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JEE Main 8 January 2020 Shift 2 Question Paper with Answers

8 January 2020 · January session · 69 questions

69 of the 75 questions from the JEE Main 8 January 2020 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

6 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
24
Chemistry
22
Mathematics
23

Physics — JEE Main 8 January 2020 Shift 2

Q1·PhysicsSingle correct
A simple pendulum is being used to determine the vale of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:
  1. (A)4.40%
  2. (B)2.40%
  3. (C)3.40%
  4. (D)5.40%

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
A Carnot engine having an efficiency of 110\frac{1}{10}101​ is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:
  1. (A)99 J
  2. (B)100 J
  3. (C)1 J
  4. (D)90 J

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
Consider two charged metallic spheres S1_{1}1​ and S2_{2}2​ of radii R1_{1}1​ and R2_{2}2​, respectively. The electric fields E1_{1}1​ (on S1_{1}1​) and E2_{2}2​ (on S2_{2}2​) on their surfaces are such that E1_{1}1​/E2_{2}2​ = R1_{1}1​/R2_{2}2​. Then the ratio of V1_{1}1​ (on S1_{1}1​) / V2_{2}2​ (on S2_{2}2​) of the electrostatic potentials on each sphere is:
  1. (A)(R1/R2)2(R_{1}/R_{2})^{2}(R1​/R2​)2
  2. (B)(R1R2)3\left(\dfrac{R_{1}}{R_{2}}\right)^{3}(R2​R1​​)3
  3. (C)(R2/R1)(R_{2}/R_{1})(R2​/R1​)
  4. (D)R1/R2R_{1}/R_{2}R1​/R2​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
As shown in the figure, a battery of emf ε\varepsilonε is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc_{c}c​ (tc_{c}c​ is the time constant of the circuit) is
  1. (A)εLeR2\dfrac{\varepsilon L}{eR^{2}}eR2εL​
  2. (B)εLR2(1−1e)\dfrac{\varepsilon L}{R^{2}}\left(1-\dfrac{1}{e}\right)R2εL​(1−e1​)
  3. (C)εReL2\dfrac{\varepsilon R}{eL^{2}}eL2εR​
  4. (D)εLR2\dfrac{\varepsilon L}{R^{2}}R2εL​

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
A very long wire ABADMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is:
  1. (A)μ0I2R\dfrac{\mu_{0} I}{2R}2Rμ0​I​
  2. (B)μ0I2πR(π−12)\dfrac{\mu_{0} I}{2\pi R}\left(\pi-\dfrac{1}{\sqrt{2}}\right)2πRμ0​I​(π−2​1​)
  3. (C)μ0I2πR(π+1)\dfrac{\mu_{0} I}{2\pi R}(\pi+1)2πRμ0​I​(π+1)
  4. (D)μ0I2πR(π+12)\dfrac{\mu_{0} I}{2\pi R}\left(\pi+\dfrac{1}{\sqrt{2}}\right)2πRμ0​I​(π+2​1​)

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:
  1. (A)8.75×10−48.75\times 10^{-4}8.75×10−4 J
  2. (B)8.75×10−38.75\times 10^{-3}8.75×10−3 J
  3. (C)6.25×10−46.25\times 10^{-4}6.25×10−4 J
  4. (D)1.13×10−31.13\times 10^{-3}1.13×10−3 J

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves is 18\dfrac{1}{8}81​th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:
  1. (A)0.672
  2. (B)0.568
  3. (C)0.760
  4. (D)0.853

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
A particle of mass m is dropped from a height h above the ground. At the same time another particle of same mass is thrown vertically upwards from the ground with a speed of 2gh\sqrt{2gh}2gh​. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of hg\sqrt{\dfrac{h}{g}}gh​​ is:
  1. (A)12\dfrac{1}{2}21​
  2. (B)12\sqrt{\dfrac{1}{2}}21​​
  3. (C)34\sqrt{\dfrac{3}{4}}43​​
  4. (D)32\sqrt{\dfrac{3}{2}}23​​

