JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 15)
A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical 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 ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :
$\frac{n^2 + 4}{n^2}$
$\frac{n + 4}{n}$
$\frac{n^2}{n^2 + 4}$
$\frac{n}{n + 4}$
Spiegazione
$$\begin{aligned}
& \mathrm{V}_{\text {Top }}=\sqrt{\mathrm{n}^2 \mathrm{gR}} \\
& \mathrm{~V}_{\text {Botom }}=\sqrt{\mathrm{n}^2 \mathrm{gR}+4 \mathrm{gR}} \\
& \text { Ratio }=\frac{\mathrm{n}^2+4}{\mathrm{n}^2}
\end{aligned}$$