JEE Exam  >  JEE Test  >  Physics Main & Advanced  >  Chapter Test: Rotational Motion - 2 - JEE MCQ

Chapter Test: Rotational Motion - 2 JEE Physics Free MCQs


MCQ Practice Test & Solutions: Chapter Test: Rotational Motion - 2 (30 Questions)

You can prepare effectively for JEE Physics for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Chapter Test: Rotational Motion - 2". These 30 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Number of Questions: 30

Sign up on EduRev for free to attempt this test and track your preparation progress.

Chapter Test: Rotational Motion - 2 - Question 1

A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is-


Detailed Solution: Question 1


Chapter Test: Rotational Motion - 2 - Question 2

A particle is moving on a circular path with constant speed v. What is the change in its velocity after it has described an angle of 60º -

Detailed Solution: Question 2


Chapter Test: Rotational Motion - 2 - Question 3

A cyclist starts from the centre O of a circular track of radius 1 km, reaches the edge A of the track and then cycles along the circumference and stops at point B as shown in the figure. If the total time taken is 10 min, what is the average velocity of the cyclist ?


Detailed Solution: Question 3


Chapter Test: Rotational Motion - 2 - Question 4


Detailed Solution: Question 4


Chapter Test: Rotational Motion - 2 - Question 5

A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is -


Detailed Solution: Question 5

Chapter Test: Rotational Motion - 2 - Question 6

When a ceiling fan is switched on, it makes 10 rotations in the first 3 second. How many rotations will it makes in the next 3 seconds? (Assume uniform angular acceleration)

Detailed Solution: Question 6


Chapter Test: Rotational Motion - 2 - Question 7

A particle is given an initial speed u inside a smooth spherical shell of radius R =1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is -


Detailed Solution: Question 7


Chapter Test: Rotational Motion - 2 - Question 8

A non-uniform thin rod of length L is placed along x-axis as such its one of ends is at the origin. The linear mass density of rod is λ = λ0x. The distance of centre of mass of rod from the origin is :

Detailed Solution: Question 8

The centre of mass of a rod is given by:

xcm = ∫ x dm / ∫ dm

The linear mass density of the rod varies with position as:

λ = λ0x

So the elemental mass is:

dm = λ dx = λ0x dx

Numerator:

0L x dm = λ0 ∫0L x2 dx = λ0 [x3/3]0L = λ0L3/3

Denominator:

0L dm = λ0 ∫0L x dx = λ0 [x2/2]0L = λ0L2/2

Centre of mass:

xcm = (L3/3) / (L2/2) = 2L/3

Therefore, the centre of mass of the rod is located at a distance 2L/3 from the origin.

Chapter Test: Rotational Motion - 2 - Question 9

A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as KE = 2t2 J, where t is in seconds. Then -


Detailed Solution: Question 9


Chapter Test: Rotational Motion - 2 - Question 10


Detailed Solution: Question 10


Chapter Test: Rotational Motion - 2 - Question 11

Assertion : When a particle is moving in a circular path, both centripetal force and centrifugal force acts on the particle.

Reason : Centripetal force and centrifugal force depend on frame of reference.


Detailed Solution: Question 11


Chapter Test: Rotational Motion - 2 - Question 12

The minimum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is-

Detailed Solution: Question 12


Chapter Test: Rotational Motion - 2 - Question 13


Detailed Solution: Question 13


Chapter Test: Rotational Motion - 2 - Question 14

A small block of mass m slides along a smooth frictional track as shown in the figure. (i) If it starts from rest at P, what is the resultant force  acting on it at Q? (ii) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight-


Detailed Solution: Question 14


Chapter Test: Rotational Motion - 2 - Question 15

A simple pendulum consisting of a mass M attached in a string of length L is released from rest at an angle α. A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in the figure. The maximum angle θ which string makes with the vertical after hitting the pin is –


Detailed Solution: Question 15


Chapter Test: Rotational Motion - 2 - Question 16

A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is –

Detailed Solution: Question 16

Chapter Test: Rotational Motion - 2 - Question 17

A wind farm generator uses a two-bladed propeller mounted on a pylon at a height of 20 m. The length of each propeller blade is 12 m. A tip of the propeller breaks off when the propeller is vertical. At that instant, the period of the motion of the propeller is 1.2 second. The fragment files off horizontally, falls and strikes the ground at point P.

