×6!2!=52(6!)5\times\frac{6!}{2!} = \frac52(6!).

Question 56

Mathematics · 4 marks · −1

Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is −12-\frac{1}{2}, then the greatest number amongst them is:

  1. A.7
  2. B.16
  3. C.212\frac{21}{2}
  4. D.27

Correct answer: B

Explanation: Middle term 5; (25−d2)(25−4d2)⋅5=2520(25-d^2)(25-4d^2)\cdot5 = 2520 ⇒ d=±112d = \pm\frac{11}{2}; terms −12,…,16-\frac12,\ldots,16.

Question 57

Mathematics · 4 marks · −1

The greatest positive integer k, for which 49k+149^k + 1 is a factor of the sum 49125+49124+…+492+49+149^{125} + 49^{124} + \ldots + 49^2 + 49 + 1, is:

  1. A.32
  2. B.60
  3. C.63
  4. D.65

Correct answer: C

Explanation: Sum =49126−148=(4963+1)(4963−1)48= \frac{49^{126}-1}{48} = \frac{(49^{63}+1)(49^{63}-1)}{48}, so k = 63.

Question 58

Mathematics · 4 marks · −1

Let xk+yk=akx^k + y^k = a^k, (a,k>0)(a, k > 0) and dydx+(yx)13=0\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac13} = 0, then k is:

  1. A.23\frac{2}{3}
  2. B.13\frac{1}{3}
  3. C.43\frac{4}{3}
  4. D.32\frac{3}{2}

Correct answer: A

Explanation: dydx=−(xy)k−1=−(yx)1−k\frac{dy}{dx} = -\left(\frac xy\right)^{k-1} = -\left(\frac yx\right)^{1-k} ⇒ 1−k=131-k = \frac13.

Question 59

Mathematics · 4 marks · −1

Let the function, f:[−7,0]→Rf: [-7, 0] \to \mathbb{R} be continuous on [−7,0][-7, 0] and differentiable on (−7,0)(-7, 0). If f(−7)=−3f(-7) = -3 and f′(x)≤2f'(x) \le 2, for all x∈(−7,0)x \in (-7, 0), then for all such functions ff, f(−1)+f(0)f(-1) + f(0) lies in the interval:

  1. A.(−∞,11](-\infty, 11]
  2. B.[−3,11][-3, 11]
  3. C.[−6,20][-6, 20]
  4. D.(−∞,20](-\infty, 20]

Correct answer: D

Explanation: MVT: f(−1)≤9f(-1) \le 9, f(0)≤11f(0) \le 11; no lower bound.

Question 60

Mathematics · 4 marks · −1

If f(a+b+1−x)=f(x)f(a + b + 1 - x) = f(x), for all x, where a and b are fixed positive real numbers, then 1a+b∫abx (f(x)+f(x+1)) dx\frac{1}{a+b}\int_a^b x\,(f(x) + f(x+1))\,dx is equal to:

  1. A.∫a−1b−1f(x) dx\int_{a-1}^{b-1} f(x)\,dx
  2. B.∫a−1b−1f(x+1) dx\int_{a-1}^{b-1} f(x+1)\,dx
  3. C.∫a+1b+1f(x) dx\int_{a+1}^{b+1} f(x)\,dx
  4. D.∫a+1b+1f(x+1) dx\int_{a+1}^{b+1} f(x+1)\,dx

Correct answer: C

Question 61

Mathematics · 4 marks · −1

The area of the region, enclosed by the circle x2+y2=2x^2 + y^2 = 2 which is not common to the region bounded by the parabola y2=xy^2 = x and the straight line y=xy = x, is:

  1. A.16(24π−1)\frac{1}{6}(24\pi - 1)
  2. B.16(12π−1)\frac{1}{6}(12\pi - 1)
  3. C.13(6π−1)\frac{1}{3}(6\pi - 1)
  4. D.13(12π−1)\frac{1}{3}(12\pi - 1)

Correct answer: B

Explanation: Region between y2=xy^2 = x and y=xy = x has area 16\frac16 and lies inside the circle: 2π−16=16(12π−1)2\pi - \frac16 = \frac16(12\pi-1).

