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JEE Main 6 September 2020 Shift 2 Question Paper with Answers

6 September 2020 · September session · 70 questions

70 of the 75 questions from the JEE Main 6 September 2020 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

5 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
22
Chemistry
23
Mathematics
25

Physics — JEE Main 6 September 2020 Shift 2

Q1·PhysicsSingle correct
In the figure shown, the current in the 10 V battery is close to:
  1. (A)0.71 A from positive to negative terminal
  2. (B)0.42 A from positive to negative terminal
  3. (C)0.21 A from positive to negative terminal.
  4. (D)0.36 A from negative to positive terminal.

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field B⃗\vec{B}B. The magnetic moment of this moving particle:
  1. (A)mv2B⃗2B2\frac{mv^{2}\vec{B}}{2B^{2}}2B2mv2B​
  2. (B)−mv2B⃗2πB2-\frac{mv^{2}\vec{B}}{2\pi B^{2}}−2πB2mv2B​
  3. (C)−mv2B⃗B2-\frac{mv^{2}\vec{B}}{B^{2}}−B2mv2B​
  4. (D)−mv2B⃗2B2-\frac{mv^{2}\vec{B}}{2B^{2}}−2B2mv2B​

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity K1K_{1}K1​, K2K_{2}K2​ and K3K_{3}K3​, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100°C and the other at 0°C (see figure). If the joints of the rod are at 70°C and 20°C in steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1K_{1}K1​, K2K_{2}K2​ and K2K_{2}K2​ is:
  1. (A)K1:K3=2:3K_{1} : K_{3} = 2 : 3K1​:K3​=2:3, K2:K3=2:5K_{2} : K_{3} = 2 : 5K2​:K3​=2:5
  2. (B)K1<K2<K3K_{1} < K_{2} < K_{3}K1​<K2​<K3​
  3. (C)K1:K2=5:2K_{1} : K_{2} = 5 : 2K1​:K2​=5:2, K1:K3=3:5K_{1} : K_{3} = 3 : 5K1​:K3​=3:5
  4. (D)K1>K2>K3K_{1} > K_{2} > K_{3}K1​>K2​>K3​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Two identical electric point dipoles have dipole moments p⃗1=pi^\vec{p}_{1} = p\hat{i}p​1​=pi^ and p⃗2=−pi^\vec{p}_{2} = -p\hat{i}p​2​=−pi^ and are held on the x axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is:
  1. (A)pa1π∈0ma\frac{p}{a}\sqrt{\frac{1}{\pi ∈_{0} ma}}ap​π∈0​ma1​​
  2. (B)pa12π∈0ma\frac{p}{a}\sqrt{\frac{1}{2\pi ∈_{0} ma}}ap​2π∈0​ma1​​
  3. (C)pa2π∈0ma\frac{p}{a}\sqrt{\frac{2}{\pi ∈_{0} ma}}ap​π∈0​ma2​​
  4. (D)pa32π∈0ma\frac{p}{a}\sqrt{\frac{3}{2\pi ∈_{0} ma}}ap​2π∈0​ma3​​

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
For a plane electromagnetic wave, the magnetic field at a point x and time t is B⃗(x,t)=[1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]\vec{B}(x, t) = \left[1.2×10^{-7} \sin(0.5×10^{3}x + 1.5×10^{11}t)\hat{k}\right]B(x,t)=[1.2×10−7sin(0.5×103x+1.5×1011t)k^] T The instantaneous electric field E⃗\vec{E}E corresponding to B⃗\vec{B}B is: (speed of light c = 3 × 10810^{8}108 ms−1ms^{-1}ms−1)
  1. (A)E⃗(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\vec{E}(x, t) = \left[-36 \sin(0.5×10^{3}x + 1.5×10^{11}t)\hat{j}\right]\frac{V}{m}E(x,t)=[−36sin(0.5×103x+1.5×1011t)j^​]mV​
  2. (B)E⃗(x,t)=[36sin⁡(1×103x+0.5×1011t)j^]Vm\vec{E}(x, t) = \left[36 \sin(1×10^{3}x + 0.5×10^{11}t)\hat{j}\right]\frac{V}{m}E(x,t)=[36sin(1×103x+0.5×1011t)j^​]mV​
  3. (C)E⃗(x,t)=[36sin⁡(0.5×103x+1.5×1011t)k^]Vm\vec{E}(x, t) = \left[36 \sin(0.5×10^{3}x + 1.5×10^{11}t)\hat{k}\right]\frac{V}{m}E(x,t)=[36sin(0.5×103x+1.5×1011t)k^]mV​
  4. (D)E⃗(x,t)=[36sin⁡(1×103x+1.5×1011t)i^]Vm\vec{E}(x, t) = \left[36 \sin(1×10^{3}x + 1.5×10^{11}t)\hat{i}\right]\frac{V}{m}E(x,t)=[36sin(1×103x+1.5×1011t)i^]mV​