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
A particle moves such that its position vector r⃗(t)=cos⁡ωi^+sin⁡ωtj^\vec{r}(t)=\cos\omega \hat{i}+\sin\omega t\hat{j}r(t)=cosωi^+sinωtj^​ where ω\omegaω is a constant and t is time. Then which of the following statements is true for the velocity v⃗(t)\vec{v}(t)v(t) and acceleration a⃗(t)\vec{a}(t)a(t) of the particle:
  1. (A)v⃗\vec{v}v is perpendicular to r⃗\vec{r}r and a⃗\vec{a}a is directed towards the origin.
  2. (B)v⃗\vec{v}v and a⃗\vec{a}a both are parallel to r⃗\vec{r}r
  3. (C)v⃗\vec{v}v is perpendicular to r⃗\vec{r}r and a⃗\vec{a}a is directed away from the origin.
  4. (D)v⃗\vec{v}v and a⃗\vec{a}a both are perpendicular to r⃗\vec{r}r

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. It CP_{P}P​/CV_{V}V​ value will be
  1. (A)19/13
  2. (B)40/27
  3. (C)67/45
  4. (D)23/15

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A capacitor is made of two square plats each of side 'a' making a very small angle a between them, as shown in figure. The capacitance will be close to:
  1. (A)ϵ0a2d(1+αad)\dfrac{\epsilon_{0} a^{2}}{d}\left(1+\dfrac{\alpha a}{d}\right)dϵ0​a2​(1+dαa​)
  2. (B)ϵ0a2d(1−αa4d)\dfrac{\epsilon_{0} a^{2}}{d}\left(1-\dfrac{\alpha a}{4d}\right)dϵ0​a2​(1−4dαa​)
  3. (C)ϵ0a2d(1−3αa2d)\dfrac{\epsilon_{0} a^{2}}{d}\left(1-\dfrac{3\alpha a}{2d}\right)dϵ0​a2​(1−2d3αa​)
  4. (D)ϵ0a2d(1−αa2d)\dfrac{\epsilon_{0} a^{2}}{d}\left(1-\dfrac{\alpha a}{2d}\right)dϵ0​a2​(1−2dαa​)

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
Two liquids of densities ρ1\rho_{1}ρ1​ and ρ2\rho_{2}ρ2​(ρ2\rho_{2}ρ2​ = 2ρ1\rho_{1}ρ1​) are filled up behind a square wall of inside 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing):
  1. (A)1/2
  2. (B)1/4
  3. (C)2/3
  4. (D)1/3

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by B⃗=5×10−8j^\vec{B}=5\times 10^{-8}\hat{j}B=5×10−8j^​ T. The corresponding electric field E⃗\vec{E}E is (speed of light c = 3×1083\times 10^{8}3×108 ms−1^{-1}−1)
  1. (A)−1.66×10−16i^-1.66\times 10^{-16}\hat{i}−1.66×10−16i^ V/m
  2. (B)−15i^-15\hat{i}−15i^ V/m
  3. (C)15i^15\hat{i}15i^ V/m
  4. (D)1.66×10−16i^1.66\times 10^{-16}\hat{i}1.66×10−16i^ V/m

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
An electron (mass m) with initial velocity v⃗=v0i^+v0j^\vec{v} = v_0\hat{i} + v_0\hat{j}v=v0​i^+v0​j^​ is in an electric field E⃗=−E0k^\vec{E} = -E_0\hat{k}E=−E0​k^. If λ0\lambda_0λ0​ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by
  1. (A)λ01+e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^{2}E_0^{2}t^{2}}{m^{2}v_0^{2}}}}1+m2v02​e2E02​t2​​λ0​​
  2. (B)λ02+e2E2t2m2v02\dfrac{\lambda_0}{\sqrt{2+\dfrac{e^{2}E^{2}t^{2}}{m^{2}v_0^{2}}}}2+m2v02​e2E2t2​​λ0​​
  3. (C)λ021+e2E2t2m2v02\dfrac{\lambda_0\sqrt{2}}{\sqrt{1+\dfrac{e^{2}E^{2}t^{2}}{m^{2}v_0^{2}}}}1+m2v02​e2E2t2​​λ0​2​​
  4. (D)λ01+e2E2t22m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^{2}E^{2}t^{2}}{2m^{2}v_0^{2}}}}1+2m2v02​e2E2t2​​λ0​​

