JEE Main 2020 Question Paper with Solution (8th January - Morning)
74 Questions MCQ Test Additional Documents and Tests for JEE | JEE Main 2020 Question Paper with Solution (8th January - Morning)
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A block of mass m is connected at one end of spring fixed at other end having natural length l0 and spring constant K. The block is rotated with constant angular speed (ω) about the fixed end in gravity free space. The elongation in spring is-
3 point charges are placed on circumference of a circle of radius 'd' as shown in figure. The electric field along x-axis at centre of circle is:
Choose the correct P-V graph of ideal gas for given V-T graph.
Find the coordinates of centre of mass of the lamina, shown in figure.
Which graph correctly represents v ariation between relaxation time 
of gas molecules with absolute temperature (T) of an ideal gas.
If two capacitors C1 & C2 are connected in parallel then equivalent capacitance is 10μF. If both capacitor are connected across 1V battery then energy stored by C2 is 4 times that of C1. Then the equivalent capacitance if they are connected in series is -
Given C1 + C2 = 10μF ........ (i)
⇒ 4C1 = C2 …(ii)
from equation (i) & (ii)
C1 = 2μF
C2 = 8μF
If they are in series
A rod of mass 4m and length L is hinged at the mid point. A ball of mass 'm' moving with speed V in the plane of rod which is perpendicular to axis of rotation, strikes at one end at an angle of 45º and sticks to it. The angular velocity of system after collision is–
Loi = Lof
Two photons of energy 4eV and 4.5eV are incident on two metals A and B respectively. Maximum kinetic energy for ejected electron is TA for A and TB = TA – 1.5eV for metal B. Relation between de-Broglie wavelengths of ejected electron of A and B are λB = 2λA. The work function of metal B is -
Relation between De-Broglie wav elength and K.E. is
⇒ TA = 2 eV
∴ KEB = 2 – 1.5 = 0.5 eV
φB = 4.5 - 0.5 = 4 eV
There is a potentiometer wire of length 1200 cm and 60 mA current is flowing in it. A battery of emf 5V and internal resistance of 20Ω is balanced on potentiometer wire with balancing length 1000 cm. The resistance of potentiometer wire is–
Potential gradient = 5 / 1000 = Vp/1200
VP = 6V
and RP = 
= 100Ω
A telescope has magnifying power 5 and length of tube is 60cm then the focal length of eye piece is–
m = fo/fe
5 = fo/fe
fo = 5fe
fo + fe = 60
6fe = 60
fe = 10
Two spherical bodies of mass m1 and m2 have radii 1 m and 2 m respectively. The gravitational field of the two bodies with the radial distance from centre is shown below. The value of 
is-
When a proton of KE = 1.0 MeV moving in South to North direction enters the magnetic field (from West to East direction), it accelerates with acceleration a = 1012 m/s2. The magnitude of magnetic field is–
∴ Bqv = ma
= 0.71× 10–3T
So, 0.71 mT
If electric field around a surface is given by 
where 'A' is the normal area of surface and Qin is the charge enclosed by the surface. This relation of gauss's law is valid when
Magnitude of electric field is constant & the surface is equipotential
Stopping potential depends on Planks constant (h), current (I), Universal gravitational constant (G) and speed of light (C). Choose the correct option for the dimension of stopping potential (V)
V = K (h)a(I)b(G)c(C)d (V is voltage)
we know [h] = ML2T-1
a – c = 1………………(1)
2a + 3c + d = 2………………(2)
–a –2c –d = –3 ………………(3)
b = –1………………(4)
on solving
c = –1
a = 0
d = 5,
b = –1
V = K (h)° (I)-1(G)-1(C)5
A cylinder of height 1m is floating in water at 0°C with 20 cm height in air. Now temperature of water is raised to 4°C, height of cylinder in air becomes 21cm. The ratio of density of water at 4°C to density of water at 0°C is– (Consider expansion of cylinder is negligible)
mg = A(80) ρ0ºc g
mg = A(79) ρ4ºc g
Number N of the α-particles deflected in Rutherford's α-scattering experiment varies with the angle of deflection θ. Then the graph between the two is best represented by.
