Q64 / 90 · JEE Main 2022 (26 Jun Shift 2) · Physics

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Solution

Under pure rolling condition: $v = ω R$

Translational $K . E = \frac{1}{2} m v^{2}$

Rotational $K . E = \frac{1}{2} \times \frac{2}{5} m R^{2} {(\frac{v}{R})}^{2} = \frac{1}{5} m v^{2}$

Total $K . E = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = \frac{7}{10} m v^{2}$

So, $\frac{Rotational K \cdot E}{Total K \cdot E} = \frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}} = \frac{2}{7}$

Hence, the correct answer is option: 'C'.