JEE Main 2025 Jan 29 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

JEE Main2025Jan 29, Shift 1Physics

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 $n g R$ , where $n \geq 1$ , then ratio of kinetic energy of the body at bottom to that at top of the circle is

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

Solution

Energy conservation between the top and bottom of the vertical circle (height difference $2 R$ ):

$\frac{1}{2} m v_{b}^{2} = \frac{1}{2} m v_{t}^{2} + 2 m g R .$

With $v_{t}^{2} = n^{2} g R$ : $v_{b}^{2} = n^{2} g R + 4 g R = (n^{2} + 4) g R .$

Ratio of kinetic energies $= \frac{v _{b}^{2}}{v _{t}^{2}} = \frac{n ^{2} + 4}{n ^{2}} .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

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