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…

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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

Aicharya · Accepted Answer

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

A body of mass ' \(m\) ' connected to a massless and unstretchable string goes in verticle circle of radius…

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