JEE Main 2020 — 7 Jan Shift 2 (Previous Year Paper)

JEE Main · 2020 · with answers and explanations

questions
72
Total marks
288
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

The dimension of B22μ0\frac{B^2}{2\mu_0}, where B is magnetic field and μ0\mu_0 is the magnetic permeability of vacuum, is:

  1. A.ML2T−2ML^2T^{-2}
  2. B.MLT−2MLT^{-2}
  3. C.ML2T−1ML^2T^{-1}
  4. D.ML−1T−2ML^{-1}T^{-2}

Correct answer: D

Explanation: It is energy per unit volume: ML−1T−2ML^{-1}T^{-2}.

Question 2

Physics · 4 marks · −1

A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}} and the rope to be massless)

  1. A.90 N
  2. B.70 N
  3. C.100 N
  4. D.75 N

Correct answer: C

Explanation: At the midpoint: Tcos⁡45°=mgT\cos45° = mg, Tsin⁡45°=FT\sin45° = F ⇒ F=mg=100F = mg = 100 N.

Question 3

Physics · 4 marks · −1

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg. The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g=10 m/s2g = 10\ \mathrm{m/s^2}) must be at least:

  1. A.56300 W
  2. B.66000 W
  3. C.48000 W
  4. D.62360 W

Correct answer: B

Explanation: F=1600×10+6000=22000F = 1600\times10 + 6000 = 22000 N; P=Fv=66000P = Fv = 66000 W.

Question 4

Physics · 4 marks · −1

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ(r)=A+Br\sigma(r) = A + Br. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

  1. A.2πa4(A4+aB5)2\pi a^4\left(\frac{A}{4} + \frac{aB}{5}\right)
  2. B.πa4(A4+aB5)\pi a^4\left(\frac{A}{4} + \frac{aB}{5}\right)
  3. C.2πa4(A4+B5)2\pi a^4\left(\frac{A}{4} + \frac{B}{5}\right)
  4. D.2πa4(aA4+B5)2\pi a^4\left(\frac{aA}{4} + \frac{B}{5}\right)

Correct answer: A

Explanation: I=∫0a(A+Br)r2 2πr dr=2π(Aa44+Ba55)I = \int_0^a (A+Br)r^2\,2\pi r\,dr = 2\pi\left(\frac{Aa^4}{4} + \frac{Ba^5}{5}\right).

Question 5

Physics · 4 marks · −1

A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}} at the north pole and the radius of the earth = 6400 km):

  1. A.194.32 N
  2. B.195.66 N
  3. C.195.32 N
  4. D.194.66 N

Correct answer: C

Explanation: m=19.6m = 19.6 kg; ω2R=(7.27×10−5)2×6.4×106≈0.034\omega^2R = (7.27\times10^{-5})^2\times6.4\times10^6 \approx 0.034; W=19.6(10−0.034)≈195.3W = 19.6(10 - 0.034) \approx 195.3 N.

Question 6

Physics · 4 marks · −1

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:

  1. A.32\frac{\sqrt3}{2}
  2. B.34\frac{3}{4}
  3. C.81256\frac{81}{256}
  4. D.916\frac{9}{16}

Correct answer: D

Explanation: v∝1/d2v \propto 1/d^2: vmin/vmax=(4.8/6.4)2=9/16v_{min}/v_{max} = (4.8/6.4)^2 = 9/16.

Question 7

Physics · 4 marks · −1

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T1T_1 and T2T_2. The temperature of the hot reservoir of the first engine is T1T_1 and the temperature of the cold reservoir of the second engine is T2T_2. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1T_1 and T2T_2, if both the engines perform equal amount of work?

  1. A.T=T1+T22T = \frac{T_1+T_2}{2}
  2. B.T=T1T2T = \sqrt{T_1T_2}
  3. C.T=2T1T2T1+T2T = \frac{2T_1T_2}{T_1+T_2}
  4. D.T=0T = 0

Correct answer: A

Explanation: Equal work ⇒ T1−T=T−T2T_1 - T = T - T_2.

Question 8

Physics · 4 marks · −1

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is ν0=1400\nu_0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:

  1. A.1 m/s
  2. B.12\frac{1}{2} m/s
  3. C.14\frac{1}{4} m/s
  4. D.18\frac{1}{8} m/s

Correct answer: C

Explanation: Beat ≈2ν0u/v\approx 2\nu_0 u/v ⇒ u=2×3502×1400=14u = \frac{2\times350}{2\times1400} = \frac14 m/s.

Question 9

Physics · 4 marks · −1

A particle of mass m and charge q has an initial velocity v⃗=v0j^\vec v = v_0\hat j. If an electric field E⃗=E0i^\vec E = E_0\hat i and magnetic field B⃗=B0i^\vec B = B_0\hat i act on the particle, its speed will double after a time:

  1. A.2mv0qE0\frac{2mv_0}{qE_0}
  2. B.2mv0qE0\frac{\sqrt2 mv_0}{qE_0}
  3. C.3mv0qE0\frac{3mv_0}{qE_0}
  4. D.3mv0qE0\frac{\sqrt3 mv_0}{qE_0}

Correct answer: D

Explanation: Speed in the y–z plane stays v0v_0; vx=qE0t/mv_x = qE_0t/m. v=2v0v = 2v_0 when vx=3v0v_x = \sqrt3 v_0 ⇒ t=3mv0/(qE0)t = \sqrt3 mv_0/(qE_0).

Question 10

Physics · 4 marks · −1

In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be:

  1. A.10 A
  2. B.20 A
  3. C.15 A
  4. D.25 A

Correct answer: B

Explanation: Total power 4325 W; I=4325/220≈19.7I = 4325/220 \approx 19.7 A ⇒ 20 A.