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firings speed needed is:
  1. (A)2GMa2\sqrt{\frac{GM}{a}}2aGM​​
  2. (B)4GMa4\sqrt{\frac{GM}{a}}4aGM​​
  3. (C)GM2ma\sqrt{\frac{GM^{2}}{ma}}maGM2​​
  4. (D)325GMa\frac{3}{2}\sqrt{\frac{5GM}{a}}23​a5GM​​

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
A particle moving in the xy plane experiences a velocity dependent force F⃗=k(vyi^+vxj^)\vec{F} = k\left(v_{y}\hat{i} + v_{x}\hat{j}\right)F=k(vy​i^+vx​j^​), where vxv_{x}vx​ and vyv_{y}vy​ are the x and y components of its velocity v⃗\vec{v}v. If a⃗\vec{a}a is the acceleration of the particle, then which of the following statements is true for the particle?
  1. (A)quantity v⃗×a⃗\vec{v} × \vec{a}v×a is constant in time
  2. (B)F⃗\vec{F}F arises due to a magnetic field.
  3. (C)kinetic energy of particle is constant in time.
  4. (D)quantity v⃗⋅a⃗\vec{v} · \vec{a}v⋅a is constant in time.

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + β)v, this angle changes to 45°. The value of β is close to:
  1. (A)0.50
  2. (B)0.41
  3. (C)0.37
  4. (D)0.73

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Given the masses of various atomic particles mpm_{p}mp​ = 1.0072 u, mnm_{n}mn​ = 1.0087 u, mem_{e}me​ = 0.000548 u, mvˉm_{\bar{v}}mvˉ​ = 0, mdm_{d}md​ = 2.0141 u, where p ≡ proton, n ≡ neutron, e ≡ electron, vˉ\bar{v}vˉ ≡ antineutrino and d ≡ deuteron. Which of the following process is allowed by momentum and energy conservation?
  1. (A)n + n → deuterium atom (electron bound to the nucleus)
  2. (B)p→n+e++vˉp → n + e^{+} + \bar{v}p→n+e++vˉ
  3. (C)n+p→d+γn + p → d + γn+p→d+γ
  4. (D)e++e−→γe^{+} + e^{-} → γe++e−→γ

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit:
  1. (A)ammeter is always used in parallel and voltmeter is series.
  2. (B)both ammeter and voltmeter must be connected in parallel.
  3. (C)ammeter is always connected in series and voltmeter in parallel.
  4. (D)both, ammeter and voltmeter must be connected in series.

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?
  1. (A)F=14π∈0QqR2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{R^{2}}F=4π∈0​1​R2Qq​ for r<Rr < Rr<R
  2. (B)14π∈0qQR2>F>0\frac{1}{4\pi ∈_{0}}\frac{qQ}{R^{2}} > F > 04π∈0​1​R2qQ​>F>0 for r<Rr < Rr<R
  3. (C)F=14π∈0Qqr2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{r^{2}}F=4π∈0​1​r2Qq​ for r>Rr > Rr>R
  4. (D)F=14π∈0Qqr2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{r^{2}}F=4π∈0​1​r2Qq​ for all r

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm ; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
  1. (A)(5.5375 ± 0.0739) mm
  2. (B)(5.5375 ± 0.0740) mm
  3. (C)(5.538 ± 0.074) mm
  4. (D)(5.54 ± 0.07) mm

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is
  1. (A)2R2R2R
  2. (B)R2\frac{R}{2}2R​
  3. (C)3R2\frac{3R}{2}23R​
  4. (D)R3\frac{R}{3}3R​

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
A square loop of side 2a and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be
  1. (A)2μ0I2a2πb\frac{2\mu_{0}I^{2}a^{2}}{\pi b}πb2μ0​I2a2​
  2. (B)2μ0I2a2bπ(a2+b2)\frac{2\mu_{0}I^{2}a^{2}b}{\pi(a^{2}+b^{2})}π(a2+b2)2μ0​I2a2b​
  3. (C)μ0I2a2b2π(a2+b2)\frac{\mu_{0}I^{2}a^{2}b}{2\pi(a^{2}+b^{2})}2π(a2+b2)μ0​I2a2b​
  4. (D)μ0I2a22πb\frac{\mu_{0}I^{2}a^{2}}{2\pi b}2πbμ0​I2a2​

