a road roller takes 800 complete revolutions to move once over to level a road. find the area of the road if the diameter of the road roller is '91 'cm and length is 1 m. also find the cost of levelling if it cost rs 150 per m² the diameter is 9q
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Answer:
Area = 2288sq.cm Cost = 343200
Step-by-step explanation:
In one revolution, area covered will be the curved surface area of the roller or cylinder.
Area = 2πrh = 2x(22/7)x(91/2)x100 (since length = 100cm = 1m)
A = 28600 sq.cm
For 800 revolutions
A = 800x28600 = 22880000 sq.cm = 2288 sq.m
Cost of leveling = 150₹/sq.m
Total cost = 2288x150 = 343200₹
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