Question 11

Physics · 4 marks · −1

A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

  1. A.2.5 s and 5.0 s
  2. B.5.0 s and 7.5 s
  3. C.5.0 s and 10.0 s
  4. D.2.5 s and 7.5 s

Correct answer: A

Explanation: ϕ=BAcos⁡ωt\phi = BA\cos\omega t, e∝sin⁡ωte \propto \sin\omega t: maximum at T/4 = 2.5 s, zero at T/2 = 5 s.

Question 12

Physics · 4 marks · −1

The figure gives experimentally measured B vs. H variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of the material are:

Figure for question 12
  1. A.1.0 T, 50 A/m and 1.5 T
  2. B.1.5 T, 50 A/m and 1.0 T
  3. C.150 A/m, 1.0 T and 1.5 T
  4. D.1.5 T, 50 A/m and 1.0 T

Correct answer: A

Explanation: Read from the loop: B at H = 0 is 1.0 T, H at B = 0 is 50 A/m, saturation 1.5 T.

Question 13

Physics · 4 marks · −1

The electric field of a plane electromagnetic wave is given by E⃗=E0i^+j^2cos⁡(kz+ωt)\vec E = E_0\frac{\hat i + \hat j}{\sqrt2}\cos(kz + \omega t). At t = 0, a positively charged particle is at the point (x,y,z)=(0,0,πk)(x, y, z) = \left(0, 0, \frac{\pi}{k}\right). If its instantaneous velocity at (t = 0) is v0k^v_0\hat k, the force acting on it due to the wave is:

  1. A.parallel to i^+j^2\frac{\hat i + \hat j}{\sqrt2}
  2. B.antiparallel to i^+j^2\frac{\hat i + \hat j}{\sqrt2}
  3. C.zero
  4. D.parallel to k^\hat k

Correct answer: B

Explanation: At that point E⃗=−E0n^\vec E = -E_0\hat n; the wave moves along −z-z, and qv⃗×B⃗q\vec v\times\vec B is also along −n^-\hat n: net force antiparallel to n^=(i^+j^)/2\hat n = (\hat i+\hat j)/\sqrt2.

Question 14

Physics · 4 marks · −1

A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is flf_l, then the ratio fl/ff_l/f is closest to the integer:

  1. A.1
  2. B.5
  3. C.9
  4. D.17

Correct answer: C

Explanation: flf=0.51.5/1.42−1≈8.9\frac{f_l}{f} = \frac{0.5}{1.5/1.42 - 1} \approx 8.9.

Question 15

Physics · 4 marks · −1

In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is:

  1. A.6.9 mm
  2. B.5.9 mm
  3. C.4.9 mm
  4. D.3.9 mm

Correct answer: B

Explanation: β=λD/d=589×10−9×1.5/0.15×10−3=5.89\beta = \lambda D/d = 589\times10^{-9}\times1.5/0.15\times10^{-3} = 5.89 mm.

Question 16

Physics · 4 marks · −1

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

  1. A.c(2mE)1/2c(2mE)^{1/2}
  2. B.1c(2Em)1/2\frac{1}{c}\left(\frac{2E}{m}\right)^{1/2}
  3. C.(E2m)1/2\left(\frac{E}{2m}\right)^{1/2}
  4. D.1c(E2m)1/2\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}

Correct answer: D

Explanation: λe=h/2mE\lambda_e = h/\sqrt{2mE}, λp=hc/E\lambda_p = hc/E; ratio =Ec2mE=1cE2m= \frac{E}{c\sqrt{2mE}} = \frac1c\sqrt{\frac{E}{2m}}.

Question 17

Physics · 4 marks · −1

The activity of a radioactive sample falls from 700 s−1^{-1} to 500 s−1^{-1} in 30 minutes. Its half life is close to:

  1. A.52 min
  2. B.62 min
  3. C.66 min
  4. D.72 min

Correct answer: B

Explanation: λ=ln⁡(7/5)/30\lambda = \ln(7/5)/30; t1/2=0.693×30/0.336≈62t_{1/2} = 0.693\times30/0.336 \approx 62 min.

Question 18

Physics · 4 marks · −1

In the figure, potential difference between A and B is:

Figure for question 18
  1. A.zero
  2. B.5 V
  3. C.10 V
  4. D.15 V

Correct answer: C

Explanation: Diode forward biased: the two 10 kΩ resistors from A are in parallel (5 kΩ) in series with 10 kΩ: VAB=30×5/15=10V_{AB} = 30\times5/15 = 10 V.

Question 19

Physics · 4 marks

The sum of two forces P⃗\vec P and Q⃗\vec Q is R⃗\vec R such that ∣R⃗∣=∣P⃗∣|\vec R| = |\vec P|. The angle θ (in degrees) that the resultant of 2P⃗2\vec P and Q⃗\vec Q will make with Q⃗\vec Q is, ________.

Correct answer: 90

Explanation: ∣P⃗+Q⃗∣2=P2⇒Q2+2P⃗⋅Q⃗=0⇒(2P⃗+Q⃗)⋅Q⃗=0|\vec P+\vec Q|^2 = P^2 ⇒ Q^2 + 2\vec P\cdot\vec Q = 0 ⇒ (2\vec P+\vec Q)\cdot\vec Q = 0: 90°.

Question 20

Physics · 4 marks

M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is ________.

Correct answer: 40

Explanation: M(540+60)=200(80+40)M(540 + 60) = 200(80 + 40) ⇒ M=40M = 40 g.

