Radius of cylinderical roller = 42 cm and height = 120 cm ⇒ Distance covered in 1 revolution by the roller = Curved surface area of the roller = 2πrh = 2×
22
7
×42×120 = 44×6×120=31680cm2=3.168m2 ⇒ Total distance covered in 500 revolutions = 500×3.168=1584m2 Now, cost of levelling the 1 m2 ground = Rs. 1.50 ∴ Total cost required = 1584×1.50=Rs.2376