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JEE Main 17 March 2021 Shift 2 Question Paper with Answers

17 March 2021 · March session · 86 questions

86 of the 90 questions from the JEE Main 17 March 2021 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
28
Chemistry
28
Mathematics
30

Physics — JEE Main 17 March 2021 Shift 2

Q1·PhysicsSingle correct
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81100\frac{81}{100}10081​ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms−2^{-2}−2)
  1. (A)3.0 ms−1^{-1}−1
  2. (B)3.50 ms−1^{-1}−1
  3. (C)2.0 ms−1^{-1}−1
  4. (D)2.50 ms−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
If one mole of the polyatomic gas is having two vibrational modes and β is the ratio of molar specific heats for polyatomic gas (β=CPCV)\left(\beta = \frac{C_{P}}{C_{V}}\right)(β=CV​CP​​) then the value of β is :
  1. (A)1.02
  2. (B)1.2
  3. (C)1.25
  4. (D)1.35

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
Which one of the following will be the output of the given circuit ?
  1. (A)NOR Gate
  2. (B)NAND Gate
  3. (C)AND Gate
  4. (D)XOR Gate

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
An object is located at 2 km beneath the surface of the water. If the fractional compression ΔVV\frac{\Delta V}{V}VΔV​ is 1.36% , the ratio of hydraulic stress to the corresponding hydraulic strain will be ___________. [Given : density of water is 1000 kg m−3^{-3}−3 and g = 9.8 ms−2^{-2}−2.]
  1. (A)1.96 × 107^{7}7 Nm−2^{-2}−2
  2. (B)1.44 × 107^{7}7 Nm−2^{-2}−2
  3. (C)2.26 × 109^{9}9 Nm−2^{-2}−2
  4. (D)1.44 × 109^{9}9 Nm−2^{-2}−2

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
A geostationary satellite is orbiting around an arbitary planet 'P' at a height of 11R above the surface of 'P' , R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is__________.'P' has the time period of 24 hours.
  1. (A)626\sqrt{2}62​
  2. (B)62\frac{6}{\sqrt{2}}2​6​
  3. (C)3
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A sound wave of frequency 245 Hz travels with the speed of 300 ms−1^{-1}−1 along the positive x-axis. Each point of the wave moves to and fro through a total distance of 6 cm. What will be the mathematical expression of this travelling wave ?
  1. (A)Y(x,t) = 0.03 [sin 5.1 x − (0.2 × 103^{3}3)t]
  2. (B)Y(x,t) = 0.06 [sin 5.1 x − (1.5 × 103^{3}3)t]
  3. (C)Y(x,t) = 0.06 [sin 0.8 x − (0.5 × 103^{3}3)t]
  4. (D)Y(x,t) = 0.03 [sin 5.1 x − (1.5 × 103^{3}3)t]

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
Which one is the correct option for the two different thermodynamic processes ?
  1. (A)(c) and (a)
  2. (B)(c) and (d)
  3. (C)(a) only
  4. (D)(b) and (c)

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
The velocity of a particle is v = v0_{0}0​ + gt + Ft2^{2}2. Its position is x = 0 at t = 0 ; then its displacement after time (t = 1) is :
  1. (A)v0_{0}0​ + g + F
  2. (B)v0_{0}0​ + g2\frac{g}{2}2g​ + F3\frac{F}{3}3F​
  3. (C)v0_{0}0​ + g2\frac{g}{2}2g​ + F
  4. (D)v0_{0}0​ + 2g + 3F

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
A carrier signal C(t) = 25 sin (2.512 × 1010^{10}10 t) is amplitude modulated by a message signal m(t) = 5 sin (1.57 × 108^{8}8 t) and transmitted through an antenna.What will be the bandwidth of the modulated signal ?
  1. (A)8 GHz
  2. (B)2.01 GHz
  3. (C)1987.5 MHz
  4. (D)50 MHz

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
Two cells of emf 2E and E with internal resistance r1_{1}1​ and r2_{2}2​ respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is
  1. (A)r1_{1}1​ + r2_{2}2​
  2. (B)r12−r2\frac{r_{1}}{2} - r_{2}2r1​​−r2​
  3. (C)r12+r2\frac{r_{1}}{2} + r_{2}2r1​​+r2​
  4. (D)r1_{1}1​ − r2_{2}2​

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle ?
  1. (A)μ0I4πr(2−π)\frac{\mu_{0}I}{4\pi r}\left(2 - \pi\right)4πrμ0​I​(2−π)
  2. (B)μ0I4πr(2+π)\frac{\mu_{0}I}{4\pi r}\left(2 + \pi\right)4πrμ0​I​(2+π)
  3. (C)μ0I2πr(2+π)\frac{\mu_{0}I}{2\pi r}\left(2 + \pi\right)2πrμ0​I​(2+π)
  4. (D)μ0I2πr(2−π)\frac{\mu_{0}I}{2\pi r}\left(2 - \pi\right)2πrμ0​I​(2−π)

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.
  1. (A)2.44 μA
  2. (B)2.44 mA
  3. (C)4.87 mA
  4. (D)4.87 μA