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
A fluid is flowing through a horizontal pipe of varying cross-section, with speed v ms−1^{-1}−1 a ta point where the pressure is P Pascal. At another point where pressure is P2\frac{P}{2}2P​ Pascal its speed is V ms−1^{-1}−1. If the density of the fluid is ρ kg m−3^{-3}−3 and the flow is streamline, then V is equal to:
  1. (A)Pρ+v\sqrt{\frac{P}{ρ}+v}ρP​+v​
  2. (B)2Pρ+v2\sqrt{\frac{2P}{ρ}+v^{2}}ρ2P​+v2​
  3. (C)P2ρ+v2\sqrt{\frac{P}{2ρ}+v^{2}}2ρP​+v2​
  4. (D)Pρ+v2\sqrt{\frac{P}{ρ}+v^{2}}ρP​+v2​

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by y(t)=y0sin⁡2ωty(t) = y_{0}\sin^{2}\omega ty(t)=y0​sin2ωt, where 'y' is measured from the lower end of unstretched spring. Then ω is:
  1. (A)12gy0\frac{1}{2}\sqrt{\frac{g}{y_{0}}}21​y0​g​​
  2. (B)gy0\sqrt{\frac{g}{y_{0}}}y0​g​​
  3. (C)g2y0\sqrt{\frac{g}{2y_{0}}}2y0​g​​
  4. (D)2gy0\sqrt{\frac{2g}{y_{0}}}y0​2g​​

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to: (Given: nitrogen molecule weight: 4.64×10−264.64 × 10^{-26}4.64×10−26 kg, Boltzman constant: 1.38×10−231.38 × 10^{-23}1.38×10−23 J/K, Planck constant: 6.63×10−346.63 × 10^{-34}6.63×10−34 Js)
  1. (A)0.24 Å
  2. (B)0.20 Å
  3. (C)0.34 Å
  4. (D)0.44 Å

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
The liner mass density of a thin rod AB of length L varies from A to B as λ(x)=λ0(1+xL)\lambda(x) = \lambda_{0}\left(1+\frac{x}{L}\right)λ(x)=λ0​(1+Lx​), where x is the distance from A. If M is the mass of the rod hen its moment of inertia about an axis passing through A and perpendicular to the rod is:
  1. (A)512ML2\frac{5}{12}ML^{2}125​ML2
  2. (B)718ML2\frac{7}{18}ML^{2}187​ML2
  3. (C)25ML2\frac{2}{5}ML^{2}52​ML2
  4. (D)37ML2\frac{3}{7}ML^{2}73​ML2

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsNumerical
The out put characteristics of a transistor is shown in the figure. When VCEV_{CE}VCE​ is 10 V and ICI_{C}IC​ = 4.0 mA, then value of βac\beta_{ac}βac​ is ______.

Correct answer: 150

Step-by-step solution →
Q20·PhysicsNumerical
The centre of mass of a solid hemisphere of radius 8 cm is x cm from the centre of the flat surface. Then value of x is ______.

Correct answer: 3.00

Step-by-step solution →
Q21·PhysicsNumerical
A Young's double-slit experiment is performed using monochromatic light of wavelength λ. The intensity of light at a point on the screen, where the path difference is λ, is K units. The intensity of the light at a point where the path difference is λ6\frac{\lambda}{6}6λ​ is given by nK12\frac{nK}{12}12nK​, where 'n' is an integer. The value of 'n' is ______.

Correct answer: 9.00

Step-by-step solution →
Q22·PhysicsNumerical
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as (n3π)\left(\frac{n}{3\pi}\right)(3πn​) μF, then value of 'n' is ______.

Correct answer: 400.00

Step-by-step solution →

Chemistry — JEE Main 6 September 2020 Shift 2

Q23·ChemistrySingle correct
For a reaction, 4M(s) + nO2_22​(g) → 2M2_22​On_nn​(s), the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which
  1. (A)the slope changes from negative to positive.
  2. (B)the free energy change shows a change from negative to positive value.
  3. (C)the slope changes from positive to negative.
  4. (D)the slope changes from positive to zero.