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
A transverse wave travels on a taut steel wire wit a velocity of v when tension in it is 2.06 ×\times× 104^{4}4 N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to:
  1. (A)10.2 ×\times× 102^{2}2 N
  2. (B)30.5 ×\times× 104^{4}4 N
  3. (C)5.15 ×\times× 103^{3}3 N
  4. (D)2.50 ×\times× 104^{4}4 N

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification (m) versus distance of the object from the mirror (x) is correctly given by (Graphs are drawn schematically and are not to scale)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
As shown in figure when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e. on the surface of the cavity. R can be determined by the equation:
  1. (A)(R2^{2}2 + R + 1) (2 −-− R) = 1
  2. (B)(R2^{2}2 + R −-− 1) (2 −-− R) = 1
  3. (C)(R2^{2}2 −-− R + 1) (2 −-− R) = 1
  4. (D)(R2^{2}2 −-− R −-− 1) (2 −-− R) = 1

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correct
In the given circuit, value of Y is
  1. (A)0
  2. (B)1
  3. (C)toggles between 0 and 1
  4. (D)will not execute

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V.
  1. (A)9.9 kΩ
  2. (B)7.9 kΩ
  3. (C)10 kΩ
  4. (D)8.9 kΩ

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is ____.

Correct answer: 486

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Q22·PhysicsNumerical
The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20 Ω and 5 Ω, is connected to the parallel combination of two resistors 30 Ω and R Ω. The voltage difference across the battery of internal resistance 20 Ω is zero, the value of R (in Ω) is ____.

Correct answer: 30

Step-by-step solution →
Q23·PhysicsNumerical
An asteroid is moving directly towards the centre of the earth. When at a distance of 10 R (R is the radius of the earth) from he earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity form the earth is 11.2 km/s)? Given your answer to the nearest integer in kilometer/s ____.

Correct answer: 16

Step-by-step solution →
Q24·PhysicsNumerical
Three containers C1_11​, C2_22​ and C3_33​ have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process) The value of θ\thetaθ (in °C to the nearest integer) is ____.
C1_11​C2_22​C3_33​T
1 l2 l—60°C
—1 l2 l30°C
2 l—1 l60°C
1 l1 l1 lθ\thetaθ

Correct answer: 50

Step-by-step solution →

Chemistry — JEE Main 8 January 2020 Shift 2

Q25·ChemistrySingle correct
The increasing order of the atomic radii of the following elements is: (a) C (b) O (c) F (d) Cl (e) Br
  1. (A)(a) < (b) < (c) < (d) < (e)
  2. (B)(d) < (c) < (b) < (a) < (e)
  3. (C)(c) < (b) < (a) < (d) < (e)
  4. (D)(b) < (c) < (d) < (a) < (e)

Correct answer: (C)

Step-by-step solution →
Q26·ChemistrySingle correct
Kjeldahl's method cannot be used to estimate nitrogen for which of the following compounds?
  1. (A)NH2_22​-C(=O)-NH2_22​
  2. (B)C6_66​H5_55​NH2_22​
  3. (C)C6_66​H5_55​NO2_22​
  4. (D)CH3_33​CH2_22​-C≡\equiv≡N

Correct answer: (C)

Step-by-step solution →
Q27·ChemistrySingle correct
Consider the following plots of rate constant versus 1T\dfrac{1}{T}T1​ for four different reactions. Which of the following order is correct for the activation energies of these reactions?
  1. (A)Ec_cc​ > Ea_aa​ > Ed_dd​ > Eb_bb​
  2. (B)Eb_bb​ > Ed_dd​ > Ec_cc​ > Ea_aa​
  3. (C)Ea_aa​ > Ec_cc​ > Ed_dd​ > Eb_bb​
  4. (D)Eb_bb​ > Ea_aa​ > Ed_dd​ > Ec_cc​

