Question

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

Answer 1

Given length  = 28 m.

          diameter = 5 cm then the radius will be r = 5/2 cm = 0.025m.

Total Radiating Surface in the system = 2pirh

                                                               = 2 * 22/7 * 0.025* 28

                                                               = 4.4 m^2.

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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