Question

Two metallic discs $A$ and $B$ of densities $d_1$ and $d_2$ respectively, possess same mass and thickness. If $I_1$ and $I_2$ are the moment of inertia of $A$ and $B$ respectively, then $I_1=$

• A. $I_2$
• B. $\frac{d_2^2}{d_1^2}{I}_2$
• C. $\frac{d_1}{d_2}{I}_2$
• D. $\frac{d_2}{d_1}{I}_2$

Answer 1

Answer: (D)

Let $R_1$ and $R_2$ be the radii of discs $A$ and $B$ respectively
If $M$ and $t$ are the mass and thickness of each disc, then
$M=\pi R_1^{2}t{d}_1=\pi R_2^{2}t{d}_2$
or $\frac{R_1^2}{R_2^2}=\frac{d_2}{d_1}\dots \left(i\right)$
As $I_1=\frac{1}{2}M{R}_1^2$ and $I_2=\frac{1}{2}M{R}_2^2$
$\therefore \frac{I_1}{I_2}=\frac{R_1^2}{R_2^2}=\frac{d_2}{d_1}$ (using (i))
or $I_1=\frac{d_2}{d_1}{I}_2$