Question

Two rings of radius \( R \) and \( nR \) made of the same material have the ratio of moment of inertia about an axis passing through their centre and perpendicular to the plane as \( 1 : 8 \). The value of \( n \) is (mass per unit length is constant).

• A. 1
• B. 3
• C. 4
• D. 2

Hint:

For rings and wires of uniform material, remember that mass is proportional to circumference. Always include this while comparing moments of inertia.

Answer 1

The Correct Option is D

Step 1: Write the formula for moment of inertia of a ring.
For a ring about an axis passing through its centre and perpendicular to its plane, the moment of inertia is \[ I = MR^2 \] where \( M \) is the mass of the ring and \( R \) is its radius.
Step 2: Express mass in terms of radius.
Since both rings are made of the same material and mass per unit length is constant, mass is directly proportional to the circumference.
\[ M \propto 2\pi R \Rightarrow M \propto R \]
Step 3: Write moments of inertia for both rings.
For the first ring (radius \( R \)):
\[ I_1 \propto R \times R^2 = R^3 \]
For the second ring (radius \( nR \)):
\[ I_2 \propto (nR) \times (nR)^2 = n^3 R^3 \]
Step 4: Use the given ratio.
\[ \frac{I_1}{I_2} = \frac{1}{8} \Rightarrow \frac{R^3}{n^3 R^3} = \frac{1}{8} \] \[ \Rightarrow n^3 = 8 \Rightarrow n = 2 \]
Step 5: Conclusion.
The value of \( n \) is \( 2 \).