Question

The figure shows two solid discs with radius $R$ and $r$ respectively. If mass per unit area is same for both, what is the ratio of $\text{MI}$ of bigger disc around axis $\text{AB}$ $($ which is $\perp$ to the plane of the disc and passing through its centre $)$ of $\text{MI}$ of smaller disc around one of its diameters lying on its plane ? Given ${}^{\prime }{M}^{\prime }$ is the mass of the larger disc. ($\text{MI}$ stands for moment of inertia)

• A. $R^2:r^2$
• B. $2R^4:r^4$
• C. $2r^4:R^4$
• D. $2R^2:r^2$

Answer 1

The correct option is B $2R^4:r^4$

Let $\sigma$ be the mass per unit area of both discs.

$\therefore$ mass of smaller disc, $m=\sigma \left(\pi r^2\right)$
and, mass of bigger disc, $M=\sigma \left(\pi R^2\right)$

$\frac{{\text{MI}}_{big}}{{\text{MI}}_{small}}=\frac{\frac{1}{2}M{R}^2}{\frac{1}{4}m{r}^2}=\frac{2M{R}^2}{m{r}^2}$

$\therefore \frac{{\text{MI}}_{big}}{{\text{MI}}_{small}}=\frac{2\left(\sigma \pi R^2\right)R^2}{\left(\sigma \pi r^2\right)r^2}=\frac{2R^4}{r^4}$

Hence, option $\left(\text{D}\right)$ is correct.