Solution:
Given diameter of the road roller, d = 0.7 m
Radius, r = d/2 = 0.7/2 = 0.35 m
Width, h = 1.2 m
Curved surface area of the road roller = 2rh
=2 ×(22/7)×0.35×1.2
= 2.64 m2
Area of the play ground = l×b
= 120×44
= 5280 m2
Number of revolutions = Area of play ground / Curved surface area
= 5280/2.64
= 2000
Hence the road roller must take 2000 revolutions to level the ground.