**Question:**

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground?

**Solution:**

Given:

Diameter of the roller $=84 \mathrm{~cm}$

$\therefore$ Radius, $r=\frac{\text { Diameter }}{2}=42 \mathrm{~cm}$

In 1 revolution, it covers the distance of its lateral surface area.

Roller is a cylinder of height, $h=120 \mathrm{~cm}$

Radius $=42 \mathrm{~cm}$

Lateral surface area of the cylinder $=2 \pi r h$

$=2 \times \frac{22}{7} \times 42 \times 120$

$=31680 \mathrm{~cm}^{2}$

It takes 500 complete revolutions to level a playground.

$\therefore$ Area of the field $=31680 \times 500=15840000 \mathrm{~cm}^{2}$ $\left(1 \mathrm{~cm}^{2}=\frac{1}{10000} \mathrm{~m}^{2}\right)$

$\therefore 15840000 \mathrm{~cm}^{2}=1584 \mathrm{~m}^{2}$.

Thus, the area of the field in $\mathrm{m}^{2}$ is $1584 \mathrm{~m}^{2}$.