If a cone of maximum volume is inscribed in a given sphere,

Solution:

(d) $\frac{2}{3}$

Let $h, r, V$ and $R$ be the height, radius of the base, volume of the cone and the radius of the sphere, respectively.

Given : $h=R+\sqrt{R^{2}-r^{2}}$

$\Rightarrow h-R=\sqrt{R^{2}-r^{2}}$

Squaring both side, we get

$h^{2}+R^{2}-2 h R=R^{2}-r^{2}$

$\Rightarrow r^{2}=2 h r-h^{2}$          .......(1)

Now,

Volume $=\frac{1}{3} \pi r^{2} h$

$\Rightarrow V=\frac{\pi}{3}\left(2 h^{2} R-h^{3}\right)$                    $[$ From eq. $(1)]$

$\Rightarrow \frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right)$

For maximum or minimum values of $V$, we must have

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

$\Rightarrow \frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0$

$\Rightarrow 4 h R-3 h^{2}=0$

$\Rightarrow 4 h R=3 h^{2}$

$\Rightarrow h=\frac{4 R}{3}$

Now,

$\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)=-\frac{4 \pi R}{3}<0$

So, volume is maximum when $h=\frac{4 R}{3}$.

$\Rightarrow h=\frac{2(2 R)}{3}$

$\Rightarrow \frac{h}{2 R}=\frac{2}{3}$

$\therefore \frac{\text { Height }}{\text { Diameter of sphere }}=\frac{2}{3}$