 # NCERT Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.2 in Chapter 13 - Surface Areas and Volumes

In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter of 5 cm. Find

the total radiating surface in the system. (Assume π=22/7)

Height of cylindrical pipe = Length of cylindrical pipe = 28 m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5 cm = 0.025 m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2 x 22/7 x 0.025 x 28 \mathrm{m}^{2}

=4.4 \mathrm{m}^{2}

The area of the radiating surface of the system is 4.4 \mathrm{m}^{2}

Video transcript
"hello students welcome to lino q a video session i am saf your match tutor and question for today is in a hot water heating system there is a cylindrical pipe of length 28 meter and diameter 5 centimeter find the total radiating surface in the system assume phi is equal to 22 by 7 height of the cylindrical pipe will be equal to the length of the cylindrical pipe which is 28 meter it is given over here that length of the cylindrical pipe is 28 meter therefore this is the height so and that is equal to let us say it is h radius of the circular end of the pipe can be found now with the help of the diameter we know that the radius is equal to half of the diameter so r is equal to radius diameter of pipe divided by 2 pi upon 2 so that is 2.5 centimeter every unit over here is in meter so let us convert this 2.5 centimeter in meter so this will give us 0.025 meter now curved surface area of the cylindrical pipe is 2pi rh pi into radius into height so it can be found by substituting values of r and h y value over here is 22 by 7 r is 0.025 and h is 28 simplify it and you will get the answer 4.4 meter square so the area of the radiating surface of the system is 4.4 meter square this is our final answer if you have any queries you can drop it in our comment section and subscribe to lido for more such q a thank you for watching "
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