A hemispherical tank is made up of an iron sheet $ 1 \mathrm{~cm} $ thick. If the inner radius is $ 1 \mathrm{~m} $, then find the volume of the iron used to make the tank.
Given:
A hemispherical tank is made up of an iron sheet $1\ cm$ thick.
The inner radius is $1\ m$.
To do:
We have to find the volume of the iron used to make the tank.
Solution:
Thickness of the hemispherical tank $= 1\ cm$
Inner radius of the tank $(r) = 1\ m$
$= 100\ cm$
This implies,
Outer radius of the tank $(\mathrm{R})=100+1$
$=101 \mathrm{~cm}$
Therefore,
Volume of the iron used to make the hemispherical tank $=\frac{2}{3} \pi[\mathrm{R}^{3}-r^{3}]$
$=\frac{2}{3} \times \frac{22}{7}[(101)^{3}-(100)^{3}]$
$=\frac{44}{21}[1030301-1000000]$
$=\frac{44}{21} \times 30301$
$=63487.81 \mathrm{~cm}^{3}$
$=\frac{63487.81}{100 \times 100 \times 100}$
$=0.06348781 \mathrm{~m}^{3}$
$=0.063487 \mathrm{~m}^{3}$
The volume of the iron used to make the tank is $0.063487 \mathrm{~m}^{3}$.
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