Question 169 of 183

A body of mass ' \(m\) ' connected to a massless and unstretchable string goes in verticle circle of radius ' \(R\) 'under gravity \(g\). The other end of the string is fixed at the center of circle. If velocity at top of circular path is \(n \sqrt{g R}\), where, \(n \geqslant 1\), then ratio of kinetic energy of the body at bottom to that at top of the circle is

A\(\frac{n^2}{n^2+4}\)B\(\frac{n^2+4}{n^2}\)C\(\frac{n+4}{n}\)D\(\frac{n}{n+4}\)

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Answer: (B) \(\frac{n^2+4}{n^2}\)

$\begin{aligned} & v_0=\sqrt{v^2+2 g(2 R)} \\ & v_0=\sqrt{n^2 g R+4 g R} \\ & \therefore \quad \frac{k_{\text {bottom }}}{k_{\text {top }}}=\frac{v_0^2}{v^2}=\frac{n^2+4}{n^2}\end{aligned}$