Given: For road roller, diameter (d) = 1.4 m, length (h) = 2.1 m
Number of rotations required for levelling the ground = 500, rate of levelling = Rs 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
∴ Curved surface area of the road roller = 2πrh
= πdh … [∵ d = 2r]
= (22/7) x 1.4 x 2.1
= 22 x 0.2 x 2.1
= 9.24 sq.m.
ii. Area of ground levelled in 1 rotation = 9.24 sq.m.
∴ Area of ground levelled in 500 rotations
= 9.24 x 500
= 4620 sq.m.
iii. Rate of levelling Rs 7 per sq.m.
∴ Total cost = Area of ground levelled x Rate of levelling
= 4620 x 7 = Rs 32340
∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is Rs 32340.