Question 62

Mathematics · 4 marks · −1

If y=y(x)y = y(x) is the solution of the differential equation, ey(dydx−1)=exe^y\left(\frac{dy}{dx} - 1\right) = e^x such that y(0)=0y(0) = 0, then y(1)y(1) is equal to:

  1. A.1+log⁡e21 + \log_e 2
  2. B.log⁡e2\log_e 2
  3. C.2+log⁡e22 + \log_e 2
  4. D.2e2e

Correct answer: A

Explanation: Let v=y−xv = y - x: evv′=1e^{v}v' = 1 ⇒ ey−x=x+1e^{y-x} = x + 1 (using y(0)=0y(0)=0) ⇒ y(1)=1+ln⁡2y(1) = 1 + \ln2.

Question 63

Mathematics · 4 marks · −1

If y=mx+4y = mx + 4 is a tangent to both the parabolas, y2=4xy^2 = 4x and x2=2byx^2 = 2by, then b is equal to:

  1. A.−128
  2. B.128
  3. C.−64
  4. D.−32

Correct answer: A

Explanation: Tangent to y2=4xy^2 = 4x: 4=1/m4 = 1/m ⇒ m=14m = \frac14. For x2=2byx^2 = 2by: x2−2bmx−8b=0x^2 - 2bmx - 8b = 0 has D = 0 ⇒ b2m2+8b=0b^2m^2 + 8b = 0 ⇒ b=−128b = -128.

Question 64

Mathematics · 4 marks · −1

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:

  1. A.32\frac{3}{\sqrt2}
  2. B.232\sqrt3
  3. C.323\sqrt2
  4. D.3\sqrt3

Correct answer: C

Explanation: ae=3ae = 3, a/e=6a/e = 6 ⇒ a2=18a^2 = 18, b2=9b^2 = 9; LR =2b2/a=32= 2b^2/a = 3\sqrt2.

Question 65

Mathematics · 4 marks · −1

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is:

  1. A.(6, 5, −2)
  2. B.(4, 3, 2)
  3. C.(6, 5, 2)
  4. D.(3, 4, −2)

Correct answer: A

Explanation: Plane: x+y−2z=3x + y - 2z = 3; image of (2,1,6) is (6,5,−2).

Question 66

Mathematics · 4 marks · −1

A vector a⃗=αi^+2j^+βk^\vec a = \alpha\hat i + 2\hat j + \beta\hat k (α,β∈R)(\alpha, \beta \in \mathbb{R}) lies in the plane of the vectors, b⃗=i^+j^\vec b = \hat i + \hat j and c⃗=i^−j^+4k^\vec c = \hat i - \hat j + 4\hat k. If a⃗\vec a bisects the angle between b⃗\vec b and c⃗\vec c, then:

  1. A.a⃗⋅i^+1=0\vec a\cdot\hat i + 1 = 0
  2. B.a⃗⋅k^+2=0\vec a\cdot\hat k + 2 = 0
  3. C.a⃗⋅i^+3=0\vec a\cdot\hat i + 3 = 0
  4. D.a⃗⋅k^+4=0\vec a\cdot\hat k + 4 = 0

Correct answer: B

Explanation: b^±c^∝(4i^+2j^+4k^)\hat b \pm \hat c \propto (4\hat i + 2\hat j + 4\hat k) or (i^+2j^−2k^)(\hat i + 2\hat j - 2\hat k); with j-coefficient 2 the latter gives a⃗⋅k^=−2\vec a\cdot\hat k = -2 (the internal bisector gives β=4\beta = 4, not offered).

Question 67

Mathematics · 4 marks · −1

Let α\alpha and β\beta be two real roots of the equation (k+1)tan⁡2x−2⋅λtan⁡x=(1−k)(k+1)\tan^2x - \sqrt2\cdot\lambda\tan x = (1-k), where k(≠−1)k (\ne -1) and λ\lambda are real numbers. If tan⁡2(α+β)=50\tan^2(\alpha + \beta) = 50, then a value of λ\lambda is:

  1. A.5
  2. B.10
  3. C.10210\sqrt2
  4. D.525\sqrt2

Correct answer: B

Explanation: tan⁡(α+β)=2λ/(k+1)1+(1−k)/(k+1)=λ2\tan(\alpha+\beta) = \frac{\sqrt2\lambda/(k+1)}{1 + (1-k)/(k+1)} = \frac{\lambda}{\sqrt2} ⇒ λ2/2=50\lambda^2/2 = 50 ⇒ λ=10\lambda = 10.