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
Two particles A and B of equal masses are suspended from two massless springs of spring constants K1_{1}1​ and K2_{2}2​ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is
  1. (A)K2K1\frac{K_{2}}{K_{1}}K1​K2​​
  2. (B)K1K2\frac{K_{1}}{K_{2}}K2​K1​​
  3. (C)K1K2\sqrt{\frac{K_{1}}{K_{2}}}K2​K1​​​
  4. (D)K2K1\sqrt{\frac{K_{2}}{K_{1}}}K1​K2​​​

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
Match List-I with List-II Choose the most appropriate answer from the options given below :
List-IList-II
a.Phase difference between current and voltage in a purely resistive AC circuiti.π2\frac{\pi}{2}2π​ ; current leads voltage
b.Phase difference between current and voltage in a pure inductive AC circuitii.zero
c.Phase difference between current and voltage in a pure capacitive AC circuitiii.π2\frac{\pi}{2}2π​ ; current lags voltage
d.Phase difference between current and voltage in an LCR series circuitiv.tan⁡−1(XC−XLR)\tan^{-1}\left(\frac{X_{C} - X_{L}}{R}\right)tan−1(RXC​−XL​​)
  1. (A)(a)−(i),(b)−(iii),(c)−(iv),(d)−(ii)
  2. (B)(a)−(ii),(b)−(iv),(c)−(iii),(d)−(i)
  3. (C)(a)−(ii),(b)−(iii),(c)−(iv),(d)−(i)
  4. (D)(a)−(ii),(b)−(iii),(c)−(i),(d)−(iv)

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
The atomic hydrogen emits a line spectrum consisting of various series.Which series of hydrogen atomic spectra is lying in the visible region ?
  1. (A)Brackett series
  2. (B)Paschen series
  3. (C)Lyman series
  4. (D)Balmer series

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
Two identical photocathodes receive the light of frequencies f1_{1}1​ and f2_{2}2​ respectively. If the velocities of the photo-electrons coming out are v1_{1}1​ and v2_{2}2​ respectively, then
  1. (A)v12−v22=2hm[f1−f2]v_{1}^{2} - v_{2}^{2} = \frac{2h}{m}\left[f_{1} - f_{2}\right]v12​−v22​=m2h​[f1​−f2​]
  2. (B)v12+v22=2hm[f1+f2]v_{1}^{2} + v_{2}^{2} = \frac{2h}{m}\left[f_{1} + f_{2}\right]v12​+v22​=m2h​[f1​+f2​]
  3. (C)v1+v2=[2hm(f1+f2)]12v_{1} + v_{2} = \left[\frac{2h}{m}\left(f_{1} + f_{2}\right)\right]^{\frac{1}{2}}v1​+v2​=[m2h​(f1​+f2​)]21​
  4. (D)v1−v2=[2hm(f1−f2)]1/2v_{1} - v_{2} = \left[\frac{2h}{m}\left(f_{1} - f_{2}\right)\right]^{1/2}v1​−v2​=[m2h​(f1​−f2​)]1/2

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved ?
  1. (A)Both, inductive reactance and current will be halved.
  2. (B)Inductive reactance will be halved and current will be doubled.
  3. (C)Inductive reactance will be doubled and current will be halved.
  4. (D)Both, inducting reactance and current will be doubled.

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
A sphere of mass 2kg and radius 0.5 m is rolling with an initial speed of 1 ms−1^{-1}−1 goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ?
  1. (A)0.60 s
  2. (B)0.52 s
  3. (C)0.57 s
  4. (D)0.80 s

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsNumerical
The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is x5\sqrt{\frac{x}{5}}5x​​E. Where the value of x =_________.

Correct answer: 3

Step-by-step solution →
Q20·PhysicsNumerical
A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 13\frac{1}{\sqrt{3}}3​1​ . It is desired to make the body move by applying the minimum possible force F N. The value of F will be ___________. (Round off to the Nearest Integer) [Take g = 10 ms−2^{-2}−2]

Correct answer: 5

Step-by-step solution →
Q21·PhysicsNumerical
A boy of mass 4 kg is standing on a piece of wood having mass 5kg . If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is _________N.(Round off to the Nearest Integer) [Take g = 10 ms−2^{-2}−2]

Correct answer: 30

Step-by-step solution →
Q22·PhysicsNumerical
Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3^{3}3 of oleic acid per cm3^{3}3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2^{2}2 by considering 100 spherical drops of radius (340π)13×10−3\left(\frac{3}{40\pi}\right)^{\frac{1}{3}} \times 10^{-3}(40π3​)31​×10−3 cm. Then the thickness of oleic acid layer will be x × 10−14^{-14}−14 m. Where x is___________.

Correct answer: 25

Step-by-step solution →
Q23·PhysicsNumerical
A particle of mass m moves in a circular orbit in a central potential field U(r) = U0_{0}0​r4^{4}4 . If Bohr's quantization conditions are applied, radii of possible orbitals rn_{n}n​ vary with n1/α^{1/\alpha}1/α , where α is___________.

