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# Ex.13.2 Q8 Surface Areas and Volumes - NCERT Maths Class 9

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## Question

In a hot water heating system, there is a cylindrical pipe of length $$28 \,\rm{m}$$ and diameter $$5 \,\rm{cm}$$. Find the total radiating surface in the system.

Video Solution
Surface Areas And Volumes
Ex 13.2 | Question 8

## Text Solution

Reasoning:

The curved surface area of a right circular cylinder of base radius  $$r$$  and height  $$h$$  is\begin{align}2\pi rh \end{align}.

What is the known?

The length and diameter of the cylindrical pipe of a water heater.

What is the unknown?

Radiating surface area in the system.

Steps:

Diameter $$= 2r = 5\,\rm{cm}$$

\begin{align}r = \frac{5}{2}\,cm &= \frac{5}{{2 \times 100}}\,\,m\end{align}

height $$= h = 28 \,m$$

Curved surface area

\begin{align}&= 2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 28 \times \frac{5}{{2 \times 100}}\\ &=\frac{{22}}{5} \\&= 4.4\,\,\rm{{m^2}} \end{align}

The total radiating surface \begin{align} = \frac{{22}}{5} = 4.4\,\,\rm{{m^2}} \end{align}

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