JEE Main 2020 — 8 Jan Shift 1 (Previous Year Paper)

JEE Main · 2020 · with answers and explanations

questions
73
Total marks
292
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length ll. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

  1. A.mlω2k+mω2\frac{ml\omega^2}{k + m\omega^2}
  2. B.mlω2k−mω2\frac{ml\omega^2}{k - m\omega^2}
  3. C.mlω2k+mω\frac{ml\omega^2}{k + m\omega}
  4. D.mlω2k−ωm\frac{ml\omega^2}{k - \omega m}

Correct answer: B

Explanation: kx=mω2(l+x)kx = m\omega^2(l + x) ⇒ x=mω2lk−mω2x = \frac{m\omega^2 l}{k - m\omega^2}.

Question 2

Physics · 4 marks · −1

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4 kg. (The coordinates of the same are shown in figure) are:

Figure for question 2
  1. A.(0.75 m, 1.75 m)
  2. B.(0.75 m, 0.75 m)
  3. C.(1 m, 1.75 m)
  4. D.(1.25 m, 1.50 m)

Correct answer: A

Explanation: Strip 1×3 (mass 3) at (0.5, 1.5) + square 1×1 (mass 1) at (1.5, 2.5): (0.75, 1.75).

Question 3

Physics · 4 marks · −1

Consider a uniform rod of mass M = 4m and length ll pivoted about its centre. A mass m moving with velocity v making angle θ=π4\theta = \frac{\pi}{4} to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:

  1. A.37vl\frac37\frac{v}{l}
  2. B.372vl\frac{3}{7\sqrt2}\frac{v}{l}
  3. C.327vl\frac{3\sqrt2}{7}\frac{v}{l}
  4. D.47vl\frac47\frac{v}{l}

Correct answer: C

Explanation: mvsin⁡45°l2=(4ml212+ml24)ω=7ml212ωmv\sin45°\frac l2 = \left(\frac{4ml^2}{12} + \frac{ml^2}{4}\right)\omega = \frac{7ml^2}{12}\omega ⇒ ω=327vl\omega = \frac{3\sqrt2}{7}\frac vl.

Question 4

Physics · 4 marks · −1

Consider two solid spheres of radii R1R_1 = 1m, R2R_2 = 2m and masses M1M_1 and M2M_2, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of M1M2\frac{M_1}{M_2} is:

Figure for question 4
  1. A.16\frac16
  2. B.13\frac13
  3. C.12\frac12
  4. D.23\frac23

Correct answer: A

Explanation: Surface fields: GM1/12=2GM_1/1^2 = 2, GM2/22=3GM_2/2^2 = 3 ⇒ M1/M2=2/12=1/6M_1/M_2 = 2/12 = 1/6.

Question 5

Physics · 4 marks · −1

Consider a solid sphere of radius R and mass density ρ(r)=ρ0(1−r2R2)\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right), 0<r≤R0 < r \le R. The minimum density of a liquid in which it will float is:

  1. A.ρ05\frac{\rho_0}{5}
  2. B.ρ03\frac{\rho_0}{3}
  3. C.2ρ05\frac{2\rho_0}{5}
  4. D.2ρ03\frac{2\rho_0}{3}

Correct answer: C

Explanation: M=4πρ0R3(13−15)M = 4\pi\rho_0R^3\left(\frac13 - \frac15\right); mean density =2ρ05= \frac{2\rho_0}{5}.

Question 6

Physics · 4 marks · −1

A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to:

  1. A.1.04
  2. B.1.01
  3. C.1.26
  4. D.1.03

Correct answer: B

Explanation: 80ρ0=79ρ480\rho_0 = 79\rho_4 ⇒ ρ4/ρ0=1.013\rho_4/\rho_0 = 1.013.

Question 7

Physics · 4 marks · −1

A thermodynamic cycle xyzx is shown on a V-T diagram. The P-V diagram that best describes this cycle is: (Diagrams are schematic and not to scale)

Figure for question 7
  1. A.P–V diagram 1 (see image)
  2. B.P–V diagram 2 (see image)
  3. C.P–V diagram 3 (see image)
  4. D.P–V diagram 4 (see image)

Correct answer: B

Question 8

Physics · 4 marks · −1

The plot that depicts the behavior of the mean free time τ (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale)

Figure for question 8
  1. A.τ vs T: straight line through the origin
  2. B.τ vs √T: straight line with negative slope
  3. C.τ vs 1/√T: straight line through the origin
  4. D.τ vs 1/T: straight line through the origin

Correct answer: C

Explanation: τ=λ/vˉ∝1/T\tau = \lambda/\bar v \propto 1/\sqrt T at fixed density.

