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

${ \bf{ \underline{ \boxed{ \mathbb{ \green{SOLUTION :}}}}}}$

$</u></strong><strong><u>{ \mathsf{Height \: ( h ) \: of \: cylindrical \: pipe = 28m}}$

${ \mathsf{Radius \: ( r ) \: of \: circular \: end \: of \: pipe = \frac{5}{2} = 2.5cm }} </p><p></p><p>$

${ \mathsf{r = 2.5 \times \frac{1}{100} m = 0.025m}}$

${ \mathsf{ \underline{We \: know \: that : }}}$

${ \mathsf{CSA \: of \: cylindrical \: pipe = 2\pi \: rh}}$

${ \mathsf{(2 \times \frac{22}{7} \times 0.025 \times 28)m {}^{2} }}$

${ \mathsf{ \boxed{ \blue{ = 4.4 \: m {}^{2} }}}}★$

The area of the radiating surface of system is 4.4 m²

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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