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
$M=\pi R_{1}^{2}td_{1}=\pi R_{2}^{2}td_{2}$
or $\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{d_{2}}{d_{1}} \ldots\left(i\right)$
As $I_{1}=\frac{1}{2}MR_{1}^{2}$ and $I_{2}=\frac{1}{2}MR_{2}^{2}$
$\therefore \frac{I_{1}}{I_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{d_{2}}{d_{1}}$ (using (i))
or $I_{1}=\frac{d_{2}}{d_{1}}I_{2}$