Explanation

(a)
Let the diameter of the sphere is $AE\text{=}2r$
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Let the radius of cone is $x$ and height is $y$
$\therefore\ AD = y$
Since, $BD^{2} = AD.DE$
$\Rightarrow \ \ x^{2} = y(2r - y)$ …(i)
Volume of cone,
$V = \frac{1}{3}\pi x^{2}y = \frac{1}{3}\pi y(2r - y)y$
$= \frac{1}{3}\pi\left( 2ry^{2} - y^{3} \right)$
On differentiating w.r.t. $y$, we get
$\frac{dV}{dy} = \frac{1}{3}\pi\left( 4ry - 3y^{2} \right)$
For maxima and minima, put $\frac{dV}{dy} = 0$
$\Rightarrow \ \frac{1}{3}\pi\ \left( 4ry - 3y^{2} \right) = 0$
$\Rightarrow \ \ y(4r - 3y) = 0$
$\Rightarrow \ \ \ y = \frac{4}{3}r,\ 0$
Again on differentiating w.r.t. $y$, we get
$\frac{d^{2}V}{dy^{2}} = \frac{1}{3}\pi(4r - 6y)$
At $y = \frac{4}{3}\ r,\ \ \frac{d^{2}V}{dy^{2}} = \frac{1}{3}\pi(4r - 8r) = - ve$
$\therefore\$ Volume of cone is maximum at $y = \frac{4}{3}r.$
Now, $required\ ratio = \frac{height\ of\ cone}{Diameter\ of\ sphere}$
$= \frac{y}{2r} = \frac{\frac{4r}{3}}{2r} = \frac{2}{3}$