JEE Main 2020 Question Paper with Solution (7th January - Evening)
72 Questions MCQ Test Additional Study Material for JEE | JEE Main 2020 Question Paper with Solution (7th January - Evening)
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If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s2 ; radius of earth = 6400 Km]
at pole, weight = mg = 196
m = 19.6 kg
at equator, weight = mg – mω2R
= 195.33 N
In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is connected to 220 V Mains supply then find minimum fuse current
Total power is (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
So current is = 4325/220
= 19.66 A
Ans is 20 Amp.
In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial relaxation time. Given 
relaxation time 
and 
A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal from midpoint. Upper half of string makes 45° with vertical, value of F is
The surface mass density of a disc of radius a varies with radial distance as σ = A + Br where A & B are positive constants then moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane
σ = A + Br
I = 
= 
(A + Br )2πr3 dr
= 2pa4 
Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high temperature for one engine is T1, low temperature for other engine is T2 (Assume work done by both engine is same) Calculate lower temperature of first engine.
Let, QH : Heat input to Ist engine
QL : Heat rejected from Ist
engine
QL': Heat rejected f rom IInd engine
Work done by Ist engine = work done by IInd engine
QH – QL = QL – QL'
2QL = QH + QL'
Activity of a substance changes from 700 s–1 to 500 s–1 in 30 minute. Find its half-life in minutes
=> ln2 = λt1/2...........(i)
(i)/(ii)
(2.06004) 30 = t1/2 = 61.8 min.
In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is
= 5.9mm
An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.
Using equation of continuity
A1V1 = A2V2
There is a electric circuit as shown in the figure. Find potential difference between points a and b.
Diode is in forward bias, so it will behave as simple wire so,
So Vab = 
= 10V
A particle of mass m and positive charge q is projected with a speed of v0 in y–direction in the presence of electric and magnetic field are in x–direction. Find the instant of time at which the speed of particle becomes double the initial speed.
As
(magnitude of velocity does not change in y–z plane)
∴ 
Two sources of sound moving with same speed v and emitting frequency of 1400 Hz are moving such that one source s1 is moving towards the observer and s2 is moving away from observer. If observer hears beat frequency of 2 Hz. Then find the speed of source. Given
Vsound >> VSource
Vsound = 350 m/s
An electron & a photon have same energy E. Find the ratio of de Broglie wavelength of electron to wavelength of photon. Given mass of electron is m & speed of light is C.
λd for electron = 
λ for photon = hC/E
Ratio = 
A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :
When ωt = π
∵ φ will have maximum.
∵ e will be minimum.
t = 
= 5 sec.
Electric field in space is given by 
A positively charged particle at (0, 0, π/K) is given velocity v0kˆ at t = 0. Direction of f orce acting on particle is
Force due to electric field is in direction 
because at t = 0, E = 
Force due to magnetic field is in direction 
∴ it is parallel to
∴ net force is antiparallel to 
Focal length of convex lens in air is 16 cm (μglass = 1.5). Now the lens is submerged in liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in air has closest value
A lift of mass 920 kg has a capacity of 10 persons. If average mass of person is 68 kg. Friction f orce between lift and lift shaft is 6000 N. The minimum power of motor required to move the lift upward with constant velocity 3 m/s is [g = 10 m/s2]
Net force on motor will be
Fm = [920 + 68(10)]g + 6000
= 22000 N
So, required power for motor
= 22000 x 3
= 66000 watt
The hysteresis curve for a material is shown in the figure. Then for the material retentivity, coercivity and saturation magnetization respectively will be
x = retentivity
y = coercivity
z = saturation magnetization
An inductor of inductance 10 mH and a resistance of 5Ω is connected to a battery of 20 V at t = 0. Find the ratio of current in circuit at t = ∞ to current at t = 40 sec.
Find the dimension of 
Energy density in magnetic field = 
A capacitor of 60 pF charged to 20 volt. Now battery is removed and then this capacitor is connected to another identical uncharged capacitor. Find heat loss in nJ.
V0 = 20 V
Heat loss = Ui – Uf
= 6 × 10–9 J = 6 nJ
When m gram of steam at 100°C is mixed with 200 gm of ice at 0°C. it results in water at 40°C. Find the value of m in gram .
