Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2020
  4. /3 Sep Shift 2

JEE Main 3 September 2020 Shift 2 Question Paper with Answers

3 September 2020 · September session · 71 questions

71 of the 75 questions from the JEE Main 3 September 2020 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

4 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
25

Physics — JEE Main 3 September 2020 Shift 2

Q1·PhysicsSingle correct
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E⃗=E0j^cos⁡(ωt−kx)\vec{E} = E_0\hat{j}\cos(\omega t - kx)E=E0​j^​cos(ωt−kx). The magnetic field B⃗\vec{B}B, at the moment t = 0 is:
  1. (A)B⃗=E0μ0ϵ0 cos⁡(kx) j^\vec{B} = E_0\sqrt{\mu_0 \epsilon_0}\,\cos(kx)\,\hat{j}B=E0​μ0​ϵ0​​cos(kx)j^​
  2. (B)B⃗=E0μ0ϵ0 cos⁡(kx) k^\vec{B} = E_0\sqrt{\mu_0 \epsilon_0}\,\cos(kx)\,\hat{k}B=E0​μ0​ϵ0​​cos(kx)k^
  3. (C)B⃗=E0μ0ϵ0cos⁡(kx) k^\vec{B} = \dfrac{E_0}{\sqrt{\mu_0 \epsilon_0}}\cos(kx)\,\hat{k}B=μ0​ϵ0​​E0​​cos(kx)k^
  4. (D)B⃗=E0μ0ϵ0cos⁡(kx) j^\vec{B} = \dfrac{E_0}{\sqrt{\mu_0 \epsilon_0}}\cos(kx)\,\hat{j}B=μ0​ϵ0​​E0​​cos(kx)j^​

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the combined system strikes the floor, is [Take g = 10 m/s2^{2}2. Assume there is no rotational motion and loss of energy after the collision is negligible.]
  1. (A)21 J
  2. (B)20 J
  3. (C)23 J
  4. (D)19 J

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to:
  1. (A)1 : 2
  2. (B)10 : 7
  3. (C)5 : 7
  4. (D)2 : 1

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) ml212ω2sin⁡θcos⁡θ\dfrac{ml^{2}}{12}\omega^{2}\sin\theta\cos\theta12ml2​ω2sinθcosθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHF_HFH​ and FVF_VFV​ about the CM. The value of θ is then such that:
  1. (A)cos⁡θ=glω2\cos\theta = \dfrac{g}{l\omega^{2}}cosθ=lω2g​
  2. (B)cos⁡θ=2g3lω2\cos\theta = \dfrac{2g}{3l\omega^{2}}cosθ=3lω22g​
  3. (C)cos⁡θ=3g2lω2\cos\theta = \dfrac{3g}{2l\omega^{2}}cosθ=2lω23g​
  4. (D)cos⁡θ=g2lω2\cos\theta = \dfrac{g}{2l\omega^{2}}cosθ=2lω2g​

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is:
  1. (A)M2L0T−1M^{2}L^{0}T^{-1}M2L0T−1
  2. (B)ML2T−2ML^{2}T^{-2}ML2T−2
  3. (C)ML0T−3ML^{0}T^{-3}ML0T−3
  4. (D)MLT−2MLT^{-2}MLT−2

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A calorimeter of water equivalent 20 g contains 180 g of water at 25∘C25^{\circ}C25∘C, 'm' grams of steam at 100∘C100^{\circ}C100∘C is mixed in it till the temperature of the mixture is 31∘C31^{\circ}C31∘C. The value of 'm' is close to (Latent heat of water = 540 cal g−1^{-1}−1, specific heat of water =1 cal g−1 ∘C−1= 1\,\mathrm{cal}\,g^{-1}\,^{\circ}C^{-1}=1calg−1∘C−1)
  1. (A)2.6
  2. (B)4
  3. (C)2
  4. (D)3.2

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
Two sources of light emit X − rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X − rays to the number density of photons of the visible light of the given wavelengths is:
  1. (A)1500\dfrac{1}{500}5001​
  2. (B)500
  3. (C)250
  4. (D)1250\dfrac{1}{250}2501​

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
Concentric metallic hollow spheres of radii R and 4R hold charges Q1Q_1Q1​ and Q2Q_2Q2​ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V (R) − V (4R) is:
  1. (A)3Q24πϵ0R\dfrac{3Q_2}{4\pi \epsilon_0 R}4πϵ0​R3Q2​​
  2. (B)Q24πϵ0R\dfrac{Q_2}{4\pi \epsilon_0 R}4πϵ0​RQ2​​
  3. (C)3Q116πϵ0R\dfrac{3Q_1}{16\pi \epsilon_0 R}16πϵ0​R3Q1​​
  4. (D)3Q14πϵ0R\dfrac{3Q_1}{4\pi \epsilon_0 R}4πϵ0​R3Q1​​

