A solid sphere of mass 500g radius 5cm is rotated about one of its...

The angular momentum about the diameter is

${L}_{\text{diameter}} = \dfrac{2}{5}M{R}^{2}\omega;$

The moment of inertia of a sphere is ${I}_{CM} = \dfrac{2}{5}M{R}^{2}$. The moment of inertia about the tangent is

${I}_{\text{tangent}} = {I}_{CM} + M{R}^{2} = \dfrac{2}{5}M{R}^{2} + M{R}^{2} = \dfrac{7}{5}M{R}^{2}$.

It is given that $\omega = 10rad{s}^{- 1}$

Using the given relation between the moment of inertia and the angular momentum,

$\dfrac{7}{5}M{R}^{2} = x \times {10}^{- 2} \times \dfrac{2}{5}M{R}^{2}\omega ⇒ x = \dfrac{3 . 5 \times {10}^{2}}{\omega} = 3 . 5 \times 10 = 35$