The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in \({m}^2\).

Given :
Height (h) of cylindrical roller = Length of roller = 120 cm
And radius (r) of the circular end roller = \(\frac{84}{2}\) = 42 cm

It can be observed that a roller is cylindrical.

Now, we know that,
CSA of rollar
= \(2{\pi}rh\)
= \(2 × \frac{22}{7} × 42 × 120\) \({cm}^2\)
= \(2 × 22 × 6 × 120\)
= 31680 \({cm}^2\)

Therefore, Area of field
= 500 x CSA of roller
= \(31680 × 500\) \({cm}^2\)
= 1584 \({m}^2\)
\(\therefore\) the area of the playground in \({m}^2\) is 1584.