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fAfAfA. The value of f is:
  1. (A)12\dfrac{1}{2}21​
  2. (B)12\dfrac{1}{\sqrt{2}}2​1​
  3. (C)2\sqrt{2}2​
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
Two light waves having the same wavelength λ in vacuum are in phase initially. Then the first wave travels path L1L_1L1​ through a medium of refractive index n1n_1n1​ while the second wave travels a path of length L2L_2L2​ through a medium of refractive index n2n_2n2​. After this the phase difference between the two waves is:
  1. (A)2πλ(L2n1−L1n2)\dfrac{2\pi}{\lambda}\left(\dfrac{L_2}{n_1} - \dfrac{L_1}{n_2}\right)λ2π​(n1​L2​​−n2​L1​​)
  2. (B)2πλ(L1n1−L2n2)\dfrac{2\pi}{\lambda}\left(\dfrac{L_1}{n_1} - \dfrac{L_2}{n_2}\right)λ2π​(n1​L1​​−n2​L2​​)
  3. (C)2πλ(n1L1−n2L2)\dfrac{2\pi}{\lambda}\left(n_1L_1 - n_2L_2\right)λ2π​(n1​L1​−n2​L2​)
  4. (D)2πλ(n2L1−n1L2)\dfrac{2\pi}{\lambda}\left(n_2L_1 - n_1L_2\right)λ2π​(n2​L1​−n1​L2​)

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
A metallic sphere cools from 50∘C50^{\circ}C50∘C to 40∘40^{\circ}40∘ in 300s. If atmospheric temperature around is 20∘C20^{\circ}C20∘C, then the sphere's temperature after the next 5 minutes will be close to:
  1. (A)35∘C35^{\circ}C35∘C
  2. (B)31∘C31^{\circ}C31∘C
  3. (C)33∘C33^{\circ}C33∘C
  4. (D)28∘C28^{\circ}C28∘C

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is: Planck's constant h=6.63×10−34h = 6.63 \times 10^{-34}h=6.63×10−34 J.s. Speed of light c=3×108c = 3 \times 10^{8}c=3×108 m / s
  1. (A)3.1eV
  2. (B)1.5eV
  3. (C)2.0eV
  4. (D)1.1eV

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
To raise the temperature of a certain mass of gas by 50∘50^{\circ}50∘ C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100∘C100^{\circ}C100∘C at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?
  1. (A)7
  2. (B)5
  3. (C)3
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic substance. The whole system is placed in a uniform magnetic field B⃗\vec{B}B. Then the field inside the paramagnetic substance is:
  1. (A)zero
  2. (B)much large than ∣B⃗∣\left|\vec{B}\right|​B​ but opposite to B⃗\vec{B}B
  3. (C)B⃗\vec{B}B
  4. (D)much large than ∣B⃗∣\left|\vec{B}\right|​B​ and parallel to B⃗\vec{B}B

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
The radius R of a nucleus of mass number A can be estimated by the formula R=(1.3×10−15)A1/3mR=\left(1.3\times 10^{-15}\right)A^{1/3}mR=(1.3×10−15)A1/3m. It follows that the mass density of a nucleus is of the order of: (Mprot=Mneut.≃1.67×10−27 kg)\left(M_{prot}=M_{neut.}\simeq 1.67\times 10^{-27}\,kg\right)(Mprot​=Mneut.​≃1.67×10−27kg)
  1. (A)103kg m−310^{3}kg\ m^{-3}103kg m−3
  2. (B)1017 kg m−310^{17}\,kg\ m^{-3}1017kg m−3
  3. (C)1010 kg m−310^{10}\,kg\ m^{-3}1010kg m−3
  4. (D)1024 kg m−310^{24}\,kg\ m^{-3}1024kg m−3

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρ0(1−r2R2)\rho\left(r\right)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right)ρ(r)=ρ0​(1−R2r2​). Then the gravitational field is maximum at:
  1. (A)r=13Rr=\frac{1}{\sqrt{3}}Rr=3​1​R
  2. (B)r=Rr=Rr=R
  3. (C)r=59Rr=\sqrt{\frac{5}{9}}Rr=95​​R
  4. (D)r=34Rr=\sqrt{\frac{3}{4}}Rr=43​​R