Question 21

Physics · 4 marks

A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.

Correct answer: 6

Explanation: Ui=12(60 pF)(20)2=12U_i = \frac12(60\,\mathrm{pF})(20)^2 = 12 nJ; equal capacitors ⇒ half is lost: 6 nJ.

Question 22

Physics · 4 marks

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is μ=0.4\mu = 0.4, the maximum possible value of 100×ba100\times\frac{b}{a} for box not to topple before moving is ________.

Figure for question 22

Correct answer: 50

Explanation: Not toppling: μmg(a2+b)≤mga2\mu mg(\frac a2 + b) \le mg\frac a2 ⇒ b≤0.75ab \le 0.75a; but the point of application cannot be above the top face, b≤a/2b \le a/2. Max 100 b/a=50100\,b/a = 50.

Question 23

Physics · 4 marks

The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the cell is N10\frac{N}{10} Ω, where N is an integer then value of N is ________.

Correct answer: 12

Explanation: r=Rl1−l2l2=10×60500=1.2r = R\frac{l_1 - l_2}{l_2} = 10\times\frac{60}{500} = 1.2 Ω ⇒ N = 12.

Question 24

Chemistry · 4 marks · −1

For the reaction 2H2_2(g) + 2NO(g) → N2_2(g) + 2H2_2O(g) the observed rate expression is, rate = kfk_f[NO]2^2[H2_2]. The rate expression for the reverse reaction is:

  1. A.kbk_b[N2_2][H2_2O]
  2. B.kbk_b[N2_2][H2_2O]2^2
  3. C.kbk_b[N2_2][H2_2O]2^2/[NO]
  4. D.kbk_b[N2_2][H2_2O]2^2/[H2_2]

Correct answer: D

Explanation: At equilibrium kf/kb=K=[N2][H2O]2[H2]2[NO]2k_f/k_b = K = \frac{[N_2][H_2O]^2}{[H_2]^2[NO]^2} ⇒ reverse rate =kb[N2][H2O]2/[H2]= k_b[N_2][H_2O]^2/[H_2].

Question 25

Chemistry · 4 marks · −1

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

  1. A.the volume of the solution increases and the volume of the solvent decreases
  2. B.the volume of the solution decreases and the volume of the solvent increases
  3. C.the volume of the solution does not change and the volume of the solvent decreases
  4. D.the volume of the solution and the solvent does not change

Correct answer: A

Question 26

Chemistry · 4 marks · −1

The equation that is incorrect is:

  1. A.(Λm0)NaBr−(Λm0)NaCl=(Λm0)KBr−(Λm0)KCl(\Lambda^0_m)_{NaBr} - (\Lambda^0_m)_{NaCl} = (\Lambda^0_m)_{KBr} - (\Lambda^0_m)_{KCl}
  2. B.(Λm0)KCl−(Λm0)NaCl=(Λm0)KBr−(Λm0)NaBr(\Lambda^0_m)_{KCl} - (\Lambda^0_m)_{NaCl} = (\Lambda^0_m)_{KBr} - (\Lambda^0_m)_{NaBr}
  3. C.(Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBr(\Lambda^0_m)_{NaBr} - (\Lambda^0_m)_{NaI} = (\Lambda^0_m)_{KBr} - (\Lambda^0_m)_{NaBr}
  4. D.(Λm0)H2O=(Λm0)HCl+(Λm0)NaOH−(Λm0)NaCl(\Lambda^0_m)_{H_2O} = (\Lambda^0_m)_{HCl} + (\Lambda^0_m)_{NaOH} - (\Lambda^0_m)_{NaCl}

Correct answer: C

Explanation: Option 3: LHS = λ(Br⁻) − λ(I⁻), RHS = λ(K⁺) − λ(Na⁺): not equal.

Question 27

Chemistry · 4 marks · −1

The ammonia (NH3_3) released on quantitative reaction of 0.6 g urea (NH2_2CONH2_2) with sodium hydroxide (NaOH) can be neutralized by:

  1. A.100 ml of 0.1 N HCl
  2. B.200 ml of 0.2 N HCl
  3. C.100 ml of 0.2 N HCl
  4. D.200 ml of 0.4 N HCl

Correct answer: C

Explanation: 0.01 mol urea → 0.02 mol NH3 = 100 mL of 0.2 N HCl.

Question 28

Chemistry · 4 marks · −1

Identify the correct labels of A, B and C in the following graph from the options given below: Root mean square speed (VrmsV_{rms}); most probable speed (VmpV_{mp}); Average speed (VavV_{av})

Figure for question 28
  1. A.A - VrmsV_{rms}; B - VmpV_{mp}; C - VavV_{av}
  2. B.A - VmpV_{mp}; B - VavV_{av}; C - VrmsV_{rms}
  3. C.A - VavV_{av}; B - VrmsV_{rms}; C - VmpV_{mp}
  4. D.A - VmpV_{mp}; B - VrmsV_{rms}; C - VavV_{av}

Correct answer: B

Question 29

Chemistry · 4 marks · −1

The bond order and the magnetic characteristics of CN−^- are:

  1. A.2122\frac12, diamagnetic
  2. B.3, paramagnetic
  3. C.2122\frac12, paramagnetic
  4. D.3, diamagnetic

Correct answer: D

Question 30

Chemistry · 4 marks · −1

Within each pair of elements F & Cl, S & Se, and Li & Na, respectively, the elements that release more energy upon an electron gain are:

  1. A.Cl, Se and Na
  2. B.F, S and Li
  3. C.F, Se and Na
  4. D.Cl, S and Li

Correct answer: D

Question 31

Chemistry · 4 marks · −1

The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is:

  1. A.distillation
  2. B.liquation
  3. C.vapour phase refining
  4. D.zone refining

Correct answer: B

Question 32

Chemistry · 4 marks · −1

Among statements (a)-(d), the correct ones are: (a) Decomposition of hydrogen peroxide gives dioxygen. (b) Like hydrogen peroxide, compounds, such as KClO3_3, Pb(NO3_3)2_2 and NaNO3_3 when heated liberate dioxygen. (c) 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide. (d) Hydrogen peroxide is used for the manufacture of sodium perborate.