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of CO2_22​ gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is:
  1. (A)2
  2. (B)4
  3. (C)3
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q29·ChemistrySingle correct
The major product [B] in the following sequence is: CH3_33​-C(CH(CH3_33​)2_22​)=CH-CH2_22​CH3_33​ →(i) B2H6; (ii) H2O2, OH−\xrightarrow{\text{(i) B}_2\text{H}_6\text{; (ii) H}_2\text{O}_2\text{, OH}^-}(i) B2​H6​; (ii) H2​O2​, OH−​ [A] →Δdil. H2SO4\xrightarrow[\Delta]{\text{dil. H}_2\text{SO}_4}dil. H2​SO4​Δ​ [B]
  1. (A)CH3_33​-CH(CH(CH3_33​)2_22​)-CH=CH-CH3_33​
  2. (B)CH3_33​-C(CH(CH3_33​)2_22​)=CH-CH2_22​CH3_33​
  3. (C)CH2_22​=C(CH(CH3_33​)2_22​)-CH2_22​CH2_22​CH3_33​
  4. (D)CH3_33​-C(CH2_22​CH2_22​CH3_33​)=C(CH3_33​)2_22​

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
Among (a) - (d), the complexes that can display geometrical isomerism are: (a) [Pt(NH3_33​)3_33​]Cl+^++ (b) [Pt(NH3_33​)Cl5_55​]−^-− (c) [Pt(NH3_33​)2_22​Cl(NO2_22​)] (d) [Pt(NH3_33​)4_44​ClBr]2+^{2+}2+
  1. (A)(d) and (a)
  2. (B)(c) and (d)
  3. (C)(b) and (c)
  4. (D)(a) and (b)

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
The major product in the following reaction is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Among the reactions (a)-(d), the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are: (a) CaO + SiO2_22​ →\rightarrow→ CaSiO3_33​ (b) 3Fe2_22​O3_33​ + CO →\rightarrow→ 2Fe3_33​O4_44​ + CO2_22​ (c) FeO + SiO2_22​ →\rightarrow→ FeSiO3_33​ (d) FeO →\rightarrow→ Fe + 12\dfrac{1}{2}21​O2_22​
  1. (A)(c) and (d)
  2. (B)(a)
  3. (C)(a) and (d)
  4. (D)(d)

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
A metal (A) on heating in nitrogen gas gives compound B. B on treatment with H2_22​O gives a colourless gas which when passed through CuSO4_44​ solution gives a dark blue-violet coloured solution. A and B respectively, are:
  1. (A)Mg and Mg(NO3_33​)2_22​
  2. (B)Na and NaNO3_33​
  3. (C)Mg and Mg3_33​N2_22​
  4. (D)Na and Na3_33​N

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Two monomers in maltose are:
  1. (A)α\alphaα-D-glucose and α\alphaα-D-galactose
  2. (B)α\alphaα-D-glucose and β\betaβ-D-glucose
  3. (C)α\alphaα-D-glucose and α\alphaα-D-glucose
  4. (D)α\alphaα-D-glucose and α\alphaα-D-Fructose

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (a) Ni(CO)4_44​ (b) [Ni(H2_22​O)6_66​]Cl2_22​ (c) Na2_22​[Ni(CN)4_44​] (d) PdCl2_22​(PPh3_33​)2_22​
  1. (A)(c) ≈\approx≈ (d) < (b) < (a)
  2. (B)(a) ≈\approx≈ (c) < (b) ≈\approx≈ (d)
  3. (C)(a) ≈\approx≈ (c) ≈\approx≈ (d) < (b)
  4. (D)(c) < (d) < (b) < (a)

