 Hi and welcome to the session. The question says, in a hot water heating system there is a cylindrical pipe of length 28 meters in diameter 5 centimeters. Find the total radiating surface in the system. Let us now start with the solution. Now here we are given the length of pipe that is denoted by h is equal to 28 meters and this is equal to 28 into 100 centimeters since 1 meter is equal to 100 centimeters therefore changing 28 meters into centimeters this gives 2800 centimeters. Also we are given the diameter of the pipe is equal to 5 centimeters therefore radius of pipe as denoted by small r is equal to 5 upon 2 centimeters. Now we have to find the total radiating surface in the system and that will be equal to the surface area of the cylindrical pipe. So we have to find the surface area of the cylindrical pipe. Now the formula to calculate the curved surface area of a cylinder is 2 pi r h. So here the curved surface area of the pipe is equal to 2 pi r h where r is the radius of the pipe and h is the length of the pipe. So 2 into 22 upon 7 into r is 2800 centimeters into h is 2800 centimeters sorry and r is 5 upon 2 centimeters. So the curved surface area is in square centimeters and simplifying this we get 7 into 4 is 28, 2 0s, 2 cancels out with 2. So we have 22 into 400 into 5 centimeters square and this is further equal to 4400 centimeters square and since 1 centimeter is equal to 1 upon 100 meters therefore 1 centimeter square is equal to 1 upon 10000 meters square. So this gives 44000 divided by 10000 meters square and this is equal to 4.4 meters square and hence our answer is the total radiating surface in the system is 4.4 meters square. So this completes the session by intake care.