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Test: Rotational Motion - 2


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30 Questions MCQ Test Physics For JEE | Test: Rotational Motion - 2

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Test: Rotational Motion - 2 - Question 1

Four masses are fixed on a massless rod as shown fig. The moment of inertia about the axis P is about :  

Detailed Solution for Test: Rotational Motion - 2 - Question 1

Moment of inertia=m X r2
Here origin is chosen at the point through which the axis passes the rod. Using the formula total moment of inertia about axis is,
=2 X 5 X (0.2)2 + 2 X 2 X (0.4)2
=1.04kgm2

Test: Rotational Motion - 2 - Question 2

An inclined plane makes an angle of 30o with the horizontal. A ring rolling down this inclinedplane from rest without slipping has a linear acceleration equal to :

Detailed Solution for Test: Rotational Motion - 2 - Question 2

We know for a rolling body acceleration down an inclined plane, a = g.sinθ / (1 + I/mR2)
Where I is body's moment of inertia. Here, I = mR2  
Thus just by putting the values to the formula we get
a = g.sin 30° / 2
= g/4

Test: Rotational Motion - 2 - Question 3

A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will be 

Detailed Solution for Test: Rotational Motion - 2 - Question 3

The difference in the potential of the body when it rolls down through a vertical height h, is mgh.
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½  mv2 + ¼  mv
Thus we get v = √gh/3

Test: Rotational Motion - 2 - Question 4

The kinetic energy T of a particle of mass m moving in a circle of radius r varies with the distance traced, S as T = KS2. The tangential acceleration of the particle is 

Detailed Solution for Test: Rotational Motion - 2 - Question 4

Test: Rotational Motion - 2 - Question 5

Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is

Detailed Solution for Test: Rotational Motion - 2 - Question 5
Circular loops (rings) of the wire:
         Radii:     r₁ =  r    ,   r₂ = n * r

Mass of the wire per unit length = d
         Mass of loop₁ = m₁ = 2 π r₁ * d = 2π r d
         Mass of loop₂ = m₂ = 2 π r₂ * d  = 2π n r d

Moment of Inertia of a circular loop about an axis perpendicular to the plane of the loop and passing through the center of the loop : 
           I = M R^2,
           M = Mass,     R = radius

MOI of given loops:
        I₁ = (2π r d) r^2 = 2 π d r^3
        I₂ = (2π n r d) (n r)^2 = 2 π d n^3 r^3

Given   I₂ = 8 I₁
      =>  2π d n^3 r^3 = 8 * 2 π d r^3
      =>  n^3 = 8
      =>  n = 2.

Test: Rotational Motion - 2 - Question 6

Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is 

Detailed Solution for Test: Rotational Motion - 2 - Question 6


IA​=0 (on the Axis .)
IB​=ID​=mr2
=m(​l​/√2)2
=ml2/​2
IC​=m(√2​l)2
IC​=2ml2
I=IA​+IB​+IC​+ID
I=0+ml2/2​+ml2/2​+2ml2
I=3ml2

Test: Rotational Motion - 2 - Question 7

A solid sphere of radius R has moment of inertia I about its diameter. It is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r and t are (respectively)

Detailed Solution for Test: Rotational Motion - 2 - Question 7

We must apply the parallel axis theorem to find moment of inertia of the disc about the given axis. On equating the moment of inertia of sphere and disc we will get a relation between "r" and "R". Also, mass remains constant and from here we can get the relation between "t" and "R" (for disc volume = area x thickness). 
Isphere=Idisc+mr2
⇒ 2/5mR2=1/2mr2+mr2
⇒ 2mR2/5=3mr2/2
⇒ r=2R/√15
On conserving mass:
⇒ (4/3πR3)ρ=(πr2t) ρ
⇒ 4/3πR3=π(4R2/15)t
⇒ t=5R

Test: Rotational Motion - 2 - Question 8

Point C is the centre of the rigid body shown in fig. Find the total torque acting on the body

about point C.(CW - clockwise ACW - anticlockwise)

