Question

In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer 1

Hello!

• Height, h = $\bf{28 \:m}$

• Radius, r = $\dfrac{5}{2\times100} = \bf{\dfrac{1}{40} \: m}$

Then,

$\underline{\bf{Total\:radiating \:surface}}$ :

$\sf 2πrh \\ \\ \implies 2 \times \dfrac{22}{7} \times \dfrac{1}{40} \times 28 \\ \\ \implies \dfrac{1232}{280} \implies \boxed{\red{\bf{4.4\:m^{2}}}}$

Cheers!

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Length of Cylindrical heating system (h)

= 28m

$\textbf{\underline{Convert\;it\;into\;cm}}$

[1m = 100 cm]

= 28 × 100

= 2800 cm

Also (Given)

$\textbf{\underline{Diameter\;of\;Pipe = 5cm}}$

Hence,

$\tt{\rightarrow Radius=\dfrac{Diameter}{2}}$

$\tt{\rightarrow Radius=\dfrac{5}{2}}$

Now,

Curved Surface area = 2πrh

$\tt{\rightarrow 2\times\dfrac{22}{7}\times\dfrac{5}{2}\times 2800}$

= 110 × 400

= 44000 cm²

$\textbf{\underline{Convert\;it\;into\;metre\;square}}$

$\tt{\rightarrow\dfrac{44000}{100\times 100}m^2}$

= 4.4 m²

Hence we get :-

Total radiating surface of the system = 4.4 m²