A hollow sphere of internal and external radii $ 2 \mathrm{~cm} $ and $ 4 \mathrm{~cm} $ respectively is melted into a cone of base radius $ 4 \mathrm{~cm} $. Find the height and slant height of the cone.

Given:

A hollow sphere of internal and external radii \( 2 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) respectively is melted into a cone of base radius \( 4 \mathrm{~cm} \).

To do:

We have to find the height and slant height of the cone.

Solution:

Internal radius of the hollow sphere  $r = 2\ m$

External radius of the hollow sphere $R = 4\ m$
Volume of the hollow sphere $=\frac{4}{3} \pi(\mathrm{R}^{3}-r^{3})$

$=\frac{4}{3} \pi(4^{3}-2^{3})$

$=\frac{4}{3} \pi(64-8)$

$=\frac{4}{3} \pi \times 56$

$=\frac{224}{3} \pi \mathrm{cm}^{3}$

Radius of the base of the cone formed $r_{1}=4 \mathrm{~cm}$
Let $h$ be the height and $l$ the slant height of the cone.

Volume of the cone $=$ Volume of the hollow sphere

Therefore,

$\frac{1}{3} \pi r_{1}^{2} h=\frac{224}{3} \pi$

$\Rightarrow \frac{1}{3} \pi(4)^{2} h=\frac{224}{3} \pi$

$\Rightarrow \frac{16}{3} \pi h=\frac{224}{3} \pi$

$\Rightarrow h=\frac{224 \pi}{3} \times \frac{3}{16 \pi}$

$\Rightarrow h=14$

The height of the cone is $14 \mathrm{~cm}$

Slant height of the cone $l=\sqrt{r_{1}^{2}+h^{2}}$

$=\sqrt{(4)^{2}+(14)^{2}}$

$=\sqrt{16+196}$

$=\sqrt{212}$

$=14.56 \mathrm{~cm}$

The height and slant height of the cone are $14\ cm$ and $14.56\ cm$ respectively.

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Updated on: 10-Oct-2022

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