A roadroller (in the shape of the cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

Given, diameter = 0.7 m

So radius = $\dfrac{\text{diameter}}{2} = \dfrac{0.7}{2} = 0.35 \text{ m}.$

Height = width = 1.2 m

Area covered in 1 revolution = Curved surface area of cylinder.

Curved surface area of cylinder = 2πrh

= $2 \times \dfrac{22}{7} \times 0.35 \times 1.2 = \dfrac{18.48}{7}$ = 2.64 m².

∴ Area covered in 1 revolution = Curved surface area of cylinder = 2.64 m².

Hence, the number of revolutions required to cover playground of size 120 m by 44 m

= $\dfrac{120 \times 44}{2.64} = \dfrac{5280}{2.64}$ = 2000.

Hence, the minimum number of revolutions required to cover playground of size 120 m by 44 m are 2000.