Discussion Forum

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

Answer:

Explanation:

According to question, Moment of inertia of disc is given by $l= \frac{MR^{2}}{2}$

 [symbols have their usual meanings]

 When the disc is remoulded into solid sphere, then volume remains same.

 i.e, Volume of disc= Volume of solid sphere

i.e,   $\pi R^{2}\times\frac{R}{6}=\frac{4}{3}\pi r^{3}$

 $\Rightarrow$    $r^{3}=\frac{R^{3}}{8}\Rightarrow r=\frac{R}{2}$

 Now,moment of inertia of solid sphere is given by $\frac{2}{5}m r^{2}$

 $=\frac{2}{5}\times m\times\frac{R^{2}}{4}=\frac{mR^{2}}{10}$

 =$\frac{l}{5}$  [l= moment of inertia of disc]