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

Step 1: Apply the law of conservation of mechanical energy.

As the stone moves from the lowest position to the horizontal position, it rises by a height $L$.

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

Step 2: Solve for the final speed squared.

Simplify the energy equation to express $v^2$ in terms of $u$, $g$, and $L$.

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

Step 3: Define the initial and final velocity vectors.

At the lowest point, the velocity is horizontal, and at the horizontal position, the velocity is vertical.

$$\vec{v}_i = (u, 0)$$ $$\vec{v}_f = (0, v)$$

Step 4: Find the change in velocity vector.

Subtract the initial velocity vector from the final velocity vector.

$$\Delta \vec{v} = \vec{v}_f - \vec{v}_i$$ $$\Delta \vec{v} = (-u, v)$$

Step 5: Calculate the magnitude of the change in velocity.

Find the magnitude of the vector $\Delta \vec{v}$.

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

Step 6: Substitute the value of $v^2$.

Substitute $v^2 = u^2 - 2gL$ into the magnitude expression.

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

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