Q 11-05-130JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
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 $\sqrt{x(u^2 - gL)}$. The value of $x$ is
Answer: (A) $2$
Energy conservation from the bottom to the horizontal position (rise $L$):
$$v^2 = u^2 - 2gL$$
At the bottom the velocity is horizontal; with the string horizontal it is vertical. The two are perpendicular, so
$$|\Delta\vec v| = \sqrt{u^2 + v^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$$
$x = 2$.
Solution by Sreeraj P, M.Sc Physics