Q 11-06-035NEETJEE MainMedium
A solid sphere rolls down from rest without slipping through a vertical height of $7$ m. Its speed at the bottom is ($g = 10\ \text{m/s}^2$)
Answer: (A) $10$ m/s
$$mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2}mv^2 \cdot \frac{7}{5} \;\Rightarrow\; v^2 = \frac{10gh}{7} = 100$$
$$v = 10\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics