Question

A hemisphere 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.

Answer 1

Given data

Inner radius $=r_1=1m$

Given thta hemispherical tank is made up to $1cm$ thick iron sheet.

So, outer radius $=$ thickness of sheet +inner radius

$=1cm+1m$

$=\frac{1}{100}m+1m$

$=0.01m+1m(\because 1cm=\frac{1}{100})$

$=1.01m$

Volume of iron used = difference between the volumes of outer and inner hemispheres

Volume of hemisphere $\frac{2}{2}\pi r^3$

$=\frac{2}{3}\pi r_2^3-\frac{2}{3}\pi r_1^3$

$=\frac{2}{3}\pi [r_2^3-r_1^3]$

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

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

$=0.06348m^3$

$\therefore$ Volume of iron used $=0.06348m^3$