Question

Two rings of radii \( R \) and \( nR \) made from the same wire have the ratio of moments of inertia about an axis passing through their centre and perpendicular to the plane of the rings is 1 : 8. The value of \( n \) is:

• A. \( \frac{1}{2} \)
• B. \( 2\sqrt{2} \)
• C. 2
• D. 4

Hint:

The moment of inertia of a ring is directly proportional to the square of its radius.

Answer 1

The Correct Option is D

Step 1: Moment of Inertia of a Ring.
The moment of inertia of a ring is given by: \[ I = m R^2 \] where \( m \) is the mass and \( R \) is the radius of the ring. If the radii of the two rings are \( R \) and \( nR \), the ratio of their moments of inertia is: \[ \frac{I_1}{I_2} = \frac{R^2}{(nR)^2} = \frac{1}{n^2} \] Given that the ratio is 1 : 8, we have: \[ \frac{1}{n^2} = \frac{1}{8} \quad \Rightarrow \quad n = 4 \] Step 2: Conclusion.
Thus, the value of \( n \) is 4.