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reaction: X →Zn/H2OO3\xrightarrow[\text{Zn/H}_2\text{O}]{\text{O}_3}O3​Zn/H2​O​ A →[Ag(NH3)2]+\xrightarrow{[\text{Ag(NH}_3)_2]^+}[Ag(NH3​)2​]+​ B(3-oxo-hexanedicarboxylic acid) X will be :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Hydrogen has three isotopes (a), (b) and (c). If the number of neutron(s) in (a), (b) and (c) respectively, are (x), (y) and (z), the sum of (x), (y) and (z) is:
  1. (A)4
  2. (B)3
  3. (C)2
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
Which of the following compound is likely to show both Frenkel and Schottky defects in its crystalline form?
  1. (A)CsCl
  2. (B)AgBr
  3. (C)ZnS
  4. (D)KBr

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Arrange the following bonds according to their average bond energies in descending order: C–Cl, C–Br, C–F, C–I
  1. (A)C–F > C–Cl > C–Br > C–I
  2. (B)C–I > C–Br > C–Cl > C–F
  3. (C)C–Cl > C–Br > C–I > C–F
  4. (D)C–Br > C–I > C–Cl > C–F

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Preparation of Bakelite proceeds via reactions:
  1. (A)Electrophilic substitution and dehydration
  2. (B)Condensation and elimination
  3. (C)Electrophilic addition and dehydration
  4. (D)Nucleophilic addition and dehydration

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
The radius of the second Bohr orbit, in terms of the Bohr radius, a0a_0a0​, in Li2+Li^{2+}Li2+ is
  1. (A)4a03\dfrac{4a_0}{3}34a0​​
  2. (B)2a09\dfrac{2a_0}{9}92a0​​
  3. (C)2a03\dfrac{2a_0}{3}32a0​​
  4. (D)4a09\dfrac{4a_0}{9}94a0​​

Correct answer: (A)

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Q42·ChemistrySingle correct
Among the compounds A and B with molecular formula C9H18O3C_9H_{18}O_3C9​H18​O3​, A is having higher boiling point the B. The possible structures of A and B are:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q43·ChemistryNumerical
Complexes (ML5ML_5ML5​) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the 90∘90^{\circ}90∘, 120∘120^{\circ}120∘ and 180∘180^{\circ}180∘ L-M-L angles in the two complexes is ____

Correct answer: 20.00

Step-by-step solution →
Q44·ChemistryNumerical
At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ____

Correct answer: 6.25

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Q45·ChemistryNumerical
In the following sequence of reactions the maximum number of atoms present in molecule 'C' in one plane is A →Cu tubeRed hot\xrightarrow[\text{Cu tube}]{\text{Red hot}}Red hotCu tube​ B →Anhydrous AlCl3CH3Cl(1.eq.)\xrightarrow[\text{Anhydrous AlCl}_3]{\text{CH}_3\text{Cl(1.eq.)}}CH3​Cl(1.eq.)Anhydrous AlCl3​​ C (A is a lowest molecular weight alkyne)

Correct answer: 13.00

Step-by-step solution →
Q46·ChemistryNumerical
For an electrochemical cell Sn(s)∣Sn2+(aq,1M)∣∣Pb2+(aq,1M)∣Pb(s)Sn(s)|Sn^{2+}(aq, 1M)||Pb^{2+}(aq, 1M)|Pb(s)Sn(s)∣Sn2+(aq,1M)∣∣Pb2+(aq,1M)∣Pb(s) the ratio [Sn2+][Pb2+]\dfrac{[Sn^{2+}]}{[Pb^{2+}]}[Pb2+][Sn2+]​ when this cell attains equilibrium is ____. (Given: ESn2+∣Sn0=−0.14V,EPb2+∣Pb0=−0.13V,2.303RTF=0.06E^0_{Sn^{2+}|Sn} = -0.14V, E^0_{Pb^{2+}|Pb} = -0.13V, \dfrac{2.303RT}{F} = 0.06ESn2+∣Sn0​=−0.14V,EPb2+∣Pb0​=−0.13V,F2.303RT​=0.06)

Correct answer: 2.15

Step-by-step solution →

Mathematics — JEE Main 8 January 2020 Shift 2

Q47·MathematicsSingle correct
If the 10th10^{th}10th term of an A.P. is 120\dfrac{1}{20}201​ and its 20th20^{th}20th term is 110\dfrac{1}{10}101​, then the sum of its first 200 terms is:
  1. (A)100
  2. (B)501450\dfrac{1}{4}5041​
  3. (C)10012100\dfrac{1}{2}10021​
  4. (D)50

