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JEE Main 2020 Question Paper with Solution (8th January - Morning) - JEE MCQ


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30 Questions MCQ Test - JEE Main 2020 Question Paper with Solution (8th January - Morning)

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 1

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- 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 1


JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 2

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:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 2


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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 3

Choose the correct P-V graph of ideal gas for given V-T graph.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 4

Find the coordinates of centre of mass of the lamina, shown in figure.  

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 4



JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 5

Which graph correctly represents v ariation between relaxation time  of gas molecules with absolute temperature (T) of an ideal gas.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 5

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 6

​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 -

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 6


Given C1 + C2 = 10μF ........ (i)

⇒ 4C1 = C2 …(ii)
from equation (i) & (ii)
C1 = 2μF
C2 = 8μF
If they are in series

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 7

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–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 7


Loi = Lof

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 8

​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 -

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 8

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

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 9

​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–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 9


Potential gradient = 5 / 1000 = Vp/1200
 VP = 6V
and RP = = 100Ω

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 10

 A telescope has magnifying power 5 and length of tube is 60cm then the focal length of eye piece is–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 10

m = fo/fe
5 = fo/fe
fo = 5fe
fo + fe = 60
6fe = 60
fe = 10

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 11

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-

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 11


JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 12

​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–

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 12


∴ Bqv = ma

= 0.71× 10–3T
So, 0.71 mT

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 13

​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

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 13

Magnitude of electric field is constant & the surface is equipotential

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 14

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)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 14

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

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 15

​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)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 15



mg = A(80) ρ0ºg
mg = A(79) ρ4ºg

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 16

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. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 16


 

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 17

If relative permittivity and relative permeability of a medium are 3 and 4/3 respectively. The critical angle for this medium is.

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 18

​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–

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 19

​Choose the correct Boolean expression for the given circuit diagram:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 19

First part of figure shown is OR gate and Second part of figure shown is NOT gate
So Yp = OR + NOT = NOR gate

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 20

A solid sphere of density just floats in a liquid. The density of liquid is – 
(r is distance from centre of sphere)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 20

 0 < r ≤ R
mg = B

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 21

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.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 21

For elastic collision KEi = KEf

34 = 32 + v2

x = 1

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 22

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).  


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 22


R1 = ∝
R2 = –30 cm


f = 60 cm

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 23

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.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 23



= 580

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 24

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.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 24


*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 25

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.


Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 25



= 15 × 4
on solving
R = 10 Ω

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 26

Number of S–O bond in S2O82– and number of S–S bond in Rhombic sulphur are respectively:

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JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 27

Following vanderwaal forces are present in ethyl acetate liquid

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 27

Ethyl acetate is polar molecule so dipole-dipole interaction will be present there.

*Multiple options can be correct
JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 28

Given, for H-atom
 
Select the correct options regarding this formula for Balmer series. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 28

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.

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 29

Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol-1 respectively are:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 29

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

JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 30

Select the correct stoichiometry and its ksp value according to given graphs. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Morning) - Question 30


Ksp = [X+] [Y]
or Ksp = 2 × 10–3 × 10–3
or Ksp = 2 × 10–6

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