A particle is connected to a string of length l and rotated in a...

Let the point where the string is horizontal be B, and the highest point of the vertical circle be C.

For the particle to just complete the vertical circle, the tension at the highest point C must be zero. The minimum speed at the highest point is given by:

$v_{C} = \sqrt{gl}$

Applying the principle of conservation of mechanical energy between point B (horizontal position) and point C (highest position):

$E_{B} = E_{C}$

$\dfrac{1}{2}mv_B^{2} = \dfrac{1}{2}mv_C^{2} + mgh$

Here, the height difference between B and C is $h = l$. Substituting $v_{C} = \sqrt{gl}$:

$\dfrac{1}{2}mv_B^{2} = \dfrac{1}{2}m(\sqrt{gl})^{2} + mgl$

$\dfrac{1}{2}mv_B^{2} = \dfrac{1}{2}mgl + mgl$

$\dfrac{1}{2}mv_B^{2} = \dfrac{3}{2}mgl$

$v_{B} = \sqrt{3gl}$

Thus, the minimum speed when the string is horizontal is $\sqrt{3gl}$.