The velocity of a particle at highest point of the vertical circle is $\sqrt{3rg}$. The tension at the lowest point, if mass of the particle is $m$, is

Let the velocity at the lowest point be $v$. then according to the law of conservation of energy.

energy at the lowest point= energy at the highest point

$\Rightarrow \dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}m(\sqrt{3rg})^2 \\ \Rightarrow v=\sqrt{7rg}$

At lowest point,

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg \\ \Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{7rg} )^{ 2 } }{ r }$

$\Rightarrow T=mg+7mg=8mg$