Let height of a right circular cone $=h cm$ and $O A = rcm$
Given, slant height of a right circular cone $=3 cm$ In $\triangle O A B$,
$ \angle B O A=90^{\circ}$
$ (O B)^2+(O A)^2=(A B)^2$
[apply pythogoras theorem]
$ (h)^2+(r)^2=(3)^2$
$ r=\sqrt{9-h^2} \ldots \text { (i) }$
We know that, Volume of cone
$=\frac{\pi}{3} \pi r^2 h $
$ V=\frac{\pi}{3}\left(9-h^2\right) \times h$
$ {\left[\text { from Eq. (i), } r=\sqrt{9-h^2}\right]}$
$ V=\frac{\pi}{3}\left(9 h-h^3\right)$
$ \frac{d V}{d h}=\frac{\pi}{3}\left(9-3 h^2\right) $
$ \therefore \quad \frac{d V}{d h}=0 \Rightarrow \frac{\pi}{3}\left(9-3 h^2\right)=0 $
$ \Rightarrow \quad \frac{h^2}{}=\frac{9}{3}=3 \Rightarrow h=\sqrt{3}$
$ \frac{d^2 V}{d h^2}=\frac{-6 \times h}{3}<0$
So, $h=\sqrt{3}$ of the cone for maximum volume