(given : Latent heat of fusion (Lf) = 80 cal/gm, Latent heat of vaporisation (Lv) = 540 cal/gm., specific heat of water (Cw)= 1 cal/gm/°C)
A solid cube of side 'a' is shown in the figure. Find maximum value of 
for which the block does
not topple before sliding.
For no toppling
For no toppling
0.2a + 0.4b ≤ 0.5a
0.4b ≤ 0.3a
b ≤ 0.75a (in limiting case)
But it is not possible as b can maximum be equal to 0.5a
∴ 
Magnitude of resultant of two vectors 
is equal to magnitude of 
. Find the angle between 
and resultant of 2
and 
So angle between (2
+ 
) and 
is 90°
Alternate solution
P2 + Q2 + 2PQcosθ = P2
Q + 2Pcosθ =θ
cosθ = 
tan α = 
= ∞
∵ [2Pcosθ + Q = 0]
α = 90°
A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of 'n' is
Let the emf of cell is e internal resistance is 'r' and potential gradient is x.
only cell connected :
ε = 560 x ................. (i)
After connecting the resistor
................... (ii)
from (1) and (2)
56 = 50 + 5r
n = 12
Which of the following reactions are possible?
(A)
(B)
(C)
(D)
Vinyl halides and aryl halides do not give Friedel craft's reaction.
A and B are in the given reaction ?
The correct statement about gluconic acid is
Gluconic acid 
is obtained by partial oxidation of glucose by Tollen's reagent or Fehling solution or Br2,H2O.
Gluconic acid can not form hemiacetal or acetal
Stability order of following alkoxide ions is
When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.
(A) Negative charge is delocalised with NO2 group
(B) Negative charge is delocalised with carbon of alkene
(C) Negative charge is localised
For the complex [Ma2b2] if M is sp3 or dsp2 hybridised respectiv ely then total number of optical isomers are respectively :
Both will not show optical isomerism.
Bond order and magnetic nature of CN– are respectively
CN– is a 14 electron system.
Which of the following is incorrect?
Theory based.
NaOH + Cl2 → A + other products
Hot & conc.
Ca(OH) + Cl2 → B + other products
Cold & dil.
A & B are respectively
6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
2Ca(OH)2 + Cl2 → Ca(OCl)2 + CaCl2 + H2O
There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non-volatile solute. If both beakers are placed together in a closed container then:
There will be lowering in vapour pressure in second beaker.
Metal with low melting point containing impurities of high melting point can be purified by
Theory based
Which of the following statements are correct ?
(I) On decomposition of H2O2, O2 gas is released .
(II) 2-ethylanthraquinol is used in preparation of H2O2
(III) On heating KClO3, Pb(NO3)2, NaNO3, O2 gas is released.
(IV) In the preparation of sodium peroxoborate, H2O2 is treated with sodium metaborate.
Theory based
Amongs the following which is redox reaction?
N2 + O2 → 2NO
3O2 → 2O3
2NaOH + H2SO4 → Na2SO4 + 2H2O
AgNO3 + NaCl → NaNO3 + AgCl
Select the correct options :
Option a is correct. The graph is for no of molecules vs speed. So don’t go for values in Vrms,Vavg and Vmp. It's a derivation according to Maxwell velocity distribution of gas. Most gas particles have the most probable velocity. Some of the particles have avg velocity while few particles have root mean square velocity.
Which one of the following amongs each pair will release maximum energy on gaining one electron (A = F, Cl), (B = S, Se), (C = Li, Na)
Theory based
Which of the following statements are incorrect ?
(A) Co+3 with strong field ligand forms high magnetic moment complex.
(B) For Co+3 if pairing energy(P) > Δo then the complex formed will have t42g, e2g configuration
(C) For [Co(en)3]3+ λabsorbed is less than λabsorbed for [CoF6]3-
(D) If Δo = 18000 cm-1 for Co+3 then with same ligands for it Δt = 16000 cm-1
Theory based
0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?