Question 9

Physics · 4 marks · −1

Three charged particles A, B and C with charges −4q, 2q and −2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is:

Figure for question 9
  1. A.3q4πϵ0d2\frac{\sqrt3 q}{4\pi\epsilon_0d^2}
  2. B.33q4πϵ0d2\frac{3\sqrt3 q}{4\pi\epsilon_0d^2}
  3. C.23qπϵ0d2\frac{2\sqrt3 q}{\pi\epsilon_0d^2}
  4. D.3qπϵ0d2\frac{\sqrt3 q}{\pi\epsilon_0d^2}

Correct answer: D

Explanation: A (−4q) at 30° above +x, C (−2q) at 30° below +x, B (2q) at 150°: Ex=kqd2(4cos⁡30°+2cos⁡30°+2cos⁡30°)=8kqd232=3qπϵ0d2E_x = \frac{kq}{d^2}(4\cos30° + 2\cos30° + 2\cos30°) = \frac{8kq}{d^2}\frac{\sqrt3}{2} = \frac{\sqrt3 q}{\pi\epsilon_0d^2}.

Question 10

Physics · 4 marks · −1

Effective capacitance of parallel combination of two capacitors C1C_1 and C2C_2 is 10 μF. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor C2C_2 is 4 times that of C1C_1. If these capacitors are connected in series, their effective capacitance will be:

  1. A.8.4 μF
  2. B.4.2 μF
  3. C.3.2 μF
  4. D.1.6 μF

Correct answer: D

Explanation: C2=4C1C_2 = 4C_1, C1+C2=10C_1 + C_2 = 10 ⇒ 2 and 8 μF; series =16/10=1.6= 16/10 = 1.6 μF.

Question 11

Physics · 4 marks · −1

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 101210^{12} m/s2^2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10−271.6\times10^{-27} kg)

  1. A.71 mT
  2. B.7.1 mT
  3. C.0.71 mT
  4. D.0.071 mT

Correct answer: C

Explanation: v=2K/m≈1.41×107v = \sqrt{2K/m} \approx 1.41\times10^7 m/s; B=maqv≈0.71B = \frac{ma}{qv} \approx 0.71 mT.

Question 12

Physics · 4 marks · −1

In finding the electric field using Gauss law the formula ∣E⃗∣=qencϵ0∣A∣|\vec E| = \frac{q_{enc}}{\epsilon_0|A|} is applicable. In the formula ϵ0\epsilon_0 is permittivity of free space, A is the area of Gaussian surface and qencq_{enc} is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

  1. A.For any choice of Gaussian surface.
  2. B.Only when the Gaussian surface is an equipotential surface.
  3. C.Only when the Gaussian surface is an equipotential surface and ∣E⃗∣|\vec E| is constant on the surface.
  4. D.Only when ∣E⃗∣|\vec E| = constant on the surface.

Correct answer: C

Question 13

Physics · 4 marks · −1

At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is:

Figure for question 13
  1. A.28 μV
  2. B.36 μV
  3. C.48 μV
  4. D.56 μV

Correct answer: D

Explanation: Area = 16×4 − 2×(½×4×2) = 56 cm²; EMF = 0.05 T × 56×10⁻⁴ m² / 5 s = 56 μV.

Question 14

Physics · 4 marks · −1

The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 43\frac43 for this wavelength, will be:

  1. A.30°
  2. B.45°
  3. C.60°
  4. D.15°

Correct answer: A

Explanation: n=3×4/3=2n = \sqrt{3\times4/3} = 2; sin⁡C=1/2\sin C = 1/2.

Question 15

Physics · 4 marks · −1

The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece?

  1. A.10 cm
  2. B.20 cm
  3. C.30 cm
  4. D.40 cm

Correct answer: A

Explanation: fo+fe=60f_o + f_e = 60, fo=5fef_o = 5f_e ⇒ fe=10f_e = 10 cm.

Question 16

Physics · 4 marks · −1

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAT_A eV and de-Broglie wavelength λA\lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB=(TA−1.5)T_B = (T_A - 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB=2λA\lambda_B = 2\lambda_A, then the work function of metal B is:

  1. A.1.5 eV
  2. B.4 eV
  3. C.3 eV
  4. D.2 eV

Correct answer: B

Explanation: TB=TA/4T_B = T_A/4 ⇒ TA=2T_A = 2, TB=0.5T_B = 0.5 ⇒ ϕB=4.5−0.5=4\phi_B = 4.5 - 0.5 = 4 eV.

Question 17

Physics · 4 marks · −1

The graph which depicts the results of Rutherford gold foil experiment with α-particles is: θ: Scattering angle; Y: Number of scattered α-particles detected (Plots are schematic and not to scale)

Figure for question 17
  1. A.Y decreases linearly from θ = 0 to π
  2. B.Y rises and saturates with θ
  3. C.Y rises to a peak and falls (bell-shaped)
  4. D.Y falls steeply from a large value at small θ and flattens towards π

Correct answer: D

Question 18

Physics · 4 marks · −1

Boolean relation at the output stage-Y for the following circuit is:

Figure for question 18
  1. A.A + B
  2. B.A · B
  3. C.Aˉ+Bˉ\bar A + \bar B
  4. D.Aˉ⋅Bˉ\bar A\cdot\bar B

Correct answer: D

Explanation: Diodes form an OR gate; the common-emitter transistor inverts: Y=A+B‾=Aˉ⋅BˉY = \overline{A+B} = \bar A\cdot\bar B.

