Question

A disc of radius 'R' and thickness $\frac{R}{6}$ has moment of inertia 'I' about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is

• A. $\frac{I}{5}$
• B. $\frac{I}{6}$
• C. $\frac{I}{32}$
• D. $\frac{I}{64}$

Answer 1

The correct option is A $\frac{I}{5}$

Moment of Inertia (MI) of disc is $I=\frac{M{R}^2}{2}$..................(1)

MI of Solid sphere $=\frac{2}{5}M{R}^2$

$Volum{e}_{disc}=Volum{e}_{solidsphere}$

$\pi \times R^2\times \frac{r}{6}=\frac{4}{3}\pi \times R^{\prime 3}$

this will give $R^{\prime }=\frac{R}{2}$

MI of Solid sphere $=\frac{2}{5}M\times (\frac{R}{2}{)}^2$

$=\frac{M{R}^2}{10}$

or from (1): ${MI}_{SolidSphere}=\frac{I}{5}$