 

Detailed Solution for Test: Rotational Motion - 2 - Question 8

Test: Rotational Motion - 2 - Question 9

A thin rod of length 4 l , mass 4 m is bent at the points as shown in the fig. What is the moment of inertia of the rod about the axis passing through point O and perpendicular to the plane of the paper :  

Detailed Solution for Test: Rotational Motion - 2 - Question 9

Since the mass of rod is 54m and length is 4l.
so mass C = m
and length of AB = BO=OC= CD = l
WE, know moment of Inertia of a rod about to end = ml2​/3
So, moment of Inertia of AB, BO, OC, CD about B, O, O, C respectively = ml2/3​
From parallel axis theorem 
Moment of Inertia of AB about O.
=ml2/3​+ml2=4ml2​/3
Similarly od CD about O   =4ml2/3​
SO moment of Inertia Rod about O
=ml2/3​+ml2/3​+4ml2/3​+4ml2/3​ 
=10ml2/3​

Test: Rotational Motion - 2 - Question 10

In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1 m respectively and the acceleration of block ‘A’ is 5 m/s2 in the downward direction , then the acceleration of block B will be :

Detailed Solution for Test: Rotational Motion - 2 - Question 10

Since both pulley are connected to each other. So, their angular acceleration are equal.
⇒α1​=α2⇒​a1/ r1=a2/ r2 because α=a/r
⇒a1/1​​=5/2(Given, a2=5m/s2)
⇒a1=5/2 ​m/s2
Therefore, D is the correct option.

Test: Rotational Motion - 2 - Question 11

An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction m . A horizontal force F is applied on the prism as shown in the fig. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :

Detailed Solution for Test: Rotational Motion - 2 - Question 11

As the force is applied towards right the prism will topple at O. Now, for the prism to topple the torque due to force "F" must be greater than or equal to the torque due to weight "Mg" about the point O.

Test: Rotational Motion - 2 - Question 12

A uniform rod of mass m and length L lies radially on a disc rotating with angular speed ω in a horizontal plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc,. Then the kinetic energy of the rod is

Detailed Solution for Test: Rotational Motion - 2 - Question 12

By theorem of parallel axes

Test: Rotational Motion - 2 - Question 13

A uniform rod of length L and mass M is bent to make a reactangle PQRS such that QR/PQ = 1/2 Moment of inertia of the rectangle about the side QR can be expressed as :

Detailed Solution for Test: Rotational Motion - 2 - Question 13

Firstly, using the relations given in the questions, we will find the length and mass of the parts

Test: Rotational Motion - 2 - Question 14

A uniform disc of radius 20 cm and mass 2 kg is fixed at its centre and can rotate about an axis through the centre and perpendicular to its plane. A massless cord is round along the rim of the disc. If a uniform force 2 newton is applied on the cord, tangential acceleration of a point on the rim of the disc will be

Detailed Solution for Test: Rotational Motion - 2 - Question 14

Test: Rotational Motion - 2 - Question 15

Fig shows a hemisphere of radius 4R A ball of radius R is released from position P. It rolls . A ball of without slipping along the inner surface of the hemisphere. Linear speed of its centre of mass when the ball is at position Q is 

Test: Rotational Motion - 2 - Question 16

Three point masses, each m, are placed at the vertices of an equilateral triangle of side ‘α" Moment of inertia of the system about the axis COD which passes through the mass at O and lies in the plane of triangle and perpendicular to OA is 

Detailed Solution for Test: Rotational Motion - 2 - Question 16

The two masses are at a distance of “a” and “a/2” so,

Test: Rotational Motion - 2 - Question 17

An object of mass m is projected with a velocity u at an angle 45° with the horizontal. When  the object is at maximum height, its angular momentum about the point of projection is 

