A) \[\frac{7v}{10g}\]
B) \[\frac{7{{v}^{2}}}{10g}\]
C) \[\frac{2{{v}^{2}}}{5g}\]
D) \[\frac{3v}{5g}\]
Correct Answer: B
Solution :
The velocity of solid sphere on the bottom of inclined \[v=\sqrt{\frac{2gh}{1+\frac{l}{M{{R}^{2}}}}}\] The moment of inertia of solid sphere about its diameter \[l=\frac{2}{5}\,M{{R}^{2}}\] \[v=\sqrt{\frac{2gh}{\left( 1+\frac{2}{5} \right)}}=\sqrt{\frac{10}{7}}gh\] \[h=\frac{7{{v}^{2}}}{10g}\]
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