A wire of length 1m fixed at one end has a sphere attached to it at the other end. The sphere is projected horizontally with a velocity of $\sqrt{9\mathrm{g}}.$ When it describes a vertical circle, the ratio of elongations of the wire when the sphere is at the top and bottom of the circle is 

 

When the sphere is at its point, its kinetic energy will be at its bottom raise and will increase potential energy.

${\mathrm{KE}}_{\mathrm{top}}={\mathrm{KE}}_{\mathrm{bottom}}-\mathrm{PE}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{b}}}^2-\mathrm{mgh}=\frac{1}{2}\mathrm{m}9\mathrm{g}-\mathrm{mg}2\mathrm{r}$

$\mathrm{as} \mathrm{r}=1$

${\mathrm{KE}}_{\mathrm{top}}=\frac{9}{2}\mathrm{mg}-2\mathrm{mg}=\frac{5}{2}\mathrm{mg}$

$$\begin{gathered} \mathrm{As}, \mathrm{KE}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{t}}}^2 \\ \Rightarrow \frac{5}{2}\mathrm{mg}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{t}}}^2 \\ \Rightarrow {\mathrm{v}}_{\mathrm{t}}=\sqrt{5\mathrm{g}} \end{gathered}$$

$\mathrm{Tension} \mathrm{in} \mathrm{top} \mathrm{position}, \mathrm{m}\frac{{{\mathrm{v}}_{\mathrm{t}}}^2}{\mathrm{r}}-\mathrm{mg}=5\mathrm{mg}-\mathrm{mg}=4\mathrm{mg}$

$\mathrm{tension} \mathrm{in} \mathrm{bottom} \mathrm{position} \mathrm{is}: \mathrm{m}\frac{{{\mathrm{v}}_{\mathrm{b}}}^2}{\mathrm{r}}+\mathrm{mg}=9\mathrm{mg}+\mathrm{mg}=10\mathrm{mg}$

$$\begin{gathered} \mathrm{Thus}, \mathrm{Y}=\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{∆l}{l}}} \\ \Rightarrow ∆l=\frac{\mathrm{F}l}{\mathrm{AY}} \end{gathered}$$

$∆l\propto \mathrm{F}$

$\frac{∆l_{\mathrm{t}}}{∆l_{\mathrm{b}}}=\frac{{\mathrm{F}}_{\mathrm{t}}}{{\mathrm{F}}_{\mathrm{b}}}=\frac{4\mathrm{mg}}{10\mathrm{mg}}=\frac{2}{5}$

Hence the correct answer is 2:5.