Q78 / 90 · JEE Main 2012 (12 May Online) · Physics
A solid sphere is rolling on a surface as shown in figure, with a translational velocity $v \mathrm{~m} \mathrm{~s}^{-1}$. If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is
- A$\sqrt{2 g h}$
- B$\sqrt{\frac{7}{5} g h}$
- C$\sqrt{\frac{7}{2} g h}$
- Correct answerD$\sqrt{\frac{10}{7} g h}$
Solution
Minimum velocity for a body rolling without slipping
$$
v=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}
$$
For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$
$$
\therefore \quad v=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}=\sqrt{\frac{10}{7} g h}
$$
Hence, the correct answer is option: 'D'.