Question 19

Physics · 4 marks · −1

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of 20 Ω, the null point on it is found to be at 1000 cm. The resistance of whole wire is:

  1. A.60 Ω
  2. B.80 Ω
  3. C.100 Ω
  4. D.120 Ω

Correct answer: C

Explanation: Gradient 5 V/1000 cm ⇒ 6 V across 1200 cm; R=6/0.06=100R = 6/0.06 = 100 Ω.

Question 20

Physics · 4 marks

A particle is moving along the x-axis with its coordinate with time 't' given by x(t)=10+8t−3t2x(t) = 10 + 8t - 3t^2. Another particle is moving along the y-axis with its coordinate as a function of time given by y(t)=5−8t3y(t) = 5 - 8t^3. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as v\sqrt v. Then v (in m/s) is ________.

Correct answer: 580

Explanation: v⃗1=(8−6t)i^=2i^\vec v_1 = (8 - 6t)\hat i = 2\hat i, v⃗2=−24t2j^=−24j^\vec v_2 = -24t^2\hat j = -24\hat j; relative speed² =4+576=580= 4 + 576 = 580.

Question 21

Physics · 4 marks

A body A, of mass m = 0.1 kg has an initial velocity of 3i^3\hat i m s−1^{-1}. It collides elastically with another body, B of the same mass which has an initial velocity of 5j^5\hat j m s−1^{-1}. After collision, A moves with a velocity v⃗=4(i^+j^)\vec v = 4(\hat i + \hat j). The energy of B after collision is written as x10\frac{x}{10} J. The value of x is ________.

Correct answer: 1

Explanation: Momentum: v⃗B=(3i^+5j^)−(4i^+4j^)=−i^+j^\vec v_B = (3\hat i + 5\hat j) - (4\hat i + 4\hat j) = -\hat i + \hat j; KB=12(0.1)(2)=0.1K_B = \frac12(0.1)(2) = 0.1 J ⇒ x = 1 (energy check: 1.7 J before and after).

Question 22

Physics · 4 marks

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz.

Correct answer: 106

Explanation: v=300/2=212.1v = 300/\sqrt2 = 212.1 m/s; open pipe fn=nv/2Lf_n = nv/2L; f2−f1=v/2L=106f_2 - f_1 = v/2L = 106 Hz.

Question 23

Physics · 4 marks

Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is ________ Ω.

Correct answer: 10

Explanation: Balance: 15×4=12×R′15\times4 = 12\times R' ⇒ R′=5R' = 5 Ω; 10∥X=510\parallel X = 5 ⇒ X = 10 Ω.

Question 24

Physics · 4 marks

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is ________.

Correct answer: 60

Explanation: 1f=(1.5−1)130\frac1f = (1.5 - 1)\frac{1}{30} ⇒ f = 60 cm.

Question 25

Chemistry · 4 marks · −1

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

Figure for question 25
  1. A.XY, 2×10−62\times10^{-6} M3^3
  2. B.XY2_2, 4×10−94\times10^{-9} M3^3
  3. C.X2_2Y, 2×10−92\times10^{-9} M3^3
  4. D.XY2_2, 1×10−91\times10^{-9} M3^3

Correct answer: B

Explanation: From the curve [Y] = 2 mM when [X] = 1 mM: Ksp=[X][Y]2=10−3(2×10−3)2=4×10−9K_{sp} = [X][Y]^2 = 10^{-3}(2\times10^{-3})^2 = 4\times10^{-9} M3^3.

Question 26

Chemistry · 4 marks · −1

The rate of a certain biochemical reaction at physiological temperature (T) occurs 10610^6 times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

  1. A.−6RT
  2. B.−6(2.303)RT
  3. C.+6RT
  4. D.+6(2.303)RT

Correct answer: B

Explanation: k2/k1=e−ΔEa/RT=106k_2/k_1 = e^{-\Delta E_a/RT} = 10^6 ⇒ ΔEa=−6(2.303)RT\Delta E_a = -6(2.303)RT.

Question 27

Chemistry · 4 marks · −1

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

  1. A.K3_3[Fe(CN)6_6] > AlCl3_3 > K2_2CrO4_4 > KBr > KNO3_3
  2. B.AlCl3_3 > K3_3[Fe(CN)6_6] > K2_2CrO4_4 > KBr = KNO3_3
  3. C.K3_3[Fe(CN)6_6] < K2_2CrO4_4 < AlCl3_3 < KBr < KNO3_3
  4. D.K3_3[Fe(CN)6_6] < K2_2CrO4_4 < KBr = KNO3_3 = AlCl3_3

Correct answer: D

Explanation: Positive sol: coagulating ions are anions; higher charge ⇒ lower flocculation value.

