Question

A hollow sphere and a solid sphere, of equal mass and equal radii roll down without slipping on an inclined plane. If the torques experienced are \( \tau_H \) and \( \tau_S \), then (A) \( \tau_H < \tau_S \)

• A. \( \tau_H > \tau_S \)
• B. \( \tau_H = \tau_S \)
• C. \( \tau_H = 0 \)
• D. \( \tau_S = 0 \)
• E.

Hint:

Higher moment of inertia ⇒ higher torque requirement.

Answer 1

The Correct Option is B

Concept: Torque due to friction
Torque produced by friction is: \[ \tau = fR \] where $f$ is frictional force and $R$ is radius.

Step 1: Relation between torque and moment of inertia
For rolling motion: \[ \tau = I\alpha \] For the same angular acceleration $\alpha$, torque required is directly proportional to moment of inertia: \[ \tau \propto I \]

Step 2: Moment of inertia comparison
- Hollow sphere: \[ I_H = \frac{2}{3}MR^2 \] - Solid sphere: \[ I_S = \frac{2}{5}MR^2 \]

Step 3: Compare values
\[ \frac{2}{3} > \frac{2}{5} \Rightarrow I_H > I_S \]

Step 4: Effect on friction and torque
Since $I_H$ is greater, more torque is required to produce the same angular acceleration. Thus, frictional force (and hence torque) must be greater for the hollow sphere. Final Result:
\[ \boxed{\tau_H > \tau_S} \] Conclusion:
Objects with larger moment of inertia require greater frictional torque for rolling motion.