A metal pipe is \( 77 \mathrm{~cm} \) long. The inner diameter of a cross section is \( 4 \mathrm{~cm} \), the outer diameter being \( 4.4 \mathrm{~cm} \) (see below figure). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

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Given:

A metal pipe is \( 77 \mathrm{~cm} \) long. The inner diameter of a cross section is \( 4 \mathrm{~cm} \), the outer diameter being \( 4.4 \mathrm{~cm} \)

To do:

We have to find its

(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Solution:

Length of the metal pipe $ l= 77\ cm$

Outer diameter $(d_1) = 4.4\ cm$

Inner diameter $(d_2) = 4\ cm$

This implies,

Outer radius $(r_1) = 2.2\ cm$

Inner radius $(r_2) = 2\ cm$

(i) Inner curved surface area of the pipe $= 2\pi r_2l$

$= 2\times\frac{22}{7} \times 2 \times 77$

$=2\times22 \times 2 \times 11$

$=968\ cm^2$

(ii) Outer curved surface area of the pipe $= 2\pi r_1l$

$= 2\times\frac{22}{7} \times 2.2 \times 77$

$=2\times22 \times 2.2 \times 11$

$=1064.8\ cm^2$

(iii) Total surface area of the pipe $=$ Inner curved surface area $+$ Outer curved surface area $+$ Areas of two bases

$= 968 + 1064.8 + 2\times\frac{22}{7} (r_1^2 - r_2^2)$

$= 968 + 1064.8 + 2\times\frac{22}{7} [(2.2)^2 - (2)^2)]$

$= 2032.8 + 2 \times \frac{22}{7} (4.84 - 4)]$

$= 2032.8 +\frac{44}{7} \times 0.84$

$= 2032.8+ 44 \times 0.12$

$= 2032.8 + 5.28\ cm^2$

$= 2038.08\ cm^2$