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.
$[Assume\pi =\frac{22}{7}]$

• A. Volume = $=0.06348m^3$
• B. Volume = $=1.06348m^3$
• C. Volume = $=0.09348m^3$
• D. Volume = $=10.06348m^3$

Answer 1

The correct option is A

Volume = $=0.06348m^3$

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 = (1 + 0.01) m = 1.01 m
Volume of the iron used to make such a tank $=\pi \times \frac{2}{3}(r_2^3-r_1^3)$
$=[\frac{2}{3}\times \frac{22}{7}\times \left\{\right(1.01{)}^3-\left(1{)}^3\right\}]m^3=[\frac{44}{21}\times (1.030301-1\left)\right]m^3$
$=0.06348m^3$ (approximately)