# 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.

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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 $0.063487 \mathrm{~m}^{3}$.

Updated on 10-Oct-2022 13:46:39