Correct answer: 3

Step-by-step solution →
Q24·PhysicsNumerical
The electric field in a region is given by E⃗=25E0i^+35E0j^\vec{E} = \frac{2}{5}E_{0}\hat{i} + \frac{3}{5}E_{0}\hat{j}E=52​E0​i^+53​E0​j^​ with E0_{0}0​ = 4.0 × 103^{3}3 NC\frac{N}{C}CN​ .The flux of this field through a rectangular surface area 0.4 m2^{2}2 parallel to the Y − Z plane is _________Nm2^{2}2C−1^{-1}−1.

Correct answer: 640

Step-by-step solution →
Q25·PhysicsNumerical
The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position x3aπ\frac{x}{3}\frac{a}{\pi}3x​πa​ , x3aπ\frac{x}{3}\frac{a}{\pi}3x​πa​ where x is ________. (Round off to the Nearest Integer) [a is an area as shown in the figure]

Correct answer: 4

Step-by-step solution →
Q26·PhysicsNumerical
The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at 2rd3\frac{2^{rd}}{3}32rd​ of the distance of the object from the surface .The wavelength of light inside the surface is 23\frac{2}{3}32​ times the wavelength in air. The radius of the curved surface is x13\frac{x}{13}13x​ m. the value of 'x' is__________.

Correct answer: 30

Step-by-step solution →
Q27·PhysicsNumerical
A 2 μF capacitor C1_{1}1​ is first charged to a potential difference of 10 V using a battery.Then the battery is removed and the capacitor is connected to an uncharged capacitor C2_{2}2​ of 8μF. The charge in C2_{2}2​ on equilibrium condition is_______μC. (Round off to the Nearest Integer)

Correct answer: 16

Step-by-step solution →
Q28·PhysicsNumerical
Seawater at a frequency f = 9 × 102^{2}2 Hz, has permittivity ε = 80ε0_{0}0​ and resistivity ρ = 0.25 Ωm. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t)=V0_{0}0​ sin (2πft). Then the conduction current density becomes 10x^{x}x times the displacement current density after time t = 1800\frac{1}{800}8001​ s. The value of x is ________ (Given : 14πε0=9×109 Nm2C−2\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}\,\mathrm{Nm}^{2}\mathrm{C}^{-2}4πε0​1​=9×109Nm2C−2)

Correct answer: 6

Step-by-step solution →

Chemistry — JEE Main 17 March 2021 Shift 2

Q29·ChemistrySingle correct
Fructose is an example of :-
  1. (A)Pyranose
  2. (B)Ketohexose
  3. (C)Aldohexose
  4. (D)Heptose

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correct
The set of elements that differ in mutual relationship from those of the other sets is :
  1. (A)Li – Mg
  2. (B)B – Si
  3. (C)Be – Al
  4. (D)Li – Na

Correct answer: (D)

Step-by-step solution →
Q31·ChemistrySingle correct
The functional groups that are responsible for the ion-exchange property of cation and anion exchange resins, respectively, are :
  1. (A)–SO3_{3}3​H and –NH2_{2}2​
  2. (B)–SO3_{3}3​H and –COOH
  3. (C)–NH2_{2}2​ and –COOH
  4. (D)–NH2_{2}2​ and –SO3_{3}3​H

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Match List-I and List-II : Choose the correct answer from the options given below :
List-IList-II
a.Haematitei.Al2_{2}2​O3_{3}3​.xH2_{2}2​O
b.Bauxiteii.Fe2_{2}2​O3_{3}3​
c.Magnetiteiii.CuCO3_{3}3​.Cu(OH)2_{2}2​
d.Malachiteiv.Fe3_{3}3​O4_{4}4​
  1. (A)(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
  2. (B)(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
  3. (C)(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
  4. (D)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
The correct pair(s) of the ambident nucleophiles is (are) : (A) AgCN/KCN (B) RCOOAg/RCOOK (C) AgNO2_{2}2​/KNO2_{2}2​ (D) AgI/KI
  1. (A)(B) and (C) only
  2. (B)(A) only
  3. (C)(A) and (C) only
  4. (D)(B) only

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
The set that represents the pair of neutral oxides of nitrogen is :
  1. (A)NO and N2_{2}2​O
  2. (B)N2_{2}2​O and N2_{2}2​O3_{3}3​
  3. (C)N2_{2}2​O and NO2_{2}2​
  4. (D)NO and NO2_{2}2​

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Match List-I with List-II : Choose the correct answer from the options given below :
List-IList-II
a.[Co(NH3_{3}3​)6_{6}6​] [Cr(CN)6_{6}6​]i.Linkage isomerism
b.[Co(NH3_{3}3​)3_{3}3​ (NO2_{2}2​)3_{3}3​]ii.Solvate isomerism
c.[Cr(H2_{2}2​O)6_{6}6​]Cl3_{3}3​iii.Co-ordination isomerism
d.cisciscis-[CrCl2_{2}2​(ox)2_{2}2​]3−^{3-}3−iv.Optical isomerism
  1. (A)(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
  2. (B)(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
  3. (C)(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
  4. (D)(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Correct answer: (A)

