Circular Motion JEE Main | KE Ratio in Vertical Circle

JEE MAIN 2025

29-01-2025 SHIFT-1

Question

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 {gR} $ , where, $n\ge1$ , then ratio of kinetic energy of the body at bottom to that at top of the circle is

Select the correct option:

$\frac{{n + 4}}{n}$

$\frac{{{{\rm{n}}^2} + 4}}{{{{\rm{n}}^2}}}$

$\frac{n}{{n + 4}}$

$\frac{{{n^2}}}{{{n^2} + 4}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

${{\rm{V}}_{{\rm{Top }}}} = \sqrt {{{\rm{n}}^2}{\rm{gR}}} $
$\;{{\rm{V}}_{{\rm{Bottom }}}} = \sqrt {{{\rm{n}}^2}{\rm{gR}} + 4{\rm{gR}}} $
${\rm{ Ratio }} = \frac{{{{\rm{n}}^2} + 4}}{{{{\rm{n}}^2}}}$

Question Tags

JEE Main

Physics

Easy

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