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JEE Main 18 March 2021 Shift 1 Question Paper with Answers

18 March 2021 · March session · 86 questions

86 of the 90 questions from the JEE Main 18 March 2021 Shift 1 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
30
Chemistry
27
Mathematics
29

Physics — JEE Main 18 March 2021 Shift 1

Q1·PhysicsSingle correct
An oil drop of radius 2 mm with a density 3g cm−3^{-3}−3 is held stationary under a constant electric field 3.55 × 105^{5}5 V m−1^{-1}−1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider g = 9.81 m/s2^{2}2)
  1. (A)48.8 × 1011^{11}11
  2. (B)1.73 × 1010^{10}10
  3. (C)17.3 × 1010^{10}10
  4. (D)1.73 × 1012^{12}12

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
Match List–I with List–II. List–I (a) 10 km height over earth's surface (b) 70 km height over earth's surface (c) 180 km height over earth's surface (d) 270 km height over earth's surface List–II (i) Thermosphere (ii) Mesosphere (iii) Stratosphere (iv) Troposphere
  1. (A)(a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
  2. (B)(a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
  3. (C)(a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
  4. (D)(a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
Imagine that the electron in a hydrogen atom is replaced by a muon (µ). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :-
  1. (A)13.6 eV
  2. (B)2815.2 eV
  3. (C)331.2 eV
  4. (D)27.2 eV

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the x-direction. At a particular point in space and time, B⃗=2.0×10−8k^ T\vec{B} = 2.0 \times 10^{-8}\hat{k}\,\text{T}B=2.0×10−8k^T. (where, k^\hat{k}k^ is unit vector along z-direction) What is E⃗\vec{E}E at this point ?
  1. (A)0.6 j^\hat{j}j^​ V/m
  2. (B)6.0 k^\hat{k}k^ V/m
  3. (C)6.0 j^\hat{j}j^​ V/m
  4. (D)0.6 k^\hat{k}k^ V/m

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
  1. (A)ωMM+m\omega\dfrac{M}{M+m}ωM+mM​
  2. (B)ωM+2mM\omega\dfrac{M+2m}{M}ωMM+2m​
  3. (C)ωMM+2m\omega\dfrac{M}{M+2m}ωM+2mM​
  4. (D)ωM−2mM+2m\omega\dfrac{M-2m}{M+2m}ωM+2mM−2m​

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
Four identical long solenoids A, B, C and D are connected to each other as shown in the figure. If the magnetic field at the center of A is 3T, the field at the center of C would be : (Assume that the magnetic field is confined with in the volume of respective solenoid).
  1. (A)12T
  2. (B)6T
  3. (C)9T
  4. (D)1T

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
The time period of a simple pendulum is given by T=2πℓgT = 2\pi\sqrt{\dfrac{\ell}{g}}T=2πgℓ​​ . The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is:-
  1. (A)2%
  2. (B)3%
  3. (C)5%
  4. (D)4%

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-
  1. (A)t2/3^{2/3}2/3
  2. (B)t3/2^{3/2}3/2
  3. (C)t
  4. (D)t1/2^{1/2}1/2

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T ? (kB_{B}B​ is Boltzmann constant)
  1. (A)12kBT\dfrac{1}{2}k_{B}T21​kB​T
  2. (B)23kBT\dfrac{2}{3}k_{B}T32​kB​T
  3. (C)32kBT\dfrac{3}{2}k_{B}T23​kB​T
  4. (D)kBTk_{B}TkB​T

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A radioactive sample disintegrates via two independent decay processes having half lives T1/2(1)T_{1/2}^{(1)}T1/2(1)​ and T1/2(2)T_{1/2}^{(2)}T1/2(2)​ respectively. The effective half-life T1/2T_{1/2}T1/2​ of the nuclei is :
  1. (A)None of the above
  2. (B)T1/2=T1/2(1)+T1/2(2)T_{1/2} = T_{1/2}^{(1)} + T_{1/2}^{(2)}T1/2​=T1/2(1)​+T1/2(2)​
  3. (C)T1/2=T1/2(1)T1/2(2)T1/2(1)+T1/2(2)T_{1/2} = \dfrac{T_{1/2}^{(1)}T_{1/2}^{(2)}}{T_{1/2}^{(1)} + T_{1/2}^{(2)}}T1/2​=T1/2(1)​+T1/2(2)​T1/2(1)​T1/2(2)​​
  4. (D)T1/2=T1/2(1)+T1/2(2)T1/2(1)−T1/2(2)T_{1/2} = \dfrac{T_{1/2}^{(1)} + T_{1/2}^{(2)}}{T_{1/2}^{(1)} - T_{1/2}^{(2)}}T1/2​=T1/2(1)​−T1/2(2)​T1/2(1)​+T1/2(2)​​

