A stone tied to a string of length L is whirled in a vertical circle… | JEE Main 2022 (27 Jun, Shift 2)

Q14JEE Main 2022 · 27 JUN · SHIFT 2Medium

Vertical Circle: Speed & Tension

A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{2(u^2 - gL)}$ . The value of x is -

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Answer: (b)

Energy: speed when the string is horizontal, $v^2 = u^2 - 2gL$
$\vec{u}$ (horizontal) and $\vec{v}$ (vertical) are perpendicular, so $|\Delta \vec{v}| = \sqrt{u^2 + v^2}$
$|\Delta \vec{v}| = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$
Comparing with $\sqrt{x(u^2 - gL)}$ : $x = 2$ , so the answer is (b).

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