A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the center. 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?

$\sqrt{u^{2} - 2 g L}$

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$\sqrt{2 g L}$

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$\sqrt{u^{2} - g L}$

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$\sqrt{2 (u^{2} - g L)}$

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Solution

The correct option is D
$\sqrt{2 (u^{2} - g L)}$

Applying energy conservation, we get final velocity, $v = \sqrt{u^{2} - 2 g L}$ .
In vector form, we can express final velocity $\to v = \sqrt{(u^{2} - 2 g L)}^j$ and initial velocity $\to u = u^i$ .
Now, the magnitude of the change in its velocity $= \sqrt{u^{2} + v^{2}} = \sqrt{2 (u^{2} - g L)}$ .

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The correct option is D √2(u2−gL) Applying energy conservation, we get final velocity, v=√u2−2gL. In vector form, we can express final velocity →v=√(u2−2gL) ^j and initial velocity →u=u ^i. Now, the magnitude of the change in its velocity =√u2+v2=√2(u2−gL).