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BITSAT Physics Test - 4 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Physics Test - 4

BITSAT Physics Test - 4 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Physics Test - 4 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Physics Test - 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Physics Test - 4 below.
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BITSAT Physics Test - 4 - Question 1

A projectile is projected with an initial velocity of  The equation of the trajectory followed by the projectile will be

Detailed Solution for BITSAT Physics Test - 4 - Question 1

Equation of projectile is given as:


This is the required solution.

BITSAT Physics Test - 4 - Question 2

Two springs with spring constant K1 and K2 are stretched by the same force. If the potential energy stored in the two springs respectively is U1 and U2 and the ratio U1: U2 is given to be 1 : 2, find the ratio of the spring constants K1 : K2.

Detailed Solution for BITSAT Physics Test - 4 - Question 2

We know,
Potential energy stored in the spring, 
Therefore,

This is the required solution.

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BITSAT Physics Test - 4 - Question 3

In a situation it is observed that the coefficient of restitution of a collision is greater than 1. This condition implies to the possibility that

Detailed Solution for BITSAT Physics Test - 4 - Question 3

The coefficient of restitution is almost every time less than 1. It is the ratio of the relative velocity after collision to that before collision. As some part of the energy is dissipated during the collision, the relative velocity after collision decreases, giving e < 1.
For e > 1, it is possible that there is some amount of energy produced via chemical reaction or explosion during the collision.
Therefore, it is the correct option.

BITSAT Physics Test - 4 - Question 4

A particle of mass 2 kg is at the point P of a circular track as shown in the figure. It is released from the rest. At point Q, the velocity of the body is 10 m/s. If the radius of the track is 40 m, what is the work done by the body against friction between points P and Q? (Take g = 10 N/m2)

Detailed Solution for BITSAT Physics Test - 4 - Question 4


Energy of the body at point P
= mgh = 2 × 10 × 40 = 800 J
Energy of the particle at point Q

Work done against friction = Change in the mechanical energy of the body
= 800 - 500
= 300 J
This is the required solution.

BITSAT Physics Test - 4 - Question 5

What will be the angular velocity of the rotation of the Earth, so that the acceleration due to gravity at the equator becomes zero? Mass of the Earth = 6400 × 103 kg and g = 9.8 m/s2

Detailed Solution for BITSAT Physics Test - 4 - Question 5

BITSAT Physics Test - 4 - Question 6

A force F(x) = -2kx is acting on a body of mass 1 kg, moving along the positive x-direction. The velocity of the particle at time t = 0 is zero and the displacement at t = 0 is 2 m. If k = 1 Nm-2, what is the velocity of the body when the displacement is 1 m?

Detailed Solution for BITSAT Physics Test - 4 - Question 6

Given: F(x) = -2kx, k = 1 Nm-2
At time t = 0, v = 0 m/s and x = 2 m
At time t = t, x = 1 m and velocity is v.
Therefore, change in kinetic energy in moving from x = 2 to x = 1 m,

This is the required solution.

BITSAT Physics Test - 4 - Question 7

A time dependent force F is acting on a block of mass 10 kg. The plot of the magnitude of force versus time is given below. The momentum gained by the body at the end of 20 seconds will be

Detailed Solution for BITSAT Physics Test - 4 - Question 7

Change in momentum = Area under F-t curve

Therefore,

Therefore, it is the correct option.

BITSAT Physics Test - 4 - Question 8

A ball of mass m, moving with a constant velocity, collides with another identical ball at rest. After the collision, the first ball acquires velocity v1 and the second ball that was at rest before collision acquires velocity v2. If the coefficient of restitution is given to be 5/7, what will be the ratio v1 : v2?

Detailed Solution for BITSAT Physics Test - 4 - Question 8

Let the velocity of the ball before the collision be u.
Velocity of the bodies after the collision is given by:
Using conservation of momentum
mu = mv1 + mv2
u = v1 + v2
Coefficient of restitution,

Hence,

Therefore, the ratio of velocities is given by:

Therefore, it is the correct option.

BITSAT Physics Test - 4 - Question 9

A circular ring of mass m is rolling without slipping along an inclined plane. The angle of inclination of the plane is given to be θ and the linear velocity of the ring at the bottom of the inclined plane is v. If a body of the same mass is slipping on the same inclined plane, what will be the velocity at the bottom of the inclined plane? Consider the plane to be smooth and frictionless.