If relative permittivity and relative permeability of a medium are 3 and 4/3 respectively. The critical angle for this medium is.
The given loop is kept in a uniform magnetic field perpendicular to plane of loop. The field changes from 1000G to 500G in 5seconds. The average induced emf in loop is–
Choose the correct Boolean expression for the given circuit diagram:
First part of figure shown is OR gate and Second part of figure shown is NOT gate
So Yp = OR + NOT = NOR gate
A solid sphere of density 
just floats in a liquid. The density of liquid is –
(r is distance from centre of sphere)
0 < r ≤ R
mg = B
Two particles each of mass 0.10kg are moving with velocities 3m/s along x-axis and 5m/s along y-axis respectively. After an elastic collision one of the mass moves with a velocity 
. The energy of other particle after collision is x/10 , then x is.
For elastic collision KEi = KEf
34 = 32 + v2
x = 1
A plano-convex lens of radius of curvature 30 cm made of refractive index 1.5 is kept in air. Find its focal length (in cm).
R1 = ∝
R2 = –30 cm
f = 60 cm
Position of two particles A and B as a function of time are given by XA = – 3t2 + 8t + c and YB = 10 – 8t3. The velocity of B with respect to A at t = 1 is √v . Find v.
= 580
An open organ pipe of length 1m contains an ideal gas whose density is twice the density of atmosphere at STP. Find the difference between fundamental and second harmonic frequencies if speed of sound in atmosphere is 300 m/s.
Four resistors of 15Ω, 12Ω, 4Ω and 10Ω are arranged in cyclic order to form a wheatstone bridge. What resistance (in Ω) should be connected in parallel across the 10Ω resistor to balance the wheatstone bridge.
= 15 × 4
on solving
R = 10 Ω
Number of S–O bond in S2O82– and number of S–S bond in Rhombic sulphur are respectively:
Following vanderwaal forces are present in ethyl acetate liquid
Ethyl acetate is polar molecule so dipole-dipole interaction will be present there.
Given, for H-atom
Select the correct options regarding this formula for Balmer series.
Correct Answer : a,c
Explanation : (a) For balmer series always started ground state n 1 = 2
(c) At the longest wavelength, the higher state energy will be minimum therefore the excited state will be n2 = 3 next to the ground always.
Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol-1 respectively are:
The electronic configurations are as follows:
11Na [Ne]3s1, 12Mg [Ne]3s2, 13Al [Ne] 3s23p1, 14Si [Ne]3s23p2
The ionization energy of Mg will be larger than that of Na due to fully filled configuration [Ne] 3s2
The ionization of Al will be smaller than that of Mg due to one electron extra than the stable configuration but smaller than Si due to the increase in effective nuclear charge of Si.
⇒ Na < Mg > Al < S
Select the correct stoichiometry and its ksp value according to given graphs.
Ksp = [X+] [Y–]
or Ksp = 2 × 10–3 × 10–3
or Ksp = 2 × 10–6
According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is :
According to hardy-schultz rule,
Coagulation value or flocculation value 
Which of the following complex exhibit facial meridional geometrical isomerism.
[Ma3b3] type complex shows facial and meridional isomerism
(A) Intermolecular force of attraction of X > Y.
(B) Intermolecular force of attraction of X < Y.
(C) Intermolecular force of attraction of Z < X.
Select the correct option(s).
At a particular temperature as intermolecular force of attraction increases vapour pressure decreases.
Rate of a reaction increases by 106 times when a reaction is carried out in presence of enzyme catalyst at same temperature. Determine change in activation energy.
or 6 ln 10 = (E - Ec ) / RT
or 
or E–EC = 2.303 × 6RT
or 
= 
Gypsum on heating at 393K produces
Theory based.
