Question

A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm.

Find its,
(i) Inner curved surface area
(ii) Outer curved surface area
(iii) Total surface area. $[Assume\pi =\frac{22}{7}]$

Answer 1

Inner radius $\left(r_1\right)$ of the cylindrical pipe $=\left[\frac{4}{2}\right]cm=2cm$
outer radius $\left(r_2\right)$ of cylindrical pipe
$=\left[\frac{4.4}{2}\right]cm=2.2cm$
Height (h) of cylindrical pipe = length of cylindrical pipe = 77 cm
CSA of inner surface of pipe = $2\pi r_{1}h$
$=[2\times \frac{22}{7}\times 2\times 77]c{m}^2$
$=968c{m}^2$
CSA of outer surface of pipe = $2\pi r_{2}h$
$=[2\times \frac{22}{7}\times 2.2\times 77]c{m}^2$
$(22\times 22\times 2.2)c{m}^2$
$=1064.8c{m}^2$
Total surface area of pipe = CSA inner surface + CSA of outer surface + Area of both circular ends of pipe.
$=2\pi r_{1}h+2\pi r_{2}h+2\pi (r_2^2-r_1^2)=[9668+1064.8+2\pi \left\{\right(2.2{)}^2-\left(2{)}^2\right\}]c{m}^2=(2032.8+5.28)c{m}^2=2038.08c{m}^2$
Therefore, the total surface area of the cylindrical pipe is $2038.08c{m}^2$.