The velocity of a body revolving in a vertical circle of radius ‘r’ at the lowest point  $\sqrt{7gr}$ . The ratio of maximum to minimum tensions in the string is 

$\mathrm{given} \mathrm{velocity} \mathrm{at} \mathrm{lowest} \mathrm{point} \mathrm{is} {{\mathrm{v}}_1}^2=7\mathrm{gr}---\left(1\right)$

$\mathrm{we} \mathrm{know} \mathrm{that} {\mathrm{v}}_{\mathrm{\theta }}=\sqrt{{{\mathrm{v}}_1}^2-2\mathrm{gr}(1-\mathrm{cos\theta })}$

$\Rightarrow {\mathrm{v}}_2=\sqrt{{{\mathrm{v}}_1}^2-2\mathrm{gr}(1+1)}$

$\mathrm{substitute} \mathrm{given} {{\mathrm{v}}_1}^2=7\mathrm{gr} \mathrm{in} \mathrm{above} \mathrm{equation}$

$$\begin{gathered} \Rightarrow {{\mathrm{v}}_2}^2=7\mathrm{gr}-4\mathrm{gr} \\ \Rightarrow {{\mathrm{v}}_2}^2=3\mathrm{gr}---\left(2\right) \end{gathered}$$

$\frac{{\mathrm{T}}_{\max }}{{\mathrm{T}}_{\min }}=\frac{{\displaystyle \frac{\mathrm{m}{{\mathrm{v}}_1}^2}{\mathrm{r}}+\mathrm{mg}}}{\frac{\mathrm{m}{{\mathrm{v}}_2}^2}{\mathrm{r}}-\mathrm{mg}}$

$\mathrm{substitute} \mathrm{eqn}\left(1\right) \& \left(2\right)$

$\frac{{\mathrm{T}}_{\max }}{{\mathrm{T}}_{\min }}=\frac{8\mathrm{mg}}{2\mathrm{mg}}$

$\frac{{\mathrm{T}}_{\max }}{{\mathrm{T}}_{\min }}=\frac{4}{1}$