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dBdt=0.032 Ts−1\frac{dB}{dt}=0.032\,Ts^{-1}dtdB​=0.032Ts−1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23×10−8 Ωm1.23\times 10^{-8}\,\Omega m1.23×10−8Ωm )
  1. (A)0.61A0.61A0.61A
  2. (B)0.43 A0.43\,A0.43A
  3. (C)0.53 A0.53\,A0.53A
  4. (D)0.34 A0.34\,A0.34A

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement(s) – time (t) graph that describes the motion of the particle is (graph are drawn schematically and are not to scale):
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
Two resistors 400Ω and 800Ω are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10kΩ across 400Ω resistor is close to:
  1. (A)2 V
  2. (B)2.05V
  3. (C)1.95 V
  4. (D)1.8 V

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsNumerical
A block starts moving up an inclined plane of inclination 30∘30^{\circ}30∘ with an initial velocity of v0v_{0}v0​. It comes back to its initial position with velocity v02\frac{v_{0}}{2}2v0​​. The value of the coefficient f kinetic friction between the block and the inclined plane is close to I1000\frac{I}{1000}1000I​, The nearest integer to I is

Correct answer: 346.00

Step-by-step solution →
Q21·PhysicsNumerical
A galvanometer coil has 500 turns and each turn has an average are of 3×10−4m23\times 10^{-4}m^{2}3×10−4m2. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is __________.

Correct answer: 20.00

Step-by-step solution →
Q22·PhysicsNumerical
If minimum possible work is done by a refrigerator in converting 100 grams of water at 0∘C0^{\circ}C0∘C to ice, how much heat (in calories) is released to the surroundings at temperature 27∘C27^{\circ}C27∘C (Latent heat of ice = 80 Cal/gram) to the nearest integer?

Correct answer: 8791

Step-by-step solution →
Q23·PhysicsNumerical
A massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG is N20ma2\frac{N}{20}ma^{2}20N​ma2 where N is an integer. The value of N is __________.

Correct answer: 25.00

Step-by-step solution →
Q24·PhysicsNumerical
When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cm s−1s^{-1}s−1, the speed (in cm s−1s^{-1}s−1) with which image moves at that instant is _______________.

Correct answer: 1.00

Step-by-step solution →

Chemistry — JEE Main 3 September 2020 Shift 2

Q25·ChemistrySingle correct
Consider the following reaction: The product ‘P’ gives positive ceric ammonium nitrate test. This is because of the presence of which of these – OH group(s)?
  1. (A)(d) only
  2. (B)(b) and (d)
  3. (C)(b) only
  4. (D)(c) and (d)

Correct answer: (C)

Step-by-step solution →
Q26·ChemistrySingle correct
An ionic micelles is formed on the addition of:
  1. (A)(A)
  2. (B)sodium stearate to pure toluene
  3. (C)liquid diethyl ether to aqueous NaCl solution
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q27·ChemistrySingle correct
The five successive ionization enthalpies of an element are 800, 2427, 3658, 25024 and 32824 kJ mol−1^{-1}−1. The number of valence electrons in the element is:
  1. (A)4
  2. (B)2
  3. (C)5
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q28·ChemistrySingle correct
The compound A in the following reactions is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correct
Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0, 1, 2,…….n + 1, where n is the principal quantum number. Then, the element with atomic number:
  1. (A)9 is the first alkali metal
  2. (B)8 is the first noble gas
  3. (C)13 has a half – filled valence subshell
  4. (D)6 has a 2p – valence subshell

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correct
The strengths of 5.6 volume hydrogen peroxide (of density 1 g/mL) in terms of mass percentage and molarity (M), respectively, are: (Take molar mass of hydrogen peroxide as 34 g/mol)
  1. (A)0.85 and 0.25
  2. (B)0.85 and 0.5
  3. (C)1.7 and 0.5
  4. (D)1.7 and 0.25

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
The incorrect statement(s) among (a) – (d) regarding acid rain is (are): (a) It can corrode water pipes. (b) It can damage structures made up of stone. (c) It cannot cause respiratory ailments in animals (d) It is not harmful for trees
  1. (A)(a), (b) and (d)
  2. (B)(c) and (d)
  3. (C)(a), (c) and (d)
  4. (D)(c) only