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

Correct answer: D

Question 33

Chemistry · 4 marks · −1

In the following reactions, products (A) and (B), respectively, are: NaOH (hot and conc.) + Cl2_2 → (A) + side products; Ca(OH)2_2 (dry) + Cl2_2 → (B) + side products

  1. A.NaOCl and Ca(OCl)2_2
  2. B.NaClO3_3 and Ca(OCl)2_2
  3. C.NaClO3_3 and Ca(ClO3_3)2_2
  4. D.NaOCl and Ca(ClO3_3)2_2

Correct answer: B

Question 34

Chemistry · 4 marks · −1

The redox reaction among the following is:

  1. A.formation of ozone from atmospheric oxygen in the presence of sunlight
  2. B.reaction of H2_2SO4_4 with NaOH
  3. C.combination of dinitrogen with dioxygen at 2000 K
  4. D.reaction of [Co(H2_2O)6_6]Cl3_3 with AgNO3_3

Correct answer: C

Question 35

Chemistry · 4 marks · −1

The number of possible optical isomers for the complexes MA2_2B2_2 with sp3^3 and dsp2^2 hybridized metal atom, respectively, is: Note: A and B are unidentate neutral and unidentate monoanionic ligands, respectively.

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

Correct answer: D

Explanation: Tetrahedral MA2B2 and square-planar MA2B2 both have a mirror plane: no optical isomers.

Question 36

Chemistry · 4 marks · −1

Among the statements (a)-(d), the incorrect ones are: (a) Octahedral Co(III) complexes with strong field ligands have very high magnetic moments (b) When Δ0<P\Delta_0 < P, the d-electron configuration of Co(III) in an octahedral complex is t2g4eg2t_{2g}^4e_g^2 (c) Wavelength of light absorbed by [Co(en)3_3]3+^{3+} is lower than that of [CoF6_6]3−^{3-} (d) If the Δ0\Delta_0 for an octahedral complex of Co(III) is 18,000 cm−1^{-1}, the Δt\Delta_t for its tetrahedral complex with the same ligand will be 16,000 cm−1^{-1}

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

Correct answer: C

Explanation: (a) strong field ⇒ low spin, diamagnetic; (d) Δt=49Δ0=8000\Delta_t = \frac49\Delta_0 = 8000 cm−1^{-1}.

Question 37

Chemistry · 4 marks · −1

A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (A) benzanilide (B) aniline and (C) acetophenone. When the column is eluted with a mixture of solvents, hexane : ethyl acetate (20:80), the sequence of obtained compounds is:

  1. A.(A), (B) and (C)
  2. B.(B), (C) and (A)
  3. C.(C), (A) and (B)
  4. D.(B), (A) and (C)

Correct answer: C

Question 38

Chemistry · 4 marks · −1

In the following reaction sequence, p-toluidine (4-methylaniline) →Ac2O\xrightarrow{\text{Ac}_2\text{O}} A →Br2/AcOH\xrightarrow{\text{Br}_2/\text{AcOH}} B, the major product B is:

Figure for question 38
  1. A.2-bromo-4-methylacetanilide (Br ortho to NHCOCH3, CH3 para)
  2. B.2-acetyl-4-methylacetanilide (COCH3 ortho to NHCOCH3, CH3 para)
  3. C.3-bromo-4-methylacetanilide (Br ortho to CH3)
  4. D.2-bromo-4-(bromomethyl)acetanilide (Br ortho to NHCOCH3, CH2Br para)

Correct answer: A

Explanation: –NHCOCH3 is the stronger o/p director; para is blocked by CH3, so Br enters ortho to the acetamido group.

Question 39

Chemistry · 4 marks · −1

The correct order of stability for the following alkoxides is: (A) CH3_3–CH(NO2_2)–O−^- (B) CH2_2=C(NO2_2)–O−^- (C) O2_2N–CH=CH–O−^-

Figure for question 39
  1. A.(C) > (A) > (B)
  2. B.(C) > (B) > (A)
  3. C.(B) > (A) > (C)
  4. D.(B) > (C) > (A)

Correct answer: B

Explanation: In (C) the negative charge is delocalised onto the nitro group through the C=C (extended conjugation); (B) is stabilised only inductively/cross-conjugated; (A) has no resonance.

Question 40

Chemistry · 4 marks · −1

In the following reaction sequence, structures of A and B, respectively will be: chromane bearing a CH2_2Br group on the benzene ring (see figure) →HBr, Δ\xrightarrow{\text{HBr},\ \Delta} A →Na, Ether\xrightarrow{\text{Na, Ether}} (Intramolecular Product) B

Figure for question 40
  1. A.A: 2-(3-bromopropyl)-phenol bearing the ring CH2Br; B: a phenol fused to a six-membered carbocycle (5,6,7,8-tetrahydronaphthol type)
  2. B.A: same bromopropyl phenol with ring CH2Br; B: a phenol fused to a five-membered ring carrying a methyl (indane type)
  3. C.A: aryl bromide with a 3-hydroxypropyl chain and ring CH2Br; B: bromo-substituted fused bicyclic compound (ring drawn with a substituent)
  4. D.A: aryl bromide with a 3-hydroxypropyl chain and ring CH2Br; B: a CH2Br-substituted benzopyran (chromane)

Correct answer: A

Explanation: HBr cleaves the alkyl–O bond of the aryl alkyl ether, giving a phenol and a primary alkyl bromide; Wurtz coupling of the two C–Br groups (alkyl Br and benzylic CH2Br) closes a new carbocycle.