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use γ = 1.4) :
  1. (A)–500 J
  2. (B)–400 J
  3. (C)400 J
  4. (D)200 J

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Å is :-
  1. (A)1178 × 10−9^{-9}−9 m
  2. (B)1178 × 10−6^{-6}−6 m
  3. (C)1178 × 10−12^{-12}−12 m
  4. (D)5890 × 10−7^{-7}−7 m

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :-
  1. (A)116\dfrac{1}{16}161​ times the mass of e−^{-}−
  2. (B)8 times the mass of e−^{-}−
  3. (C)16 times the mass of e−^{-}−
  4. (D)18\dfrac{1}{8}81​ times the mass of e−^{-}−

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :-
  1. (A)3.9
  2. (B)8.4
  3. (C)7.5
  4. (D)3.0

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
  1. (A)The bandwidth of resonance circuit will increase.
  2. (B)The resonance frequency will increase.
  3. (C)The quality factor will increase.
  4. (D)The quality factor and the resonance frequency will remain constant.

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :
  1. (A)2.5 ms
  2. (B)25 ms
  3. (C)2.5 s
  4. (D)0.25 ms

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as non-uniform and distorted. The doctor diagnosed the problem as :
  1. (A)Astigmatism
  2. (B)Myopia with Astigmatism
  3. (C)Presbyopia with Astigmatism
  4. (D)Myopia and hypermetropia

Correct answer: (B)

Step-by-step solution →
Q19·Physics·Magnetic Field of CurrentSingle correct
A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.
  1. (A)Loop assumes circular shape with its plane normal to the field.
  2. (B)Loop assumes circular shape with its plane parallel to the field.
  3. (C)Wire gets stretched to become straight.
  4. (D)Shape of the loop remains unchanged.

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :
  1. (A)9 T
  2. (B)27 T
  3. (C)12 T
  4. (D)3 T

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1a\dfrac{1}{a}a1​ s. The value of 'a' to the nearest integer is _______ .

Correct answer: 6

Step-by-step solution →
Q22·PhysicsNumerical
The circuit shown in the figure consists of a charged capacitor of capacity 3 µF and a charge of 30 µC. At time t = 0, when the key is closed, the value of current flowing through the 5 MΩ resistor is 'x' µ-A. The value of 'x to the nearest integer is _______.

Correct answer: 2

Step-by-step solution →
Q23·PhysicsNumerical
The voltage across the 10Ω resistor in the given circuit is x volt. The value of 'x' to the nearest integer is _____.

Correct answer: 70

Step-by-step solution →
Q24·PhysicsNumerical
Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a : b. Then a/b can be expressed as 1/x where x is _______ .

Correct answer: 32

Step-by-step solution →
Q25·PhysicsNumerical
A person is swimming with a speed of 10 m/s at an angle of 120° with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is 'x' m/s. The value of 'x' to the nearest integer is ______.

Correct answer: 5

Step-by-step solution →
Q26·PhysicsNumerical
A parallel plate capacitor has plate area 100 m2^{2}2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF. The value of ε0_{0}0​ = 8.85 × 10−12^{-12}−12 F.m−1^{-1}−1. The value of 'x' to the nearest integer is_____.

Correct answer: 161

Step-by-step solution →
Q27·PhysicsNumerical
A ball of mass 10 kg moving with a velocity 10310\sqrt{3}103​ m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x-axis. The velocity of the ball moving at 30° with x-axis is x m/s. The configuration of pieces after collision is shown in the figure below. The value of x to the nearest integer is _______ .

Correct answer: 20

Step-by-step solution →
Q28·PhysicsNumerical
As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s2^{2}2) The value of 'x' to the nearest integer is_____.

