A bob attached to a light string of length $10\text{m}$ is given an initial velocity of $20\text{m/s}$ at the bottom-most point to perform vertical circular motion. What will be the velocity of bob just before slacking of string?
(Take $g=10m/s^2$)

The correct option is B $10\sqrt{\frac{2}{3}}\text{m/s}$
The FBD of the bob at the position where the slackness happen is shown in the figure

The slackness of the string happens when the tension in the string becomes zero.
Hence the component of $mg$ along the string ($mg\cos \theta$) will balance the centrifugal force which is $\frac{m{v}^2}{R}$ in the radially outward direction.

$\Rightarrow mg\cos \theta =\frac{m{v}^2}{R}...\left(i\right)$,
Height of the bob, $H=R+R\cos \theta$

Now by the conservation of mechanical energy we get,
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+mg(R+R\cos \theta )...\left(ii\right)$, where $u$ is the initial velocity given to the bob at the bottom most point and $v$ is the velocity at the point where the slackness happen.

Putting the value of , $\cos \theta =\frac{v^2}{Rg}$ from equation $\left(i\right)$ in the equation $\left(ii\right)$ we get

$v^2=\frac{u^2-2gR}{3}$ , where $u=20\text{m/s},R=10\text{m}$

$v=\sqrt{\frac{200}{3}}=10\sqrt{\frac{2}{3}}\text{m/s}$

Hence option B is the correct answer