Answer
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Hint: Here the solution is only to find the radiating surface in the system not the total surface of the system. So we will find only the curved surface area of the cylindrical pipe with the help of given dimensions.
Step by step solution:
Given that,
Length of the cylinder is 28m. This length of pipe is nothing but the height of the system.
So h=28m=0.28cm
Diameter is 5cm. so radius r=2.5cm
Now the curved surface area of the cylindrical pipe that is radiating is given by,
\[CS{A_{cylinder}} = 2\pi rh\]
Putting the values,
\[ \Rightarrow 2 \times \dfrac{{22}}{7} \times \dfrac{{2.5}}{{100}} \times 28\]
On doing the calculations
\[ \Rightarrow 44 \times 4 \times \dfrac{{2.5}}{{100}}\]
\[ \Rightarrow 44 \times \dfrac{{2.5}}{{25}}\]
\[ \Rightarrow 44 \times 0.1\]
\[ \Rightarrow 4.4c{m^2}\]
This is the area of the cylindrical pipe that radiates.
Note:
Note that the length is given in meters and diameter is given in centimeters. So first convert the dimensions to any one measuring system either meters or in centimeters and then start solving. Also note that we are asked to find the radiating surface only!
Step by step solution:
Given that,
Length of the cylinder is 28m. This length of pipe is nothing but the height of the system.
So h=28m=0.28cm
Diameter is 5cm. so radius r=2.5cm
Now the curved surface area of the cylindrical pipe that is radiating is given by,
\[CS{A_{cylinder}} = 2\pi rh\]
Putting the values,
\[ \Rightarrow 2 \times \dfrac{{22}}{7} \times \dfrac{{2.5}}{{100}} \times 28\]
On doing the calculations
\[ \Rightarrow 44 \times 4 \times \dfrac{{2.5}}{{100}}\]
\[ \Rightarrow 44 \times \dfrac{{2.5}}{{25}}\]
\[ \Rightarrow 44 \times 0.1\]
\[ \Rightarrow 4.4c{m^2}\]
This is the area of the cylindrical pipe that radiates.
Note:
Note that the length is given in meters and diameter is given in centimeters. So first convert the dimensions to any one measuring system either meters or in centimeters and then start solving. Also note that we are asked to find the radiating surface only!
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