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Q36·ChemistrySingle correct
Primary, secondary and tertiary amines can be separated using :-
  1. (A)Para-Toluene sulphonyl chloride
  2. (B)Chloroform and KOH
  3. (C)Benzene sulphonic acid
  4. (D)Acetyl amide

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
The common positive oxidation states for an element with atomic number 24, are :
  1. (A)+2 to +6
  2. (B)+1 and +3 to +6
  3. (C)+1 and +3
  4. (D)+1 to +6

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List-I with List-II : Choose the correct match :
List-I (Chemical Compound)List-II (Used as)
a.Sucralosei.Synthetic detergent
b.Glyceryl ester of stearic acidii.Artificial sweetener
c.Sodium benzoateiii.Antiseptic
d.Bithionoliv.Food preservative
  1. (A)(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
  2. (B)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  3. (C)(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
  4. (D)(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Given below are two statements : Statement-I : 2-methylbutane on oxidation with KMnO4_{4}4​ gives 2-methylbutan-2-ol. Statement-II : n-alkanes can be easily oxidised to corresponding alcohol with KMnO4_{4}4​. Choose the correct option :
  1. (A)Both statement I and statement II are correct
  2. (B)Both statement I and statement II are incorrect
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Statement I is incorrect but Statement II is correct

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Nitrogen can be estimated by Kjeldahl's method for which of the following compound ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
Amongst the following, the linear species is :
  1. (A)NO2_{2}2​
  2. (B)Cl2_{2}2​O
  3. (C)O3_{3}3​
  4. (D)N3−_{3}^{-}3−​

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
C12_{12}12​H22_{22}22​O11_{11}11​ (Sucrose) + H2_{2}2​O →Enzyme A\xrightarrow{\text{Enzyme A}}Enzyme A​ C6_{6}6​H12_{12}12​O6_{6}6​ (Glucose) + C6_{6}6​H12_{12}12​O6_{6}6​ (Fructose) C6_{6}6​H12_{12}12​O6_{6}6​ (Glucose) →Enzyme B\xrightarrow{\text{Enzyme B}}Enzyme B​ 2C2_{2}2​H5_{5}5​OH + 2CO2_{2}2​ In the above reactions, the enzyme A and enzyme B respectively are :-
  1. (A)Amylase and Invertase
  2. (B)Invertase and Amylase
  3. (C)Invertase and Zymase
  4. (D)Zymase and Invertase

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
One of the by-products formed during the recovery of NH3_{3}3​ from Solvay process is :
  1. (A)Ca(OH)2_{2}2​
  2. (B)NaHCO3_{3}3​
  3. (C)CaCl2_{2}2​
  4. (D)NH4_{4}4​Cl

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
In the above reaction, the structural formula of (A), "X" and "Y" respectively are :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
For the coagulation of a negative sol, the species below, that has the highest flocculating power is :
  1. (A)SO42−_{4}^{2-}42−​
  2. (B)Ba2+^{2+}2+
  3. (C)Na+^{+}+
  4. (D)PO43−_{4}^{3-}43−​

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
Which of the following statement(s) is (are) incorrect reason for eutrophication ? (A) excess usage of fertilisers (B) excess usage of detergents (C) dense plant population in water bodies (D) lack of nutrients in water bodies that prevent plant growth Choose the most appropriate answer from the options given below :
  1. (A)(A) only
  2. (B)(C) only
  3. (C)(B) and (D) only
  4. (D)(D) only

Correct answer: (D)

Step-by-step solution →
Q47·Chemistry·Electronic Effects and StabilitySingle correct
Choose the correct statement regarding the formation of carbocations A and B given :-
  1. (A)Carbocation B is more stable and formed relatively at faster rate
  2. (B)Carbocation A is more stable and formed relatively at slow rate
  3. (C)Carbocation B is more stable and formed relatively at slow rate
  4. (D)Carbocation A is more stable and formed relatively at faster rate

Correct answer: (A)

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Q48·ChemistrySingle correct
During which of the following processes, does entropy decrease ? (A) Freezing of water to ice at 0°C (B) Freezing of water to ice at –10°C (C) N2_{2}2​(g) + 3H2_{2}2​(g) →\rightarrow→ 2NH3_{3}3​(g) (D) Adsorption of CO(g) and lead surface (E) Dissolution of NaCl in water
  1. (A)(A), (B), (C) and (D) only
  2. (B)(B) and (C) only
  3. (C)(A) and (E) only
  4. (D)(A), (C) and (E) only

Correct answer: (A)

Step-by-step solution →
Q49·ChemistryNumerical
A KCl solution of conductivity 0.14 S m−1^{-1}−1 shows a resistance of 4.19 Ω\OmegaΩ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω\OmegaΩ. The conductivity of the HCl solution is ____ ×\times× 10−2^{-2}−2 S m−1^{-1}−1. (Round off to the Nearest Integer).

Correct answer: 57

Step-by-step solution →
Q50·ChemistryNumerical
On complete reaction of FeCl3_{3}3​ with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is ____. (Round off to the Nearest Integer).