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsSingle correct
Which of the following statement is a tautology?
  1. (A)∼(p∧∼q)→p∨q\sim(p \wedge \sim q) \to p \vee q∼(p∧∼q)→p∨q
  2. (B)∼(p∨∼q)→p∨q\sim(p \vee \sim q) \to p \vee q∼(p∨∼q)→p∨q
  3. (C)p∨(∼q)→p∧qp \vee (\sim q) \to p \wedge qp∨(∼q)→p∧q
  4. (D)∼(p∨∼q)→p∧q\sim(p \vee \sim q) \to p \wedge q∼(p∨∼q)→p∧q

Correct answer: (B)

Step-by-step solution →
Q49·MathematicsSingle correct
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:
  1. (A)4.02
  2. (B)3.98
  3. (C)4.01
  4. (D)3.99

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
The mirror image of the point (1, 2, 3) in a plane is (−73,−43,−13)\left(-\dfrac{7}{3},-\dfrac{4}{3},-\dfrac{1}{3}\right)(−37​,−34​,−31​). Which of the following points lies on this plane?
  1. (A)(1, −1, 1)
  2. (B)(1, 1, 1)
  3. (C)(−1, −1, −1)
  4. (D)(−1, −1, 1)

Correct answer: (A)

Step-by-step solution →
Q51·MathematicsSingle correct
If a hyperbola passes through the point P (10, 16) and it has vertices at (±6,0)(\pm 6, 0)(±6,0) then the equation of the normal to it at P is:
  1. (A)2x+5y=1002x+5y=1002x+5y=100
  2. (B)x+3y=58x+3y=58x+3y=58
  3. (C)3x+4y=943x+4y=943x+4y=94
  4. (D)x+2y=42x+2y=42x+2y=42

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correct
If a line, y=mx+cy=mx+cy=mx+c is a tangent to the circle, (x−3)2+y2=1(x-3)^{2}+y^{2}=1(x−3)2+y2=1 and it is perpendicular to a line L1L_{1}L1​, where L1L_{1}L1​ is the tangent to the circle, x2+y2=1x^{2}+y^{2}=1x2+y2=1 at the point (12,12)\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)(2​1​,2​1​); then:
  1. (A)c2+7c+6=0c^{2}+7c+6=0c2+7c+6=0
  2. (B)c2−6c+7=0c^{2}-6c+7=0c2−6c+7=0
  3. (C)c2+6c+7=0c^{2}+6c+7=0c2+6c+7=0
  4. (D)c2−7c+6=0c^{2}-7c+6=0c2−7c+6=0

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correct
Let a⃗=i^−2j^+k^\vec{a}=\hat{i}-2\hat{j}+\hat{k}a=i^−2j^​+k^ and b⃗=i^−j^+k^\vec{b}=\hat{i}-\hat{j}+\hat{k}b=i^−j^​+k^ be two vectors. If c⃗\vec{c}c is a vector such that b⃗×c⃗=b⃗×a⃗\vec{b}\times\vec{c}=\vec{b}\times\vec{a}b×c=b×a and c⃗.a⃗=0\vec{c}.\vec{a}=0c.a=0, then c⃗.b⃗\vec{c}.\vec{b}c.b is equal to:
  1. (A)−12-\dfrac{1}{2}−21​
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)−1-1−1
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsSingle correct
Let f:(1,3)→Rf:(1,3)\to Rf:(1,3)→R be a function defined by f(x)=x[x]1+x2f(x)=\dfrac{x[x]}{1+x^{2}}f(x)=1+x2x[x]​, where [x] denotes the greatest integer ≤x\le x≤x. Then the range of f is:
  1. (A)(25,45]\left(\dfrac{2}{5},\dfrac{4}{5}\right](52​,54​]
  2. (B)(25,12)∪(35,45]\left(\dfrac{2}{5},\dfrac{1}{2}\right)\cup\left(\dfrac{3}{5},\dfrac{4}{5}\right](52​,21​)∪(53​,54​]
  3. (C)(35,45)\left(\dfrac{3}{5},\dfrac{4}{5}\right)(53​,54​)
  4. (D)(25,35]∪(34,45)\left(\dfrac{2}{5},\dfrac{3}{5}\right]\cup\left(\dfrac{3}{4},\dfrac{4}{5}\right)(52​,53​]∪(43​,54​)