Detailed Solution for BITSAT Physics Test - 4 - Question 9

Let the velocity of the object sliding down the plane be v'.
Therefore,  where h is the height of the inclined plane.
If l is the length of the plane, then
h = I sin θ
For pure rolling without slipping of the ring,
v2 = 2al; where a = linear acceleration of the ring

Therefore,

Therefore,

BITSAT Physics Test - 4 - Question 10


In the graph, the stress-strain curve for two different materials P and Q are plotted. Which of the following statements are correct?
I. Tensile strength of P is more than that of Q.
II. Young's modulus of P is more than that of Q.
III. Tensile strength of P is less than that of Q.
IV. Young's modulus of P is less than that of Q.

Detailed Solution for BITSAT Physics Test - 4 - Question 10

From the graph,


For same stress,

Therefore, Young's modulus of P > Young's modulus of Q
Also, the breaking stress in P is more than breaking stress in Q. Therefore, tensile stress of P is greater than that of Q. Therefore, it is the correct option.

BITSAT Physics Test - 4 - Question 11

Refer to the system shown in figure. The acceleration of the masses is

Detailed Solution for BITSAT Physics Test - 4 - Question 11

5g − T2 = 5a        ....(i)
T2 − T1 − 3g = 3a      ...(ii)
T1 − g = a    ...(iii)
Adding Eqs (i) and (iii),
−T2 + T1 + 4g = 6a
Adding this to Eq. (ii), we get
g = 9a or a = g/9

BITSAT Physics Test - 4 - Question 12

Two blocks of masses m and 2m are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass m is placed on a smooth inclined plane of inclination 30° and 2m hangs vertically. If the system is released, the blocks move with an acceleration equal to

Detailed Solution for BITSAT Physics Test - 4 - Question 12

As 2m > m
The acceleration of the two block system will be as shown in figure.

The equation for block of mass 2m is
2mg − T = 2ma  ...(i)
Similarly, for block of mass m
T − mg sin30° = ma ...(ii)
From Eqs. (i) and (ii), we have

⟹ a = g/2

BITSAT Physics Test - 4 - Question 13

A wooden box of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with a constant acceleration of 0.4 ms−2. What is the force of friction between the box and the inclined plane? [g = 10 ms−2]

Detailed Solution for BITSAT Physics Test - 4 - Question 13

ma = mg sin θ − f
or f = mg sinθ − ma

BITSAT Physics Test - 4 - Question 14

A Gardner waters the plants by a pipe of diameter 1 mm. The water comes out at the rate of 10 cm3s−1. The reactionary force exerted on the hand of the Gardner is

Detailed Solution for BITSAT Physics Test - 4 - Question 14


So, we get 

BITSAT Physics Test - 4 - Question 15

Which of the following statements is absolutely correct about mass?

Detailed Solution for BITSAT Physics Test - 4 - Question 15

Explanation:


  • Statement A: This statement is correct because according to Hooke's law, the elongation in a spring is directly proportional to the mass connected with the spring balance. Therefore, more mass will result in more elongation in the spring balance.
  • Statement B: This statement is correct as well. When more mass is added to one pan of a beam balance, more mass is required on the other pan to keep the beam horizontal and balanced. This is based on the principle of torque and the concept of moments in physics.
  • Statement D: This statement is also true. According to Newton's second law of motion, the acceleration of an object is inversely proportional to its mass for a given force. Therefore, the greater the mass of a body, the lesser its acceleration will be when the same force is applied.

Therefore, all of the statements are absolutely correct about mass.

BITSAT Physics Test - 4 - Question 16

If the block P as shown in the figure below were to be at rest, what should the magnitude of force F be?

Detailed Solution for BITSAT Physics Test - 4 - Question 16

BITSAT Physics Test - 4 - Question 17

A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Detailed Solution for BITSAT Physics Test - 4 - Question 17

The force which is trying to accelerate the block on the inclined plane in the downward direction is mg sinθ, while the force opposing this is the force of friction f. The force of friction f⁡ can take any value for 0 to μmg cosθ. Here



As the dragging force is 9.8 N, the force of friction will grow up to this value to stop the block from sliding.

BITSAT Physics Test - 4 - Question 18

A pendulum consisting of a small heavy bob suspended from a rigid rod oscillates in a vertical plane as shown in the figure. When the bob passes through the position of equilibrium, the rod is subjected to a tension equal to twice the weight of the bob. Through what maximum angle α from the vertical will the pendulum be deflected? Disregard the weight of the rod and the resistance of the air.

