the diameter of road roller is 84cm and length 1m how many revolution will it take to level playground 1980 sq m in area
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750 revolution it will do
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Assume π=22/7
The roller is in the shape of a cylinder. The area covered by the roller is the curved surface area of the cylinder.
r=84cm/2=42cm or 0.42m. h=1m.
Curved surface area =
=22/7 ×2×0.42×1 m^2
=2.64m^2
Revolutions taken to cover the playground =area of playground ÷area covered in 1 revolution
=1980 m^2÷ 2.64m^2
=750
So the road roller will take 750 Revolutions to level the playground.
The roller is in the shape of a cylinder. The area covered by the roller is the curved surface area of the cylinder.
r=84cm/2=42cm or 0.42m. h=1m.
Curved surface area =
=22/7 ×2×0.42×1 m^2
=2.64m^2
Revolutions taken to cover the playground =area of playground ÷area covered in 1 revolution
=1980 m^2÷ 2.64m^2
=750
So the road roller will take 750 Revolutions to level the playground.
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