A sphere is rolling on a frictionless surface as shown in figure with translational velocity V ms^-1. if it is to climb the inclined surface then V should be

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$\frac{10}{7} gh$

$\geq \sqrt{2 gh}$

2 gh

$\geq \sqrt{\frac{10}{7} gh}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {MV}^{2} + \frac{1}{2} I ω^{2} \geq Mgh \\ V \geq \sqrt{\frac{10}{7} gh} \end{array}$

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