Question 28

Chemistry · 4 marks · −1

The predominant intermolecular forces present in ethyl acetate, a liquid, are:

  1. A.hydrogen bonding and London dispersion
  2. B.London dispersion, dipole-dipole and hydrogen bonding
  3. C.Dipole-dipole and hydrogen bonding
  4. D.London dispersion and dipole-dipole

Correct answer: D

Question 29

Chemistry · 4 marks · −1

For the Balmer series in the spectrum of H atom, νˉ=RH{1n12−1n22}\bar\nu = R_H\left\{\frac{1}{n_1^2} - \frac{1}{n_2^2}\right\}, the correct statements among (I) to (IV) are: (I) As wavelength decreases, the lines in the series converge (II) The integer n1n_1 is equal to 2 (III) The lines of longest wavelength corresponds to n2=3n_2 = 3 (IV) The ionization energy of hydrogen can be calculated from wave number of these lines

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

Correct answer: A

Question 30

Chemistry · 4 marks · −1

A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below: The following inferences are made: (A) X has higher intermolecular interactions compared to Y. (B) X has lower intermolecular interactions compared to Y. (C) Z has lower intermolecular interactions compared to Y. The correct inference(s) is/are:

Figure for question 30
  1. A.(A)
  2. B.(B)
  3. C.(C)
  4. D.(A) and (C)

Correct answer: B

Explanation: X reaches a given vapour pressure at the lowest temperature: most volatile, weakest interactions.

Question 31

Chemistry · 4 marks · −1

The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are:

  1. A.496, 577, 737, 786
  2. B.786, 737, 577, 496
  3. C.496, 737, 577, 786
  4. D.496, 577, 786, 737

Correct answer: C

Question 32

Chemistry · 4 marks · −1

The strength of an aqueous NaOH solution is most accurately determined by titrating: (Note: consider that an appropriate indicator is used)

  1. A.Aq. NaOH in a volumetric flask and concentrated H2_2SO4_4 in a conical flask
  2. B.Aq. NaOH in a burette and concentrated H2_2SO4_4 in a conical flask
  3. C.Aq. NaOH in a burette and aqueous oxalic acid in a conical flask
  4. D.Aq. NaOH in a pipette and aqueous oxalic acid in a burette

Correct answer: C

Question 33

Chemistry · 4 marks · −1

When gypsum is heated to 393 K, it forms:

  1. A.Anhydrous CaSO4_4
  2. B.Dead burnt plaster
  3. C.CaSO$_4\cdot0.5H0.5 H_2$O
  4. D.CaSO$_4\cdot5H5 H_2$O

Correct answer: C

Question 34

Chemistry · 4 marks · −1

The number of bonds between sulphur and oxygen atoms in S2_2O82−_8^{2-} and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are:

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

Correct answer: D

Question 35

Chemistry · 4 marks · −1

The third ionization enthalpy is minimum for:

  1. A.Ni
  2. B.Mn
  3. C.Fe
  4. D.Co

Correct answer: C

Question 36

Chemistry · 4 marks · −1

The complex that can show fac- and mer- isomers is:

  1. A.[Co(NH3_3)3_3(NO2_2)3_3]
  2. B.[CoCl2_2(en)2_2]
  3. C.[Co(NH3_3)4_4Cl2_2]+^+
  4. D.[Pt(NH3_3)2_2Cl2_2]

Correct answer: A

Question 37

Chemistry · 4 marks · −1

Among the gases (a) - (e), the gases that cause greenhouse effect are: (a) CO2_2 (b) H2_2O (c) CFCs (d) O2_2 (e) O3_3

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

Correct answer: C

Question 38

Chemistry · 4 marks · −1

The most suitable reagent for the given conversion is: a decalin-type bicyclic compound bearing CONH2_2, COCH3_3 (acetyl), CN and CO2_2H groups is converted into the same compound with the CO2_2H group reduced to CH2_2OH (other groups unchanged).

Figure for question 38
  1. A.LiAlH4_4
  2. B.B2_2H6_6
  3. C.H2_2/Pd
  4. D.NaBH4_4

Correct answer: B

Explanation: Diborane reduces –COOH to –CH2OH selectively, leaving amide, ketone (slow) and nitrile largely intact.

Question 39

Chemistry · 4 marks · −1

A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63 °C while the other boils at 60 °C. What is the best way to separate the two liquids and which one will be distilled out first?

  1. A.fractional distillation, isohexane
  2. B.fractional distillation, 3-methylpentane
  3. C.simple distillation, isohexane
  4. D.simple distillation, 3-methylpentane

Correct answer: A

Question 40

Chemistry · 4 marks · −1

Arrange the following compounds in increasing order of C−OH bond length: methanol, phenol, p-ethoxyphenol

  1. A.methanol < phenol < p-ethoxyphenol
  2. B.phenol < p-ethoxyphenol < methanol
  3. C.phenol < methanol < p-ethoxyphenol
  4. D.methanol < p-ethoxyphenol < phenol

Correct answer: B

Question 41

Chemistry · 4 marks · −1

Which of the following statement is not true for glucose?

