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JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2021 July 20 Shift 1 Question Paper & Solutions

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JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 1

The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 10-4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is:
(Mass of electron = 9.1 × 10-31 kg)

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 1


3 → 2 ⇒ 1.89 eV
B = 5 × 10-4 T
r = 7 mm

= 1.077 eV
We know, work function = energy incident - (KE)electron
ϕ = 1.89 - 1.077 = 0.813 eV

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 2

If  and  are two vectors satisfying the relation  . Then the value of  will be:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 2


AB cosθ = AB sinθ ⇒ θ = 45°

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JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 3

The value of current in the 6Ω resistance is:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 3


Applying KCL at point P,

⇒ 10V + 12V - 1080 + 3V - 420 = 0
⇒ V = 60
⇒ Current in 6 Ω = V - 0/6 = 10 A

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 4

A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let rd and rα  be their respective radii of circular path. The value of  is equal to

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 4

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 5

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years, respectively. What will be the time after which one third of the material remains? (Take In 3 = 1.1)

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 5


Given:


Now, let the initial number of radioactive nuclei be N0.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 6

A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in the sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 m/s2 and 4 m/s2, respectively. Identify the curve that fits best for the weight of the passenger as a function of time from the below given figures.

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 6

Acceleration due to gravity on Earth,
gE = 10 m/s2
Acceleration due to gravity on Mars,
gM = 4 m/s2
We know that,

∴ Weight at Earth,
mgE = 100 × 10 = 1000 N
As the spaceship moves far away from Earth, the value of gE decreases to zero at a point where gE + gM = 0.
Hence, weight will also be zero.
This point is called neutral point and is shown by graph (c) in the given figure.
Then, gE increases 4ms-2 at the surface of Mars and weight becomes 400N.
This is exhibited by graph (c).
So, option C is correct.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 7

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is ______. (R is the universal gas constant.)

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 7

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 8

The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are I1 and I2, respectively. The actual length of the metal wire is

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 8


∴  F = kx

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 9

A butterfly is flying with a velocity of 4√2m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 9

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 10

A certain charge Q is divided into two parts q and (Q - q). How should the charges Q and q be related so that q and (Q - q) placed at a certain distance apart experience maximum electrostatic repulsion?

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 10



⇒ Q - 2q = 0
⇒ Q = 2q

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 11

The entropy of any system is given by:

When α and β are the constants,μ , J, K and R are number of moles, mechanical equivalent of heat, Boltzmann constant and gas constant, respectively. [Take S = dQ/T ]
Choose the incorrect option from the following.

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 11

Entropy of the system is given by:


∵ Dimensions of Q = [ML2T-2]
Dimensions of T = [K]
Boltzman constant, k = energy / T
[∵ Dimensions of Energy = [ML2T-2]

From Eqs. (ii) and (iii), we can write,

∵ 
and mechanical equivalent of heat

Using Eqs. (i), we can write,


So, from Eqs. (iii) and (viii), we can say that α and k have different dimensions.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 12

The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0, as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph showing the variation of different quantities with distance.

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 12

As the rectangular conductor moves in field area, flux will increase up to x = b. Then, flux is generated on return journey from x = b to x = 0. The flux is shown by plot A of the graph.
⇒ e = - dϕ/dt
This is shown by curve B.
⇒ Power dissipated = VI
This is shown by curve C.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 13

A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration of 0.2 m/s2. The normal reaction 'R' by the floor, if mass of the iron cylinders is equal, i.e. 20 kg each, is __________ N.
[Take g = 10 m/s2 and μs = 0.2]

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 13


Force along vertical axis:
Mg - N = Ma
70g - N = 70 × 0.2
N = 70[g - 0.2] = 70 × 9.8
∴ N = 686 Newton

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 14

For the circuit shown below, calculate the value of Iz:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 14

I = 50/1000  = 50 mA
R = 1000Ω

I = 50/2000  = 25 mA
Iz = I1000 - I2000
= 50 - 25 = 25 mA

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 15

A nucleus of mass M emits γ-ray photon of frequency 'ν'. The loss of internal energy by the nucleus is:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 15

Energy of γ-ray [Eγ] = hν
Momentum of γ-ray [Pγ]

Total momentum is conserved.
, where  = Momentum of decayed nuclei

⇒ 
⇒ K.E. of nuclei

Loss in internal energy = 
​​​​​​​

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 16

Regions I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 16

As we know that, the equation of reflection at spherical surface is:


μII = refractive index of region II = 1.4
μI = refractive index of region I = 1.25
R = radius of curvature = -25 cm
u = object distance = -40 cm

⇒ v = -37.58 cm
where, negative sign indicate real image.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 17

AC voltage V(t) = 20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is ____________.
[Take  = 8.85 × 10-12 F/m]

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 17


Separation between the plates, d = 2 mm = 2 x  10-3 m
Area, A = 1 m2

Where ε0 = absolute electrical permitivity of free space = 8.854 x 10-12N-1kg2m-2

Capacitive Resiatance


From Eqs. (i) and (ii), we get,

By using Ohm's Law

As,


⇒ I0 = 27.79μA
∴ The amplitude of the oscillating displacement current for applied AC voltage will approximately be 27.79 μA.

