Explanation

(b) The kinetic energy of a rolling body is
$\frac{1}{2}mv^{2}\left( 1 + \frac{k^{2}}{R^{2}} \right)$
According to law of conservation of energy, we get
$\frac{1}{2}mv^{2}\left( 1 + \frac{k^{2}}{R^{2}} \right) = mgh$
Where $h$ is the height of the inclined plane
$\therefore v = \sqrt{\frac{2gh}{1 + \frac{k^{2}}{R^{2}}}}$
For a solid sphere $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Substituting the given values, we get
$v = \sqrt{\frac{2 \times 10 \times 7}{\left( 1 + \frac{2}{5} \right)}} = \sqrt{\frac{2 \times 10 \times 7 \times 5}{7}} = 10\ ms^{- 1}$