Explanation

(c) Given that,
Mass of solid sphere, $m = 2$ kg
Velocity, $v = 10\ ms^{- 1}$
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Let the sphere attained a height $h$.
When the sphere is at point $A$, it possesses kinetic energy and rotational kinetic energy, and when it is at point $B$ it possesses only potential energy.
So, from law of conservation of energy
$\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = mgh$
or $\frac{1}{2}mv^{2} + \frac{1}{2}mK^{2} \times \frac{v^{2}}{R^{2}} = mgh$
or $\frac{1}{2}mv^{2}\left\lbrack 1 + \frac{K^{2}}{R^{2}} \right\rbrack = mgh$
or $\frac{1}{2} \times (10)^{2}\left\lbrack 1 + \frac{2}{5} \right\rbrack = 9.8 \times h$
(for solid sphere, $\frac{K^{2}}{R^{2}} = \frac{2}{5})\ $
or $\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 \times h$
or $h = 7.1$ m