The height of a cone is $60 \mathrm{~cm}$. A small cone is cut off at the top by a plane parallel to the base and its volume is $\frac{1}{64} t h$ the volume of original cone. The height from the base at which the section is made is:
It is given that the height of a cone = 60 cm
The volume of a small cone cut on the top by plane parallel to base of given cone
= volume of given cone
We have to find the height from the base at which section is made.
Let R be the radius of base of given cone
Let H be the height of given cone
H = 60 cm
Let r be the radius of base of small cone
Let h be the height of small cone
In ΔAPC and ΔAQE
PC || QE
Therefore
$\angle \mathrm{APC}=\angle \mathrm{AQE} \quad[$ Corresponding angles $]$
$\angle \mathrm{ACP}=\angle \mathrm{AEQ} \quad[$ Corresponding angles $]$
Also $\angle \mathrm{PAC}=\angle \mathrm{QAE}$ [Same angle]
Hence ΔAPC ~ ΔAQE
$\frac{A P}{A Q}=\frac{P C}{Q E}$
$\frac{h}{H}=\frac{r}{R}$.......(1)
Volume of cone $\mathrm{ADE}=\frac{1}{3} \pi R^{2} H$
Volume of cone $\mathrm{ABC}=\frac{1}{3} \pi r^{2} h$
$\frac{\text { Volume of cone } \mathrm{ABC}}{\text { Volume of cone } \mathrm{ADE}}=\frac{\frac{1}{3} \pi r^{2} h}{\frac{1}{3} \pi R^{2} H}$
$\frac{1}{64}=\left(\frac{r}{R}\right)^{2} \times \frac{h}{H}$
$\frac{1}{64}=\left(\frac{h}{H}\right)^{2} \times \frac{h}{H}[$ From $(1)]$
$\frac{1}{64}=\left(\frac{h}{H}\right)^{3}$
$\sqrt[3]{\frac{1}{64}}=\frac{h}{H}$
$\frac{1}{4}=\frac{h}{60}$
$h=15$
The height from the base at which section is made $=H-h$
$=60-15 \mathrm{~cm}$
$=45 \mathrm{~cm}$
Hence Option (c) is correct.
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