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

## 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.

## 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 \(\begin{align} = \frac{{22}}{5} = 4.4\,\,\rm{{m^2}} \end{align}\)