Correct answer: (B)

Step-by-step solution →
Q24·ChemistrySingle correct
The average molar mass of chlorine is 35.5 g mol−1^{-1}−1. The ratio of 35^{35}35Cl to 37^{37}37Cl in naturally occurring chlorine is close to:
  1. (A)4 : 1
  2. (B)3 : 1
  3. (C)2 : 1
  4. (D)1 : 1

Correct answer: (B)

Step-by-step solution →
Q25·ChemistrySingle correct
The value of KC_CC​ is 64 at 800 K for the reaction N2_22​(g) + 3H2_22​(g) ⇌ 2NH3_33​(g) The value of KC_CC​ for the following reaction is: NH3_33​(g) ⇌ 12\frac{1}{2}21​N2_22​(g) + 32\frac{3}{2}23​H2_22​(g)
  1. (A)164\frac{1}{64}641​
  2. (B)8
  3. (C)14\frac{1}{4}41​
  4. (D)18\frac{1}{8}81​

Correct answer: (D)

Step-by-step solution →
Q26·ChemistrySingle correct
Dihydrogen of high purity (>99.95%) is obtained through:
  1. (A)the reaction of Zn with dilute HCl.
  2. (B)the electrolysis of acidified water using Pt electrodes
  3. (C)the electrolysis of brine solution.
  4. (D)the electrolysis of warm Ba(OH)2_22​ solution using Ni electrodes.

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
The reaction of NO with N2_22​O4_44​ at 250 K gives:
  1. (A)N2_22​O
  2. (B)NO2_22​
  3. (C)N2_22​O3_33​
  4. (D)N2_22​O5_55​

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correct
The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is:
Item-IItem-II
I.BenzeneP.HCl and SnCl2_22​, H3_33​O+^++
II.BenzonitrileQ.H2_22​, Pd-BaSO4_44​, S and quinoline
III.Benzoyl ChlorideR.CO, HCl and AlCl3_33​
  1. (A)(I) − (Q), (II) − (R) and (III) − (P)
  2. (B)(I) − (P), (II) − (Q) and (III) − (R)
  3. (C)(I) − (R), (II) − (P) and (III) − (Q)
  4. (D)(I) − (R), (II) − (Q) and (III) − (P)

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
A crystal is made up of metal ions 'M1_11​' and 'M2_22​' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1_11​' occupies 50% of octahedral voids and the cation 'M2_22​' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1_11​' and 'M2_22​' are respectively:
  1. (A)+2, +4
  2. (B)+1, +3
  3. (C)+3, +1
  4. (D)+4, +2

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
The element that can be refined by distillation is:
  1. (A)nickel
  2. (B)zinc
  3. (C)tin
  4. (D)gallium

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
For a d4^44 metal ion in an octahedral field, the correct electronic configuration is:
  1. (A)t2g3eg1t_{2g}^{3}e_{g}^{1}t2g3​eg1​ when Δo_oo​ < P
  2. (B)t2g3eg1t_{2g}^{3}e_{g}^{1}t2g3​eg1​ when Δo_oo​ > P
  3. (C)t2g4eg0t_{2g}^{4}e_{g}^{0}t2g4​eg0​ when Δo_oo​ < P
  4. (D)tg2e2g2t_{g}^{2}e_{2g}^{2}tg2​e2g2​ when Δo_oo​ < P

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Match the following:
Test / MethodReagent
i.Lucas Testa.C6_66​H5_55​SO2_22​Cl / aq. KOH
ii.Dumas methodb.HNO3_33​ / AgNO3_33​
iii.Kjeldahl's methodc.CuO / CO2_22​
iv.Hinsberg Testd.Conc. HCl and ZnCl2_22​
e.H2_22​SO4_44​
  1. (A)(i)-(d), (ii)-(c), (iii)-(b), (iv)-(e)
  2. (B)(i)-(b), (ii)-(d), (iii)-(e), (iv)-(a)
  3. (C)(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)
  4. (D)(i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Match the following compounds (Column-I) with their uses (Column-II):
Column-IColumn-II
i.Ca(OH)2_22​A.casts of statues
ii.NaClB.white wash
iii.CaSO4_44​ · 12\frac{1}{2}21​H2_22​OC.antacid
iv.CaCO3_33​D.Washing soda preparation
  1. (A)(i)-(D), (ii)-(A), (iii)-(C), (iv)-(B)
  2. (B)(i)-(B), (ii)-(D), (iii)-(A), (iv)-(A)
  3. (C)(i)-(B), (ii)-(C), (iii)-(D), (iv)-(A)
  4. (D)(i)-(C), (ii)-(D), (iii)-(B), (iv)-(A)

