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

From the question,

${\mathrm{T}}_{\max }=\frac{{\mathrm{mv}}^2}{\mathrm{r}}+\mathrm{mg}=8\mathrm{mg}$

By conserving energy,

$-\frac{1}{2}{\mathrm{mu}}^2+\mathrm{mg}.2\mathrm{r}=\frac{1}{2}{\mathrm{mV}}^2$

$\Rightarrow \frac{{\mathrm{mu}}^2}{\mathrm{r}}=\frac{\mathrm{m}{\left(\sqrt{7\mathrm{gr}}\right)}^2}{\mathrm{r}}-4\mathrm{mg}=3\mathrm{mg}$

$$\begin{gathered} {\mathrm{T}}_{\min }+\mathrm{mg}=\frac{{\mathrm{mu}}^2}{\mathrm{r}} \\ \Rightarrow {\mathrm{T}}_{\min }=3\mathrm{mg}-\mathrm{mg}=2\mathrm{mg} \end{gathered}$$

$\frac{{\mathrm{T}}_{\max }}{{\mathrm{T}}_{\min }}=\frac{8\mathrm{mg}}{2\mathrm{mg}}=\frac{4}{1}$

Hence the correct answer is 4:1.