Question 41

Chemistry · 4 marks · −1

Which of the following statements is correct?

  1. A.Gluconic acid can form cyclic (acetal/hemiacetal) structure
  2. B.Gluconic acid is a partial oxidation product of glucose
  3. C.Gluconic acid is a dicarboxylic acid
  4. D.Gluconic acid is obtained by oxidation of glucose with HNO3_3

Correct answer: B

Question 42

Chemistry · 4 marks · −1

Consider the following reactions: (a) benzene + chlorobenzene →anhyd. AlCl3\xrightarrow{\text{anhyd. AlCl}_3} biphenyl (b) benzene + Cl2_2 (excess) →anhyd. AlCl3, dark\xrightarrow{\text{anhyd. AlCl}_3,\ \text{dark}} hexachlorobenzene (c) benzene + CH2_2=CH–Cl →anhyd. AlCl3\xrightarrow{\text{anhyd. AlCl}_3} styrene (C6_6H5_5–CH=CH2_2) (d) benzene + CH2_2=CH–CH2_2Cl →anhyd. AlCl3\xrightarrow{\text{anhyd. AlCl}_3} allylbenzene (C6_6H5_5–CH2_2–CH=CH2_2). Which of these reactions are possible?

Figure for question 42
  1. A.(b), (c) and (d)
  2. B.(b) and (d)
  3. C.(a) and (b)
  4. D.(a) and (d)

Correct answer: B

Explanation: Aryl and vinyl halides do not form carbocations under Friedel–Crafts conditions.

Question 43

Chemistry · 4 marks · −1

For the following reactions: CH3_3CH2_2CH2_2Br + Z−^- →ks, Substitution\xrightarrow{k_s,\ \text{Substitution}} CH3_3CH2_2CH2_2Z + Br−^- and CH3_3CH2_2CH2_2Br + Z−^- →ke, elimination\xrightarrow{k_e,\ \text{elimination}} CH3_3CH=CH2_2 + HZ + Br−^-, where, Z−^- = CH3_3CH2_2O−^- (A) or (CH3_3)3_3C–O−^- (B), ksk_s and kek_e, are, respectively, the rate constants for substitution and elimination, and μ=kske\mu = \frac{k_s}{k_e}, the correct option is ________.

  1. A.μA>μB\mu_A > \mu_B and ke(A)>ke(B)k_e(A) > k_e(B)
  2. B.μA>μB\mu_A > \mu_B and ke(B)>ke(A)k_e(B) > k_e(A)
  3. C.μB>μA\mu_B > \mu_A and ke(A)>ke(B)k_e(A) > k_e(B)
  4. D.μB>μA\mu_B > \mu_A and ke(B)>ke(A)k_e(B) > k_e(A)

Correct answer: B

Explanation: Bulky tert-butoxide favours elimination (larger k_e, smaller k_s/k_e).

Question 44

Chemistry · 4 marks

3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution 12\frac{1}{2} mL of 5 M NaOH is added. The pH of the solution is ________. [Given: pKa of acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.

Correct answer: 5.23

Explanation: In 20 mL: CH3_3COOH 2 mmol, HCl 1 mmol; NaOH 2.5 mmol neutralises HCl and converts 1.5 mmol acid to acetate: pH = 4.75 + log(1.5/0.5) = 5.23.

Question 45

Chemistry · 4 marks

The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L−1^{-1}. If H2_2SO4_4 is used for the flocculation of arsenic sulphide, the amount, in grams, of H2_2SO4_4 in 250 ml required for the above purpose is ________. (molecular mass of H2_2SO4_4 = 98 g/mol)

Correct answer: 0.37

Explanation: The coagulating ion is H+^+: 30 mmol L−1^{-1} of H+^+ needs 15 mmol L−1^{-1} H2_2SO4_4; in 250 mL: 3.75 mmol × 98 = 0.3675 g.

Question 46

Chemistry · 4 marks

Consider the following reactions: NaCl + K2_2Cr2_2O7_7 + H2_2SO4_4 (Conc.) → (A) + side products; (A) + NaOH → (B) + Side products; (B) + H2_2SO4_4 (dilute) + H2_2O2_2 → (C) + Side products. The sum of the total number of atoms in one molecule each of (A), (B) and (C) is ________.

Correct answer: 18

Explanation: A = CrO2_2Cl2_2 (5), B = Na2_2CrO4_4 (7), C = CrO5_5 (6): 18.

Question 47

Chemistry · 4 marks

The number of sp2^2 hybridised carbons present in "Aspartame" is ________.

Correct answer: 9

Explanation: 6 aromatic carbons + 3 carbonyl carbons (acid, amide, ester) = 9.

Question 48

Mathematics · 4 marks · −1

Let α and β be the roots of the equation x2−x−1=0x^2 - x - 1 = 0. If pk=(α)k+(β)kp_k = (\alpha)^k + (\beta)^k, k≥1k \ge 1, then which one of the following statements is not true?