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
The IUPAC name of the following compound is [structure shown]
  1. (A)2-nitrogen-4-hydroxymethyl-5-amine benzaldehyde.
  2. (B)3-amino-4-hydroxymethyl 1-5-nitrobvenzaldehyde
  3. (C)5-amino-4hydroxymethyl-2-nitrobenzaldehyde
  4. (D)4-amino-2formyl-5hydroxymethyl nitrobenzene

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Which of the following compounds can e prepared in good yield by Gabriel phthalimide synthesis?
  1. (A)(A)
  2. (B)CH3_33​–CH2_22​–NHCH3_33​
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A = 100 g mol−1^{-1}−1 ; B = 200 g mol−1^{-1}−1 ; C = 10,000 g mol−1^{-1}−1]
  1. (A)B > C > A
  2. (B)C > B > A
  3. (C)A > B > C
  4. (D)A > C > B

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
For the given cell; Cu(s) | Cu2+^{2+}2+ (C1_11​M) || Cu2+^{2+}2+ (C2_22​M) | Cu(s) Change in Gibbs energy (ΔG) is negative, if
  1. (A)C1_11​ = C2_22​
  2. (B)C2=C12\frac{C_2 = C_1}{\sqrt{2}}2​C2​=C1​​
  3. (C)C1_11​ = 2C2_22​
  4. (D)C2_22​ = 2\sqrt{2}2​C1_11​

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Reaction of an inorganic sulphite X with dilute H2_22​SO4_44​ generates compound Y. Reaction of Y with NaOH gives X. Further, the reaction of X. Further, the reaction of X with Y and water affords compound Z. Y and Z respectively, are:
  1. (A)SO2_22​ and Na2_22​SO3_33​
  2. (B)SO3_33​ and NaHSO3_33​
  3. (C)SO2_22​ and NaHSO3_33​
  4. (D)S and Na2_22​SO3_33​

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
The increasing order of the boiling points of the major products A, B and C of the following reactions will be:
  1. (A)B < C < A
  2. (B)C < A < B
  3. (C)A < B < C
  4. (D)A < C < B

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Mischmetal is an alloy consisting mainly of:
  1. (A)lanthanoid metals
  2. (B)actinoid and transition metals
  3. (C)lanthanoid and actinoid metals
  4. (D)actionoid metals

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
The correct match between Item-I and Item-II is:
Item-IItem-II
a.Natural rubberI.1, 3-butadiene + styrene
b.NeopreneII.1, 3-butadiene + acrylonitrile
c.Buna-NIII.Chloroprene
d.Buna-SIV.Isoprene
  1. (A)(a) − (III), (b) − (IV), (c) − (I), (d) − (II)
  2. (B)(a) − (III), (b) − (IV), (c) − (II), (d) − (I)
  3. (C)(a) − (IV), (b) − (III), (c) − (II), (d) − (I)
  4. (D)(a) − (IV), (b) − (III), (c) − (I), (d) − (II)

Correct answer: (C)

Step-by-step solution →
Q42·ChemistryNumerical
If the solubility product of AB2_22​ is 3.20 × 10−11^{-11}−11 M3^33, then the solubility of AB2_22​ in pure water is __________ × 10−4^{-4}−4 mol L−1^{-1}−1. [Assuming that neither kind of ion reacts with water]

Correct answer: 02.00

Step-by-step solution →
Q43·ChemistryNumerical
For Freudlich adsorption isotherm, a plot of log (x/m) (y-axis) and log p (x-axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is __________ × 10−4^{-4}−4 g. (log 3 = 0.4771)

Correct answer: 48.00

Step-by-step solution →
Q44·ChemistryNumerical
A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is __________. (to the nearest integer) (Atomic mass: C = 12 ; H = 1 ; O = 16)

Correct answer: 69.00

Step-by-step solution →
Q45·ChemistryNumerical
The atomic number of Unnilunium is __________.

Correct answer: 101.00

Step-by-step solution →

Mathematics — JEE Main 6 September 2020 Shift 2

Q46·MathematicsSingle correct
The integral ∫12ex⋅x2(2+log⁡ex)dx\int_{1}^{2} e^{x} \cdot x^{2}(2+\log_{e} x)dx∫12​ex⋅x2(2+loge​x)dx equals:
  1. (A)e(4e+1)e(4e + 1)e(4e+1)
  2. (B)4e2−14e^{2} - 14e2−1
  3. (C)e(4e−1)e(4e - 1)e(4e−1)
  4. (D)e(2e−1)e(2e - 1)e(2e−1)

Correct answer: (C)