  1. A.The pentaacetate of glucose does not react with hydroxylamine to give oxime
  2. B.Glucose reacts with hydroxylamine to form oxime
  3. C.Glucose gives Schiff's test for aldehyde
  4. D.Glucose exists in two crystalline forms α and β

Correct answer: C

Question 42

Chemistry · 4 marks · −1

The major products A and B in the following reactions are: (CH3_3)2_2CH–CN →Peroxide, Heat\xrightarrow{\text{Peroxide, Heat}} [A]; [A] + CH2_2=CH–CH2_2CH3_3 (but-1-ene) → B

Figure for question 42
  1. A.A = (CH3)2C•–CN; B = the radical adduct drawn with C(CH3)2CN attached at the internal carbon of the alkene
  2. B.A = (CH3)2C•–CN; B = zig-zag chain ending in C(CH3)2–CN on the other side (see image)
  3. C.A = (CH3)2C•–CN; B = CH3CH2CH2CH2–C(CH3)2–CN
  4. D.A = (CH3)2CH–CH•–CN type radical; B = corresponding chain nitrile (see image)

Correct answer: C

Explanation: Peroxide abstracts the tertiary H to give the stabilised (CH3)2C•CN radical, which adds to the terminal carbon of but-1-ene; H-abstraction then gives CH3CH2CH2CH2C(CH3)2CN.

Question 43

Chemistry · 4 marks · −1

The decreasing order of reactivity towards dehydrohalogenation (E1) reaction of the following compounds is: (A) 1-chlorobutane (B) 4-chlorobut-1-ene (C) 2-chlorobutane (D) 3-chlorobut-1-ene

Figure for question 43
  1. A.B > D > C > A
  2. B.D > B > C > A
  3. C.B > A > D > C
  4. D.B > D > A > C

Correct answer: B

Question 44

Chemistry · 4 marks · −1

The major product of the following reaction is: (CH3_3)2_2C=CH–CH2_2–CH2_2–C(CH3_3)=CH–CH2_2OH (geraniol) →dil. H2SO4\xrightarrow{\text{dil. H}_2\text{SO}_4}

Figure for question 44
  1. A.Open-chain triol (hydration of both C=C and retention of CH2OH)
  2. B.α-Terpineol: 1-methylcyclohexene ring bearing a C(CH3)2OH group
  3. C.Cyclopentane bearing CH3, a vinyl group and C(CH3)2OH
  4. D.Open-chain diol with a C(CH3)2OH terminus

Correct answer: B

Question 45

Chemistry · 4 marks

Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is ________. Atomic weight: Fe = 55.85; S = 32.00; O = 16.00

Correct answer: 4.98

Explanation: 10 ppm of 100 kg = 1 g Fe; FeSO$_4\cdot7H7H_2$O (M ≈ 278.0) needed: 278.0/55.85 ≈ 4.98 g.

Question 46

Chemistry · 4 marks

What would be the electrode potential for the given half cell reaction at pH = 5? ________. 2H2_2O → O2_2 + 4H+^+ + 4e−^-; Ered0E^0_{red} = 1.23 V (R = 8.314 J mol−1^{-1} K−1^{-1}; Temp = 298 K; oxygen under std. atm. pressure of 1 bar)

Correct answer: 0.93

Explanation: Ered=1.23−0.0591 pH=1.23−0.296=0.93E_{red} = 1.23 - 0.0591\,\mathrm{pH} = 1.23 - 0.296 = 0.93 V.

Question 47

Chemistry · 4 marks

The volume (in mL) of 0.125 M AgNO3_3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3_3)6_6]Cl3_3 is ________. M[Co(NH3)6]Cl3M_{[Co(NH_3)_6]Cl_3} = 267.46 g/mol, MAgNO3M_{AgNO_3} = 169.87 g/mol

Correct answer: 26.92

Explanation: Cl−^- = 3 × 0.3/267.46 = 3.365 mmol; V = 3.365/0.125 = 26.92 mL.

Question 48

Chemistry · 4 marks

The number of chiral centres in penicillin is ________.

Correct answer: 3

Explanation: The penam core carries three stereocentres (C2, C5, C6).

Question 49

Mathematics · 4 marks · −1

The inverse function of f(x)=82x−8−2x82x+8−2xf(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}, x∈(−1,1)x \in (-1, 1), is ________.

  1. A.14log⁡e(1+x1−x)\frac14\log_e\left(\frac{1+x}{1-x}\right)
  2. B.14(log⁡8e)log⁡e(1−x1+x)\frac14(\log_8 e)\log_e\left(\frac{1-x}{1+x}\right)
  3. C.14log⁡e(1−x1+x)\frac14\log_e\left(\frac{1-x}{1+x}\right)
  4. D.14(log⁡8e)log⁡e(1+x1−x)\frac14(\log_8 e)\log_e\left(\frac{1+x}{1-x}\right)

Correct answer: D

Explanation: 84x=1+y1−y8^{4x} = \frac{1+y}{1-y} ⇒ x=14log⁡81+y1−yx = \frac14\log_8\frac{1+y}{1-y}.

Question 50

Mathematics · 4 marks · −1

If the equation, x2+bx+45=0x^2 + bx + 45 = 0 (b∈R)(b \in \mathbb{R}) has conjugate complex roots and they satisfy ∣z+1∣=210|z + 1| = 2\sqrt{10}, then:

  1. A.b2+b=12b^2 + b = 12
  2. B.b2−b=42b^2 - b = 42
  3. C.b2−b=30b^2 - b = 30
  4. D.b2+b=72b^2 + b = 72

Correct answer: C

Explanation: ∣z∣2=45|z|^2 = 45, ∣z+1∣2=40|z+1|^2 = 40 ⇒ Re z=−3\mathrm{Re}\,z = -3 ⇒ b = 6 ⇒ b2−b=30b^2 - b = 30.

