A body is moving in a vertical circle of radius ‘r’ by a string. If the ratio of maximum to minimum speed is $\sqrt{3} :1$, the ratio of maximum to minimum tensions in the string is

Let the speeds at the lowest and highest positions be ${\mathrm{v}}_1$ and ${\mathrm{v}}_2$, respectively.

$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\sqrt{3}}{1}$

Between the lowest and highest locations, energy conservation gives

${{\mathrm{v}}_1}^2={{\mathrm{v}}_2}^2+2\mathrm{gh}$

$$\begin{gathered} \Rightarrow 3{{\mathrm{v}}_2}^2={{\mathrm{v}}_2}^2+2\mathrm{gh} \\ \Rightarrow {{\mathrm{v}}_2}^2=\mathrm{gh} \\ \Rightarrow {{\mathrm{v}}_1}^2=3\mathrm{gh} \end{gathered}$$

h = 2r, which is the distance between the lowest and tallest locations.

At its lowest and hottest, tensions

${\mathrm{T}}_1=\frac{\mathrm{m}{{\mathrm{v}}_1}^2}{\mathrm{r}}+\mathrm{mg}$

${\mathrm{T}}_2=\frac{\mathrm{m}{{\mathrm{v}}_2}^2}{\mathrm{r}}+\mathrm{mg}$

$\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{{\displaystyle \frac{{{\mathrm{v}}_1}^2}{\mathrm{r}}+\mathrm{g}}}{{\displaystyle \frac{{{\mathrm{v}}_2}^2}{\mathrm{r}}-\mathrm{g}}}$

$\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{{\displaystyle \frac{6\mathrm{gr}}{\mathrm{r}}+\mathrm{g}}}{{\displaystyle \frac{2\mathrm{gr}}{\mathrm{r}}-\mathrm{g}}} = \frac{7}{1}$

Hence the correct answer is 7:1.