Correct answer: 6

Step-by-step solution →
Q51·ChemistryNumerical
The reaction 2A + B2_{2}2​ →\rightarrow→ 2AB is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____. (Round off to the Nearest Integer).

Correct answer: 27

Step-by-step solution →
Q52·ChemistryNumerical
The total number of C–C sigma bond/s in mesityl oxide (C6_{6}6​H10_{10}10​O) is____. (Round off to the Nearest Integer).

Correct answer: 5

Step-by-step solution →
Q53·ChemistryNumerical
A 1 molal K4_{4}4​Fe(CN)6_{6}6​ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is____ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm−3^{-3}−3]

Correct answer: 85

Step-by-step solution →
Q54·ChemistryNumerical
In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ______ ×\times× 10−1^{-1}−1 BM. (Round off to the Nearest Integer). [Given : 3\sqrt{3}3​ = 1.73, 2\sqrt{2}2​ = 1.41]

Correct answer: 49

Step-by-step solution →
Q55·ChemistryNumerical
Consider the reaction N2_{2}2​O4_{4}4​(g) ⇌\rightleftharpoons⇌ 2NO2_{2}2​(g). The temperature at which KC_{C}C​ = 20.4 and KP_{P}P​ = 600.1, is_____K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 354

Step-by-step solution →
Q56·ChemistryNumerical
Consider the above reaction. The percentage yield of amide product is ______. (Round off to the Nearest Integer). (Given : Atomic mass : C : 12.0 u, H : 1.0u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)

Correct answer: 77

Step-by-step solution →

Mathematics — JEE Main 17 March 2021 Shift 2

Q57·MathematicsSingle correct
Let f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)=e−xsin⁡xf(x) = e^{-x}\sin xf(x)=e−xsinx. If F:[0,1]→RF : [0, 1] \rightarrow RF:[0,1]→R is a differentiable function such that F(x)=∫0xf(t) dtF(x) = \int_{0}^{x} f(t)\,dtF(x)=∫0x​f(t)dt, then the value of ∫01(F′(x)+f(x))exdx\int_{0}^{1}\left(F'(x) + f(x)\right)e^{x}dx∫01​(F′(x)+f(x))exdx lies in the interval
  1. (A)[327360,329360]\left[\frac{327}{360}, \frac{329}{360}\right][360327​,360329​]
  2. (B)[330360,331360]\left[\frac{330}{360}, \frac{331}{360}\right][360330​,360331​]
  3. (C)[331360,334360]\left[\frac{331}{360}, \frac{334}{360}\right][360331​,360334​]
  4. (D)[335360,336360]\left[\frac{335}{360}, \frac{336}{360}\right][360335​,360336​]

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
If the integral ∫010[sin⁡2πx]ex−[x]dx=αe−1+βe−12+γ\int_{0}^{10}\frac{[\sin 2\pi x]}{e^{x-[x]}}dx = \alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma∫010​ex−[x][sin2πx]​dx=αe−1+βe−21​+γ, where α,β,γ\alpha, \beta, \gammaα,β,γ are integers and [x][x][x] denotes the greatest integer less than or equal to xxx, then the value of α+β+γ\alpha + \beta + \gammaα+β+γ is equal to :
  1. (A)000
  2. (B)202020
  3. (C)252525
  4. (D)101010

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation cos⁡x (3sin⁡x+cos⁡x+3)dy=(1+ysin⁡x (3sin⁡x+cos⁡x+3))dx\cos x\,(3\sin x + \cos x + 3)dy = (1 + y\sin x\,(3\sin x + \cos x + 3))dxcosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx, 0≤x≤π20 \le x \le \frac{\pi}{2}0≤x≤2π​, y(0)=0y(0) = 0y(0)=0. Then, y(π3)y\left(\frac{\pi}{3}\right)y(3π​) is equal to:
  1. (A)2log⁡e(23+96)2\log_{e}\left(\frac{2\sqrt{3}+9}{6}\right)2loge​(623​+9​)
  2. (B)2log⁡e(23+1011)2\log_{e}\left(\frac{2\sqrt{3}+10}{11}\right)2loge​(1123​+10​)
  3. (C)2log⁡e(3+72)2\log_{e}\left(\frac{\sqrt{3}+7}{2}\right)2loge​(23​+7​)
  4. (D)2log⁡e(33−84)2\log_{e}\left(\frac{3\sqrt{3}-8}{4}\right)2loge​(433​−8​)

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
The value of ∑r=06(6Cr⋅6C6−r)\sum_{r=0}^{6}\left({}^{6}C_{r}\cdot{}^{6}C_{6-r}\right)∑r=06​(6Cr​⋅6C6−r​) is equal to :
  1. (A)112411241124
  2. (B)132413241324
  3. (C)102410241024
  4. (D)924924924