Question 51

Mathematics · 4 marks · −1

For which of the following ordered pairs (μ, δ), the system of linear equations x+2y+3z=1x + 2y + 3z = 1, 3x+4y+5z=μ3x + 4y + 5z = \mu, 4x+4y+4z=δ4x + 4y + 4z = \delta is inconsistent?

  1. A.(1, 0)
  2. B.(3, 4)
  3. C.(4, 3)
  4. D.(4, 6)

Correct answer: C

Explanation: det = 0; 2E2−2E1=4x+4y+4z=2μ−22E_2 - 2E_1 = 4x + 4y + 4z = 2\mu - 2, so consistent iff δ = 2μ − 2; (4, 3) violates it.

Question 52

Mathematics · 4 marks · −1

Let two points be A(1, −1) and B(0, 2). If a point P(x', y') be such that the area of ΔPAB = 5 sq. units and it lies on the line, 3x+y−4λ=03x + y - 4\lambda = 0, then a value of λ is:

  1. A.1
  2. B.4
  3. C.−3
  4. D.3

Correct answer: D

Explanation: Area =12∣3x′+y′−2∣=5= \frac12|3x' + y' - 2| = 5 ⇒ 3x′+y′=123x' + y' = 12 or −8-8 ⇒ 4λ=124\lambda = 12 ⇒ λ = 3.

Question 53

Mathematics · 4 marks · −1

If a, b and c are the greatest values of 19Cp{}^{19}C_p, 20Cq{}^{20}C_q and 21Cr{}^{21}C_r respectively, then:

  1. A.a11=b22=c21\frac{a}{11} = \frac{b}{22} = \frac{c}{21}
  2. B.a10=b11=c42\frac{a}{10} = \frac{b}{11} = \frac{c}{42}
  3. C.a10=b11=c21\frac{a}{10} = \frac{b}{11} = \frac{c}{21}
  4. D.a11=b22=c42\frac{a}{11} = \frac{b}{22} = \frac{c}{42}

Correct answer: D

Explanation: a=19C9a = {}^{19}C_9, b=20C10=2ab = {}^{20}C_{10} = 2a, c=21C10=2111b=4211ac = {}^{21}C_{10} = \frac{21}{11}b = \frac{42}{11}a.

Question 54

Mathematics · 4 marks · −1

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be such that for all x∈Rx \in \mathbb{R} (21+x+21−x)(2^{1+x} + 2^{1-x}), f(x)f(x) and (3x+3−x)(3^x + 3^{-x}) are in A.P., then the minimum value of f(x) is:

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

Correct answer: C

Explanation: 2f=2(2x+2−x)+(3x+3−x)≥4+22f = 2(2^x + 2^{-x}) + (3^x + 3^{-x}) \ge 4 + 2 ⇒ f≥3f \ge 3.

Question 55

Mathematics · 4 marks · −1

lim⁡x→0(3x2+27x2+2)1/x2\lim_{x\to0}\left(\frac{3x^2+2}{7x^2+2}\right)^{1/x^2} is equal to:

  1. A.e
  2. B.e²
  3. C.1e\frac1e
  4. D.1e2\frac{1}{e^2}

Correct answer: D

Explanation: exp⁡lim⁡−4x2(7x2+2)x2=e−2\exp\lim\frac{-4x^2}{(7x^2+2)x^2} = e^{-2}.

Question 56

Mathematics · 4 marks · −1

Let f(x)=xcos⁡−1(−sin⁡∣x∣)f(x) = x\cos^{-1}(-\sin|x|), x∈[−π2,π2]x \in \left[-\frac\pi2, \frac\pi2\right], then which of the following is true?

  1. A.f is not differentiable at x = 0
  2. B.f′(0)=−π2f'(0) = -\frac\pi2
  3. C.f' is decreasing in (−π2,0)\left(-\frac\pi2, 0\right) and increasing in (0,π2)\left(0, \frac\pi2\right)
  4. D.f' is increasing in (−π2,0)\left(-\frac\pi2, 0\right) and decreasing in (0,π2)\left(0, \frac\pi2\right)

Correct answer: C

Explanation: f=x(π2+∣x∣)f = x\left(\frac\pi2 + |x|\right): f′=π2+2∣x∣f' = \frac\pi2 + 2|x|, decreasing then increasing.

Question 57

Mathematics · 4 marks · −1

If c is a point at which Rolle's theorem holds for the function, f(x)=log⁡e(x2+α7x)f(x) = \log_e\left(\frac{x^2+\alpha}{7x}\right) in the interval [3, 4], where α∈R\alpha \in \mathbb{R}, then f''(c) is equal to:

  1. A.37\frac{\sqrt3}{7}
  2. B.112\frac{1}{12}
  3. C.−124-\frac{1}{24}
  4. D.−112-\frac{1}{12}

Correct answer: B

Explanation: f(3) = f(4) ⇒ α = 12; f'(c) = 0 ⇒ c2=12c^2 = 12; f′′(c)=24−2c2(c2+12)2+1c2=112f''(c) = \frac{24 - 2c^2}{(c^2+12)^2} + \frac{1}{c^2} = \frac{1}{12}.

