A solid sphere is rolling on a surface as shown in figure, with a translational velocity $$v$$ ms$$^{-1}$$. If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is

Total mechanical energy is conserved during pure rolling, where total kinetic energy equals translational kinetic energy plus rotational kinetic energy, satisfying $$v = \omega R$$.

Given: $$I = \frac{2}{5}MR^2$$

From conservation of energy:

$$E_i = E_f$$

$$\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = Mgh$$

$$\frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2 = Mgh$$

$$\frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = Mgh$$

$$\frac{7}{10}Mv^2 = Mgh$$

$$v^2 = \frac{10}{7}gh \implies v = \sqrt{\frac{10}{7}gh}$$