Question

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
$[Assume\pi =\frac{22}{7}]$

Answer 1

Inner radius $\left(r_1\right)$ of the hemispherical tank = 1 m
Thickness of the hemispherical tank = 1 cm = 0.01 m
Outer radius$\left(r_2\right)$ of the hemispherical tank = Inner radius + Thickness = (1 + 0.01) m = 1.01 m

Volume of the iron used to make such a tank = Outer volume - Inner volume of hemispherical tank
$=\frac{2}{3}\times \pi (r_2^3-r_1^3)$

$=[\frac{2}{3}\times \frac{22}{7}\times \left\{\right(1.01{)}^3-\left(1{)}^3\right\}]m^3$

$=[\frac{44}{21}\times (1.030301-1\left)\right]m^3$

$=0.06348m^3$ (approximately)