A road roller has length 2.1 m and diameter 1 m . How .many rotations will be required to level a ground of area 660 sq.cm. ? *
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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.Read more on Sarthaks.com - https://www.sarthaks.com/853199/length-road-roller-diameter-levelling-ground-500-rotations-the-road-roller-were-required