Diameter and length of a roller are 84 cm and 120 cm respectively. In how many revolutions, can the roller level the playground of area 1584 m^2
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r=42cm
l=120 cm
surface area=2πrh
ground area=revolution*surface area
1584=x*2*22/7*1.2*4.2
x=50
l=120 cm
surface area=2πrh
ground area=revolution*surface area
1584=x*2*22/7*1.2*4.2
x=50
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Answer:
Roller is in the shape of a cylinder.
Diameter = 84 cm
Radius = 84/2 = 42 cm
Area of roller = Curved surface area of cylinder = 2πrh
⇒ (2*22*42*120)/7
Curved surface area of cylinder = 31680 cm²
Area to be covered = 1584 sq m or 1584*10000
= 15840000 cm² (1 sq m = 10000 sq cm)
Total number of revolution to cover the area of 15840000 sq cm =
15840000/31680
500 Revolutions.
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