The minimum velocity required for a body at the bottom of a vertical circle to complete the circle of radius 'R' is $\sqrt{5 gR}$ . A body is released from the top edge of an inclined plane as shown in the figure. The minimum value of 'h' for which the body can complete the vertical circle is

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$\frac{2 R}{5}$

$\frac{R}{2}$

$\frac{R}{5}$

$\frac{5 R}{2}$

answer is D.

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Detailed Solution

After sliding through a height of h the velocity of the body is given by $\sqrt{2 gh}$ .

So, $\sqrt{2 gh} = \sqrt{5 gR} \Rightarrow h = \frac{5 R}{2}$

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