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Complex A has a composition of H12_{12}12​O6_{6}6​Cl3_{3}3​Cr. If the complex on treatment with conc. H2_{2}2​SO4_{4}4​ loses 13.5% of its original mass, the correct molecular formula of A is: [Given : atomic mass of Cr = 52 amu and Cl = 35 amu]
  1. (A)[Cr(H2O)4Cl2]Cl⋅2H2O\left[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{4}\mathrm{Cl}_{2}\right]\mathrm{Cl}\cdot 2\mathrm{H}_{2}\mathrm{O}[Cr(H2​O)4​Cl2​]Cl⋅2H2​O
  2. (B)[Cr(H2O)6]Cl3\left[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}\right]\mathrm{Cl}_{3}[Cr(H2​O)6​]Cl3​
  3. (C)[Cr(H2O)5Cl]Cl2⋅H2O\left[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}\right]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}[Cr(H2​O)5​Cl]Cl2​⋅H2​O
  4. (D)[Cr(H2O)3Cl3]⋅3H2O\left[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{3}\mathrm{Cl}_{3}\right]\cdot 3\mathrm{H}_{2}\mathrm{O}[Cr(H2​O)3​Cl3​]⋅3H2​O

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
The incorrect statement is
  1. (A)In manganate and permanganateions, the π - bounding takes place by overlap of p –orbitals of oxygen and d – orbitals of manganese
  2. (B)Manganate ion is green in colour and permanganateion is purple in colour
  3. (C)Manganate and permanganate ions are paramagnetic
  4. (D)Manganate and permanganate ions are tetrahedral

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
The decreasing order or reactivity of the following compounds towards nucleophilic substitution (SN_{N}N​2) is:
  1. (A)(II) > (III) > (I) > (IV)
  2. (B)(III) > (II) > (IV) > (I)
  3. (C)(II) > (III) > (IV) > (I)
  4. (D)(IV) > (II) > (III) > (I)

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
A mixture of one mole each of H2_{2}2​, He and O2_{2}2​ each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2_{2}2​ is 2 atm, the total pressure of the gases in the cylinder is:
  1. (A)14 atm
  2. (B)6 atm
  3. (C)22 atm
  4. (D)38 atm

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
Among the statements (I – IV), the correct ones are: (I) Be has smaller atomic radius compared to Mg. (II) Be has higher ionization enthalpy than Al. (III) Charge/radius ratio of Be is greater than that of Al. (IV) Both Be and Al form mainly covalent compounds.
  1. (A)(I), (III) and (IV)
  2. (B)(II), (III) and (IV)
  3. (C)(I), (II) and (III)
  4. (D)(I), (II) and (IV)

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Three isomers A, B and C (mol. formula C8_{8}8​H11_{11}11​N) give the following results: R has lower boiling point than S B →C6H5SO2Cl\xrightarrow{\text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{Cl}}C6​H5​SO2​Cl​ alkali – insoluble product A, B and C, respectively are:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
For the reaction 2A+3B+32C→3P2A + 3B + \frac{3}{2}C \rightarrow 3P2A+3B+23​C→3P, which statement is correct?
  1. (A)dnAdt=23dnBdt=34dnCdt\frac{dn_{A}}{dt} = \frac{2}{3}\frac{dn_{B}}{dt} = \frac{3}{4}\frac{dn_{C}}{dt}dtdnA​​=32​dtdnB​​=43​dtdnC​​
  2. (B)dnAdt=dnBdt=dnCdt\frac{dn_{A}}{dt} = \frac{dn_{B}}{dt} = \frac{dn_{C}}{dt}dtdnA​​=dtdnB​​=dtdnC​​
  3. (C)dnAdt=32dnBdt=34dnCdt\frac{dn_{A}}{dt} = \frac{3}{2}\frac{dn_{B}}{dt} = \frac{3}{4}\frac{dn_{C}}{dt}dtdnA​​=23​dtdnB​​=43​dtdnC​​
  4. (D)dnAdt=23dnBdt=43dnCdt\frac{dn_{A}}{dt} = \frac{2}{3}\frac{dn_{B}}{dt} = \frac{4}{3}\frac{dn_{C}}{dt}dtdnA​​=32​dtdnB​​=34​dtdnC​​