Step-by-step solution →
Q47·MathematicsSingle correct
The area (in sq. units) of the region enclosed by the curves y=x2−1y = x^{2} - 1y=x2−1 and y=1−x2y = 1 - x^{2}y=1−x2 is equal to
  1. (A)43\frac{4}{3}34​
  2. (B)83\frac{8}{3}38​
  3. (C)72\frac{7}{2}27​
  4. (D)163\frac{16}{3}316​

Correct answer: (B)

Step-by-step solution →
Q48·MathematicsSingle correct
The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is:
  1. (A)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  2. (B)3+13−1\frac{\sqrt{3}+1}{\sqrt{3}-1}3​−13​+1​
  3. (C)13−1\frac{1}{\sqrt{3}-1}3​−11​
  4. (D)13+1\frac{1}{\sqrt{3}+1}3​+11​

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correct
The set of all real values of λ\lambdaλ for which the function f(x)=(1−cos⁡2x)⋅(λ+sin⁡x)f(x) = (1 - \cos^{2} x)\cdot(\lambda + \sin x)f(x)=(1−cos2x)⋅(λ+sinx), x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)x∈(−2π​,2π​), has exactly one maxima and exactly one minima, is
  1. (A)(−12,12)−{0}\left(-\frac{1}{2}, \frac{1}{2}\right) - \{0\}(−21​,21​)−{0}
  2. (B)(−32,32)\left(-\frac{3}{2}, \frac{3}{2}\right)(−23​,23​)
  3. (C)(−12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)(−21​,21​)
  4. (D)(−32,32)−{0}\left(-\frac{3}{2}, \frac{3}{2}\right) - \{0\}(−23​,23​)−{0}

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
If α\alphaα and β\betaβ are the roots of the equation 2x(2x+1)=12x(2x + 1) = 12x(2x+1)=1, then β\betaβ is equal to:
  1. (A)2α(α+1)2\alpha(\alpha + 1)2α(α+1)
  2. (B)−2α(α+1)-2\alpha(\alpha + 1)−2α(α+1)
  3. (C)2α(α−1)2\alpha(\alpha - 1)2α(α−1)
  4. (D)2α22\alpha^{2}2α2

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correct
For all twice differentiable functions f:R→Rf : R \to Rf:R→R, with f(0)=f(1)=f′(0)=0f(0) = f(1) = f'(0) = 0f(0)=f(1)=f′(0)=0,
  1. (A)f′′(x)≠0f''(x) \neq 0f′′(x)=0, at every point x∈(0,1)x \in (0,1)x∈(0,1)
  2. (B)f′′(x)=0f''(x) = 0f′′(x)=0, for some x∈(0,1)x \in (0,1)x∈(0,1)
  3. (C)f′′(0)=0f''(0) = 0f′′(0)=0
  4. (D)f′′(x)=0f''(x) = 0f′′(x)=0, at every point x∈(0,1)x \in (0, 1)x∈(0,1)

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
If y=(2πx−1)cosec⁡xy = \left(\frac{2}{\pi}x - 1\right)\operatorname{cosec} xy=(π2​x−1)cosecx is the solution of the differential equation, dydx+p(x)y=2πcosec⁡ x,0<x<π2\frac{dy}{dx} + p(x)y = \frac{2}{\pi}\operatorname{cosec}\ x, 0 < x < \frac{\pi}{2}dxdy​+p(x)y=π2​cosec x,0<x<2π​, then the function p(x) is equal to:
  1. (A)cot⁡x\cot xcotx
  2. (B)cosec⁡x\operatorname{cosec} xcosecx
  3. (C)sec⁡x\sec xsecx
  4. (D)tan⁡x\tan xtanx

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correct
Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (−1,−4)(-1, -4)(−1,−4) in this line is:
  1. (A)(115,285)\left(\frac{11}{5}, \frac{28}{5}\right)(511​,528​)
  2. (B)(295,85)\left(\frac{29}{5}, \frac{8}{5}\right)(529​,58​)
  3. (C)(85,295)\left(\frac{8}{5}, \frac{29}{5}\right)(58​,529​)
  4. (D)(295,115)\left(\frac{29}{5}, \frac{11}{5}\right)(529​,511​)