Correct answer: 10

Step-by-step solution →
Q29·PhysicsNumerical
An npn transistor operates as a common emitter amplifier with a power gain of 106^{6}6. The input circuit resistance is 100Ω and the output load resistance is 10 KΩ. The common emitter current gain 'β' will be ______. (Round off to the Nearest Integer)

Correct answer: 100

Step-by-step solution →
Q30·PhysicsNumerical
A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is ______ .

Correct answer: 10

Step-by-step solution →

Chemistry — JEE Main 18 March 2021 Shift 1

Q31·ChemistrySingle correct
Considering the above reaction, X and Y respectively are :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
The ionic radius of Na+^{+}+ ions is 1.02 Å. The ionic radii (in Å) of Mg2+^{2+}2+ and Al3+^{3+}3+, respectively, are-
  1. (A)1.05 and 0.99
  2. (B)0.72 and 0.54
  3. (C)0.85 and 0.99
  4. (D)0.68 and 0.72

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Reaction of Grignard reagent, C2_{2}2​H5_{5}5​MgBr with C8_{8}8​H8_{8}8​O followed by hydrolysis gives compound "A" which reacts instantly with Lucas reagent to give compound B, C10_{10}10​H13_{13}13​Cl. The Compound B is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Reagent, 1-naphthylamine and sulphanilic acid in acetic acid is used for the detection of
  1. (A)N2_{2}2​O
  2. (B)NO3−_{3}^{-}3−​
  3. (C)NO
  4. (D)NO2−_{2}^{-}2−​

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
A non-reducing sugar "A" hydrolyses to give two reducing mono saccharides. Sugar A is-
  1. (A)Fructose
  2. (B)Galactose
  3. (C)Glucose
  4. (D)Sucrose

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Match the list -I with list - II
List-I (Class of Drug)List-II (Example)
a.Antacidi.Novestrol
b.Artificial sweetenerii.Cimetidine
c.Antifertilityiii.Valium
d.Tranquilizersiv.Alitame
  1. (A)(a) – (ii), (b) – (iv), (c) – (i), (d) – (iii)
  2. (B)(a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
  3. (C)(a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
  4. (D)(a) – (ii), (b) – (iv), (c) – (iii), (d) – (i)

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
Consider the above chemical reaction and identify product "A"
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List-I with List-II Choose the most appropriate answer from the options given below :
List-IList-II
a.Chlorophylli.Ruthenium
b.Vitamin-B12_{12}12​ii.Platinum
c.Anticancer drugiii.Cobalt
d.Grubbs catalystiv.Magnesium
  1. (A)a-iii, b-ii, c-iv, d-i
  2. (B)a-iv, b-iii, c-ii, d-i
  3. (C)a-iv, b-iii, c-i, d-ii
  4. (D)a-iv, b-ii, c-iii, d-i

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Match List-I with List-II : Choose the most appropriate match :
List-I (Chemicals)List-II (Use / Preparation / Constituent)
a.Alcoholic potassium hydroxidei.Electrodes in batteries
b.Pd/ BaSO4_{4}4​ii.Obtained by addition reaction
c.BHC (Benzene hexachloride)iii.Used for β - elimination reaction
d.Polyacetyleneiv.Lindlar's catalyst
  1. (A)a-ii, b-i, c-iv, d-iii
  2. (B)a-iii, b-iv, c-ii, d-i
  3. (C)a-iii, b-i, c-iv, d-ii
  4. (D)a-ii, b-iv, c-i, d-iii

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
The satements that are TRUE : (A) Methane leads to both global warming and photochemical smog (B) Methane is generated from paddy fields (C) Methane is a stronger global warming gas than CO2_{2}2​ (D) Methane is a part of reducing smog Choose the most appropriate answer from the options given below :
  1. (A)(A), (B), (C) only
  2. (B)(A) and (B) only
  3. (C)(B), (C), (D) only
  4. (D)(A), (B), (D) only

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Match List-I with List-II Choose the most appropriate answer from the options given below :
List-IList-II
a.Ca(OCl)2_{2}2​i.Antacid
b.CaSO4_{4}4​.12\frac{1}{2}21​H2_{2}2​Oii.Cement
c.CaOiii.Bleach
d.CaCO3_{3}3​iv.Plaster of paris
  1. (A)a-i, b-iv, c-iii, d-ii
  2. (B)a-iii, b-ii, c-iv, d-i
  3. (C)a-iii, b-iv, c-ii, d-i
  4. (D)a-iii, b-ii, c-i, d-iv

