Question

Two iron solid discs of negligible thickness have radii \( R_1 \) and \( R_2 \) and moment of inertia \( I_1 \) and \( I_2 \), respectively. For \( R_2 = 2R_1 \), the ratio of \( I_1 \) and \( I_2 \) would be \( \frac{1}{x} \), where \( x \) is:

Hint:

Remember that the moment of inertia scales with the square of the radius for simple geometric bodies like discs and spheres.

Answer 1

The moment of inertia for a disc is expressed as \( I = \frac{1}{2} M R^2 \).Step 1: Consider two discs:\[I_1 = \frac{1}{2} M R_1^2, \quad I_2 = \frac{1}{2} M R_2^2\]Step 2: Given \( R_2 = 2R_1 \), substitute this into the equation for \( I_2 \):\[I_2 = \frac{1}{2} M (2R_1)^2 = 4 \times \frac{1}{2} M R_1^2 = 4 I_1\]Step 3: The ratio of \( I_1 \) to \( I_2 \) is \( \frac{1}{4} \), where \( x = 4 \).Final Conclusion:The determined value for \( x \) is 4, which aligns with Option (2).