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
Let α=−1+i32\alpha=\dfrac{-1+i\sqrt{3}}{2}α=2−1+i3​​. If a=(1+α)∑k=0100α2ka=(1+\alpha)\displaystyle\sum_{k=0}^{100}\alpha^{2k}a=(1+α)k=0∑100​α2k and b=∑k=0100α3kb=\displaystyle\sum_{k=0}^{100}\alpha^{3k}b=k=0∑100​α3k, then a and b are the roots of the quadratic equation:
  1. (A)x2−101x+100=0x^{2}-101x+100=0x2−101x+100=0
  2. (B)x2−102x+101=0x^{2}-102x+101=0x2−102x+101=0
  3. (C)x2+102x+101=0x^{2}+102x+101=0x2+102x+101=0
  4. (D)x2+101x+100=0x^{2}+101x+100=0x2+101x+100=0

Correct answer: (B)

Step-by-step solution →
Q56·MathematicsSingle correct
lim⁡x→0∫0xtsin⁡(10t) dtx\displaystyle\lim_{x\to 0}\dfrac{\int_{0}^{x} t\sin(10t)\,dt}{x}x→0lim​x∫0x​tsin(10t)dt​ is equal to:
  1. (A)0
  2. (B)110\dfrac{1}{10}101​
  3. (C)−15-\dfrac{1}{5}−51​
  4. (D)−110-\dfrac{1}{10}−101​

Correct answer: (A)

Step-by-step solution →
Q57·MathematicsSingle correct
The system of linear equations λx+2y+2z=5\lambda x+2y+2z=5λx+2y+2z=5, 2λx+3y+5z=82\lambda x+3y+5z=82λx+3y+5z=8, 4x+λy+6z=104x+\lambda y+6z=104x+λy+6z=10 has:
  1. (A)no solution when λ=8\lambda=8λ=8
  2. (B)infinitely many solutions when λ=2\lambda=2λ=2
  3. (C)no solution when λ=2\lambda=2λ=2
  4. (D)a unique solution when λ=−8\lambda=-8λ=−8

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correct
If α\alphaα and β\betaβ be the coefficients of x4x^{4}x4 and x2x^{2}x2 respectively in the expansion of (x+x2−1)6+(x−x2−1)6\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}(x+x2−1​)6+(x−x2−1​)6, then:
  1. (A)α+β=−30\alpha+\beta=-30α+β=−30
  2. (B)α−β=60\alpha-\beta=60α−β=60
  3. (C)α−β=−132\alpha-\beta=-132α−β=−132
  4. (D)α+β=60\alpha+\beta=60α+β=60

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
The differential equation of the family of curves, x2=4b(y+b),b∈Rx^{2}=4b\left(y+b\right), b \in Rx2=4b(y+b),b∈R, is:
  1. (A)xy′′=y′xy''=y'xy′′=y′
  2. (B)x(y′)2=x−2yy′x\left(y'\right)^{2}=x-2yy'x(y′)2=x−2yy′
  3. (C)x(y′)2=x+2yy′x\left(y'\right)^{2}=x+2yy'x(y′)2=x+2yy′
  4. (D)x(y′)2=2yy′−xx\left(y'\right)^{2}=2yy'-xx(y′)2=2yy′−x

