1. Identify the motion

This is a conical pendulum. The particle moves in a horizontal circle while the string makes an angle $\theta$ with the vertical.

Given: $r=\frac{L}{\sqrt{2}}$

For a conical pendulum, $r=L\sin\theta$ So, $L\sin\theta=\frac{L}{\sqrt{2}}$ $\sin\theta=\frac{1}{\sqrt{2}}$ Hence, $\theta=45^\circ$

So, $\cos\theta=\frac{1}{\sqrt{2}}, \qquad \tan\theta=1$

2. Resolve tension

Let tension in the string be $T$.

• Vertical balance: $T\cos\theta=mg$

• Horizontal component provides centripetal force: $T\sin\theta=\frac{mv^2}{r}$

3. Divide the two equations

$\frac{T\sin\theta}{T\cos\theta}=\frac{mv^2/r}{mg}$

$\tan\theta=\frac{v^2}{rg}$

Since $\theta=45^\circ$, we have $\tan\theta=1$ Therefore, $\frac{v^2}{rg}=1$ $v^2=rg$ $v=\sqrt{rg}$

4. Match with options

$v=\sqrt{rg}$ So the correct option is A.