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