2 × mole of Urea = mole of NH3...(1)
mole of NH3 = mole of HCl ........(2)
∴ mole of HCl = 0.02 mole
Number of sp2 hybrid carbon atoms in aspartame is –
All stared carbon atoms of aspartame are sp2 hybrid. Aspartame is methyl ester of dipeptide formed from aspartic acid and phenylalanine.
3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container 
mL of 5M NaOH is added to this. Find the pH of this solution. (log 3 = 0.4771, pKa = 4.74)
m mole of acidic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5
pH = PKa + log 
= 4.74 + log 3
= 4.74 + 0.48 = 5.22
Flocculation value for As2S3 sol by HCl is 30 m mole L–1. Calcualte mass of H2SO4 required in gram for 250 mL sol.
For 1L sol 30 m mol of HCl is required
∴ For 1L sol 15 m mol H2SO4 is required
For 250 mL of sol
m mol H2SO4 = 0.3675 g
Determine total number of atoms in per unit formula of (A), (B) & (C)
(A) = CrO2Cl2
(B) = Na2CrO4
(C) = CrO5
Calculate ΔfH° (In kJ/mol) for C2H6(g), if ΔcH° [C(graphite)] = -393.5 kJ/mol,
ΔCHº [H2(g)] = -286 kJ/mol and
ΔCHº [C6H6(g)] = -1560 kJ/mol
Let A = [aij], B=[bij] are two 3 × 3 matrices such that bij = λi + j - 2aij & |B| = 81. Find |A| if λ = 3.
=>81 = 33.3. 32 |A|
=> 34 = 36 |A|
=> |A|= 1/9
From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find the locus of mid point of PQ.
slope of PQ = 
= -1
⇒ k – α = - h + 2α
⇒ α = 
......... (1)
Also 2h = 2α + β
2k = α + β
⇒ 2h = α + 2k
⇒ a = 2h - 2k .....(2)
from (1) & (2)
= 2(h - k)
so locus is 6x –6y = x + y
⇒ 5x = 7y
Pair of tangents are drawn from origin to the circle x2 + y2 – 8x – 4y + 16 = 0 then square of length of chord of contact is
L = √S1 = √16 = 4
R = 
= 2
Length of Chord of contact = 
square of length of chord of contact = 64/5
Contrapositive of if A ⊂ B and B ⊂ C then C ⊂ D
Let P = A ⊂ B, Q = B ⊂ C, R = C ⊂ A
Contrapositive of (P ∧ Q) → R is ~ R → ~ (P ∧ Q)
R ν (~ P ν ~ Q)
Let y(x) is solution of differential equation (y2 – x) dx/dy = 1 and y(0) = 1, then find the value of x where curve cuts the x-axis
c = –e
y = 0, x = 0 – 0 + 2 + (–e) (e–0)
x = 2 –e
Let θ1 and θ2 (whereθ1 < θ2)are two solutions of 2cot2θ 
θ ∈ [0, 2π) then 
is equal to
2 cos2θ - 5sinθ + 4sin2θ = 0, sinθ ≠0
Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =
S = 7 + 17 + 27 + 37 + 47 +......20 terms
= 10[2040] = (102) (20)
⇒ m = 20
If 
6 , then number of ordered pairs (r, k) are –(where k ∈ I).
Hence number of order pair = 4
Let 4α 
= 5 then a =
⇒ 4(2 - e-αa - e-2a) = 5
Put e-α = t
⇒ 4t2 + 4t – 3 = 0
⇒(2t + 3) (2t - 1) = 0
⇒ 
⇒ a = ln2
Let f(x) is a five degree polynomial which has critical points x = ±1 and 
= 4 then which one is incorrect.
f(x) = ax5 +bx4 + cx3
= 4
⇒ 2 + c = 4
⇒ c = 2
f'(x) = 5ax4 + 4bx3 + 6x2
= 
f'(1) = 0 ⇒ 5a + 4b + 6 = 0
f'(–1) = 0 ⇒ 5a – 4b + 6 = 0
b = 0,
f'(x) = –6x4 + 6x2
= 6x2 (–x2 +1)
= –6x2 (x+1) (x–1)
Minimal at x = –1
Maxima at x = 1
x = –1
x = 1
If 
are unit vectors such that 
and 
and 
then 
⇒ λ = -3/2
Coefficient of x7 in 
is
Coefficient of x7 is 11C7 = 11C4 = 330
Let α and β are the roots of x2 - x - 1 = 0 such that Pk = αk + βk, k ≥ 1 then which one is incorrect?
