Given, height of inclined plane, $h=7 m$
$g=10 m / s ^{2}$
By conservation of energy, Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.
$m g h =\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} \cdot \frac{2}{5} m R^{2} \cdot\left(\frac{v}{R}\right)^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2} $
$mg h =\frac{7}{10} m v^{2}$
$\therefore v^{2}=\frac{10}{7} g h $
$\Rightarrow v=\sqrt{\frac{10}{7} g h}=\sqrt{\frac{10}{7} \times 10 \times 7}$
$=\sqrt{100} m / s $