Question

A uniform solid sphere of mass $M$ , radius $R$ and having moment of inertia about an axis passing through center of mass is $I$ , is recast into a uniform circular disc of thickness $t$ , whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$ . Then radius of the disc will be

• A. $R\sqrt{\frac{2}{15}}$
• B. $\frac{2R}{\sqrt{15}}$
• C. $\frac{4R}{\sqrt{15}}$
• D. $\frac{R}{4}$

Answer 1

Answer: (B)

$I=\frac{2}{5}M{R}^2$ and $\rho =\frac{3M}{4\pi R^3}$
For circular disc $\pi r^2\times t\rho =M\Rightarrow \frac{\pi r^2\cdot t\cdot 3M}{4\pi R^3}=M$
$\Rightarrow t=\frac{4\pi R^3}{3\pi r^2}=\frac{4R^3}{3r^2}\Rightarrow r^2=\frac{4R^3}{3t}$
$\frac{2}{5}M{R}^2=\frac{3}{2}M{r}^2$
$\Rightarrow \frac{2}{5}{R}^2=\frac{3}{2}\cdot \frac{4R^3}{3t}\Rightarrow t=5R$
$\Rightarrow r^2\cdot 3\cdot 5R=4R^3\Rightarrow r^2=\frac{4R^2}{15}$
$\Rightarrow r=\frac{2R}{\sqrt{15}}$