A frictionless track ABCDE ends in a circular loop of radius R. A body slides down the track from rest at point A which is at a height h = 5 cm. Maximum value of R for the body to successfully complete the loop is

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5cm

2cm

$\frac{10}{3} cm$

$\frac{15}{4} cm$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = mgh \\ (∵ v = \sqrt{5 Rg}) \\ \frac{1}{2} m (5 Rg) = mgh \\ h = \frac{5 R}{2} \\ R = \frac{2}{5} \times h = 2 cm \end{array}$

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