= 5(1) + 6
P5 = 11 and P5 = α2 + β2 = α + 1 + β + 1
P2 = 3 and
P3 = α3 + β3 = 2α + 1 + 2β + 1
= 2(1) +2 = 4
P2 × P3 = 12 and P5 = 11
⇒ P5 ≠ P2 × P3
Let f(x) = x3 –4x2 + 8x + 11, if LMVT is applicable on f(x) in [0, 1], value of c is :
f(x) is a polynomial function
∴ it is continuous and differentiable in [0, 1]
Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16 f '(x) = 3x2 – 8x + 8
∴ 
= 3c2 – 8c + 8
⇒ 3c2 – 8c + 3 = 0
∴ 
The area bounded by 4x2 ≤ y ≤ 8x + 12 is -
4x2 = y
y = 8x + 12
4x2 = 8x + 12
x2 – x – 3 = 0
x2 – 2x – 3 = 0]
x2 – 3x + x – 3 = 0
(x + 1)(x – 3) = 0
x = –1
= 128/3
There are 5 machines. Probability of a machine being faulted is 1/4 Probability of atmost two machines s faulted, is 
k then value of k is
Required probability = when no. machine has fault + when only one machine has fault + when only two machines have fault.
∴ k = 17/8
3x + 4y = 12√2 is the tangent to the ellipse 
then the distance between focii of ellipse is-
condition of tangency c2 = a2m2 + b2
a2 = 16
1a = 4
∴ focus are (± 7 , 0)
∴ distance between foci = 2√7
If z = 
is purely real and θ = 
then arg(sinθ + I cosθ) is -
as z is purely real
⇒ 3 cosθ + 4sinθ = 0
⇒ tanθ = 
arg(sinθ + icosθ) = 
a1, a2, a3 …..a9 are in GP where a1 < 0,
a1 + a2 = 4, a3 + a4 = 16, if 
= 4λ, then λ is equal to
a1 + a2 = 4
⇒ a1 + a1r = 4 ................... (i)
⇒ r2 = 4
r = ±2
r = 2, a1(1+2) = 4
⇒ a1= 4/3
r = –2, a1(1–2) = 4
⇒ a1 = - 4
λ = –171
If 
and 
.Then dy/dx at x = 1/2 is
Let X = {x : 1 ≤ x ≤ 50, x ∈ N}
A = {x: x is multiple of 2}
B = {x: x is multiple of 7}
Then find number of elements in the smallest subset of X which contain elements of both A and B
If Q
is foot of perpendicular drawn from P(1, 0, 3) on a line L and if line L is passing through (α, 7, 1), then value of α is
Since PQ is perpendicular to L, therefore
If f(x) is defined in x ∈
f(x) = 
Find k such that f(x) is continuous
= 3 + 2 = 5
∴ f(x) will be continuous if f(0) = 
If system of equation
x + y + z = 6
x + 2y + 3z = 10
3x + 2y + λz = μ has more than two solutions. Find (μ – λ2)
x + y + z = 6 …….. (1)
x + 2y + 3z = 10 …….. (2)
3x + 2y + λz = μ ..... (3)
from (1) and (2)
if z = 0
⇒ x + y = 6 and x + 2y = 10
⇒ y = 4, x = 2
(2, 4, 0)
if y = 0 ⇒ x + z = 6 and x + 3z = 10
⇒ z = 2 and x = 4
(4, 0, 2)
so, 3x + 2y + λz = μ must pass through (2, 4, 0) and (4, 0, 2)
so, 6 + 8 = μ
⇒ μ= 14 and 12 + 2λ = μ
12 + 2λ = 14
⇒ λ = 1
so μ – λ2 = 14 - 1 = 13
If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively then find xy
mean = 
⇒ x + y = 19.............(i)
variance 
= 25
⇒ x2 + y2 = 113………(ii)
(x+y)2 = (19)2
⇒ x2 + y2 + 2xy = 361 ⇒ xy = 124 (exact data is not retrieved so ans. can vary)
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