Torques of equal magnitude are applied t | Class 11 Physics Chapter System of Particles and Rotational Motion, System of Particles and Rotational Motion NCERT Solutions

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Academic session 2026-27, Chapter 6 in the current curriculum.

Welcome to the NCERT Solutions for Class 11 Physics - Chapter System of Particles and Rotational Motion. This page offers a step-by-step solution to the specific question from Exercise 1, Question 11:

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 11:

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Answer:

Let m and r be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis,II = mr2

The moment of inertia of the solid sphere about an axis passing through its centre, III  =  2/5 mr2

We have the relation:

τ  =  I α

Where, α = Angular acceleration

τ = Torque

I = Moment of inertia

For the hollow cylinder, τI   =  II αI

For the solid sphere, τII  = III αII

As an equal torque is applied to both the bodies, τI   =  τ2

∴ αII / αI   =  II  / III    =  mr2 /  2/5 mr2    =  2/5

αII > αI              ....  (i)

Now, using the relation:

ω  = ω0 + αt

Where, ω0 = Initial angular velocity

t = Time of rotation

ω = Final angular velocity

For equal ω0 and t, we have:

ω ∝ α … (ii)

From equations (i) and (ii), we can write:

ωII > ωI

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about System of Particles and Rotational Motion.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 11: Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same....

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