Detailed Solution for Test: Rotational Motion - 2 - Question 17
max height ,H= u^2sin^2θ/2g

velocity at max height = ux= ucosθ

angular momentum= muxH

and putting θ =45degree

angular momentum = mu^3/g4√2
Test: Rotational Motion - 2 - Question 18

A spherical ball of mass 2 kg is rolling without slipping with a speed 4 m/s on a rough floor. Its rotational kinetic energy is :

Detailed Solution for Test: Rotational Motion - 2 - Question 18

Test: Rotational Motion - 2 - Question 19

Fig shows a flywheel of radius 10 cm. Its moment of inertia about the rotation axis is 0.4 kg m2. A massless string passes over the flywheel and a mass 2 kg is attached at its lower end. Angular acceleration of the mass in rad/s2  is nearly. 

Detailed Solution for Test: Rotational Motion - 2 - Question 19


I took g = 10 but if g = 9.8 is taken then we get the exact answer.

Test: Rotational Motion - 2 - Question 20

Fig. shows a uniform rod of length  l and mass M which is pivoted at end A such that it can rotate in a vertical plane. The free end of the rod B is initially vertically above the pivot and then released. As the rod rotates about A, its angular acceleration when it is inclined to horizontal at angel q is : 

Test: Rotational Motion - 2 - Question 21

A thin wire of length L and uniform linear mass density r is bent into a circular loop with centre O as shown. The moment of inertia of the loop about the axis XX is

Test: Rotational Motion - 2 - Question 22

A metal wire of mass m and length l is rotated in a circle by holding it at one end. The breaking stress of wire is F and its area of cross-section is a. calculate the maximum frequency of rotation at which it breaks  

Test: Rotational Motion - 2 - Question 23

A cylinder of mass m radius R is spinned in a clockwise direction with angular velocity ωo and then gently placed on an inclined plane for which coefficient of friction μ = tanθ, θ is the angle of inclined plane with the horizontal. The cylinder continues to spin without failing for time

Test: Rotational Motion - 2 - Question 24

A solid sphere of mass M, radius R and having moment of inertia about an axis passing  through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. then, radius of the disc will be : 

Detailed Solution for Test: Rotational Motion - 2 - Question 24

Test: Rotational Motion - 2 - Question 25

A cylinder is placed on a rough inclined surface of inclination q. Minimum value of coefficient of static friction between the cylinder and the surface so that the cylinder rolls without slipping is : 

Detailed Solution for Test: Rotational Motion - 2 - Question 25

Test: Rotational Motion - 2 - Question 26

A tube of length L is filled completely with an incompressible liquid of mass M and is closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity w. Force exerted by  the liquid at the other end is :  

Detailed Solution for Test: Rotational Motion - 2 - Question 26

The force exerted by liquid at the other end is the centripetal force due to rotation. The case can be treated as rotation of the center of mass of the liquid at distance L/2 from the axis of rotation.
Hence force exerted by liquid at the end = 

Test: Rotational Motion - 2 - Question 27

A small object of uniform density rolls up a curved surface with an initial velocity v. it reaches up to a maximum height of   with respect to the initial position. The object is 

Detailed Solution for Test: Rotational Motion - 2 - Question 27

Test: Rotational Motion - 2 - Question 28

A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10 m/s. it then rolls up a smooth inclined plane of inclination 30° with the horizontal. The height attained by the sphere before it stops is  

Detailed Solution for Test: Rotational Motion - 2 - Question 28

Using the energy conservation
we know ωr = v = 10

Test: Rotational Motion - 2 - Question 29

A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is  

Detailed Solution for Test: Rotational Motion - 2 - Question 29

Test: Rotational Motion - 2 - Question 30

A body of mass 1.5 kg rotating about an axis with angular velocity of 0.3 rad s-1 has an angular momentum of 1.8 kg m2s-1. The radius of gyration of the body about an axis is :  

Detailed Solution for Test: Rotational Motion - 2 - Question 30

Angular momentum is given by 

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