Among the following least 3rd ionization energy is for
Accurate measurement of concentration of NaOH can be performed by following titration:
Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.
Arrange the following compounds in order of dehydrohalogenation (E1) reaction.
(A)
(B) 
(C) 
(D) 
E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the E1 reaction.
Product A and B are respectively :
[A] is more stable radical and undergoes Markovnikov addition to form [B].
Major product in the following reaction is
Arrange the order of C—OH bond length of the following compounds.
CH3–OH
A B C
There is not any resonance in CH3–OH. Resonance is poor in p-Ethoxyphenol than phenol
Which of the following are "green house gases"?
(a) CO2
(b) O2
(c) O3
(d) CFC
(e) H2O
CO2, O3, H2O vapours and CFC's are green house gases.
Two liquids isohexane and 3-methylpentane has boiling point 60°C and 63°C. They can be separated by
Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can't be used as boiling point difference is very small.
Which of the given statement is incorrect about glucose?
Open chain form of glucose is very very small, hence does not gives Schiff's test.
Reagent used for the given conversion is:
B2H6 is very selective and usually used to reduce acid to alcohol.
0.3 g [ML6]Cl3 of molar mass 267.46 g/mol is reacted with 0.125 M AgNO3(aq) solution, calculate volume of AgNO3 required in ml.
Given : 2H2O → O2 + 4H+ + 4e–
Eº = –1.23 V Calculate electrode potential at pH = 5.
Calculated the mass of FeSO4.7H2O, which must be added in 100 kg of wheat to get 10 PPM of Fe.
A gas undergoes expansion according to the following graph. Calculate work done by the gas.
Number of chiral centres in Pencillin is
Star marked atoms are chiral centers.
If f(x) = 
g(x) = 
then find the area bounded by f(x) and g(x) from x = 1/2 to x = 
Required area = Area of trapezium ABCD – 
z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be
z = x + iy
|x| + |y| = 4
Minimum value of |z| = 2√2
Maximum value of |z| = 4
So |z| can't be √7
If f(x) = 
and a – 2b + c = 1 then
Apply R1 = R1 + R3 – 2R2
⇒ f(x) = 1
⇒ f(50) = 1
Let an is a positive term of a GP and 
Let GP is a, ar, ar2 ……..
= 200
................. (i)
................... (ii)
Form (1) and (2) r = 2
add both
⇒ a2 + a3 + ........... a200 + a201 = 300 ⇒ r(a1 +...........a200) = 300
= 300/r
= 150
If 
y(1) = 1 and y(x) = e then x = ?
Put y = vx
When x = 1, y = 1 then
x2 = y2(1 + 2lny)
x2 = e2(3)
x = ± √3 e
So x = √3e
Let probability distribution is 
then value of p(x > 2) is
⇒ 6k2 + 5k = 1
6k2 + 5k – 1 = 0
6k2 + 6k – k – 1 = 0
(6k – 1) (k + 1) = 0 ⇒ k = - 1
(rejected) ; k = 1/6
P(x > 2) = k + 2k + 5k2
= λtanθ + 2log f(x) + c then ordered pair (λ,f(x)) is
= – t + 2 log (1+t) + C
= –tanθ + 2 log (1 + tanθ) + C
⇒ λ = -1 and f(x) = 1 + tanθ
If p → (p ∧ ~ q) is false. Truth value of p & q will be
If 
= A then the value of x at which f(x) = [x2] sinπx is discontinuous
(where [.] denotes greatest integer function)
4 – 0 = A check when
(A) x = 
⇒ x = √5 ⇒ discontinuos
