A hemispherical tank is made up of an iron sheet $1 c m$ thick. If the inner radius is $1 m,$ then find the volume of the iron used to make the tank.

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Solution

In hemispherical tank,

Inner radius, $r_{1} = 1 m$
Outer radius, is sum of the inner radius and the thickness.
$r_{2} = 1 m + 1 c m$
$= 1 m + \frac{1}{100} m$ [Since, $1 c m = \frac{1}{100} m$ ]
$= 1 m + 0.01 m$
$= 1.01 m$
Volume of iron used $=$ Volume of outer hemisphere $-$ Volume of inner hemisphere
$= \frac{2}{3} π r_{2}^{3} - \frac{2}{3} π r_{1}^{3}$ [Since, Volume of the hemisphere $= \frac{2}{3} π r^{3}$ ]
$= \frac{2}{3} \times \frac{22}{7} (r_{2}^{3} - r_{1}^{3})$
$= \frac{2}{3} \times \frac{22}{7} ({1.01}^{3} - 1^{3}) m^{3}$
$= \frac{44}{21} \times (1.030301 - 1) m^{3}$
$= \frac{44}{21} \times 0.030301 m^{3}$
$= 0.06348 m^{3}$
Hence, the volume of iron is $0.06348 m^{3}$ .

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