In figure, the distance from the base of the pylon to the point where the fragment strikes the ground is closest to -

Detailed Solution: Question 17


Chapter Test: Rotational Motion - 2 - Question 18

Detailed Solution: Question 18

Chapter Test: Rotational Motion - 2 - Question 19

A small sphere is given vertical velocity of magnitude v0 = 5m/s and it swings in vertical plane about end of massless string. If max.tension string can withstand is equal to twice of weight of sphere, then angle 'θ' with vertical at which string will break is -


Detailed Solution: Question 19


Chapter Test: Rotational Motion - 2 - Question 20

Detailed Solution: Question 20


Chapter Test: Rotational Motion - 2 - Question 21

The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed ω varies with time t as –

Detailed Solution: Question 21


Chapter Test: Rotational Motion - 2 - Question 22

The minimum velocity (in m/s) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is-

Detailed Solution: Question 22


Chapter Test: Rotational Motion - 2 - Question 23

A block of mass m is placed at the top of a smooth wedge ABC. The wedge is rotated about an axis passing through C as shown in the figure. The minimum value of angular speed ω such that the block does not slip on the wedge is-


Detailed Solution: Question 23


Chapter Test: Rotational Motion - 2 - Question 24

A particle of mass m1 is fastened to one end of a string and one of m2 to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. The particle are then projected, so that the two portions of the string are always in the same straight line and describes horizontal circles. Find the ratio of tensions in the two parts of the string


Detailed Solution: Question 24


Chapter Test: Rotational Motion - 2 - Question 25

A body of mass m slides down an incline and reaches the bottom with a velocity v. If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

Detailed Solution: Question 25


Chapter Test: Rotational Motion - 2 - Question 26

The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is

Detailed Solution: Question 26

We know that for a disc of mass m and radius r
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10

Chapter Test: Rotational Motion - 2 - Question 27

Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be

                                     

Detailed Solution: Question 27

For ring 1
(MOI)1 = MOI about diameter + PQ2
MOI about diameter = ½ PQ2 
(MOI)1 = 3/2 PQ2
Similarly, (MOI)2 = 3/2 PQ2
(MOI)3 = ½ PQ2
Total MOI = 7/2 PQ2

Chapter Test: Rotational Motion - 2 - Question 28

Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not.

Detailed Solution: Question 28

If the axes are parallel, we use the formula:
Io = Icm + md2
For the first body the distance between the axis and the axis passing through C.O.M is 0 
Therefore, Io = Icm + m(0)2 = lcm since it is mentioned in question
Whereas for the second body it is not passing through the C.O.M therefore, there is some distance between the axes, say 'd'
So, Io = Icm + md2

Chapter Test: Rotational Motion - 2 - Question 29

Moment of inertia of a thin semicircular disc (mass = M & radius = R) about an axis through point O and perpendicular to plane of disc, is given by :

                                   

Detailed Solution: Question 29

Mass of semicircular disc = M
Suppose there is a circular disc of mass 2M, then
Moment of intertia of circular disc = ½ (2M)R2
Moment of intertia of circular disc = ½ (2M)R2 = MR2
=> So, Moment of intertia of semi-circular disc = ½ MR2

Chapter Test: Rotational Motion - 2 - Question 30

A rod of length 'L' is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in vertical position is

Detailed Solution: Question 30

258 videos|856 docs|206 tests
Information about Chapter Test: Rotational Motion - 2 Page
In this test you can find the Exam questions for Chapter Test: Rotational Motion - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Chapter Test: Rotational Motion - 2, EduRev gives you an ample number of Online tests for practice
258 videos|856 docs|206 tests
Download as PDF