A) \[\frac{3R}{10}\]
B) \[\frac{9R}{7}\]
C) \[\frac{8R}{5}\]
D) \[\frac{9R}{5}\]
Correct Answer: B
Solution :
[b] Apply C.O.M.E. \[\frac{1}{2}m{{V}^{2}}=\frac{mgh}{\left( 1+\frac{h}{R} \right)}\] \[V=\frac{3}{4}{{V}_{e}}=\frac{3}{4}\sqrt{2gR}\,({{V}_{e}}=\sqrt{2gR})\] \[\frac{1}{2}\times \frac{9}{16}\times 2gR=\frac{gh}{1+\frac{h}{R}}\Rightarrow h=\frac{9R}{7}\]
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