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
The d – electron configuration of [Ru(en)3]Cl2\left[\text{Ru}(\text{en})_{3}\right]\text{Cl}_{2}[Ru(en)3​]Cl2​ and [Fe(H2O)6]Cl2\left[\text{Fe}(\text{H}_{2}\text{O})_{6}\right]\text{Cl}_{2}[Fe(H2​O)6​]Cl2​, respectively are:
  1. (A)t2g6eg0t_{2g}^{6}e_{g}^{0}t2g6​eg0​ and t2g4eg2t_{2g}^{4}e_{g}^{2}t2g4​eg2​
  2. (B)t2g4eg2t_{2g}^{4}e_{g}^{2}t2g4​eg2​ and t2g4eg2t_{2g}^{4}e_{g}^{2}t2g4​eg2​
  3. (C)t2g6ee0t_{2g}^{6}e_{e}^{0}t2g6​ee0​ and t2g6eg0t_{2g}^{6}e_{g}^{0}t2g6​eg0​
  4. (D)t2g4eg2t_{2g}^{4}e_{g}^{2}t2g4​eg2​ and t2g6eg0t_{2g}^{6}e_{g}^{0}t2g6​eg0​

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Match the following drugs with their therapeutic actions:
Column-IColumn-II
(i).Ranitidine(a).Antidepressant
(ii).Nardil (Phenelzine)(b).Antibiolic
(iii).Chloramphenicol(c).Antihistamine
(iv).Dimetane (Brompheniramine)(d).Antacid
(e).Analgesic
  1. (A)(i) – (e); (ii) – (a); (iii) – (c); (iv) – (d)
  2. (B)(i) – (a); (ii) – (c); (iii) – (b); (iv) – (e)
  3. (C)(i) – (d); (ii) – (c); (iii) – (a); (iv) – (e)
  4. (D)(i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)

Correct answer: (D)

Step-by-step solution →
Q42·ChemistryNumerical
The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ______.

Correct answer: 10

Step-by-step solution →
Q43·ChemistryNumerical
The number of [structure shown] groups present in a tripeptide Asp – Glu – Lys is ______.

Correct answer: 5

Step-by-step solution →
Q44·ChemistryNumerical
If 250 cm3^{3}3 of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular masses of A and B is ______ ×10−2\times 10^{-2}×10−2 (to the nearest integer).

Correct answer: 177

Step-by-step solution →
Q45·ChemistryNumerical
An acidic solution of dichromate is electrolyzed for 8 minutes using 2A current. As per the following equation Cr2_{2}2​O72−_{7}^{2-}72−​ + 14H+^{+}+ + 6e−^{-}− → 2Cr3+^{3+}3+ + 7H2_{2}2​O. The amount of Cr3+^{3+}3+ obtained was 0.104 g. The efficiency of the process (in %) is (Take: F = 96000 C, At. Mass of chromium = 52) ______.

Correct answer: 60

Step-by-step solution →
Q46·ChemistryNumerical
6.023 ×1022\times 10^{22}×1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing 5 g of substance 'x' in 2 L solution is ______ ×10−3\times 10^{-3}×10−3.

Correct answer: 25

Step-by-step solution →

Mathematics — JEE Main 3 September 2020 Shift 2

Q47·MathematicsSingle correct
If the sum of the series 20+1935+1915+1845+....20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + ....20+1953​+1951​+1854​+.... upto nthn^{th}nth term is 488 and the nthn^{th}nth term is negative, then:
  1. (A)nthn^{th}nth term is −425-4\frac{2}{5}−452​
  2. (B)n=41n = 41n=41
  3. (C)nthn^{th}nth term is −4-4−4
  4. (D)n=60n = 60n=60

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsSingle correct
If z1,z2z_1, z_2z1​,z2​ are complex numbers such that Re(z1)=∣z1−1∣,Re(z2)=∣z2−1∣\mathrm{Re}(z_1) = |z_1 - 1|, \mathrm{Re}(z_2) = |z_2 - 1|Re(z1​)=∣z1​−1∣,Re(z2​)=∣z2​−1∣ and arg⁡(z1−z2)=π6\arg(z_1 - z_2) = \frac{\pi}{6}arg(z1​−z2​)=6π​, then Im(z1+z2)\mathrm{Im}(z_1 + z_2)Im(z1​+z2​) is equal to:
  1. (A)232\sqrt{3}23​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)13\frac{1}{\sqrt{3}}3​1​
  4. (D)23\frac{2}{\sqrt{3}}3​2​

Correct answer: (A)

