Question

in a water heating system there is a cylindrical pipe of length 28 metre and diameter 5 cm find the total rotating surface in the system ​

Answer 1

Answer:

4.4m

Step-by-step explanation:

length of the pipe=28m

2r=5

r=5/2cm

r=5/2×1/100=0.025m

=2(22/7×0.025×28)

=4.4m

Answer 2

Correct Question :-

In a water heating system there is a cylindrical pipe of length 28 metre and diameter 5 cm find the total radiating surface in the system.

Answer :-

Total radiating surface in the system is 4.4 m².

Explanation :-

Length i.e height of the cylindrical pipe (h) = 20 m

Diameter of the cylindrical pipe (d) = 5 cm = 5/100 m = 0.05

Radius of the cylinder = d/2 = 0.05/2 = 0.025 m

Total radiating surface is the curved surface area of the cylindrical pipe

Curved surface area of the cylindrical pipe = 2πrh

= 2 * (22/7) * 0.025 * 28

= 56 * (22/7) * 0.025

= 1.4 * (22/7)

= 0.2 * 22

= 4.4 m²

i.e Curved surface of the cylindrical pipe = 4.4 m²

∴ total radiating surface in the system is 4.4m².