Question

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotating in similar way, is \( n \) times higher than the moment of inertia of the given ring. Here, \( n = \):

Hint:

In rotational motion, the moment of inertia depends on the mass distribution relative to the axis of rotation.

Answer 1

For a disc, the moment of inertia about its diameter is \( I_{\text{disc}} = \frac{1}{4} M R^2 \), and for a ring, the moment of inertia is \( I_{\text{ring}} = M R^2 \).

Step 1: The moment of inertia of a solid sphere about its center is \( I_{\text{sphere}} = \frac{2}{5} M R^2 \).

Step 2: The relationship between the sphere and the ring's moment of inertia is:

\[ I_{\text{sphere}} = \frac{2}{5} I_{\text{ring}} \]

Step 3: Substituting values, we find that the ratio \( n \) is 1, so \( n = 1 \).

Final Conclusion: The value of \( n \) is 1, which corresponds to Option (1).