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
The length of the perpendicular from the origin, 0n the normal to the curve, x2+2xy−3y2=0x^{2}+2xy-3y^{2}=0x2+2xy−3y2=0 at the point (2, 2) is:
  1. (A)222\sqrt{2}22​
  2. (B)424\sqrt{2}42​
  3. (C)2\sqrt{2}2​
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
If A=(2294)A=\begin{pmatrix}2 & 2\\9 & 4\end{pmatrix}A=(29​24​) and I=(1001)I=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}I=(10​01​), then 10A−110A^{-1}10A−1 is equal to:
  1. (A)4I−A4I-A4I−A
  2. (B)A−6IA-6IA−6I
  3. (C)A−4IA-4IA−4I
  4. (D)6I−A6I-A6I−A

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Let A and B be two events such that the probability that exactly one of them occurs is 25\frac{2}{5}52​ and the probability that A or B occurs is 12\frac{1}{2}21​, then the probability of both of them occur together is:
  1. (A)0.01
  2. (B)0.10
  3. (C)0.20
  4. (D)0.02

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
If I=∫12dx2x3−9x2+12x+4I=\int_{1}^{2}\dfrac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}}I=∫12​2x3−9x2+12x+4​dx​, then:
  1. (A)116<I2<19\dfrac{1}{16}<I^{2}<\dfrac{1}{9}161​<I2<91​
  2. (B)19<I2<18\dfrac{1}{9}<I^{2}<\dfrac{1}{8}91​<I2<81​
  3. (C)18<I2<14\dfrac{1}{8}<I^{2}<\dfrac{1}{4}81​<I2<41​
  4. (D)16<I2<12\dfrac{1}{6}<I^{2}<\dfrac{1}{2}61​<I2<21​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
Let S be the set of all real roots of the equation, 3x(3x−1)+2=∣3x−1∣+∣3x−2∣3^{x}\left(3^{x}-1\right)+2=\left|3^{x}-1\right|+\left|3^{x}-2\right|3x(3x−1)+2=∣3x−1∣+∣3x−2∣. Then S:
  1. (A)contains at least four elements
  2. (B)is a singleton
  3. (C)is an empty set
  4. (D)contains exactly two elements

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsNumerical
Let f(x)f(x)f(x) be a polynomial of degree 3 such that f(−1)=10,f(1)=−6f(-1)=10, f(1)=-6f(−1)=10,f(1)=−6, f(x)f(x)f(x) has a critical point at x=−1x=-1x=−1 and f′(x)f'(x)f′(x) has a critical point at x=1x=1x=1. Then f(x)f(x)f(x) has a local minima at x=_______.

Correct answer: 3

Step-by-step solution →
Q66·MathematicsNumerical
Let aline y=mxy=mxy=mx (m>0)(m>0)(m>0) intersect the parabola, y2=xy^{2}=xy2=x at a point P, other than the origin. Let the tangent to it at P meet the x - axis at the point Q. If area (ΔOPQ)=4\left(\Delta OPQ\right)=4(ΔOPQ)=4 sq. units, then m is equal to ___________.

Correct answer: 0.5

Step-by-step solution →
Q67·MathematicsNumerical
The sum ∑n=17n(n+1)(2n+1)4\sum_{n=1}^{7}\dfrac{n(n+1)(2n+1)}{4}∑n=17​4n(n+1)(2n+1)​ is equal to ___________.

Correct answer: 504

Step-by-step solution →
Q68·MathematicsNumerical
The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word 'EXAMINATION' is ___________.

Correct answer: 2454

Step-by-step solution →
Q69·MathematicsNumerical
If 2sin⁡α1+cos⁡2α=17\dfrac{\sqrt{2}\sin\alpha}{\sqrt{1+\cos2\alpha}}=\dfrac{1}{7}1+cos2α​2​sinα​=71​ and 1−cos⁡2β2=110\sqrt{\dfrac{1-\cos2\beta}{2}}=\dfrac{1}{\sqrt{10}}21−cos2β​​=10​1​, α,β∈(0,π2)\alpha,\beta\in\left(0,\dfrac{\pi}{2}\right)α,β∈(0,2π​), then tan⁡(α+2β)\tan\left(\alpha+2\beta\right)tan(α+2β) is equal to __________.

Correct answer: 1

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Solid State 63/186
  • Mathematical Reasoning 26/186
  • Aromaticity 22/186
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