  1. A.p5=11p_5 = 11
  2. B.p3=p5−p4p_3 = p_5 - p_4
  3. C.(p1+p2+p3+p4+p5)=26(p_1 + p_2 + p_3 + p_4 + p_5) = 26
  4. D.p5=p2⋅p3p_5 = p_2\cdot p_3

Correct answer: D

Explanation: pk=pk−1+pk−2p_k = p_{k-1} + p_{k-2}: 1, 3, 4, 7, 11; p2p3=12≠11p_2p_3 = 12 \ne 11.

Question 49

Mathematics · 4 marks · −1

If 3+isin⁡θ4−icos⁡θ\frac{3 + i\sin\theta}{4 - i\cos\theta}, θ∈[0,2π]\theta \in [0, 2\pi], is a real number, then an argument of sin⁡θ+icos⁡θ\sin\theta + i\cos\theta is:

  1. A.−tan⁡−1(34)-\tan^{-1}\left(\frac34\right)
  2. B.π−tan⁡−1(34)\pi - \tan^{-1}\left(\frac34\right)
  3. C.π−tan⁡−1(43)\pi - \tan^{-1}\left(\frac43\right)
  4. D.tan⁡−1(43)\tan^{-1}\left(\frac43\right)

Correct answer: C

Explanation: Real ⇒ 3cos⁡θ+4sin⁡θ=03\cos\theta + 4\sin\theta = 0, tan⁡θ=−34\tan\theta = -\frac34; taking sin⁡θ=−35\sin\theta = -\frac35, cos⁡θ=45\cos\theta = \frac45: point (−35,45)(-\frac35, \frac45) has argument π−tan⁡−143\pi - \tan^{-1}\frac43.

Question 50

Mathematics · 4 marks · −1

Let A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}] be two 3 × 3 real matrices such that bij=(3)(i+j−2)ajib_{ij} = (3)^{(i+j-2)}a_{ji}, where i, j = 1, 2, 3. If the determinant of B is 81, then the determinant of A is:

  1. A.1/3
  2. B.1/9
  3. C.1/81
  4. D.3

Correct answer: B

Explanation: B=DATDB = DA^TD with D=diag(1,3,9)D = \mathrm{diag}(1,3,9): ∣B∣=272∣A∣=729∣A∣=81|B| = 27^2|A| = 729|A| = 81 ⇒ ∣A∣=1/9|A| = 1/9.

Question 51

Mathematics · 4 marks · −1

The number of ordered pairs (r, k) for which 6⋅35Cr=(k2−3)⋅36Cr+16\cdot{}^{35}C_r = (k^2 - 3)\cdot{}^{36}C_{r+1}, where k is an integer, is:

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

Correct answer: B

Explanation: 36Cr+1=36r+135Cr{}^{36}C_{r+1} = \frac{36}{r+1}{}^{35}C_r ⇒ r+1=6(k2−3)≤36r + 1 = 6(k^2-3) \le 36: k=±2k = \pm2 (r = 5), k=±3k = \pm3 (r = 35): 4 pairs.

Question 52

Mathematics · 4 marks · −1

The coefficient of x7x^7 in the expression (1+x)10+x(1+x)9+x2(1+x)8+…+x10(1+x)^{10} + x(1+x)^9 + x^2(1+x)^8 + \ldots + x^{10} is:

  1. A.120
  2. B.210
  3. C.330
  4. D.420

Correct answer: C

Explanation: ∑k=0710−kC7−k=11C7=330\sum_{k=0}^{7}{}^{10-k}C_{7-k} = {}^{11}C_7 = 330.

Question 53

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a G. P. such that a1<0a_1 < 0, a1+a2=4a_1 + a_2 = 4 and a3+a4=16a_3 + a_4 = 16. If ∑i=19ai=4λ\sum_{i=1}^{9} a_i = 4\lambda, then λ is equal to:

  1. A.−513
  2. B.5113\frac{511}{3}
  3. C.−171
  4. D.171

Correct answer: C

Explanation: r2=4r^2 = 4, a1<0a_1 < 0 ⇒ r=−2r = -2, a1=−4a_1 = -4; S9=−41+5123=−684=4λS_9 = -4\frac{1+512}{3} = -684 = 4\lambda ⇒ λ=−171\lambda = -171.

Question 54

Mathematics · 4 marks · −1

If the sum of the first 40 terms of the series, 3+4+8+9+13+14+18+19+…3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \ldots is (102)m, then m is equal to:

  1. A.25
  2. B.20
  3. C.10
  4. D.5

Correct answer: B

Explanation: Pair sums 7, 17, 27, ... (20 terms): 202(14+19×10)=2040=102×20\frac{20}{2}(14 + 19\times10) = 2040 = 102\times20.

Question 55

Mathematics · 4 marks · −1

Let y=y(x)y = y(x) be a function of x satisfying y1−x2=k−x1−y2y\sqrt{1-x^2} = k - x\sqrt{1-y^2} where k is a constant and y(12)=−14y\left(\frac12\right) = -\frac14. Then dydx\frac{dy}{dx} at x=12x = \frac12, is equal to:

  1. A.25\frac{2}{\sqrt5}
  2. B.−54-\frac{\sqrt5}{4}
  3. C.−52-\frac{\sqrt5}{2}
  4. D.52\frac{\sqrt5}{2}

Correct answer: C

Explanation: $\sin^{-1}x + \sin^{-1}y = $ const ⇒ y′=−1−y21−x2=−15/43/2=−52y' = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = -\frac{\sqrt{15}/4}{\sqrt3/2} = -\frac{\sqrt5}{2}.