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
The value of lim⁡n→∞[r]+[2r]+.....+[nr]n2\lim_{n\rightarrow\infty}\frac{[r]+[2r]+.....+[nr]}{n^{2}}limn→∞​n2[r]+[2r]+.....+[nr]​, where rrr is non-zero real number and [r][r][r] denotes the greatest integer less than or equal to rrr, is equal to :
  1. (A)r2\frac{r}{2}2r​
  2. (B)rrr
  3. (C)2r2r2r
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
The number of solutions of the equation sin⁡−1[x2+13]+cos⁡−1[x2−23]=x2\sin^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos^{-1}\left[x^{2}-\frac{2}{3}\right] = x^{2}sin−1[x2+31​]+cos−1[x2−32​]=x2, for x∈[−1,1]x \in [-1, 1]x∈[−1,1], and [x][x][x] denotes the greatest integer less than or equal to xxx, is :
  1. (A)222
  2. (B)000
  3. (C)444
  4. (D)Infinite

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be 12\frac{1}{2}21​ and probability of occurrence of 0 at the odd place be 13\frac{1}{3}31​. Then the probability that '10' is followed by '01' is equal to :
  1. (A)118\frac{1}{18}181​
  2. (B)13\frac{1}{3}31​
  3. (C)16\frac{1}{6}61​
  4. (D)19\frac{1}{9}91​

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
The number of solutions of the equation x+2tan⁡x=π2x + 2\tan x = \frac{\pi}{2}x+2tanx=2π​ in the interval [0,2π][0, 2\pi][0,2π] is :
  1. (A)333
  2. (B)444
  3. (C)222
  4. (D)555

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
Let S1S_{1}S1​, S2S_{2}S2​ and S3S_{3}S3​ be three sets defined as S1={z∈C:∣z−1∣≤2}S_{1} = \left\{z \in \mathbb{C} : |z-1| \le \sqrt{2}\right\}S1​={z∈C:∣z−1∣≤2​} S2={z∈C:Re((1−i)z)≥1}S_{2} = \left\{z \in \mathbb{C} : \mathrm{Re}\left((1-i)z\right) \ge 1\right\}S2​={z∈C:Re((1−i)z)≥1} S3={z∈C:Im(z)≤1}S_{3} = \left\{z \in \mathbb{C} : \mathrm{Im}(z) \le 1\right\}S3​={z∈C:Im(z)≤1} Then the set S1∩S2∩S3S_{1} \cap S_{2} \cap S_{3}S1​∩S2​∩S3​
  1. (A)is a singleton
  2. (B)has exactly two elements
  3. (C)has infinitely many elements
  4. (D)has exactly three elements

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
If the curve y=y(x)y = y(x)y=y(x) is the solution of the differential equation 2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx2(x^{2} + x^{5/4})dy - y(x + x^{1/4})dx = 2x^{9/4}dx2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx , x>0x > 0x>0 which passes through the point (1,1−43log⁡e2)\left(1, 1-\frac{4}{3}\log_{e}2\right)(1,1−34​loge​2), then the value of y(16)y(16)y(16) is equal to :
  1. (A)4(313+83log⁡e3)4\left(\frac{31}{3}+\frac{8}{3}\log_{e}3\right)4(331​+38​loge​3)
  2. (B)(313+83log⁡e3)\left(\frac{31}{3}+\frac{8}{3}\log_{e}3\right)(331​+38​loge​3)
  3. (C)4(313−83log⁡e3)4\left(\frac{31}{3}-\frac{8}{3}\log_{e}3\right)4(331​−38​loge​3)
  4. (D)(313−83log⁡e3)\left(\frac{31}{3}-\frac{8}{3}\log_{e}3\right)(331​−38​loge​3)

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :
  1. (A)364364364
  2. (B)240240240
  3. (C)333333333
  4. (D)360360360

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
If x,y,zx, y, zx,y,z are in arithmetic progression with common difference ddd, x≠3dx \neq 3dx=3d, and the determinant of the matrix [342x452y5kz]\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}​345​42​52​k​xyz​​ is zero, then the value of k2k^{2}k2 is
  1. (A)727272
  2. (B)121212
  3. (C)363636
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
Let O be the origin. Let OP→=xi^+yj^−k^\overrightarrow{OP} = x\hat{i}+y\hat{j}-\hat{k}OP=xi^+yj^​−k^ and OQ→=−i^+2j^+3xk^\overrightarrow{OQ} = -\hat{i}+2\hat{j}+3x\hat{k}OQ​=−i^+2j^​+3xk^, x,y∈Rx, y \in Rx,y∈R, x>0x > 0x>0, be such that ∣PQ→∣=20\left|\overrightarrow{PQ}\right| = \sqrt{20}​PQ​​=20​ and the vector OP→\overrightarrow{OP}OP is perpendicular to OQ→\overrightarrow{OQ}OQ​. If OR→=3i^+zj^−7k^\overrightarrow{OR} = 3\hat{i}+z\hat{j}-7\hat{k}OR=3i^+zj^​−7k^, z∈Rz \in Rz∈R, is coplanar with OP→\overrightarrow{OP}OP and OQ→\overrightarrow{OQ}OQ​, then the value of x2+y2+z2x^{2}+y^{2}+z^{2}x2+y2+z2 is equal to
  1. (A)777
  2. (B)999
  3. (C)222
  4. (D)111