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 18

The normal reaction 'N' for a vehicle of 800 kg mass negotiating a turn on a 30° banked road at maximum possible speed without skidding is ___________ × 103 kg m/s2.
[Given: cos 30° = 0.87,μs = 0.2]

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 18


Equating forces perpendiluar to the inclined plane,

Equating forces along the inclined plane,

On dividing E.(i) by Eq. (ii), we get,

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 19

A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point, the direction of current density is at an angle of 60° with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is:
(Resistivity of magnesium, ρ = 44 × 10−8 Ωm)

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 19


J = 5 × 50 = 250 A/m2
Now, 
= 44 × 10–8 × 250 = 11 × 10–5 V/m

JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 20

Consider a mixture of gas molecule of types A, B and C having masses mA < mB < mC. The ratio of their root mean square speeds at normal temperature and pressure is:

Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 20


mA < mB < mC
⇒ VA > VB > VC

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 21

A body having specific charge 8 μC/g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ______________ s.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 21

q = 8 μC/g = 8 × 10−6 C/g
d = 10 cm = 0.1 m
Electric field = E = 100 V/mF = ma
qE = ma


Now, d = 1/2at2 (∵ u = 0)

∴ Time period = 2 x 1/2 = 1 s
If the collision of the body is perfectly elastic, the time period of motion will be 1 s.

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 22

In a spring gun having spring constant 100 N/m, a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground, the value of d is _______________ m. (g = 10 m/s2)


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 22



kx2 = mv2

From h = 1/2gt2

d = vt = 0.5 
= 1 m

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 23

A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses (m/M) is 1/x . The value of 'x' will be ____________.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 23


Just before collision

Just after collision
From momentum conservation,

mu = Mv

From angular momentum conservation about O,

Coefficient of restitution = e = Relative velocity after collsion / Relative velocity before collision


From equations (ii) and (iii), we get

x = 4

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 24

An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 24

For simple microscope,
m = 1 + D/f0
6 = 1 + D/f0
5 =25/f0
f0 = 5 cm
For compound microscope,

(where f0 = fe = focal lengths of the objective lens and eyepiece respectively and l is the length of the given tube = 0.6 m = 60 cm)

fe = 25 cm

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 25

In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by 45° is 1/x  × 10-3 F. Then the value of x is


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 25


XC - XL = R

X = 3

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 26

The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m/s, then the speed of car is ______ km/hr.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 26


Wall as an observer
Frequency received by wall,

Wall as a source
Frequency received by observer on car,

From equations (i) and (ii),

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 27

A carrier wave VC(t) = 160 sin(2 π × 106 t) volts is made to vary between Vmax = 200 V and Vmin = 120 V by a message signal Vm(t) = Am sin(2 π × 103t) volts. The peak voltage Am of the modulating signal is __________.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 27

Amax = Am + AC
⇒ vmax = Vm + VC
200 = Vm + 160
Vm = 40
∴  Peak voltage, Am = 40

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 28

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it takes time 't1' when it rolls down the plane, it takes time t2. The value of . The value of x will be _________.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 28


If the disc slips on inclined plane, then its acceleration is:
a1 = g sinθ

If the disc rolls on inclined plane, then its acceleration is:

Now,


Dividing equation (ii) by (i),

⇒ x = 2

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 29

The amplitude of wave disturbance propagating in the positive x-direction is given by  at time t = 0 and y =  at t = 1 s, where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be _________ m/s.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 29

As per question,
at time t = 0 and y =  
and at t = 1s, 
As we know,

So, at t = 1s,

On comparing Eqs. (i) and (ii), we get
v = 2 ms-1
Hence, the velocity of the wave will be 2 m/s.

*Answer can only contain numeric values
JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 30

In the given figure, heat energy absorbed by a system in going through a cyclic process is _________ πJ.


Detailed Solution for JEE Main 2021 July 20 Shift 1 Question Paper & Solutions - Question 30


For complete cyclic process,
ΔU = 0
ΔQ = U + W
= 0 + W
ΔQ = W
W = Area = πr1.r2
= π × (10 × 103) (10 × 10-3) = 100
∴ Q = 100π

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