Question 56

Mathematics · 4 marks · −1

The value of c in the Lagrange's mean value theorem for the function f(x)=x3−4x2+8x+11f(x) = x^3 - 4x^2 + 8x + 11, when x∈[0,1]x \in [0, 1] is:

  1. A.4−73\frac{4-\sqrt7}{3}
  2. B.7−23\frac{\sqrt7-2}{3}
  3. C.4−53\frac{4-\sqrt5}{3}
  4. D.23\frac{2}{3}

Correct answer: A

Explanation: f′(c)=f(1)−f(0)=5f'(c) = f(1) - f(0) = 5 ⇒ 3c2−8c+3=03c^2 - 8c + 3 = 0 ⇒ c=4−73c = \frac{4-\sqrt7}{3}.

Question 57

Mathematics · 4 marks · −1

Let f(x) be a polynomial of degree 5 such that x = ±1 are its critical points. If lim⁡x→0(2+f(x)x3)=4\lim_{x\to0}\left(2 + \frac{f(x)}{x^3}\right) = 4, then which one of the following is not true?

  1. A.f is an odd function.
  2. B.x = 1 is a point of maxima and x = −1 is a point of minimum of f.
  3. C.f(1) − 4f(−1) = 4.
  4. D.x = 1 is a point of minima and x = −1 is a point of maxima of f.

Correct answer: D

Explanation: f=2x3−65x5f = 2x^3 - \frac65x^5; f′′(1)=−12<0f'' (1) = -12 < 0 ⇒ maximum at x = 1.

Question 58

Mathematics · 4 marks · −1

The value of α for which 4α∫−12e−α∣x∣ dx=54\alpha\int_{-1}^{2} e^{-\alpha|x|}\,dx = 5, is:

  1. A.log⁡e(43)\log_e\left(\frac43\right)
  2. B.log⁡e2\log_e 2
  3. C.log⁡e2\log_e\sqrt2
  4. D.log⁡e(32)\log_e\left(\frac32\right)

Correct answer: B

Explanation: 4(2−e−α−e−2α)=54(2 - e^{-\alpha} - e^{-2\alpha}) = 5 ⇒ t2+t−34=0t^2 + t - \frac34 = 0, t=e−α=12t = e^{-\alpha} = \frac12.

Question 59

Mathematics · 4 marks · −1

The area (in sq. units) of the region {(x,y)∈R2∣4x2≤y≤8x+12}\{(x, y) \in \mathbb{R}^2 \mid 4x^2 \le y \le 8x + 12\} is:

  1. A.1243\frac{124}{3}
  2. B.1253\frac{125}{3}
  3. C.1273\frac{127}{3}
  4. D.1283\frac{128}{3}

Correct answer: D

Explanation: ∫−13(8x+12−4x2)dx=1283\int_{-1}^{3}(8x + 12 - 4x^2)dx = \frac{128}{3}.

Question 60

Mathematics · 4 marks · −1

Let y=y(x)y = y(x) be the solution curve of the differential equation, (y2−x)dydx=1(y^2 - x)\frac{dy}{dx} = 1, satisfying y(0)=1y(0) = 1. This curve intersects the x-axis at a point whose abscissa is:

  1. A.2
  2. B.2 + e
  3. C.2 − e
  4. D.−e

Correct answer: C

Explanation: dxdy+x=y2\frac{dx}{dy} + x = y^2 ⇒ x=y2−2y+2+Ce−yx = y^2 - 2y + 2 + Ce^{-y}, C = −e; at y = 0: x = 2 − e.

Question 61

Mathematics · 4 marks · −1

Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0x^2 + y^2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2(AB)^2 is equal to:

  1. A.325\frac{32}{5}
  2. B.525\frac{52}{5}
  3. C.565\frac{56}{5}
  4. D.645\frac{64}{5}

Correct answer: D

Explanation: Tangent length L = 4, r = 2: AB=2LrL2+r2=1620AB = \frac{2Lr}{\sqrt{L^2+r^2}} = \frac{16}{\sqrt{20}}, AB2=645AB^2 = \frac{64}{5}.

Question 62

Mathematics · 4 marks · −1

The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is:

  1. A.3x − 2y = 0
  2. B.2x − 3y = 0
  3. C.7x − 5y = 0
  4. D.5x − 7y = 0

Correct answer: D

Explanation: From (2t, t) the foot on y = x is (3t/2, 3t/2); midpoint (7t/4, 5t/4) ⇒ 5x = 7y.

Question 63

Mathematics · 4 marks · −1

If 3x+4y=1223x + 4y = 12\sqrt2 is a tangent to the ellipse x2a2+y29=1\frac{x^2}{a^2} + \frac{y^2}{9} = 1 for some a∈Ra \in \mathbb{R}, then the distance between the foci of the ellipse is:

  1. A.222\sqrt2
  2. B.272\sqrt7
  3. C.252\sqrt5
  4. D.4

Correct answer: B

Explanation: c2=a2m2+b2c^2 = a^2m^2 + b^2: 18=9a216+918 = \frac{9a^2}{16} + 9 ⇒ a2=16a^2 = 16; foci distance 216−9=272\sqrt{16-9} = 2\sqrt7.