Correct answer: (C)

Step-by-step solution →
Q42·Chemistry·IsomerismSingle correct
Compound with molecular formula C3_{3}3​H6_{6}6​O can show :
  1. (A)Positional isomerism
  2. (B)Both positional isomerism and metamerism
  3. (C)Metamerism
  4. (D)Functional group isomerism

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
The correct structures of trans-[NiBr2_{2}2​(PPh3_{3}3​)2_{2}2​] and meridonial-[Co(NH3_{3}3​)3_{3}3​(NO2_{2}2​)3_{3}3​], respectively, are
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
A certain orbital has no angular nodes and two radial nodes. The orbital is :
  1. (A)2s
  2. (B)3s
  3. (C)3p
  4. (D)2p

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Considering the above chemical reaction, identify the product "X" :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correct
Match List-I with List-II Choose the most appropriate answer from the options given below -
List-I (process)List-II (catalyst)
a.Deacron's processi.ZSM-5
b.Contact processii.CuCl2_{2}2​
c.Cracking of hydrocarbonsiii.Particles 'Ni'
d.Hydrogenation of vegetable oilsiv.V2_{2}2​O5_{5}5​
  1. (A)a-ii, b-iv, c-i, d-iii
  2. (B)a-i, b-iii, c-ii, d-iv
  3. (C)a-iii, b-i, c-iv, d-ii
  4. (D)a-iv, b-ii, c-i, d-iii

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
The number of ionisable hydrogens present in the product obtained from a reaction of phosphorus trichloride and phosphonic acid is:
  1. (A)3
  2. (B)0
  3. (C)2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is :
  1. (A)M2_{2}2​A3_{3}3​
  2. (B)M4_{4}4​A3_{3}3​
  3. (C)M4_{4}4​A
  4. (D)MA3_{3}3​

Correct answer: (B)

Step-by-step solution →
Q49·ChemistrySingle correct
The chemical that is added to reduce the melting point of the reaction mixture during the extraction of aluminium is :
  1. (A)Cryolite
  2. (B)Bauxite
  3. (C)Calamine
  4. (D)Kaolite

Correct answer: (A)

Step-by-step solution →
Q50·ChemistryNumerical
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ________ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]

Correct answer: 10

Step-by-step solution →
Q51·ChemistryNumerical
2 NO(g) + Cl2_{2}2​(g) ⇌ 2 NOCl(s) This reaction was studied at −10°C and the following data was obtained run [NO]0_{0}0​ [Cl2_{2}2​]0_{0}0​ r0_{0}0​ 1 0.10 0.10 0.18 2 0.10 0.20 0.35 3 0.20 0.20 1.40 [NO]0_{0}0​ and [Cl2_{2}2​]0_{0}0​ are the initial concentrations and r0_{0}0​ is the initial reaction rate. The overall order of the reaction is ______. (Round off to the Nearest Integer).

Correct answer: 3

Step-by-step solution →
Q52·ChemistryNumerical
For the reaction C2_{2}2​H6_{6}6​ → C2_{2}2​H4_{4}4​ + H2_{2}2​ the reaction enthalpy Δr_{r}r​H = ______ kJ mol−1^{-1}−1. (Round off to the Nearest Integer). [Given : Bond enthalpies in kJ mol−1^{-1}−1 : C–C : 347, C=C : 611; C–H : 414, H–H : 436]

Correct answer: 128

Step-by-step solution →
Q53·ChemistryNumerical
______ grams of 3-Hydroxy propanal (MW=74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C3_{3}3​H4_{4}4​O) if the percentage yield is 64. (Round off to the Nearest Integer). [Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]

Correct answer: 16

Step-by-step solution →
Q54·ChemistryNumerical
A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n × 10−1^{-1}−1, when n = ______. (Round off to the Nearest Integer). (Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]

Correct answer: 3

Step-by-step solution →
Q55·ChemistryNumerical
The total number of unpaired electrons present in the complex K3_{3}3​[Cr(oxalate)3_{3}3​] is ______.

