A solid sphere & a hollow sphere of radius R are rolling down an inclined plane of height h. The ratio of velocities of Solid sphere to Hollow sphere on reaching the bottom is

Correct option is B

​$$\text{The total energy at the start is just the gravitational potential energy:} \\P.E = K_{\text{translational}} + K_{\text{rotational}} \\mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \\\text{Where:} \\m = \text{mass} \\v = \text{linear velocity} \\I = \text{moment of inertia} \\\omega = \frac{v}{R} \text{ (because it is rolling without slipping)} \\2mgh = mv^2 \left( 1 + \frac{I}{mR^2} \right) \\v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}$$

$$\text{For the Solid Sphere:} \\\text{The moment of inertia } I = \frac{2}{5} mR^2. \text{ Substituting into the equation:} \\v_{\text{solid}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{5}{7} gh} \\\text{For the Hollow Sphere:} \\\text{The moment of inertia } I = \frac{2}{3} mR^2. \text{ Substituting into the equation:} \\v_{\text{hollow}} = \sqrt{\frac{2gh}{1 + \frac{2}{3}}} = \sqrt{\frac{3}{5} gh} \\\text{Ratio of Velocities:} \\\text{The ratio of the velocities of the solid sphere to the hollow sphere is:} \\\frac{v_{\text{solid}}}{v_{\text{hollow}}} = \frac{\sqrt{\frac{5}{7} gh}}{\sqrt{\frac{3}{5} gh}} = \sqrt{\frac{5}{7} \times \frac{5}{3}} = \frac{\sqrt{25}}{\sqrt{21}} = \frac{5}{\sqrt{21}}$$​​​