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsSingle correct
If the tangent to the curve, y=f(x)=xlog⁡exy = f(x) = x\log_{e} xy=f(x)=xloge​x, (x>0)(x > 0)(x>0) at a point (c,f(c))(c, f(c))(c,f(c)) is parallel to the line – segment joining the points (1,0)(1, 0)(1,0) and (e,e)(e, e)(e,e) then c is equal to:
  1. (A)e−1e\frac{e-1}{e}ee−1​
  2. (B)1e−1\frac{1}{e-1}e−11​
  3. (C)e(1e−1)e^{\left(\frac{1}{e-1}\right)}e(e−11​)
  4. (D)e(11−e)e^{\left(\frac{1}{1-e}\right)}e(1−e1​)

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correct
Let f:R→Rf : R \to Rf:R→R be a function defined by f(x)=max⁡{x,x2}f(x) = \max\{x, x^{2}\}f(x)=max{x,x2}. Let S denote the set of all points in R, where f is not differentiable. Then:
  1. (A){0,1}\{0, 1\}{0,1}
  2. (B){0}\{0\}{0}
  3. (C)ϕ\phiϕ(an empty set)
  4. (D){1}\{1\}{1}

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
Let θ=π5\theta = \frac{\pi}{5}θ=5π​ and A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}A=[cosθ−sinθ​sinθcosθ​]. If B=A+A4B = A + A^{4}B=A+A4, then det (B):
  1. (A)is one.
  2. (B)lies in (2, 3).
  3. (C)is zero.
  4. (D)lies in (1, 2).

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
A plane P meets the coordinate axes at A, B and C respectively. The centroid of △ABC\triangle ABC△ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is:
  1. (A)x−12=y−11=z−21\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-2}{1}2x−1​=1y−1​=1z−2​
  2. (B)x−11=y−11=z−22\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}1x−1​=1y−1​=2z−2​
  3. (C)x−12=y−12=z−21\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}2x−1​=2y−1​=1z−2​
  4. (D)x−12=y−12=z−22\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{2}2x−1​=2y−1​=2z−2​

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correct
The common difference of the A.P. b1,b2,…,bmb_{1}, b_{2}, \ldots, b_{m}b1​,b2​,…,bm​ is 2 more than the common difference of A.P. a1,a2,…,ana_{1}, a_{2}, \ldots, a_{n}a1​,a2​,…,an​. If a40=−159a_{40} = -159a40​=−159, a100=−399a_{100} = -399a100​=−399 and b100=a70b_{100} = a_{70}b100​=a70​, then b1b_{1}b1​ is equal to:
  1. (A)81
  2. (B)−127
  3. (C)−81
  4. (D)127

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:
  1. (A)e4+2e2−1=0e^4 + 2e^2 - 1 = 0e4+2e2−1=0
  2. (B)e2+e−1=0e^2 + e - 1 = 0e2+e−1=0
  3. (C)e2+2e−1=0e^2 + 2e - 1 = 0e2+2e−1=0
  4. (D)e4+e2−1=0e^4 + e^2 - 1 = 0e4+e2−1=0

Correct answer: (D)

Step-by-step solution →
Q60·MathematicsSingle correct
For a suitably chosen real constant a, let a function, f:R−{−a}→Rf : R - \{-a\} \to Rf:R−{−a}→R be defined by f(x)=a−xa+xf(x) = \frac{a - x}{a + x}f(x)=a+xa−x​. Further suppose that for any real number x≠−ax \neq -ax=−a and f(x)≠−af(x) \neq -af(x)=−a, (fof)(x)=x(fof)(x) = x(fof)(x)=x. Then f(−12)f\left(-\frac{1}{2}\right)f(−21​) is equal to:
  1. (A)13\frac{1}{3}31​
  2. (B)−13-\frac{1}{3}−31​
  3. (C)−3-3−3
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
If the constant term in the binomial expansion of (x−kx2)10\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}(x​−x2k​)10 is 405, then ∣k∣|k|∣k∣ equals:
  1. (A)999
  2. (B)111
  3. (C)333
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
The centre of the circle passing through the point (0,1)(0, 1)(0,1) and touching the parabola y=x2y = x^2y=x2 at the point (2,4)(2, 4)(2,4) is:
  1. (A)(−5310,165)\left(\frac{-53}{10}, \frac{16}{5}\right)(10−53​,516​)
  2. (B)(65,5310)\left(\frac{6}{5}, \frac{53}{10}\right)(56​,1053​)
  3. (C)(310,165)\left(\frac{3}{10}, \frac{16}{5}\right)(103​,516​)
  4. (D)(−165,5310)\left(\frac{-16}{5}, \frac{53}{10}\right)(5−16​,1053​)

