JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Work, Power & Energy

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 gR

, 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) $\frac{n ^{2} + 4}{n ^{2}}$ ✓ Correct

Answer: (D)

Solution

Let the velocity at the top be

v_{top} = n g R

. Using conservation of energy between the top and bottom of the circle (height difference

2 R

\frac{1}{2} m v_{bottom}^{2} = \frac{1}{2} m v_{top}^{2} + m g (2 R)

v_{bottom}^{2} = n^{2} g R + 4 g R = (n^{2} + 4) g R

The ratio of kinetic energies (same mass at both points) is the ratio of

v^{2}

values:

\frac{K E _{bottom}}{K E _{top}} = \frac{v _{bottom}^{2}}{v _{top}^{2}} = \frac{( n ^{2} + 4 ) g R}{n ^{2} g R} = \frac{n ^{2} + 4}{n ^{2}}

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Work, Power & Energy

Solution

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