Question

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

Answer 1

In hemispherical tank,

Inner radius, $r_1=1m$

Outer radius, is sum of the inner radius and the thickness.

$r_2=1m+1cm$

$=1m+\frac{1}{100}m$ [Since, $1cm=\frac{1}{100}m$]

$=1m+0.01m$

$=1.01m$

Volume of iron used $=$ Volume of outer hemisphere$-$Volume of inner hemisphere

$=\frac{2}{3}\pi r_2^3-\frac{2}{3}\pi r_1^3$ [Since, Volume of the hemisphere $=\frac{2}{3}\pi r^3$]

$=\frac{2}{3}\times \frac{22}{7}(r_2^3-r_1^3)$

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

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

$=\frac{44}{21}\times 0.030301m^3$

$=0.06348m^3$

Hence, the volume of iron is $0.06348m^3$.