Explanation

(c)
Volume of cone, $V = \frac{\pi}{3}r^{2}h$
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$\Rightarrow \ \ \ \ \ V = \frac{\pi}{3}r^{2}\sqrt{l^{2} - r^{2}}$
On differentiating w.r.t. $r,$ we get
$\frac{dV}{dr} = \frac{\pi}{3}\left\lbrack 2r\sqrt{l^{2} - r^{2}} + \frac{r^{2}}{2\sqrt{l^{2} - r^{2}}}( - 2r) \right\rbrack$
Put $\frac{dV}{dr} = 0$
$\Rightarrow \ \ \ 2r\sqrt{{(l}^{2} - r^{2}}) - \frac{r^{3}}{\sqrt{l^{2} - r^{2}}} = 0$
$\Rightarrow \ \ \ r\left\lbrack 2\left( l^{2} - r^{2} \right) - r^{2} \right\rbrack = 0$
$\Rightarrow \ \ r = \pm \ \ l\sqrt{\frac{2}{3}}$
$\therefore\ \$At $r = l\sqrt{\frac{2}{3}\ },\frac{d^{2}V}{dr^{2}} < 0,$ maxima
$\therefore\ \ \ h = \sqrt{l^{2} - \frac{2}{3}l^{2} = \frac{l}{\sqrt{3}}}$
In $\mathrm{\Delta}\ ABC$, tan$\theta = \frac{r}{h}$=$\frac{l\sqrt{\frac{2}{3}}}{\frac{l}{\sqrt{3}}} = \sqrt{2}$