Correct answer: 3

Step-by-step solution →
Q56·ChemistryNumerical
For the reaction 2Fe3+^{3+}3+(aq) + 2I−^{-}−(aq) → 2Fe2+^{2+}2+(aq) + I2_{2}2​(s) the magnitude of the standard molar free energy change, ΔrGm∘\Delta_{r}G^{\circ}_{m}Δr​Gm∘​ = – ______ kJ (Round off to the Nearest Integer). [EFe2+/Fe(s)∘E^{\circ}_{Fe^{2+}/Fe(s)}EFe2+/Fe(s)∘​ = −0.440 V; EFe3+/Fe(s)∘E^{\circ}_{Fe^{3+}/Fe(s)}EFe3+/Fe(s)∘​ = −0.036 V; EI2/2I−∘E^{\circ}_{I_{2}/2I^{-}}EI2​/2I−∘​ = 0.539 V; F = 96500 C]

Correct answer: 45

Step-by-step solution →
Q57·ChemistryNumerical
Complete combustion of 3 g of ethane gives x × 1022^{22}22 molecules of water. The value of x is ______. (Round off to the Nearest Integer). [Use : NA_{A}A​ = 6.023 × 1023^{23}23; Atomic masses in u : C : 12.0 ; O : 16.0 ; H : 1.0]

Correct answer: 18

Step-by-step solution →

Mathematics — JEE Main 18 March 2021 Shift 1

Q58·MathematicsSingle correct
The differential equation satisfied by the system of parabolas y2=4a(x+a)y^{2} = 4a(x + a)y2=4a(x+a) is :
  1. (A)y(dydx)2−2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right)^{2} - 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)2−2x(dxdy​)−y=0
  2. (B)y(dydx)2−2x(dydx)+y=0y\left(\dfrac{dy}{dx}\right)^{2} - 2x\left(\dfrac{dy}{dx}\right) + y = 0y(dxdy​)2−2x(dxdy​)+y=0
  3. (C)y(dydx)2+2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right)^{2} + 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)2+2x(dxdy​)−y=0
  4. (D)y(dydx)+2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right) + 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)+2x(dxdy​)−y=0

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
The number of integral values of m so that the abscissa of point of intersection of lines 3x+4y=93x + 4y = 93x+4y=9 and y=mx+1y = mx + 1y=mx+1 is also an integer, is :
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)000

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
Let (1+x+2x2)20=a0+a1x+a2x2+...+a40x40(1 + x + 2x^{2})^{20} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{40}x^{40}(1+x+2x2)20=a0​+a1​x+a2​x2+...+a40​x40. then a1+a3+a5+...+a37a_{1} + a_{3} + a_{5} + ... + a_{37}a1​+a3​+a5​+...+a37​ is equal to
  1. (A)220(220−21)2^{20}(2^{20} - 21)220(220−21)
  2. (B)219(220−21)2^{19}(2^{20} - 21)219(220−21)
  3. (C)219(220+21)2^{19}(2^{20} + 21)219(220+21)
  4. (D)220(220+21)2^{20}(2^{20} + 21)220(220+21)

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correct
The solutions of the equation ∣1+sin⁡2xsin⁡2xsin⁡2xcos⁡2x1+cos⁡2xcos⁡2x4sin⁡2x4sin⁡2x1+4sin⁡2x∣=0,(0<x<π)\begin{vmatrix} 1+\sin^{2}x & \sin^{2}x & \sin^{2}x \\ \cos^{2}x & 1+\cos^{2}x & \cos^{2}x \\ 4\sin 2x & 4\sin 2x & 1+4\sin 2x \end{vmatrix} = 0,\left(0 < x < \pi\right)​1+sin2xcos2x4sin2x​sin2x1+cos2x4sin2x​sin2xcos2x1+4sin2x​​=0,(0<x<π), are
  1. (A)π12,π6\dfrac{\pi}{12}, \dfrac{\pi}{6}12π​,6π​
  2. (B)π6,5π6\dfrac{\pi}{6}, \dfrac{5\pi}{6}6π​,65π​
  3. (C)5π12,7π12\dfrac{5\pi}{12}, \dfrac{7\pi}{12}125π​,127π​
  4. (D)7π12,11π12\dfrac{7\pi}{12}, \dfrac{11\pi}{12}127π​,1211π​

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
Choose the correct statement about two circles whose equations are given below : x2+y2−10x−10y+41=0x^{2} + y^{2} - 10x - 10y + 41 = 0x2+y2−10x−10y+41=0 x2+y2−22x−10y+137=0x^{2} + y^{2} - 22x - 10y + 137 = 0x2+y2−22x−10y+137=0
  1. (A)circles have same centre
  2. (B)circles have no meeting point
  3. (C)circles have only one meeting point
  4. (D)circles have two meeting points

