A solid sphere at rests rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of plane is ($g=$ acceleration due to gravity)

Applying conservation of mechanical energy for the solid sphere rolling without slipping down the inclined plane, the initial potential energy $mgh$ converts to both translational and rotational kinetic energy at the bottom.

For a solid sphere, $I = \frac{2}{5}mr^2$ and $\omega = v/r$. The rotational kinetic energy becomes $\frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{5}mv^2$.

The total kinetic energy is then $\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.

Equating $mgh = \frac{7}{10}mv^2$ and solving for velocity yields $v = \sqrt{\frac{10gh}{7}}$.

$\boxed{\left(\frac{10 g h}{7}\right)^{\frac{1}{2}}}$