The sum of the legs of a right triangle is $9\mathrm{ }\mathrm{c}\mathrm{m}$. When the triangle rotates about one of the the legs, a cone results which has the maximum volume. Then,

Let the given situation be,

Let $\mathrm{BC}=\mathrm{r} \mathrm{cm} , \mathrm{AB}=\mathrm{h} \mathrm{cm} , \angle \mathrm{x}=$Semi vertical angle.

Let $BC$ be the side about which the triangle is revolved.

As it is given that $\mathrm{r}+\mathrm{h}=9 \mathrm{cm}$

Volume of the cone formed will be given as $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Substituting $\mathrm{r}=9-\mathrm{h}$ in $V$,

$$\begin{gathered} \Rightarrow \mathrm{V}=\frac{1}{3}\mathrm{\pi }(9-\mathrm{h}{)}^2\times \mathrm{h} \\ \Rightarrow \mathrm{V}=\frac{1}{3}\times \mathrm{\pi }\times \left(81\mathrm{h}-18{\mathrm{h}}^2+{\mathrm{h}}^3\right) \end{gathered}$$

As the volume of the cone must be Maximum so put $\frac{\mathrm{dV}}{\mathrm{dh}}=0$

Differentiating $V=\frac{1}{3}\times \mathrm{\pi }\times \left(81\mathrm{h}-18{\mathrm{h}}^2+{\mathrm{h}}^3\right)$ with respect to $h$ we get,

$$\begin{gathered} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dh}}=\frac{1}{3}\mathrm{\pi }\times \left(81-36\mathrm{h}+3{\mathrm{h}}^2\right) \\ \Rightarrow \frac{\mathrm{dV}}{\mathrm{dh}}=\frac{3}{3}\mathrm{\pi }\times \left(27-12\mathrm{h}+{\mathrm{h}}^2\right) \end{gathered}$$

Putting $\frac{\mathrm{dV}}{\mathrm{dh}}=0$

$$\begin{gathered} \Rightarrow \frac{3}{3}\mathrm{\pi }\times \left(27-12\mathrm{h}+{\mathrm{h}}^2\right)=0 \\ \Rightarrow 27-12\mathrm{h}+{\mathrm{h}}^2=0 \\ \Rightarrow {\mathrm{h}}^2-9\mathrm{h}-3\mathrm{h}+27=0 \\ \Rightarrow (\mathrm{h}-3)(\mathrm{h}-9)=0 \\ \Rightarrow \Rightarrow \mathrm{h}=3,9 \end{gathered}$$

Let $h=9$

$$\begin{gathered} \Rightarrow \mathrm{r}=9-\mathrm{h} \\ \Rightarrow \mathrm{r}=9-9 \\ \Rightarrow \mathrm{r}=0 \end{gathered}$$

$\Rightarrow h=9$ is not possible.

So, $h=3$

$$\begin{gathered} \Rightarrow \mathrm{r}=9-\mathrm{h} \\ \Rightarrow \mathrm{r}=9-3=6 \end{gathered}$$

Applying Pythagoras theorem to find the slant height$\left(AC\right)$ of cone.

So, Slant height $\mathrm{BC}=\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}$

$$\begin{gathered} \Rightarrow \mathrm{AC}=\sqrt{3^2+6^2} \\ \Rightarrow \mathrm{AC}=\sqrt{45} \\ \Rightarrow \mathrm{AC}=\sqrt{5\times 3^2} \\ \Rightarrow \mathrm{AC}=3\sqrt{5} \mathrm{cm} \end{gathered}$$

Thus the maximum volume of the cone is $\mathrm{V}=\frac{1}{3}\pi {\left(6\right)}^2\times 3$

$\Rightarrow \mathrm{V}=36\mathrm{\pi } {\mathrm{cm}}^3$

As we know the curved surface of the cone is given as $S=\mathrm{\pi rl}$ where $\mathrm{l}=AC$(Slant height)

$\Rightarrow S=\pi \times 6\times 3\sqrt{5}$

$\Rightarrow \mathrm{S}=18\sqrt{5}\times \mathrm{\pi } {\mathrm{cm}}^2$

In $△\mathrm{ABC}$ $\mathrm{tanx}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{r}}{\mathrm{h}}$

$$\begin{gathered} \Rightarrow \mathrm{x}={\tan }^{-1}\frac{3}{6} \\ \Rightarrow \mathrm{x}={\tan }^{-1}\frac{1}{2} \end{gathered}$$

So, Semi vertical angle of the cone is $\mathrm{x}={\tan }^{-1}\frac{1}{2}$.

Hence all options-$1,2,3,4$ are correct answers.