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let α,β,γ\alpha, \beta, \gammaα,β,γ be the real roots of the equation, x3+ax2+bx+c=0x^{3} + ax^{2} + bx + c = 0x3+ax2+bx+c=0, (a,b,c∈R(a, b, c \in R(a,b,c∈R and a,b≠0)a, b \neq 0)a,b=0). If the system of equations (in, u, v, w) given by αu+βv+γw=0\alpha u + \beta v + \gamma w = 0αu+βv+γw=0, βu+γv+αw=0\beta u + \gamma v + \alpha w = 0βu+γv+αw=0; γu+αv+βw=0\gamma u + \alpha v + \beta w = 0γu+αv+βw=0 has non-trivial solution, then the value of a2b\dfrac{a^{2}}{b}ba2​ is
  1. (A)555
  2. (B)333
  3. (C)111
  4. (D)000

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
The integral ∫(2x−1)cos⁡(2x−1)2+54x2−4x+6 dx\displaystyle\int \dfrac{(2x-1)\cos\sqrt{(2x-1)^{2}+5}}{\sqrt{4x^{2}-4x+6}}\,dx∫4x2−4x+6​(2x−1)cos(2x−1)2+5​​dx is equal to (where c is a constant of integration)
  1. (A)12sin⁡(2x−1)2+5+c\dfrac{1}{2}\sin\sqrt{\left(2x-1\right)^{2}+5} + c21​sin(2x−1)2+5​+c
  2. (B)12cos⁡(2x+1)2+5+c\dfrac{1}{2}\cos\sqrt{\left(2x+1\right)^{2}+5} + c21​cos(2x+1)2+5​+c
  3. (C)12cos⁡(2x−1)2+5+c\dfrac{1}{2}\cos\sqrt{\left(2x-1\right)^{2}+5} + c21​cos(2x−1)2+5​+c
  4. (D)12sin⁡(2x+1)2+5+c\dfrac{1}{2}\sin\sqrt{\left(2x+1\right)^{2}+5} + c21​sin(2x+1)2+5​+c

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
The equation of one of the straight lines which passes through the point (1,3)(1,3)(1,3) and makes an angles tan⁡−1(2)\tan^{-1}\left(\sqrt{2}\right)tan−1(2​) with the straight line, y+1=32 xy + 1 = 3\sqrt{2}\,xy+1=32​x is
  1. (A)42x+5y−(15+42)=04\sqrt{2}x + 5y - \left(15 + 4\sqrt{2}\right) = 042​x+5y−(15+42​)=0
  2. (B)52x+4y−(15+42)=05\sqrt{2}x + 4y - \left(15 + 4\sqrt{2}\right) = 052​x+4y−(15+42​)=0
  3. (C)42x+5y−42=04\sqrt{2}x + 5y - 4\sqrt{2} = 042​x+5y−42​=0
  4. (D)42x−5y−(5+42)=04\sqrt{2}x - 5y - \left(5 + 4\sqrt{2}\right) = 042​x−5y−(5+42​)=0

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
If lim⁡x→0sin⁡−1x−tan⁡−1x3x3\lim_{x \to 0} \dfrac{\sin^{-1}x - \tan^{-1}x}{3x^{3}}limx→0​3x3sin−1x−tan−1x​ is equal to L, then the value of (6L+1)(6L + 1)(6L+1) is
  1. (A)16\dfrac{1}{6}61​
  2. (B)12\dfrac{1}{2}21​
  3. (C)666
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
A vector a⃗\vec{a}a has components 3p3p3p and 111 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a⃗\vec{a}a has components p+1p + 1p+1 and 10\sqrt{10}10​, then a value of p is equal to:
  1. (A)111
  2. (B)−54-\dfrac{5}{4}−45​
  3. (C)45\dfrac{4}{5}54​
  4. (D)−1-1−1