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
Two tangents are drawn from a point P to the circle x2+y2−2x−4y+4=0x^{2}+y^{2}-2x-4y+4=0x2+y2−2x−4y+4=0, such that the angle between these tangents is tan⁡−1(125)\tan^{-1}\left(\frac{12}{5}\right)tan−1(512​), where tan⁡−1(125)∈(0,π)\tan^{-1}\left(\frac{12}{5}\right) \in (0, \pi)tan−1(512​)∈(0,π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB\Delta PABΔPAB and ΔCAB\Delta CABΔCAB is :
  1. (A)11:411 : 411:4
  2. (B)9:49 : 49:4
  3. (C)3:13 : 13:1
  4. (D)2:12 : 12:1

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Consider the function f:R→Rf : R \rightarrow Rf:R→R defined by f(x)={(2−sin⁡(1x))∣x∣,x≠00,x=0f(x) = \begin{cases} \left(2-\sin\left(\frac{1}{x}\right)\right)|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}f(x)={(2−sin(x1​))∣x∣,0,​x=0x=0​. Then fff is :
  1. (A)monotonic on (−∞,0)∪(0,∞)(-\infty, 0) \cup (0, \infty)(−∞,0)∪(0,∞)
  2. (B)not monotonic on (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞)
  3. (C)monotonic on (0,∞)(0, \infty)(0,∞) only
  4. (D)monotonic on (−∞,0)(-\infty, 0)(−∞,0) only

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
Let L be a tangent line to the parabola y2=4x−20y^{2} = 4x - 20y2=4x−20 at (6,2)(6, 2)(6,2). If L is also a tangent to the ellipse x22+y2b=1\frac{x^{2}}{2}+\frac{y^{2}}{b} = 12x2​+by2​=1, then the value of bbb is equal to :
  1. (A)111111
  2. (B)141414
  3. (C)161616
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
The value of the limit lim⁡θ→0tan⁡(πcos⁡2θ)sin⁡(2πsin⁡2θ)\lim_{\theta\rightarrow 0}\frac{\tan\left(\pi\cos^{2}\theta\right)}{\sin\left(2\pi\sin^{2}\theta\right)}limθ→0​sin(2πsin2θ)tan(πcos2θ)​ is equal to :
  1. (A)−12-\frac{1}{2}−21​
  2. (B)−14-\frac{1}{4}−41​
  3. (C)000
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let the tangent to the circle x2+y2=25x^{2}+y^{2}=25x2+y2=25 at the point R(3,4)R(3, 4)R(3,4) meet x-axis and y-axis at point P and Q, respectively. If rrr is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2r^{2}r2 is equal to
  1. (A)52964\frac{529}{64}64529​
  2. (B)12572\frac{125}{72}72125​
  3. (C)62572\frac{625}{72}72625​
  4. (D)58566\frac{585}{66}66585​

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
If the Boolean expression (p∧q)⊛(p⊗q)(p \wedge q) \circledast (p \otimes q)(p∧q)⊛(p⊗q) is a tautology, then ⊛\circledast⊛ and ⊗\otimes⊗ are respectively given by
  1. (A)→,→\rightarrow, \rightarrow→,→
  2. (B)∧,∨\wedge, \vee∧,∨
  3. (C)∨,→\vee, \rightarrow∨,→
  4. (D)∧,→\wedge, \rightarrow∧,→

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
If the equation of plane passing through the mirror image of a point (2,3,1)(2, 3, 1)(2,3,1) with respect to line x+12=y−31=z+2−1\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}2x+1​=1y−3​=−1z+2​ and containing the line x−23=1−y2=z+11\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}3x−2​=21−y​=1z+1​ is αx+βy+γz=24\alpha x + \beta y + \gamma z = 24αx+βy+γz=24, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to :
  1. (A)202020
  2. (B)191919
  3. (C)181818
  4. (D)212121

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsNumerical
If 111, log⁡10(4x−2)\log_{10}(4^{x}-2)log10​(4x−2) and log⁡10(4x+185)\log_{10}\left(4^{x}+\frac{18}{5}\right)log10​(4x+518​) are in arithmetic progression for a real number xxx, then the value of the determinant ∣2(x−12)x−1x210xx10∣\begin{vmatrix} 2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{vmatrix}​2(x−21​)1x​x−101​x2x0​​ is equal to :

Correct answer: 2

Step-by-step solution →
Q78·MathematicsNumerical
Let f:[−1,1]→Rf : [-1, 1] \rightarrow Rf:[−1,1]→R be defined as f(x)=ax2+bx+cf(x) = ax^{2}+bx+cf(x)=ax2+bx+c for all x∈[−1,1]x \in [-1, 1]x∈[−1,1], where a,b,c∈Ra, b, c \in Ra,b,c∈R such that f(−1)=2f(-1) = 2f(−1)=2, f′(−1)=1f'(-1) = 1f′(−1)=1 and for x∈(−1,1)x \in (-1, 1)x∈(−1,1) the maximum value of f′′(x)f''(x)f′′(x) is 12\frac{1}{2}21​. If f(x)≤αf(x) \le \alphaf(x)≤α, x∈[−1,1]x \in [-1, 1]x∈[−1,1], then the least value of α\alphaα is equal to ______.

