NCERT Solutions for Class 11 Physics Chapter 7 System of particles and Rotational Motion

Chapter 7 System of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 7 System of particles and Rotational Motion is designed and prepared by the best teachers across India. All the important topics are covered in the exercises and each answer comes with a detailed explanation to help students understand concepts better. These NCERT solutions play a crucial role in your preparation for all exams conducted by the CBSE, including the JEE.

NCERT TEXTBOOK QUESTIONS SOLVED

Ans. In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

Ans. When the child gets up and runs about on the trolley, the speed of the centre of mass of the trolley and child remains unchanged irrespective of the manner of motion of child. It is because here child and trolley constitute one single system and forces involved are purely internal forces. As there is no external force, there is no change in momentum of the system and velocity remains unchanged.

Ans. (a) Moment of inertia of sphere about any diameter = 2/5 MR2
Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.

Ans. Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are I1 = MR2 and I2 = 2/5 MR2
Let τ be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations α1and α2 respectively. Then τ=I1 α1 = I2 α2
The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time.

Ans. M = 20 kg
Angular speed, w = 100 rad s-1; R = 0.25 m
Moment of inertia of the cylinder about its axis
=1/2 MR2 = 1/2 x 20 (0.25)2 kg m2 = 0.625 kg m2
Rotational kinetic energy,
Er = 1/2 Iw2 = 1/2 x 0.625 x (100)2 J = 3125 J
Angular momentum,
L = Iw = 0.625 x 100 Js= 62.5 Js.

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