B

To find the minimum velocity at the bottom of the circle for a body attached to a string of length \( \ell \), we need to consider the forces acting on the body at the highest point of the vertical circle.

Step 1: At the highest point of the circle, the centripetal force required to keep the body moving in a circular path is provided by the weight of the body and the tension in the string. For the minimum speed to cross the highest point, the tension can be considered to be zero. This gives us the equation:
\[ mg = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the body,
- \( g \) is acceleration due to gravity,
- \( v \) is the velocity at the highest point,
- \( r = \ell \) is the radius of the circular path.

Step 2: Simplifying the equation:
\[ g = \frac{v^2}{\ell} \]
So, \[ v = \sqrt{g\ell} \] at the highest point.

Step 3: To find the minimum velocity at the bottom of the circle, we apply the principle of conservation of energy. The potential energy at the top is converted into kinetic energy at the bottom.
The height difference between the top and bottom of the circle is \( 2\ell \) (since the height at the top is \( 2\ell \) above the lowest point):
\[ mgh = \frac{1}{2}mv^2 \] where \( h = 2\ell \).

Step 4: Plugging in the values:
\[ mg(2\ell) = \frac{1}{2}mv^2 \]
\[ 2g\ell = \frac{1}{2}v^2 \]
\[ v^2 = 4g\ell \]

Step 5: Therefore,\[ v = \sqrt{4g\ell} = 2\sqrt{g\ell} \] at the bottom of the circle.

Consequently, the minimum velocity at the bottom is \( 2\sqrt{g\ell} \). Thus, the correct answer is option B.