Question

A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.

Answer 1

Thickness of steel $=0.25cm=\frac{1}{4}cm$
Inside radius of the hemispherical bowl (r) = 5 cm
$\therefore$ Outer radius (R) = 5 + 0.25 = 5.25 cm
and volume of the steel used $=\frac{2}{3}\pi (R^3-r^3)=\frac{2}{3}\times \frac{22}{7}\times \left[\right(5.25{)}^3-\left(5.00{)}^3\right]c{m}^3=\frac{44}{21}[144.703125-125.000000]c{m}^3=\frac{44}{21}\times 19.703125c{m}^3=41.28c{m}^3$