The height of a cone is $ 20 \mathrm{~cm} $. A small cone is cut off from the top by a plane parallel to the base. If its volume be $ 1 / 125 $ of the volume of the original cone, determine at what height above the base the section is made.

Given:

The height of a cone is \( 20 \mathrm{~cm} \).

A small cone is cut off from the top by a plane parallel to the base.

Its volume is $\frac{1}{125}$ of the volume of the original cone.

To do:

We have to find the height above the base the section is made.

Solution:

Total height of the cone $H = 20\ cm$

Let a cone whose height is $h_2$ is cut off.

This implies,

The height of the remaining cone which is a frustum $h_{1}=20-h_{2} \mathrm{cm}$

Let $r_{1}$ and $r_{2}$ be the radii of the bigger cone and smaller cone respectively.

Therefore,

Volume of the bigger cone $=\frac{1}{3} \pi r_{1}^{2} h_{1}$

Volume of the smaller cone $=\frac{1}{3} \pi r_{2}^{2} h_{2}$

$\Rightarrow \frac{\frac{1}{3} \pi r_{2}^{2} h_{2}}{\frac{1}{3} \pi r_{1}^{2} H}=\frac{1}{125}$

$\Rightarrow \frac{r_{2}^{2} \times h_{2}}{r_{1}^{2} \times H} = \frac{1}{125}$

$=\frac{1}{5} \times \frac{1}{25}$

$\Rightarrow \frac{r_{2}^{2}}{r_{1}^{2}} \times \frac{h_{2}}{H}=(\frac{1}{5})^{2} \times \frac{1}{5}$

$\Rightarrow (\frac{r_{2}}{r_{1}})^{2}(\frac{h_{2}}{H})=(\frac{1}{5})^{2} \times \frac{1}{5}$

Comparing both sides, we get,

$\Rightarrow (\frac{r_{2}}{r_{1}})^{2}=(\frac{1}{5})^{2}$

$\frac{h_{2}}{H}=\frac{1}{5}$

$\Rightarrow 5 h_{2}=H$

$\Rightarrow 5 h_{2}=20$

$\Rightarrow h_{2}=4 \mathrm{~cm}$

$h_{1}=H-h_{2}$

$=20-4$

$=16 \mathrm{~cm}$

The height from the base of the section is $16 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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