A

In order for the block to complete a vertical circle, it must have a minimum speed at the highest point of the circle to ensure that the gravitational force is enough to provide the necessary centripetal force.
Step 1: Identify the forces at the top of the circle. At the top, the gravitational force acts downward, which provides the required centripetal force for circular motion. Hence,
$$mg = \frac{mv^2}{\ell}$$
where:
$m$ = mass of the block,
$v$ = speed at the highest point,
$\ell$ = length of the rope.
Step 2: Solve for the speed at the highest point. Simplifying the above equation results in:
$$g = \frac{v^2}{\ell}$$
$$v^2 = g\ell$$
$$v = \sqrt{g\ell}$$
Step 3: Determine the speed at the lowest point. For the block to reach the top, it must start with an initial speed at the bottom. Using the conservation of energy principle between the bottom and top of the circle, we can equate kinetic energy and potential energy:
At the bottom:
$$E_{initial} = \frac{1}{2}mv_0^2$$
At the top:
$$E_{final} = \frac{1}{2}mv^2 + mg(2\ell)$$
Hence,
$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mg(2\ell)$$
Step 4: Substitute and solve for $v_0$ (initial velocity). Substituting $v = \sqrt{g\ell}$ gives:
$$\frac{1}{2}mv_0^2 = \frac{1}{2}m(g\ell) + mg(2\ell)$$
Cancelling the mass $m$ and simplifying yields:
$$v_0^2 = g\ell + 4g\ell = 5g\ell$$
$$v_0 = \sqrt{5g\ell}$$
Therefore, the minimum horizontal velocity $v_0$ required for the block to move in a complete vertical circle is $$v_0 = \sqrt{5g\ell}$$.