The diameter of a road roller is 42cm and its length is 100cm.If it
takes 100 complete revolutions to level a playground, find the cost of
levelling it at Rs 2 per square metr
Answers
Answered by
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Step-by-step explanation:
Area of the roller = CSA (curved surface area)
= 2πrh
r = d ÷ 2
r = 42 ÷ 2
r = 21cm
h = 100cm
2πrh = 2 × 22/7 × 21 × 100
( 21÷7 = 3)
44 × 300
CSA = 13200m^2
Revolutions = 100
Area it covers in 100 revolutions = 13200 × 100
= 1320000m^2
Cost of levelling at ₹2 per m^2
= 1320000 × 2
= ₹2,640,000
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