Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moment of intertia $I_1$ and $I_2$, respectively. For $R_2=2 R_1$, the ratio of $I_1$ and $I_2$ would be $1 / x$, where $\mathrm{x}=$ ________ -

Solution

Given $R_2=2 R_1$ $\begin{aligned} & M_1=\sigma \times \pi R_1^2=M_o \\ & M_2=\sigma \times \pi R_2^2=M_o \\ & M_2=\sigma \times \pi R_2^2=\sigma \times \pi\left[2 R_1\right]^2=\sigma \times 4 \pi R_1^2=4 M_o \end{aligned}$ $\frac{I_1}{I_2}=\frac{\frac{M_1 R_1^2}{2}}{\frac{M_2 R_2^2}{2}}=\frac{M_1 R_1^2}{M_2 R_2^2}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}$

Hence, the correct answer is 16.