Detailed Solution for BITSAT Physics Test - 4 - Question 18

Newton's second law to find the velocity when the bob passes through the position of equilibrium

From the equation of the law of conservation of energy,  it is possible to find the height from which the bob dropped ;

BITSAT Physics Test - 4 - Question 19

A solid disc of mass M is just held in air horizontal by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 ms−1. If the mass of each stone is 0.05 kg 0.05 kg . What is the mass of the disc (g = 10 ms−2 )

Detailed Solution for BITSAT Physics Test - 4 - Question 19

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
F = n m v

BITSAT Physics Test - 4 - Question 20

A L shaped rod whose one end is horizontal and the other is vertical is rotating about a vertical axis as shown with the angular speed ω. The sleeve has mass m and friction coefficient between rod and sleeve is μ. The minimum angular speed ω for which sleeve cannot slip on rod -

Detailed Solution for BITSAT Physics Test - 4 - Question 20


BITSAT Physics Test - 4 - Question 21

One quarter of the disc of mass mis removed. If r be the radius of the disc, the new moment of inertia is

Detailed Solution for BITSAT Physics Test - 4 - Question 21

Moment of inertia of whole disc about an axis through centre of disc and perpendicular to its plane is 
As one quarter of disc is removed, new mass,

BITSAT Physics Test - 4 - Question 22

A spool is pulled horizontally by two equal and opposite forces as shown in Figure. Which of the following statements are correct?

Detailed Solution for BITSAT Physics Test - 4 - Question 22


net torque about bottommost point

then for accelaration of com ,
Friction force will act towards right.

BITSAT Physics Test - 4 - Question 23

The moment of inertia about an axis of a body which is rotating with angular velocity 1 rads−1 is numerically equal to

Detailed Solution for BITSAT Physics Test - 4 - Question 23

Rotational kinetic energy,

ie, moment of inertia about an axis of a body is twice the rotational kinetic energy.

BITSAT Physics Test - 4 - Question 24

A particle at rest is to reach an angular velocity of 36 rad s−1 in 6 s, with a constant angular acceleration. The total angle turned during this interval is

Detailed Solution for BITSAT Physics Test - 4 - Question 24

The angular displacement in a given time is,

BITSAT Physics Test - 4 - Question 25

A fly wheel of radius R and of moment of inertia I can be rotated in vertical plane without friction about its natural axis. A light string is wound over the wheel and a body of weight W is suspended at the other end of string. The angular acceleration of wheel when suspended body is released is

Detailed Solution for BITSAT Physics Test - 4 - Question 25


If the angular acceleration of the wheel is α, acceleration of the body a = Rα.
Using torque equation
TR = Iα
And for the block,

By solving the above equations,

Therefore, the angular acceleration is less than 

BITSAT Physics Test - 4 - Question 26

A solid sphere roll down two different inclined planes of same height but of different inclinations. In both cases

Detailed Solution for BITSAT Physics Test - 4 - Question 26

In pure rolling mechanical energy remains conserved,
therefore speed will be same in both cases.
Time of descent on an inclined plane for a rolling body is,

The higher the inclination, the smaller the time of descent.

BITSAT Physics Test - 4 - Question 27

At any instant, a rolling body may be considered to be in pure rotation about an axis through the point of contact. This axis is translating forward with speed

Detailed Solution for BITSAT Physics Test - 4 - Question 27

Since, in this case, instantaneous axis of rotation is always below the centre of mass. This is possible only when point of contact moves with a velocity equal to centre of mass.

BITSAT Physics Test - 4 - Question 28

The moment of inertia of a body about a given axis is 1.2 kg m2 . Initially, the body is at rest. In order to produce rotational kinetic energy of 1500 J , and angular acceleration of 25 rad  s−2 must be applied about that axis for a duration of

Detailed Solution for BITSAT Physics Test - 4 - Question 28

BITSAT Physics Test - 4 - Question 29

A flywheel rotates with uniform angular acceleration. Its angular velocity increases from 20π rads−1 to 40 π rads−1 in 10 s. How many rotations did it make in this period?

Detailed Solution for BITSAT Physics Test - 4 - Question 29

As ω2 = ω+ αt  ∴40π = 20π + α × 10 
or α = 2π rads−1

Number of rotations completed

BITSAT Physics Test - 4 - Question 30

A L shaped rod of mass M is free to rotate in a vertical plane about axis AA'  as shown in figure. Maximum angular acceleration of rod is-

Detailed Solution for BITSAT Physics Test - 4 - Question 30

Moment of inertia about axis AA' is


Torque of Mg will maximum when ‘OC’ is horizontal

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