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
If the equation a∣z∣2+α‾z+αz‾+d=0a\left|z\right|^{2} + \overline{\alpha}z + \alpha\overline{z} + d = 0a∣z∣2+αz+αz+d=0 represents a circle where a,d are real constants then which of the following condition is correct ?
  1. (A)∣α∣2−ad≠0\left|\alpha\right|^{2} - ad \neq 0∣α∣2−ad=0
  2. (B)∣α∣2−ad>0|\alpha|^{2} - ad > 0∣α∣2−ad>0 and a∈R−{0}a \in R - \{0\}a∈R−{0}
  3. (C)∣α∣2−ad≥0|\alpha|^{2} - ad \geq 0∣α∣2−ad≥0 and a∈Ra \in Ra∈R
  4. (D)α=0\alpha = 0α=0, a,d∈R+a, d \in R^{+}a,d∈R+

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
For the four circles M, N, O and P, following four equations are given : Circle M : x2+y2=1x^{2} + y^{2} = 1x2+y2=1 Circle N : x2+y2−2x=0x^{2} + y^{2} - 2x = 0x2+y2−2x=0 Circle O : x2+y2−2x−2y+1=0x^{2} + y^{2} - 2x - 2y + 1 = 0x2+y2−2x−2y+1=0 Circle P : x2+y2−2y=0x^{2} + y^{2} - 2y = 0x2+y2−2y=0 If the centre of circle M is joined with centre of the circle N, further centre of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a :
  1. (A)Rhombus
  2. (B)Square
  3. (C)Rectangle
  4. (D)Parallelogram

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
If α,β\alpha, \betaα,β are natural numbers such that 100α−199β=(100)(100)+(99)(101)+(98)(102)+.....+(1)(199)100^{\alpha} - 199\beta = (100)(100) + (99)(101) + (98)(102) + ..... + (1)(199)100α−199β=(100)(100)+(99)(101)+(98)(102)+.....+(1)(199), then the slope of the line passing through (α,β)(\alpha, \beta)(α,β) and origin is :
  1. (A)540540540
  2. (B)550550550
  3. (C)530530530
  4. (D)510510510

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
The real valued function f(x)=cosec−1xx−[x]f\left(x\right) = \dfrac{\mathrm{cosec}^{-1}x}{\sqrt{x - \left[x\right]}}f(x)=x−[x]​cosec−1x​, where [x][x][x] denotes the greatest integer less than or equal to x, is defined for all x belonging to :
  1. (A)all reals except integers
  2. (B)all non-integers except the interval [−1,1][-1,1][−1,1]
  3. (C)all integers except 0,−1,10, -1, 10,−1,1
  4. (D)all reals except the Interval [−1,1][-1,1][−1,1]

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
132−1+152−1+172−1+...+1(201)2−1\dfrac{1}{3^{2}-1} + \dfrac{1}{5^{2}-1} + \dfrac{1}{7^{2}-1} + ... + \dfrac{1}{\left(201\right)^{2}-1}32−11​+52−11​+72−11​+...+(201)2−11​ is equal to
  1. (A)101404\dfrac{101}{404}404101​
  2. (B)25101\dfrac{25}{101}10125​
  3. (C)101408\dfrac{101}{408}408101​
  4. (D)99400\dfrac{99}{400}40099​

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
If the functions are defined as f(x)=xf\left(x\right) = \sqrt{x}f(x)=x​ and g(x)=1−xg\left(x\right) = \sqrt{1 - x}g(x)=1−x​, then what is the common domain of the following functions : f+gf + gf+g, f−gf - gf−g, f/gf/gf/g, g/fg/fg/f, g−fg - fg−f where (f±g)(x)=f(x)±g(x)(f \pm g)(x) = f(x) \pm g(x)(f±g)(x)=f(x)±g(x), (f/g)(x)=f(x)g(x)(f/g)(x) = \dfrac{f\left(x\right)}{g\left(x\right)}(f/g)(x)=g(x)f(x)​
  1. (A)0≤x≤10 \leq x \leq 10≤x≤1
  2. (B)0≤x<10 \leq x < 10≤x<1
  3. (C)0<x<10 < x < 10<x<1
  4. (D)0<x≤10 < x \leq 10<x≤1