(B) x = 
⇒ x = 5 ⇒ continuous
(C) x = √A ⇒ x = 2 ⇒ continuous
(D) x = 
⇒ x = 3 ⇒ continuous
Let one end of focal chord of parabola y2 = 8x is 
, then equation of tangent at other end of this focal chord is
Let
is (2t2, 4t) ⇒ t = -1/2
Parameter of other end of focal chord is 2
⇒ point is (8, 8)
⇒ equation of tangent is 8y - 4(x+8) = 0
⇒ 2y - x = 8
Let x + 6y = 8 is tangent to standard ellipse where minor axis is 
, then eccentricity of ellipse is
, 
Equation of tangent ≡ y = mx ± 
comparing with ≡ 
m = -1/6 and a2m2 + b2 = 16/9
a2 = 16
if f(x) and g(x) are continuous functions, fog is identity function, g'(b) = 5 and g(b) = a then f'(a) is
If 7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0
then which option is correct
= 7(–20) – 6(–20) – 2(–10)
= – 140 + 120 + 20 = 0
so infinite non-trivial solution exist
now equation (1) + 3 equation (3)
10x – 20z = 0
x = 2z
Let x = 2sinθ - sin2θ and y = 2 cosθ - cos2θ
find 
θ = π
g(x) = 
dt then correct choice is
F'(x) = x2g(x)
⇒ F'(1) = 1.g(1) = 0 ......(1)
(∵ g(1) = 0)
Now F''(x) = 2xg(x) + x2g'(x)
Let both root of equation ax2 – 2bx + 5 = 0 are α and root of equation x2 - 2bx - 10 = 0 are α and β. Find the value of α2 + β2
⇒ b2 = 5a ..... (i) (a ≠ 0)
α + β = 2b ............ (ii)
and αβ = – 10 ............. (iii)
is also root of x2 – 2bx – 10 = 0
⇒ b2 - 2ab2 - 10a2 = 0 by (i)
⇒ 5a – 10a2 – 10a2 = 0
⇒ 20a2 = 5a
⇒ a = 1/4 and b2 = 5/4
α2 = 20 and β2 = 5
Now α2 + β2
= 5 + 20
= 25
Let A = 
and B = 
then
A = { x : x ∈ (-2, 2) }
B = { x : x ∈ (-∞, -1] ∪ [5, ∞) }
A ∩ B = { x : x ∈ (-2, -1] }
A ∪ B = { x : x ∈ (-∞, 2) ∪ [5, ∞) }
A - B = { x : x ∈ (-1, 2) }
B - A = { x : x ∈ (-∞, -2] ∪ [5, ∞) }
Let x = 
and y = 
where θ ∈ (0,π/4), then
y = 1 + cos2θ + cos4θ + .......
cos2θ = x
y (1 – x) = 1
Let the distance between plane passing through lines 
and 
= 
and plane 23x – 10y – 2z + 48 = 0 is 
then k is equal to
Lines must be intersecting
distance of plane contains given lines from given plane is same as distance between point (–3, –2,1) from given plane.
Required distance equal to 
= 
= 
= k = 3
If 25C0 + 525C1 + 925C2 ……… 101 25C25 = 225k find k = ?
= 100 .224 + 225 = 225(50 + 1) = 51.225
So k = 51
Let circles (x – 0)2 + (y – 4)2 = k and (x – 3)2 + (y – 0)2 = 12 touches each other than find the maximum value of 'k'
Two circles touches each other if C1 C2 = |r1 ± r2|
Distance between C2(3, 0) and C1(0, 4) is either 
(C1C2 = 5)
=> √k + 1 = 5 or |k – 1| = 5
⇒ k = 16 or k = 36
⇒ maximum value of k is 36
Let 
angle between 
equal to π/3
If vector a is perpendicular to vector b x c then find the value of |vector a x (b x c)|
⇒ 
Also,
Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is
First common term = 23
common difference = 7 × 4 = 28
Last term ≤ 407
So n = 14
If minimum value of term free from x for (x/sinθ + 1/xcosθ) is L1 in θ ∈ [π/8, π/4] and L2 in θ ∈ [π/16, π/8]
find L2/L1.
for r = 8 term is free from 'x'
L1 = 16C828
∵ {Min value of L1 at θ = π/4}
{∵ min value of L2 at θ = π/8]
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