Correct answer: 5

Step-by-step solution →
Q79·MathematicsNumerical
Let f:[−3,1]→Rf : [-3, 1] \rightarrow Rf:[−3,1]→R be given as f(x)={min⁡{(x+6),x2},−3≤x≤0max⁡{x,x2},0≤x≤1.f(x) = \begin{cases} \min\left\{(x+6), x^{2}\right\}, & -3 \le x \le 0 \\ \max\left\{\sqrt{x}, x^{2}\right\}, & 0 \le x \le 1. \end{cases}f(x)={min{(x+6),x2},max{x​,x2},​−3≤x≤00≤x≤1.​ If the area bounded by y=f(x)y = f(x)y=f(x) and x-axis is A, then the value of 6A6A6A is equal to ______.

Correct answer: 41

Step-by-step solution →
Q80·MathematicsNumerical
Let tan⁡α\tan\alphatanα, tan⁡β\tan\betatanβ and tan⁡γ\tan\gammatanγ; α,β,γ≠(2n−1)π2\alpha, \beta, \gamma \neq \frac{(2n-1)\pi}{2}α,β,γ=2(2n−1)π​, n∈Nn \in Nn∈N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of ΔABC\Delta ABCΔABC coincides with origin and its orthocentre lies on y-axis, then the value of (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha\cos\beta\cos\gamma}\right)^{2}(cosαcosβcosγcos3α+cos3β+cos3γ​)2 is equal to :

Correct answer: 144

Step-by-step solution →
Q81·MathematicsNumerical
Consider a set of 3n3n3n numbers having variance 4. In this set, the mean of first 2n2n2n numbers is 6 and the mean of the remaining nnn numbers is 3. A new set is constructed by adding 1 into each of first 2n2n2n numbers, and subtracting 1 from each of the remaining nnn numbers. If the variance of the new set is kkk, then 9k9k9k is equal to ______.

Correct answer: 68

Step-by-step solution →
Q82·MathematicsNumerical
Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax2)n\left(x+\frac{a}{x^{2}}\right)^{n}(x+x2a​)n, x≠0x \neq 0x=0, be in the ratio 12:8:312 : 8 : 312:8:3. Then the term independent of xxx in the expansion, is equal to ________.

Correct answer: 4

Step-by-step solution →
Q83·MathematicsNumerical
Let A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​] and B=[αβ]≠[00]B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}B=[αβ​]=[00​] such that AB=BAB = BAB=B and a+d=2021a + d = 2021a+d=2021, then the value of ad−bcad - bcad−bc is equal to ________.

Correct answer: 2020

Step-by-step solution →
Q84·MathematicsNumerical
Let x⃗\vec{x}x be a vector in the plane containing vectors a⃗=2i^−j^+k^\vec{a} = 2\hat{i}-\hat{j}+\hat{k}a=2i^−j^​+k^ and b⃗=i^+2j^−k^\vec{b} = \hat{i}+2\hat{j}-\hat{k}b=i^+2j^​−k^. If the vector x⃗\vec{x}x is perpendicular to (3i^+2j^−k^)\left(3\hat{i}+2\hat{j}-\hat{k}\right)(3i^+2j^​−k^) and its projection on a⃗\vec{a}a is 1762\frac{17\sqrt{6}}{2}2176​​, then the value of ∣x⃗∣2\left|\vec{x}\right|^{2}∣x∣2 is equal to ________.

Correct answer: 486

Step-by-step solution →
Q85·MathematicsNumerical
Let In=∫1ex19(log⁡∣x∣)ndxI_{n} = \int_{1}^{e} x^{19}\left(\log|x|\right)^{n}dxIn​=∫1e​x19(log∣x∣)ndx, where n∈Nn \in Nn∈N. If (20)I10=αI9+βI8(20)I_{10} = \alpha I_{9} + \beta I_{8}(20)I10​=αI9​+βI8​, for natural numbers α\alphaα and β\betaβ, then α−β\alpha - \betaα−β equal to _______.

Correct answer: 1

Step-by-step solution →
Q86·MathematicsNumerical
Let P be an arbitrary point having sum of the squares of the distance from the planes x+y+z=0x+y+z=0x+y+z=0, lx−nz=0lx-nz=0lx−nz=0 and x−2y+z=0x-2y+z=0x−2y+z=0, equal to 9. If the locus of the point P is x2+y2+z2=9x^{2}+y^{2}+z^{2}=9x2+y2+z2=9, then the value of l−nl-nl−n is equal to _______.

Correct answer: 0

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
  • 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
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • 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
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Carboxylic Acids and Derivatives 54/186
  • Chemistry in Everyday Life 60/186
  • Diazonium Salts and Reactions 53/186
  • IUPAC Nomenclature 37/186
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