Question 58

Mathematics · 4 marks · −1

If ∫cos⁡x dxsin⁡3x(1+sin⁡6x)2/3=f(x)(1+sin⁡6x)1/λ+c\int\frac{\cos x\,dx}{\sin^3x(1 + \sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\lambda} + c where c is a constant of integration, then λf(π3)\lambda f\left(\frac\pi3\right) is equal to:

  1. A.98\frac98
  2. B.−98-\frac98
  3. C.2
  4. D.−2

Correct answer: D

Explanation: With t=sin⁡xt = \sin x: ∫t−7(1+t−6)−2/3dt=−12(1+t−6)1/3\int t^{-7}(1 + t^{-6})^{-2/3}dt = -\frac12(1 + t^{-6})^{1/3} ⇒ f=−12sin⁡2xf = -\frac{1}{2\sin^2x}, λ = 3 ⇒ 3×(−23)=−23\times\left(-\frac23\right) = -2.

Question 59

Mathematics · 4 marks · −1

For a > 0, let the curves C1:y2=axC_1: y^2 = ax and C2:x2=ayC_2: x^2 = ay intersect at origin O and a point P. Let the line x = b (0 < b < a) intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, C1C_1 and C2C_2, and the area of ΔOQR=12\Delta OQR = \frac12, then 'a' satisfies the equation:

  1. A.x6+6x3−4=0x^6 + 6x^3 - 4 = 0
  2. B.x6−12x3+4=0x^6 - 12x^3 + 4 = 0
  3. C.x6−12x3−4=0x^6 - 12x^3 - 4 = 0
  4. D.x6−6x3+4=0x^6 - 6x^3 + 4 = 0

Correct answer: B

Question 60

Mathematics · 4 marks · −1

Let y = y(x) be a solution of the differential equation, 1−x2dydx+1−y2=0\sqrt{1-x^2}\frac{dy}{dx} + \sqrt{1-y^2} = 0, ∣x∣<1|x| < 1. If y(12)=32y\left(\frac12\right) = \frac{\sqrt3}{2}, then y(−12)y\left(\frac{-1}{\sqrt2}\right) is equal to:

  1. A.−12-\frac{1}{\sqrt2}
  2. B.12\frac{1}{\sqrt2}
  3. C.32\frac{\sqrt3}{2}
  4. D.−32-\frac{\sqrt3}{2}

Correct answer: B

Explanation: sin⁡−1y+sin⁡−1x=π2\sin^{-1}y + \sin^{-1}x = \frac\pi2 ⇒ y=1−x2y = \sqrt{1-x^2} ⇒ y=12y = \frac{1}{\sqrt2}.

Question 61

Mathematics · 4 marks · −1

The locus of a point which divides the line segment joining the point (0, −1) and a point on the parabola, x2=4yx^2 = 4y, internally in the ratio 1 : 2, is:

  1. A.4x2−3y=24x^2 - 3y = 2
  2. B.x2−3y=2x^2 - 3y = 2
  3. C.9x2−12y=89x^2 - 12y = 8
  4. D.9x2−3y=29x^2 - 3y = 2

Correct answer: C

Explanation: Point (2t3,t2−23)\left(\frac{2t}{3}, \frac{t^2 - 2}{3}\right) ⇒ 9x2−12y=89x^2 - 12y = 8.

Question 62

Mathematics · 4 marks · −1

Let the line y = mx and the ellipse 2x2+y2=12x^2 + y^2 = 1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at (−132,0)\left(-\frac{1}{3\sqrt2}, 0\right) and (0, β), then β is equal to:

  1. A.23\frac{\sqrt2}{3}
  2. B.23\frac23
  3. C.23\frac{2}{\sqrt3}
  4. D.223\frac{2\sqrt2}{3}

Correct answer: A

Question 63

Mathematics · 4 marks · −1

The shortest distance between the lines x−33=y−8−1=z−31\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} and x+3−3=y+72=z−64\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} is:

  1. A.2302\sqrt{30}
  2. B.3303\sqrt{30}
  3. C.3
  4. D.7230\frac72\sqrt{30}

Correct answer: B

Explanation: b⃗1×b⃗2=(−6,−15,3)\vec b_1\times\vec b_2 = (-6, -15, 3), ∣⋅∣=330|\cdot| = 3\sqrt{30}; (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−6)(−6)+(−15)(−15)+3(3)=270(\vec a_2 - \vec a_1)\cdot(\vec b_1\times\vec b_2) = (-6)(-6) + (-15)(-15) + 3(3) = 270; SD =270/330=330= 270/3\sqrt{30} = 3\sqrt{30}.

