A particle suspended by a thread of length ‘l’ is projected horizontally with a velocity $\sqrt{3\mathrm{gl}}$ at the lowest point. The height from the bottom at which the tension in the string becomes zero is

When the particle is not being affected by any external forces,

$∆\mathrm{KE}+∆\mathrm{PE}=0$

Assume that at height h the thread's tension is zero, and that the height at which a particle forms an angle is provided as well as the radial force operating at that height.

$\mathrm{h}=\mathrm{l}+\mathrm{lcos\theta }$

${\mathrm{F}}_{\mathrm{net}}=\mathrm{T}+\mathrm{mgcos\theta }$

let ${\mathrm{v}}_{\mathrm{B}}$ be the speed of the particle,

${\mathrm{F}}_{\mathrm{r}}=\frac{\mathrm{m}{{\mathrm{v}}_{\mathrm{B}}}^2}{\mathrm{l}}$

On equating the above equation,

${{\mathrm{v}}_{\mathrm{B}}}^2=\mathrm{glcos\theta }$

By using the law of conservation,

$\frac{1}{2}\mathrm{mglcos\theta }+\frac{3}{2}\mathrm{mgl}-\mathrm{mgl}(1+\mathrm{cos\theta })=0$

$\mathrm{lcos\theta }=\frac{-1}{3}$

$$\begin{gathered} \Rightarrow \mathrm{h}=\mathrm{l}+\mathrm{lcos\theta } \\ \Rightarrow \mathrm{h}=\frac{2\mathrm{l}}{3} \end{gathered}$$

Hence the correct answer is $\frac{2l}{3}$.