Step-by-step solution →
Q49·MathematicsSingle correct
Let a,b,c∈Ra, b, c \in Ra,b,c∈R be such that a2+b2+c2=1a^2 + b^2 + c^2 = 1a2+b2+c2=1. If acos⁡θ=bcos⁡(θ+2π3)=ccos⁡(θ+4π3)a\cos\theta = b\cos\left(\theta + \frac{2\pi}{3}\right) = c\cos\left(\theta + \frac{4\pi}{3}\right)acosθ=bcos(θ+32π​)=ccos(θ+34π​), where θ=π9\theta = \frac{\pi}{9}θ=9π​, then the angle between the vectors ai^+bj^+ck^a\hat{i} + b\hat{j} + c\hat{k}ai^+bj^​+ck^ and bi^+cj^+ak^b\hat{i} + c\hat{j} + a\hat{k}bi^+cj^​+ak^ is:
  1. (A)π9\frac{\pi}{9}9π​
  2. (B)2π3\frac{2\pi}{3}32π​
  3. (C)000
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
Let the latus rectum of the parabola y2=4xy^2 = 4xy2=4x be the common chord to the circles C1C_1C1​ and C2C_2C2​ each of them having radius 252\sqrt{5}25​. Then, the distance between the centres of the circles C1C_1C1​ and C2C_2C2​ is:
  1. (A)454\sqrt{5}45​
  2. (B)858\sqrt{5}85​
  3. (C)888
  4. (D)121212

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Let R1R_1R1​ and R2R_2R2​ be two relations defined as follows: R1={(a,b)∈R2:a2+b2∈Q}R_1 = \{(a,b) \in R^2 : a^2 + b^2 \in Q\}R1​={(a,b)∈R2:a2+b2∈Q} and R2={(a,b)∈R2:a2+b2∉Q}R_2 = \{(a,b) \in R^2 : a^2 + b^2 \notin Q\}R2​={(a,b)∈R2:a2+b2∈/Q}, where QQQ is the set of all rational numbers. Then:
  1. (A)Neither R1R_1R1​ nor R2R_2R2​ is transitive.
  2. (B)R1R_1R1​ is transitive but R2R_2R2​ is not transitive.
  3. (C)R1R_1R1​ and R2R_2R2​ are both transitive
  4. (D)R2R_2R2​ is transitive but R1R_1R1​ is not transitive.

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correct
If the value of the integral ∫01/2x2(1−x2)3/2dx\int_{0}^{1/2} \frac{x^2}{\left(1 - x^2\right)^{3/2}} dx∫01/2​(1−x2)3/2x2​dx is k6\frac{k}{6}6k​, then kkk is equal to:
  1. (A)23+π2\sqrt{3} + \pi23​+π
  2. (B)32−π3\sqrt{2} - \pi32​−π
  3. (C)23−π2\sqrt{3} - \pi23​−π
  4. (D)32+π3\sqrt{2} + \pi32​+π

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correct
If x3dy+xy dx=x2dy+2y dx;y(2)=ex^3 dy + xy\, dx = x^2 dy + 2y\, dx; y(2) = ex3dy+xydx=x2dy+2ydx;y(2)=e and x>1x > 1x>1, then y(4)y(4)y(4) is equal to:
  1. (A)e2\frac{\sqrt{e}}{2}2e​​
  2. (B)12+e\frac{1}{2} + \sqrt{e}21​+e​
  3. (C)32e\frac{3}{2}\sqrt{e}23​e​
  4. (D)32+e\frac{3}{2} + \sqrt{e}23​+e​

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
The probability that a randomly chosen 5 - digit number is made from exactly two digits is:
  1. (A)134104\frac{134}{10^4}104134​
  2. (B)121104\frac{121}{10^4}104121​
  3. (C)150104\frac{150}{10^4}104150​
  4. (D)135104\frac{135}{10^4}104135​

Correct answer: (D)

Step-by-step solution →
Q55·MathematicsSingle correct
If the surface area of a cube is increasing at a rate of 3.6 cm2cm^2cm2/sec, retaining its shape; then the rate of change of its volume (in cm3cm^3cm3/sec), when the length of a side of the cube is 10 cm, is:
  1. (A)999
  2. (B)181818
  3. (C)101010
  4. (D)202020

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
The set of all real values of λ\lambdaλ for which the quadratic equations, (λ2+1)x2−4λx+2=0\left(\lambda^2 + 1\right)x^2 - 4\lambda x + 2 = 0(λ2+1)x2−4λx+2=0 always have exactly one root in the interval (0, 1) is:
  1. (A)(0,2)(0, 2)(0,2)
  2. (B)(2,4](2, 4](2,4]
  3. (C)(−3,−1)(-3, -1)(−3,−1)
  4. (D)(1,3](1, 3](1,3]