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
Let z=x+iyz = x + iyz=x+iy be a non-zero complex number such that z2=i∣z∣2z^2 = i|z|^2z2=i∣z∣2, where i=−1i = \sqrt{-1}i=−1​, then z lies on the:
  1. (A)line, y=−xy = -xy=−x
  2. (B)imaginary axis
  3. (C)line, y=xy = xy=x
  4. (D)real axis

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Consider the statement: “For an integer n, if n3−1n^3 - 1n3−1 is even, then n is odd.” The contrapositive statement of this statement is:
  1. (A)For an integer n, if n is even, then n3−1n^3 - 1n3−1 is odd.
  2. (B)For an integer n, if n3−1n^3 - 1n3−1 is not even then n is not odd.
  3. (C)For an integer n, if n is even, then n3−1n^3 - 1n3−1 is even.
  4. (D)For an integer n, if n is odd, then n3−1n^3 - 1n3−1 is even.

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
The probabilities of three events A, B and C are given by P(A)=0.6P(A) = 0.6P(A)=0.6, P(B)=0.4P(B) = 0.4P(B)=0.4 and P(C)=0.5P(C) = 0.5P(C)=0.5. If P(A∪B)=0.8P(A \cup B) = 0.8P(A∪B)=0.8, P(A∩C)=0.3P(A \cap C) = 0.3P(A∩C)=0.3, P(A∩B∩C)=0.2P(A \cap B \cap C) = 0.2P(A∩B∩C)=0.2, P(B∩C)=βP(B \cap C) = \betaP(B∩C)=β and P(A∪B∪C)=αP(A \cup B \cup C) = \alphaP(A∪B∪C)=α, where 0.85≤α≤0.950.85 \leq \alpha \leq 0.950.85≤α≤0.95, then β\betaβ lies in the interval:
  1. (A)[0.35,0.36][0.35, 0.36][0.35,0.36]
  2. (B)[0.25,0.35][0.25, 0.35][0.25,0.35]
  3. (C)[0.20,0.25][0.20, 0.25][0.20,0.25]
  4. (D)[0.36,0.40][0.36, 0.40][0.36,0.40]

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsNumerical
Suppose that a function f:R→Rf : R \to Rf:R→R satisfies f(x+y)=f(x)f(y)f(x + y) = f(x)f(y)f(x+y)=f(x)f(y) for all x,y∈Rx, y \in Rx,y∈R and f(1)=3f(1) = 3f(1)=3. If ∑i=1nf(i)=363\sum_{i=1}^{n} f(i) = 363∑i=1n​f(i)=363, then n is equal to __________.

Correct answer: 5

Step-by-step solution →
Q67·MathematicsNumerical
The sum of distinct values of λ\lambdaλ for which the system of equations (λ−1)x+(3λ+1)y+2λz=0(\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0(λ−1)x+(3λ+1)y+2λz=0 (λ−1)x+(4λ−2)y+(λ+3)z=0(\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0(λ−1)x+(4λ−2)y+(λ+3)z=0 2x+(3λ+1)y+3(λ−1)z=0,2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0,2x+(3λ+1)y+3(λ−1)z=0, has non-zero solutions, is __________.

Correct answer: 3

Step-by-step solution →
Q68·MathematicsNumerical
If x⃗\vec{x}x and y⃗\vec{y}y​ be two non-zero vectors such that ∣x⃗+y⃗∣=∣x⃗∣|\vec{x} + \vec{y}| = |\vec{x}|∣x+y​∣=∣x∣ and 2x⃗+λy⃗2\vec{x} + \lambda\vec{y}2x+λy​ is perpendicular to y⃗\vec{y}y​, then the value of λ\lambdaλ is

Correct answer: 1

Step-by-step solution →
Q69·MathematicsNumerical
Consider the data on x taking the values 0,2,4,8,…,2n0, 2, 4, 8, \ldots, 2^n0,2,4,8,…,2n with frequencies nC0^{n}C_0nC0​, nC1^{n}C_1nC1​, nC2^{n}C_2nC2​, …\ldots…, nCn^{n}C_nnCn​ respectively. If the mean of this data is 7282n\frac{728}{2^n}2n728​, then n is equal to __________.

Correct answer: 6

Step-by-step solution →
Q70·MathematicsNumerical
The number of words (with or without meaning) that can be formed from all the letters of the word “LETTER” in which vowels never come together is __________.

Correct answer: 120

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
  • Application of Derivatives 139/186
  • d- and f-Block Elements 126/186
  • 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
  • Trigonometric Functions 144/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • 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
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Hydrogen 81/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Solid State 63/186
  • IUPAC Nomenclature 37/186
  • Mathematical Reasoning 26/186
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