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.

Given,

Inner radius of the tank (r) = 1 m

Thickness of iron = 1 cm = $\dfrac{1}{100}$ m = 0.01 m

Outer radius of the tank (R) = Inner radius + Thickness = 1 m + 0.01 m = 1.01 m

Volume of iron used (V) = Volume of tank with outer radius - Volume of tank with inner radius

$V = \dfrac{2}{3}πR^3 - \dfrac{2}{3}πr^3 \\[1em] = \dfrac{2}{3}π(R^3 - r^3) \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [(1.01)^3 - (1)^3] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [1.030301 - 1 ] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 0.030301 \\[1em] = \dfrac{44}{21} \times 0.030301 \\[1em] = \dfrac{1.333244}{21} \\[1em] = \text{0.06348 m}^3.$

Hence, volume of iron used to make the tank = 0.06348 m³.