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
If a ΔABC\Delta ABCΔABC have vertices A (−1,7)(-1, 7)(−1,7), B (−7,1)(-7, 1)(−7,1) and C (5,−5)(5, -5)(5,−5), then its orthocenter has coordinates:
  1. (A)(35,−35)\left( \frac{3}{5}, -\frac{3}{5} \right)(53​,−53​)
  2. (B)(−3,3)(-3, 3)(−3,3)
  3. (C)(3,−3)(3, -3)(3,−3)
  4. (D)(−35,35)\left( -\frac{3}{5}, \frac{3}{5} \right)(−53​,53​)

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
Let p, q, r be three statements such that the truth value of (p∧q)→(∼q∨r)\left( p \wedge q \right) \rightarrow \left( \sim q \vee r \right)(p∧q)→(∼q∨r) is F. Then the truth values of p, q,r are respectively:
  1. (A)T, F, T
  2. (B)T, T, F
  3. (C)F, T, F
  4. (D)T, T, T

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsSingle correct
The plane which bisects the line joining the points (4,−2,3)(4, -2, 3)(4,−2,3) and (2,4,−1)(2, 4, -1)(2,4,−1) at right angles also passes through the point:
  1. (A)(0,−1,1)(0, -1, 1)(0,−1,1)
  2. (B)(4,0,1)(4, 0, 1)(4,0,1)
  3. (C)(4,0,−1)(4, 0, -1)(4,0,−1)
  4. (D)(0,1,−1)(0, 1, -1)(0,1,−1)

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
Let e1e_1e1​ and e2e_2e2​ be the eccentricities of the ellipse, x225+y2b2=1(b<5)\frac{x^2}{25} + \frac{y^2}{b^2} = 1 \left( b < 5 \right)25x2​+b2y2​=1(b<5) and the hyperbola, x216−y2b2=1\frac{x^2}{16} - \frac{y^2}{b^2} = 116x2​−b2y2​=1 respectively satisfying e1e2=1e_1 e_2 = 1e1​e2​=1. If α\alphaα and β\betaβ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α,β)\left( \alpha, \beta \right)(α,β) is equal to :
  1. (A)(8,10)(8, 10)(8,10)
  2. (B)(245,10)\left( \frac{24}{5}, 10 \right)(524​,10)
  3. (C)(203,12)\left( \frac{20}{3}, 12 \right)(320​,12)
  4. (D)(8,12)(8, 12)(8,12)

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A be a 3×33 \times 33×3 matrix such that adj A=[2−11−1021−2−1]\mathrm{adj}\,A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{bmatrix}adjA=​2−11​−10−2​12−1​​ and B=adj(adj A)B = \mathrm{adj}\left( \mathrm{adj}\,A \right)B=adj(adjA). If ∣A∣=λ\left| A \right| = \lambda∣A∣=λ and ∣(B−1)T∣=μ\left| \left( B^{-1} \right)^{T} \right| = \mu​(B−1)T​=μ, then the ordered pair, (∣λ∣.μ)\left( \left| \lambda \right| . \mu \right)(∣λ∣.μ) is equal to:
  1. (A)(9,19)\left( 9, \frac{1}{9} \right)(9,91​)
  2. (B)(3,81)(3, 81)(3,81)
  3. (C)(9,181)\left( 9, \frac{1}{81} \right)(9,811​)
  4. (D)(3,181)\left( 3, \frac{1}{81} \right)(3,811​)

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
If ∫sin⁡−1(x1+x)dx=A(x)tan⁡−1(x)+B(x)+C,\int \sin^{-1} \left( \sqrt{\frac{x}{1+x}} \right) dx = A(x) \tan^{-1} \left( \sqrt{x} \right) + B(x) + C,∫sin−1(1+xx​​)dx=A(x)tan−1(x​)+B(x)+C, where C is a constant of integration, then the ordered pair (A(x),B(x))\left( A(x), B(x) \right)(A(x),B(x)) can be:
  1. (A)(x−1,−x)\left( x - 1, -\sqrt{x} \right)(x−1,−x​)
  2. (B)(x−1,x)\left( x - 1, \sqrt{x} \right)(x−1,x​)
  3. (C)(x+1,−x)\left( x + 1, -\sqrt{x} \right)(x+1,−x​)
  4. (D)(x+1,x)\left( x + 1, \sqrt{x} \right)(x+1,x​)

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let xi(1≤i≤10)x_i \left( 1 \le i \le 10 \right)xi​(1≤i≤10) be ten observations of a random variable X. If ∑i=110(xi−p)=3\sum_{i=1}^{10} \left( x_i - p \right) = 3∑i=110​(xi​−p)=3 and ∑i=110(xi−p)2=9\sum_{i=1}^{10} \left( x_i - p \right)^2 = 9∑i=110​(xi​−p)2=9 where 0≠p∈R0 \ne p \in R0=p∈R, then the standard deviation of these observations is:
  1. (A)35\sqrt{\frac{3}{5}}53​​
  2. (B)45\frac{4}{5}54​
  3. (C)710\frac{7}{10}107​
  4. (D)910\frac{9}{10}109​

