Solid sphere A is rotating about an axis PQ . If the radius of the...

Using parallel axis theorem,

${I}_{PQ} = {I}_{\text{COM}} + M{d}^{2}$

$= \dfrac{2}{5}M{R}^{2} + M{d}^{2}$

$= M\left[\dfrac{2}{5} \times {5}^{2} + {10}^{2}\right]$

$= M [ 110 ] = M{k}^{2}$

$⇒ k = \sqrt{110}cm$

Therefore, $x = 110$.