Question

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

Answer 1

Answer:

4.4m or 440 cm

Step-by-step explanation:

Given,

h = 28m

d = 5cm = 0.05m

So,

r = (0.05/2) m

Total radiating surface = CSA of Cylinder ===>

2πrh = 2 × 22 × 0.05 × 28

---- ------

7 2

= 22 × 0.05 × 4 = 4.4m or 440 cm

Hope this helps!!!!!

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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