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
If the term independent of x in the expansion of (32x2−13x)9\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9(23​x2−3x1​)9 is k, then 18k is equal to:
  1. (A)9
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Suppose f(x)f(x)f(x) is a polynomial of degree four, having critical points at −1,0,1-1, 0, 1−1,0,1. If T={x∈R | f(x)=f(0)}T = \left\{ x \in R \,\middle|\, f(x) = f(0) \right\}T={x∈R∣f(x)=f(0)}, then the sum of squares of all the elements of T is:
  1. (A)8
  2. (B)6
  3. (C)2
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
lim⁡x→a(a+2x)13−(3x)13(3a+x)13−(4x)13(a≠0)\lim_{x \to a} \frac{\left( a + 2x \right)^{\frac{1}{3}} - \left( 3x \right)^{\frac{1}{3}}}{\left( 3a + x \right)^{\frac{1}{3}} - \left( 4x \right)^{\frac{1}{3}}} \left( a \ne 0 \right)limx→a​(3a+x)31​−(4x)31​(a+2x)31​−(3x)31​​(a=0) is equal to:
  1. (A)(23)43\left( \frac{2}{3} \right)^{\frac{4}{3}}(32​)34​
  2. (B)(29)43\left( \frac{2}{9} \right)^{\frac{4}{3}}(92​)34​
  3. (C)(29)(23)13\left( \frac{2}{9} \right) \left( \frac{2}{3} \right)^{\frac{1}{3}}(92​)(32​)31​
  4. (D)(23)(29)13\left( \frac{2}{3} \right) \left( \frac{2}{9} \right)^{\frac{1}{3}}(32​)(92​)31​

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsNumerical
If the tangent to the curve, y=exy = e^xy=ex at a point (c,ec)\left( c, e^c \right)(c,ec) and the normal to the parabola, y2=4xy^2 = 4xy2=4x at the point (1,2)(1, 2)(1,2) intersect at the same point on the x −-− axis, then the value of c is __________.

Correct answer: 04.00

Step-by-step solution →
Q68·MathematicsNumerical
The total number of 3 −-− digit numbers, whose sum of digits is 10, is ____________.

Correct answer: 54.00

Step-by-step solution →
Q69·MathematicsNumerical
Let S be the set of all integer solutions, (x, y, z), of the system of equation x−2y+5z=0x - 2y + 5z = 0x−2y+5z=0 −2x+4y+z=0-2x + 4y + z = 0−2x+4y+z=0 −7x+14y+9z=0-7x + 14y + 9z = 0−7x+14y+9z=0 such that 15≤x2+y2+z2≤15015 \le x^2 + y^2 + z^2 \le 15015≤x2+y2+z2≤150. Then, the number of elements in the set S is equal to _______.

Correct answer: 08.00

Step-by-step solution →
Q70·MathematicsNumerical
Let a plane P contain two lines r⃗=i^+λ(i^+j^),λ∈R\vec{r} = \hat{i} + \lambda \left( \hat{i} + \hat{j} \right), \lambda \in Rr=i^+λ(i^+j^​),λ∈R and r⃗=−j^+μ(j^−k^),μ∈R\vec{r} = -\hat{j} + \mu \left( \hat{j} - \hat{k} \right), \mu \in Rr=−j^​+μ(j^​−k^),μ∈R. If Q(α,β,γ)Q\left( \alpha, \beta, \gamma \right)Q(α,β,γ) is the foot of the perpendicular drawn from the point M (1,0,1)(1, 0, 1)(1,0,1) to P, then 3(α+β+γ)3 \left( \alpha + \beta + \gamma \right)3(α+β+γ) equals ____________ .

Correct answer: 05.00

Step-by-step solution →
Q71·MathematicsNumerical
If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th4^{th}4th A.M. is equal to 2nd2^{nd}2nd G.M., then m is equal to ________________.

Correct answer: 39.00

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
  • 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
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
  • Magnetism and Matter 50/186
  • Mathematical Reasoning 26/186
← 3 Sep Shift 1 2020All papers4 Sep Shift 1 2020 →

Attempt this paper under exam timing.

Take the 3 September 2020 Shift 2 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS