A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring 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 -

Circular Motion PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q49 with Solution

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Correct answer: B

A stone is moving in a vertical circle of radius $L$ .

At the lowest point, its speed is $u$ .
We need the change in velocity when it reaches the position where the string is horizontal.

The change in velocity depends on the initial and final velocity vectors.

From the lowest point to the horizontal point, the stone rises by a height $L$ .

Using conservation of mechanical energy:

\frac12 m u^2 = \frac12 m v^2 + mgL

So,

v^2 = u^2 - 2gL

Thus the speed at the horizontal position is

v = \sqrt{u^2 - 2gL}

Therefore, the two velocity vectors are perpendicular.

Let:

So magnitude of change in velocity is

|\Delta \vec v| = \sqrt{u^2 + (u^2-2gL)}

because for perpendicular vectors,

|\vec v_2 - \vec v_1| = \sqrt{v_1^2 + v_2^2}

Hence,

|\Delta \vec v| = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2-gL)}

Comparing with the given form:

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

we get

x=2

So the correct option is B.