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(D) The height $(h)$ of the cylindrical pipe is equal to its length,so $h = 28 \, m$.The radius $(r)$ of the circular end of the pipe is half of the d

Vedclass · Accepted Answer

$4.4$

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(D) The height $(h)$ of the cylindrical pipe is equal to its length,so $h = 28 \, m$.The radius $(r)$ of the circular end of the pipe is half of the diameter,so $r = \frac{5}{2} \, cm = 2.5 \, cm$.To maintain consistent units,convert the radius into meters: $r = \frac{2.5}{100} \, m = 0.025 \, m$.The total radiating surface of the pipe is its Curved Surface Area $(CSA)$,which is given by the formula $CSA = 2 \pi r h$.Substituting the values: $CSA = 2 	imes \frac{22}{7} 	imes 0.025 	imes 28$.$CSA = 2 	imes 22 	imes 0.025 	imes 4 = 44 	imes 0.1 = 4.4 \, m^2$.Thus,the total radiating surface in the system is $4.4 \, m^2$.

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. $\left[ \text{Assume } \pi = \frac{22}{7} \right]$ (in $, m^2$)

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