Question 64

Mathematics · 4 marks · −1

Let a⃗\vec a, b⃗\vec b and c⃗\vec c be three unit vectors such that a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0. If λ=a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\lambda = \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a and d⃗=a⃗×b⃗+b⃗×c⃗+c⃗×a⃗\vec d = \vec a\times\vec b + \vec b\times\vec c + \vec c\times\vec a, then the ordered pair, (λ,d⃗)(\lambda, \vec d) is equal to:

  1. A.(−32,3a⃗×b⃗)\left(-\frac32, 3\vec a\times\vec b\right)
  2. B.(32,3a⃗×c⃗)\left(\frac32, 3\vec a\times\vec c\right)
  3. C.(−32,3c⃗×b⃗)\left(-\frac32, 3\vec c\times\vec b\right)
  4. D.(32,3b⃗×c⃗)\left(\frac32, 3\vec b\times\vec c\right)

Correct answer: A

Explanation: ∣a⃗+b⃗+c⃗∣2=0⇒λ=−32|\vec a+\vec b+\vec c|^2 = 0 ⇒ \lambda = -\frac32; with c⃗=−a⃗−b⃗\vec c = -\vec a - \vec b each cross product equals a⃗×b⃗\vec a\times\vec b.

Question 65

Mathematics · 4 marks · −1

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 14\frac14. If the probability that at most two machines will be out of service on the same day is (34)3k\left(\frac34\right)^3k, then k is equal to:

  1. A.174\frac{17}{4}
  2. B.172\frac{17}{2}
  3. C.4
  4. D.178\frac{17}{8}

Correct answer: D

Explanation: P=(34)3[916+1516+1016]=(34)3178P = \left(\frac34\right)^3\left[\frac{9}{16} + \frac{15}{16} + \frac{10}{16}\right] = \left(\frac34\right)^3\frac{17}{8}.

Question 66

Mathematics · 4 marks · −1

If θ1\theta_1 and θ2\theta_2 be respectively the smallest and the largest values of θ in (0,2π)−{π}(0, 2\pi) - \{\pi\} which satisfy the equation, 2cot⁡2θ−5sin⁡θ+4=02\cot^2\theta - \frac{5}{\sin\theta} + 4 = 0, then ∫θ1θ2cos⁡23θ dθ\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta is equal to:

  1. A.π3+16\frac{\pi}{3} + \frac16
  2. B.π9\frac{\pi}{9}
  3. C.π3\frac{\pi}{3}
  4. D.2π3\frac{2\pi}{3}

Correct answer: C

Explanation: 2 cosec2θ−5 cosec θ+2=02\,\mathrm{cosec}^2\theta - 5\,\mathrm{cosec}\,\theta + 2 = 0 ⇒ sin⁡θ=12\sin\theta = \frac12: θ = π/6, 5π/6; integral =π3= \frac{\pi}{3}.

Question 67

Mathematics · 4 marks · −1

Let A, B, C and D be four non-empty sets. The contrapositive statement of \"If A⊆BA \subseteq B and B⊆DB \subseteq D, then A⊆CA \subseteq C\" is:

  1. A.If A⊈CA \not\subseteq C, then A⊈BA \not\subseteq B and B⊆DB \subseteq D
  2. B.If A⊈CA \not\subseteq C, then A⊈BA \not\subseteq B or B⊈DB \not\subseteq D
  3. C.If A⊈CA \not\subseteq C, then A⊆BA \subseteq B and B⊆DB \subseteq D
  4. D.If A⊆CA \subseteq C, then B⊂AB \subset A or D⊂BD \subset B

Correct answer: B

Question 68

Mathematics · 4 marks

Let X={n∈N:1≤n≤50}X = \{n \in \mathbb{N} : 1 \le n \le 50\}. If A={n∈X:nA = \{n \in X : n is a multiple of 2}2\} and B={n∈X:nB = \{n \in X : n is a multiple of 7}7\}, then the number of elements in the smallest subset of X containing both A and B is ________.

Correct answer: 29

Explanation: ∣A∪B∣=25+7−3=29|A\cup B| = 25 + 7 - 3 = 29.

Question 69

Mathematics · 4 marks

If the system of linear equations, x+y+z=6x + y + z = 6, x+2y+3z=10x + 2y + 3z = 10, 3x+2y+λz=μ3x + 2y + \lambda z = \mu has more than two solutions, then μ−λ2\mu - \lambda^2 is equal to ________.

Correct answer: 13

Explanation: Third equation = 4(first) − (second): λ = 1, μ = 14 ⇒ μ − λ² = 13.

Question 70

Mathematics · 4 marks

If the function f defined on (−13,13)\left(-\frac13, \frac13\right) by f(x)={1xlog⁡e(1+3x1−2x),x≠0k,x=0f(x) = \begin{cases}\frac1x\log_e\left(\frac{1+3x}{1-2x}\right), & x \ne 0\\ k, & x = 0\end{cases} is continuous, then k is equal to ________.

Correct answer: 5

Explanation: lim⁡ln⁡(1+3x)−ln⁡(1−2x)x=3+2=5\lim\frac{\ln(1+3x) - \ln(1-2x)}{x} = 3 + 2 = 5.

Question 71

Mathematics · 4 marks

If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through (α, 7, 1) is (53,73,173)\left(\frac53, \frac73, \frac{17}{3}\right), then α is equal to ________.

Correct answer: 4

Explanation: (1,0,3)−F∝(2,7,8)(1,0,3) - F \propto (2, 7, 8) must be ⟂ to (α−53,143,−143)(\alpha - \frac53, \frac{14}{3}, -\frac{14}{3}): 2α−103−143=02\alpha - \frac{10}{3} - \frac{14}{3} = 0 ⇒ α = 4.

Question 72

Mathematics · 4 marks

If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, then x·y is equal to ________.

Correct answer: 54

Explanation: x + y = 16; ∑xi2=8(25+100)=1000\sum x_i^2 = 8(25 + 100) = 1000 ⇒ x2+y2=148x^2 + y^2 = 148 ⇒ xy = 54.

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