A heavy particle is tied to the end $\mathrm{A}$ of a string of length $1.6 \mathrm{m}$. Its other end $\mathrm{O}$ is fixed. It revolves as a conical pendulum with the string making $60°$ with the vertical. Then

As

$$\begin{gathered} \mathrm{T} \sin \left(60°\right)=\frac{{\mathrm{mv}}^2}{\mathrm{r}} ...\left(1\right) \\ \mathrm{T} \cos \left(60°\right)=\mathrm{mg} ...\left(2\right) \\ \therefore \tan \left(60°\right)=\frac{{\mathrm{v}}^2}{\mathrm{rg}} \\ \mathrm{v}=\sqrt{\mathrm{rg} \tan \left(60°\right)} \end{gathered}$$

Also, $\mathrm{r}=\mathrm{l} \sin \left(60°\right)$

So v = $\sqrt{0.8 \times \sqrt{3}\times 10\times \sqrt{3}} m/s =2.8\sqrt{3} m/s$

Time period$=\frac{2\mathrm{\pi r}}{\mathrm{v}}=\frac{2\mathrm{\pi r}}{\sqrt{\mathrm{rgtan}\left(60°\right)}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{r}}{\mathrm{gtan}\left(60°\right)}}$

$=2\mathrm{\pi }\sqrt{\frac{\mathrm{lcos}\left(60°\right)}{\mathrm{g}}}$= $2\mathrm{\pi }\sqrt{\frac{0.8}{10}}\mathrm{s}=\frac{4\mathrm{\pi }}{7} \mathrm{s}$

From equation $\left(1\right)$, $\mathrm{T}=2\mathrm{mg}$

$\therefore$Centripetal acceleration${\mathrm{a}}_{\mathrm{C}}=\frac{{\mathrm{v}}^2}{\mathrm{r}}=\frac{\mathrm{rgtan}\left(60°\right)}{\mathrm{r}}=\mathrm{gtan}\left(60°\right)$ = 10$\sqrt{3} m/s^2$