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

A stone tied to a string of length $$L$$ is whirled in a vertical circle. At the lowest position, it has speed $$u$$. We need to find the magnitude of change in velocity when the string becomes horizontal.

Find the speed at the horizontal position.

Using energy conservation from the lowest point to the point where the string is horizontal (height = $$L$$):

$$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$

$$v^2 = u^2 - 2gL$$

Find the change in velocity vector.

At the lowest point, the velocity is horizontal (say, to the right): $$\vec{v_1} = u\hat{i}$$

At the horizontal position (string horizontal means the stone is at the same height as the center), the velocity is directed vertically (upward or downward). Since the stone moves in a circle, at the 90° position the velocity is vertical: $$\vec{v_2} = v\hat{j}$$

Calculate $$|\Delta \vec{v}|$$.

$$|\Delta \vec{v}| = |\vec{v_2} - \vec{v_1}| = \sqrt{u^2 + v^2}$$

Substituting $$v^2 = u^2 - 2gL$$:

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

Comparing with $$\sqrt{x(u^2 - gL)}$$, we get $$x = 2$$.

The correct answer is Option A.