Question

Solid sphere $A$ is rotating about an axis $PQ$. If the radius of the sphere is $5 cm$ then its radius of gyration about $PQ$ will be $\sqrt{x} cm$. The value of $x$ is __

Answer 1

$ I _{ cm }=\frac{2}{5} MR ^2 $
$ I _{ PQ }= I _{ cm }+ md ^2$
$ I _{ PQ }=\frac{2}{5} mR ^2+ m (10 cm )^2$
For radius of gyration
$ I _{ PQ }= mk ^2$
$ k ^2=\frac{2}{5} R ^2+(10 cm )^2$
$ =\frac{2}{5}(5)^2+100$
$ =10+100=110 $
$ k =\sqrt{110} cm $
$ x =110$