Question

An iron pipe 20 cm long has exterior
diameter equal to 25 cm. If the thickness of
the pipe is 1 cm, find the whole surface of
the pipe.
​

Answer 1

Answer: 24

Step-by-step explanation:

Outer Diameter = inner diameter + thickness of the  pipe

25 cm  = inner diameter + 1 cm

25 cm - 1 cm = inner diameter

24 cm = inner diameter

Answer 2

Answer :-

Total Surface Area of the Cylinder = 3168 $cm^{2}$

Given :-

Height = 20 cm

External Diameter = 25 cm

Thickness = 1 cm

To find :-

The whole surface of  the pipe.

Solution :-

First we have to find the external radius and internal radius of the pipe.

So, External radius (R) = $\frac{External \:\:Diameter }{2} =\frac{25 \:\:cm}{2} = 12.5\:\:cm$

Internal Radius (r) $= External \:\:radius - Thickness = 12.5\:\:cm - 1\:\:cm = 11.5\:\:cm$

Outer Curved Surface Area = 2 x π  x External radius (R) x height $= 2 \times \pi \times 12.5\:\:cm \times 20 \:\:cm = 500\:\pi\:cm^{2}$

Inner Curved Surface Area = 2 x π  x Internal Radius (r)  x height  $= 2 \times \pi \times 11.5\:\:cm \times 20 \:\:cm = 460\:\pi\:cm^{2}$

∴ Area of one end of the cylinder

$=\pi \times R^{2} - \pi \times r^{2} \\ = \pi \times (R^{2} - r^{2}) \\ = \pi \times (R+r)(R-r)\\= \pi \times (12.5\:cm+11.5\:cm)(12.5\:cm-11.5\:cm)\\=\pi \times (24.0 \:cm^{2} )(1.0 \:cm^{2} )\\= \pi \times 24.0 \:cm^{2} \\ =24.0\: \pi \:cm^{2}$

Area of both the ends

$= 2 \times Area\:\:of\:\:one\:\:end\:\:of\:\:the\:\:cylinder\\= 2 \times 24.0 \: \pi \: cm^{2} \\= 48.0 \: \pi \: cm^{2}$

∴ Total Surface Area of the Cylinder

$= Outer\:\:Curved\:\:Surface\:\:Area + Inner\:\:Curved\:\:Surface\:\:Area + 2(Area\:\:of\:\:one\:\:end) \\ \\Note:- \: 2(Area\:\:of\:\:one\:\:end) = Area of both the end of the cylinder$

$\\ = 500\: \pi \: cm^{2} +460\: \pi \: cm^{2}+48\: \pi\: cm^{2}\\=1008\: \pi \: cm^{2}\\=1008 \times \frac{22}{7} \:\:cm^{2}$

$= 3168 \: cm^{2}$