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A hollow sphere of internal and external diameters $ 4 \mathrm{~cm} $ and $ 8 \mathrm{~cm} $ respectively is melted into a cone of base diameter $ 8 \mathrm{~cm} $. Calculate the height of the cone.
Given:
A hollow sphere of internal and external diameters \( 4 \mathrm{~cm} \) and \( 8 \mathrm{~cm} \) respectively is melted into a cone of base diameter \( 8 \mathrm{~cm} \).
To do:
We have to find the height of the cone.
Solution:
Outer diameter of hollow sphere $=8 \mathrm{~cm}$
Inner diameter of hollow sphere $=4 \mathrm{~cm}$
This implies,
Outer radius $R=\frac{8}{2}$
$=4 \mathrm{~cm}$
Inner radius $r=\frac{4}{2}$
$=2 \mathrm{~cm}$
Volume of the hollow sphere $=\frac{4}{3} \pi(\mathrm{R}^{3}-r^{3})$
$=\frac{4}{3} \pi[4^{3}-2^{3}]$
$=\frac{4}{3} \pi[64-8]$
$=\frac{4 \pi}{3} \times 56$
$=\frac{224}{3} \pi \mathrm{cm}^{3}$
Diameter of the base of the solid cone $=8 \mathrm{~cm}$
This implies,
Radius of the base of the solid cone $r_1=\frac{8}{2}$
$=4 \mathrm{~cm}$
Let the height of the cone be $h$.
Therefore,
$\frac{1}{3} \pi r_{1}^{2} h=\frac{224}{3} \pi$
$\Rightarrow \frac{1}{3} \pi(4)^{2} h=\frac{224}{3} \pi$
$\Rightarrow \frac{16}{3} \pi h=\frac{224}{3} \pi$
$\Rightarrow h=\frac{224 \pi \times 3}{3 \times 16 \pi}$
$\Rightarrow h=14\ cm$
The height of the cone is $14 \mathrm{~cm}$.
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