Question

8.
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.

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Answer 1

Answer:

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Answer 2

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${\huge{\red{\underline{\overline{\mathit{\red{ANSWER↓}}}}}}}$

h = 28m

2r=5cm

$\therefore$ r = $\large\dfrac{5}{2}$ cm = $\large\dfrac{5}{2×100}$

= $\large\dfrac{5}{200}$ m = $\large\dfrac{1}{40}$ m

$\therefore$ Total radiating surface in the system

= ${\small{\boxed{\sf{\pink{2πrh}}}}}$

= 2 × $\large\dfrac{27}{7}$ × $\large\dfrac{1}{40}$ × 28 = ${\underline{\underline{4.4m²}}}$

The area of the radiating surface of system is 4.4 m²

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