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 of a disc is given by \( I = \frac{1}{2} M R^2 \). Step 1: For the two discs, we have: \[ I_1 = \frac{1}{2} M R_1^2, \quad I_2 = \frac{1}{2} M R_2^2 \] Step 2: For \( R_2 = 2R_1 \), we substitute 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: Thus, the ratio of \( I_1 \) and \( I_2 \) is \( \frac{1}{4} \), meaning \( x = 4 \). Final Conclusion: The value of \( x \) is 4, which corresponds to Option (2).