A body of mass ‘m’ connected to a massless and unstretchable string… | JEE Main 2025 (29 Jan, Shift 1)

Q6JEE Main 2025 · 29 JAN · SHIFT 1Medium

Critical Speed & Looping

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 $n\sqrt{gR}$ , where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is:

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Answer: (a)

$v_{top}^2 = n^2 gR$
Energy: $v_{bottom}^2 = v_{top}^2 + 2g(2R) = (n^2 + 4)gR$
$\frac{K_{bottom}}{K_{top}} = \frac{n^2 + 4}{n^2}$ , so the answer is (a).

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Critical Speed & Looping question from JEE Main 2025 (29 Jan, Shift 1)

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