Explanation

(b) When a body of mass $m$ and radius $R$ rolls down on inclined plane of height $h$ and angle of inclination $\theta$, it losses potential energy. However, it acquires both linear and angular speeds.
Velocity at the lowest point $v = \sqrt{\frac{2gh}{1 + \frac{K^{2}}{R^{2}}}}$
For solid sphere $\frac{K^{2}}{R^{2}} = \frac{2}{5}$
$\therefore$ $v = \sqrt{\frac{2 \times 10 \times 7}{1 + \frac{2}{5}}} = 10$ m/s