Question 68

Mathematics · 4 marks · −1

An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value −1. Then the expected value of X, is:

  1. A.316\frac{3}{16}
  2. B.−316-\frac{3}{16}
  3. C.−18-\frac{1}{8}
  4. D.18\frac{1}{8}

Correct answer: D

Explanation: P(X=3) = 5/32, P(X=4) = 2/32, P(X=5) = 1/32, P(X=−1) = 24/32: E = (15+8+5−24)/32 = 1/8.

Question 69

Mathematics · 4 marks · −1

If y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2αy(\alpha) = \sqrt{2\left(\frac{\tan\alpha + \cot\alpha}{1 + \tan^2\alpha}\right) + \frac{1}{\sin^2\alpha}}, α∈(3π4,π)\alpha \in \left(\frac{3\pi}{4}, \pi\right), then dydα\frac{dy}{d\alpha} at α=5π6\alpha = \frac{5\pi}{6} is:

  1. A.43\frac{4}{3}
  2. B.−14-\frac{1}{4}
  3. C.−4
  4. D.4

Correct answer: D

Explanation: y=∣1+cot⁡α∣=−(1+cot⁡α)y = |1 + \cot\alpha| = -(1+\cot\alpha) on the interval; y′=cosec2α=4y' = \mathrm{cosec}^2\alpha = 4.

Question 70

Mathematics · 4 marks · −1

The logical statement (p⇒q)∧(q⇒∼p)(p \Rightarrow q) \wedge (q \Rightarrow \sim p) is equivalent to:

  1. A.p
  2. B.q
  3. C.~p
  4. D.~q

Correct answer: C

Question 71

Mathematics · 4 marks

If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to ________.

Correct answer: 18

Explanation: n2−112=10⇒n=11\frac{n^2-1}{12} = 10 ⇒ n = 11; 4(m2−1)12=16⇒m=7\frac{4(m^2-1)}{12} = 16 ⇒ m = 7; m + n = 18.

Question 72

Mathematics · 4 marks

If the sum of the coefficients of all even powers of x in the product (1+x+x2+…+x2n)(1−x+x2−x3+…+x2n)(1 + x + x^2 + \ldots + x^{2n})(1 - x + x^2 - x^3 + \ldots + x^{2n}) is 61, then n is equal to ________.

Correct answer: 30

Explanation: Sum =P(1)+P(−1)2=(2n+1)+(2n+1)2=2n+1=61⇒n=30= \frac{P(1)+P(-1)}{2} = \frac{(2n+1)+(2n+1)}{2} = 2n+1 = 61 ⇒ n = 30.

Question 73

Mathematics · 4 marks

lim⁡x→23x+33−x−123−x/2−31−x\lim_{x\to2}\frac{3^x + 3^{3-x} - 12}{3^{-x/2} - 3^{1-x}} is equal to ________.

Correct answer: 36

Explanation: With u=3xu = 3^x: numerator (u−3)(u−9)u\frac{(u-3)(u-9)}{u}, denominator u−3u\frac{\sqrt u - 3}{u}; ratio (u−3)(u+3)→6×6=36(u-3)(\sqrt u+3) \to 6\times6 = 36.

Question 74

Mathematics · 4 marks

Let S be the set of points where the function, f(x)=∣2−∣x−3∣∣f(x) = |2 - |x - 3||, x∈Rx \in \mathbb{R}, is not differentiable. Then ∑x∈Sf(f(x))\sum_{x\in S} f(f(x)) is equal to ________.

Correct answer: 3

Explanation: S = {1, 3, 5}; f(1) = f(5) = 0, f(3) = 2; f(0) = 1, f(2) = 1; sum = 1 + 1 + 1 = 3.

Question 75

Mathematics · 4 marks

Let A(1, 0), B(6, 2) and C(32,6)C\left(\frac{3}{2}, 6\right) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point (−76,−13)\left(-\frac{7}{6}, -\frac{1}{3}\right), is ________.

Correct answer: 5

Explanation: P is the centroid (176,83)\left(\frac{17}{6}, \frac83\right); PQ = 42+32=5\sqrt{4^2 + 3^2} = 5.

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