Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 the change in its velocity as it reaches a position where the string is horizontal is:

$\sqrt{u^2-2gL}$

$\sqrt{2gL}$

$\sqrt{u^2-gL}$

$\sqrt{2(u^2-gL)}$

Step-by-Step Solution

Let the initial velocity at the lowest point be $\vec{v}_i = u \hat{i}$ . When the string is horizontal, the stone has reached a height $L$ . Let its velocity at this position be $\vec{v}_f = v \hat{j}$ . Using the principle of conservation of mechanical energy: $\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgL$ $v^2 = u^2 - 2gL$ $v = \sqrt{u^2 - 2gL}$ So, $\vec{v}_f = \sqrt{u^2 - 2gL} \hat{j}$ . The change in velocity is $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = \sqrt{u^2 - 2gL} \hat{j} - u \hat{i}$ . The magnitude of the change in velocity is: $|\Delta \vec{v}| = \sqrt{(\sqrt{u^2 - 2gL})^2 + (-u)^2} = \sqrt{u^2 - 2gL + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$ .

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