A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes a half-circle, reaching the topmost position B. The ratio of kinetic energies

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Solution

Q.
A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes a half-circle, reaching the topmost position B. The ratio of kinetic energies $\frac{{(K . E)}_{A}}{{(K . E)}_{B}}$ is ______. [2024]

1 3 : 2

2 5 : 1

3 2 : 5

4 1 : 5

Ans.
(2)

Apply energy conservation

$\frac{1}{2} m V_{L}^{2} = \frac{1}{2} m V_{H}^{2} + m g (2 L)$

$∵$ $V_{L} = \sqrt{5 g L}$

So, $V_{H} = \sqrt{g L}$

$\frac{{(K . E)}_{A}}{{(K . E)}_{B}} = \frac{\frac{1}{2} m {(\sqrt{5 g L})}^{2}}{\frac{1}{2} m {(\sqrt{g L})}^{2}} = \frac{5}{1}$

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