Choose the correct answer of the following question:

Solution:

Let the radii of the smaller and given cones be $r$ and $R$, respectively; and their heights be $h$ and $H$, respectively.

We have,

$H=2 h \quad \ldots \ldots$ (i)

In $\Delta \mathrm{AQD}$ and $\Delta \mathrm{APC}$,

$\angle \mathrm{QAD}=\angle \mathrm{PAC} \quad($ Common angle $)$

$\angle \mathrm{AQD}=\angle \mathrm{APC}=90^{\circ}$

So, by AA criteria

$\Delta \mathrm{AQD} \sim \Delta \mathrm{APC}$

$\Rightarrow \frac{\mathrm{AQ}}{\mathrm{AP}}=\frac{\mathrm{QD}}{\mathrm{PC}}$

$\Rightarrow \frac{h}{H}=\frac{r}{R}$

$\Rightarrow \frac{h}{2 h}=\frac{r}{R} \quad[$ Using (i) $]$

$\Rightarrow \frac{1}{2}=\frac{r}{R}$

$\Rightarrow R=2 r \quad \ldots$ (ii)

Now,

The ratio of the volume of the smaller cone to the whole cone $=\frac{\text { Volume of the smaller cone }}{\text { Volume of the whole cone }}$

$=\frac{\left(\frac{1}{3} \pi r^{2} h\right)}{\left(\frac{1}{3} \pi R^{2} H\right)}$

$=\left(\frac{r}{R}\right)^{2} \times\left(\frac{h}{H}\right)$

$=\left(\frac{r}{2 r}\right)^{2} \times\left(\frac{h}{2 h}\right)$   [Using (i) and (ii)]

$=\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{2}\right)$

$=\frac{1}{8}$

$=1: 8$

Hence, the correct answer is option (d).