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
If f(x)={1∣x∣;    ∣x∣≥1ax2+b;    ∣x∣<1f\left(x\right) = \begin{cases} \dfrac{1}{\left|x\right|} & ; \;\; \left|x\right| \geq 1 \\ ax^{2} + b & ; \;\; \left|x\right| < 1 \end{cases}f(x)=⎩⎨⎧​∣x∣1​ax2+b​;∣x∣≥1;∣x∣<1​ is differentiable at every point of the domain, then the values of a and b are respectively :
  1. (A)12,12\dfrac{1}{2}, \dfrac{1}{2}21​,21​
  2. (B)12,−32\dfrac{1}{2}, -\dfrac{3}{2}21​,−23​
  3. (C)52,−32\dfrac{5}{2}, -\dfrac{3}{2}25​,−23​
  4. (D)−12,32-\dfrac{1}{2}, \dfrac{3}{2}−21​,23​

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Let A+2B=[1206−33−531]A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}A+2B=​16−5​2−33​031​​ and 2A−B=[2−152−16012]2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}2A−B=​220​−1−11​562​​. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) −-− Tr(B) has value equal to
  1. (A)111
  2. (B)222
  3. (C)000
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:
  1. (A)266642666426664
  2. (B)122664122664122664
  3. (C)122234122234122234
  4. (D)222642226422264

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsSingle correct
The value of 3+14+13+14+13+...∞3 + \cfrac{1}{4 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{3 + ...\infty}}}}3+4+3+4+3+...∞1​1​1​1​ is equal to
  1. (A)1.5+31.5 + \sqrt{3}1.5+3​
  2. (B)2+32 + \sqrt{3}2+3​
  3. (C)3+233 + 2\sqrt{3}3+23​
  4. (D)4+34 + \sqrt{3}4+3​

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsNumerical
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is

Correct answer: 300

Step-by-step solution →
Q79·MathematicsNumerical
Let the plane ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0 bisect the line joining the points (4,−3,1)(4, -3, 1)(4,−3,1) and (2,3,−5)(2, 3, -5)(2,3,−5) at the right angles. If a, b, c, d are integers, then the minimum value of (a2+b2+c2+d2)(a^{2} + b^{2} + c^{2} + d^{2})(a2+b2+c2+d2) is

Correct answer: 28

Step-by-step solution →
Q80·MathematicsNumerical
The missing value in the following figure is

Correct answer: 4

Step-by-step solution →
Q81·MathematicsNumerical
Let z1,z2z_{1}, z_{2}z1​,z2​ be the roots of the equation z2+az+12=0z^{2} + az + 12 = 0z2+az+12=0 and z1,z2z_{1}, z_{2}z1​,z2​ form an equilateral triangle with origin. Then, the value of ∣a∣|a|∣a∣ is

Correct answer: 6

Step-by-step solution →
Q82·MathematicsNumerical
The equation of the planes parallel to the plane x−2y+2z−3=0x - 2y + 2z - 3 = 0x−2y+2z−3=0 which are at unit distance from the point (1,2,3)(1, 2, 3)(1,2,3) is ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0. If (b−d)=K(c−a)(b - d) = K(c - a)(b−d)=K(c−a), then the positive value of K is

Correct answer: 4

Step-by-step solution →
Q83·MathematicsNumerical
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is_.

Correct answer: 35

Step-by-step solution →
Q84·MathematicsNumerical
If f(x)=∫5x8+7x6(x2+1+2x7)2 dx,(x≥0),f(0)=0f\left(x\right) = \displaystyle\int \dfrac{5x^{8} + 7x^{6}}{\left(x^{2} + 1 + 2x^{7}\right)^{2}}\,dx, \left(x \geq 0\right), f\left(0\right) = 0f(x)=∫(x2+1+2x7)25x8+7x6​dx,(x≥0),f(0)=0 and f(1)=1Kf\left(1\right) = \dfrac{1}{K}f(1)=K1​, then the value of K is

Correct answer: 4

Step-by-step solution →
Q85·MathematicsNumerical
A square ABCD has all its vertices on the curve x2y2=1x^{2}y^{2} = 1x2y2=1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is

Correct answer: 80

Step-by-step solution →
Q86·MathematicsNumerical
The number of solutions of the equation ∣cot⁡x∣=cot⁡x+1sin⁡x\left|\cot x\right| = \cot x + \dfrac{1}{\sin x}∣cotx∣=cotx+sinx1​ in the interval [0,2π][0, 2\pi][0,2π] is

Correct answer: 1

Step-by-step solution →

Chapters tested in this paper

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

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • 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
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • 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
  • 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
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
  • Solid State 63/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • Diazonium Salts and Reactions 53/186
  • Isomerism 51/186
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