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


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30 Questions MCQ Test Additional Study Material for JEE - JEE Main 2020 Question Paper with Solution (9th January - Morning)

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 1

Kinetic energy of the particle is E and it's De–Broglie wavelength is λ. On increasing it's KE by ΔE, it's new De–Broglie wavelength becomes λ/2 . Then ΔE is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 1

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 2

The dimensional formula of is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 2

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 3

Two immiscible liquids of refractive index √2 and 2√2 are filled with equal height h in a vessel. Then apparent depth of bottom surface of the container given that outside medium is air:

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 3



JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 4

Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing through point P and B as shown in figure. Given P is centroid of triangle ABC.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 4

M.I about P = 
M.I about B =

Now ratio = 13 / 23

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 5

A solid sphere having radius R and Uniform charge density ρ has a cavity of radius R/2 as shown in figure. Find the ratio of magnitude of electric field at point A and B i.e.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 5

For a solid sphere



Electric field at point B = EB = E1A + E2A 
E1A = Electric Field Due to solid sphere of radius R at point B = 
E2A = Electric Field Due to solid sphere of radius R/2 (which having charge density –ρ)
E2A = R/2

EB =  E1A + E2A  = 

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 6

Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of magnetic field at distance a/3 and 2a from axis of wire is.
 

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 7

Find current in the wire BC.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 7


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

Two electromagnetic waves are moving in free space whose electric field vectors are given by  . A charge q is moving with velocity .Find the 
net Lorentz force on this charge at t = 0 and when it is at origin.

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

Magnetic field vectors associated with this electromagnetic wave are given by


by putting the value of 
The net Lorentz force on the charged particle is 
at t = 0 and at x = y = 0
t = 0 , x = y = 0

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

Two ideal di-atomic gases A and B. A is rigid, B has an extra degree of freedom due to vibration. Mass of A is m and mass of B is m/4. The ratio of molar specific heat of A to B at constant volume is :  

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

Molar heat capacity of A at constant volume = 5R/2
Molar heat capacity of B at constant volume = 7R/2
Dividing both 

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 10

An ideal liquid (water) flowing through a tube of non uniform cross section area at A and B are 40 cm2 and 20 cm2 respectively. If  pressure difference between A & B is 700 N/m2 then volume flow rate is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 10

using equation of continuity
40 VA = 20 VB
2VA =  VB
Using Bernoullies equation


Volume flow rate = 20 × 100 × VB = 2732 cm3/s

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 11

A screw gauge adv ances by 3mm in 6 rotations. There are 50 divisions on circular scale. Find least  count of screw gauge ? 

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 11

Pitch = 3/6 = 0.5 mm
L.C. = =  = 0.01 mm = 0.001 cm

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 12

A telescope of aperture diameter 5m is used to observe the moon from the earth. Distance between the moon and earth is 4 × 105 km. Determine the minimum distance between two points on the moon's surface which can be resolved using this telescope. (Wave length of light is 5893 Å).

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 12



distance = O1O2 = dθ

distance = O1O2≈ 57.5 m
∴ answer from options = 60m
(minimum distance)

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 13

A particle of mass m is revolving around a planet in a circular orbit of radius R. At the instant the particle has velocity another particle of mass m/2 moving at velocity collides perfectly in-elastically with the first particle. The new path of the combined body will take is    

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 13

Conserving momentum


vf < vorb (= v) thus the combined mass will go on to an elliptical path.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 14

Two particles of same mass 'm' moving with velocities and  collide in-elastically. Find the loss in kinetic energy. 

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 14

Conserving momentum 
on solving

Change in K.E

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 15

Three wav es of same intensity (I0) having initial phases rad respectively interfere at a point.Find the resultant Intensity

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 15


JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 16

Particle moves from point A to point B along the line shown in figure under the action of force. . Determine the work done on the particle by in moving the particle from point A to point B

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 16

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 17

For the given P-V graph for an ideal gas, chose the correct V- T graph. Process BC is adiabatic.

 

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 17

For process A – B Volume is constant
PV = nRT ; as P increases T increases
For process B – C
PVγ = Constant
TVγ-1 = Constant
For process C – A pressure is constant
V = kT 

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 18

Given. Find vector parallel to electric field at position
[Note that ]

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 18

Since
must be antiparallel to 
So ,
where λ is a arbitrary positive constant
Now 


so

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 19

Which of the following statements are correct for moving charge as shown in figure.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 19

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 20

Photons of wav elength 6556 Å falls on a metal surface. If ejected electrons with maximum K.E. moves in magnetic field of 3 × 10–4 T in circular orbit of radius 10–2m, then work function of metal surface is 

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 20



= 1.1 eV

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 21

A rod of length 1 m is released from rest as shown in the figure below.

If ω of rod is √n at the moment it hits the ground, then find n.


Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 21

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 22

If reversible voltage of 100 V is applied across an inductor, current in it reduces from 0.25A to 0A in 0.025ms. Find inductance of inductor (in mH).  


Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 22


∴ L = 100 × 10–4 H
= 10 mH 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 23

A wire of length l = 3m and area of cross section 10–2cm2 and breaking stress 48×107N/m2 is attached with block of mass 10kg. Find the maximum possible value of angular velocity with which block can be moved in circle with string fixed at one end.


Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 23



Solving
ω = 4 rad/s 

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 24

Position of a particle as a function of time is given as x2 = at2 + 2bt + c, where a, b, c are constants. Acceleration of particle varies with x–n then value of n is.  


Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 24


*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 25

In the given circuit both diodes having zero forward resistance and built-in potential of 0.7 V. Find the potential of point E in volts.


Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 25


Let VB = 0
Right diode is reversed biased and left diode is forward biased
∴ VE = 12.7 – 0.7
= 12 Volt

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 26

Determine wavelength of electron in 4th Bohr's orbit ?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 26

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 27

Which of the following species have one unpaired electron each?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 27

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 28

For Br2(l) Enthalpy of atomisation = x kJ/mol Bond dissociation enthalpy of bromine = y kJ/mole
then

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 28


ΔHatomisation = ΔHvap + Bond energy
Hence x > y

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 29

Which of the following oxides are acidic, Basic Amphoteric Respectively.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 29

Non-metal oxides are acidic in nature alkali metal oxides are basic in nature Al2O3 is amphoteric.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 30

Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution.
Given : Spin only magnetic moment = 3.8 B.M.
What is the IUPAC name of the complex.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 30

Cr(H2O)6Clncomplex)spin = 3.8 B.M.
From data of magnetic moment oxidation number of Cr should be +3
Hence complex is Cr(H2O)6Cl3.
Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl . 2H2O.
It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate.

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