Question 64

Mathematics · 4 marks · −1

Let the volume of a parallelopiped whose coterminous edges are given by u⃗=i^+j^+λk^\vec u = \hat i + \hat j + \lambda\hat k, v⃗=i^+j^+3k^\vec v = \hat i + \hat j + 3\hat k and w⃗=2i^+j^+k^\vec w = 2\hat i + \hat j + \hat k be 1 cu. unit. If θ be the angle between the edges u⃗\vec u and w⃗\vec w, then cos θ can be:

  1. A.57\frac57
  2. B.766\frac{7}{6\sqrt6}
  3. C.763\frac{7}{6\sqrt3}
  4. D.533\frac{5}{3\sqrt3}

Correct answer: C

Explanation: [u⃗ v⃗ w⃗]=λ−3=±1[\vec u\,\vec v\,\vec w] = \lambda - 3 = \pm1 ⇒ λ = 4 or 2; λ = 4: cos⁡θ=7186=763\cos\theta = \frac{7}{\sqrt{18}\sqrt6} = \frac{7}{6\sqrt3}.

Question 65

Mathematics · 4 marks · −1

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p ≠ 0 and q ≠ 0. If the new mean and new s.d. become half of their original values, then q is equal to:

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

Correct answer: D

Explanation: ∣p∣=12|p| = \frac12; 20p−q=1020p - q = 10; p = 1/2 gives q = 0 (excluded), so p = −1/2, q = −20.

Question 66

Mathematics · 4 marks · −1

Let A and B be two independent events such that P(A)=13P(A) = \frac13 and P(B)=16P(B) = \frac16. Then, which of the following is TRUE?

  1. A.P(A/(A∪B))=14P(A/(A\cup B)) = \frac14
  2. B.P(A/B)=23P(A/B) = \frac23
  3. C.P(A/B′)=13P(A/B') = \frac13
  4. D.P(A′/B′)=13P(A'/B') = \frac13

Correct answer: C

Explanation: Independence: P(A|B') = P(A) = 1/3.

Question 67

Mathematics · 4 marks · −1

Let f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1f(x) = (\sin(\tan^{-1}x) + \sin(\cot^{-1}x))^2 - 1, ∣x∣>1|x| > 1. If dydx=12ddx(sin⁡−1(f(x)))\frac{dy}{dx} = \frac12\frac{d}{dx}(\sin^{-1}(f(x))) and y(3)=π6y(\sqrt3) = \frac\pi6, then y(−3)y(-\sqrt3) is equal to:

  1. A.−π6-\frac\pi6
  2. B.2π3\frac{2\pi}{3}
  3. C.5π6\frac{5\pi}{6}
  4. D.π3\frac\pi3

Correct answer: C

Question 68

Mathematics · 4 marks · −1

Which one of the following is a tautology?

  1. A.P∧(P∨Q)P \wedge (P \vee Q)
  2. B.P∨(P∧Q)P \vee (P \wedge Q)
  3. C.(P∧(P→Q))→Q(P \wedge (P \to Q)) \to Q
  4. D.Q→(P∧(P→Q))Q \to (P \wedge (P \to Q))

Correct answer: C

Question 69

Mathematics · 4 marks

The least positive value of 'a' for which the equation, 2x2+(a−10)x+332=2a2x^2 + (a - 10)x + \frac{33}{2} = 2a has real roots is ________.

Correct answer: 8

Explanation: D=(a−10)2−8(332−2a)=a2−4a−32≥0D = (a-10)^2 - 8\left(\frac{33}{2} - 2a\right) = a^2 - 4a - 32 \ge 0 ⇒ a ≥ 8 (for a > 0).

Question 70

Mathematics · 4 marks

The number of all 3 × 3 matrices A, with enteries from the set {−1, 0, 1} such that the sum of the diagonal elements of AATAA^T is 3, is ________.

Correct answer: 672

Explanation: tr(AAT)=∑aij2=3(AA^T) = \sum a_{ij}^2 = 3 ⇒ exactly three non-zero entries: 9C3×23=672{}^9C_3\times2^3 = 672.

Question 71

Mathematics · 4 marks

The sum ∑k=120(1+2+3+…+k)\sum_{k=1}^{20}(1 + 2 + 3 + \ldots + k) is ________.

Correct answer: 1540

Explanation: ∑k(k+1)2=20⋅21⋅226=1540\sum\frac{k(k+1)}{2} = \frac{20\cdot21\cdot22}{6} = 1540.

Question 72

Mathematics · 4 marks

Let the normal at a point P on the curve y2−3x2+y+10=0y^2 - 3x^2 + y + 10 = 0 intersect the y-axis at (0,32)\left(0, \frac32\right). If m is the slope of the tangent at P to the curve, then |m| is equal to ________.

Correct answer: 4

Explanation: m=6x2y+1m = \frac{6x}{2y+1}; normal meets y-axis at y0+2y0+16=32y_0 + \frac{2y_0+1}{6} = \frac32 ⇒ y0=1y_0 = 1, x02=4x_0^2 = 4 ⇒ |m| = 12/3 = 4.

Question 73

Mathematics · 4 marks

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ________.

Correct answer: 490

Explanation: 12C4−5C4=495−5=490{}^{12}C_4 - {}^5C_4 = 495 - 5 = 490.

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