a road roller in the shape of a cylinder has a diameter 0.7 M and its width is 1.2 M Find the least number of revolutions that the ruler must make in the order to level a playground of size 120 m by 44 m
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CSA of roller=2 pi r h
=2×22/7×70/2×120
2&2 ,7 & 70 will be cancelled
so ,22×10×120
so CSA =26400cm^2
=2.6400m^2
Minimum no of revolutions =
area of playground / CSA of roller
120/2.6400
= 45.454545....
I hope this would help u
Do thank me .if it helped u
=2×22/7×70/2×120
2&2 ,7 & 70 will be cancelled
so ,22×10×120
so CSA =26400cm^2
=2.6400m^2
Minimum no of revolutions =
area of playground / CSA of roller
120/2.6400
= 45.454545....
I hope this would help u
Do thank me .if it helped u
Rushikesh17:
area of playground is not given only dimensions are given.
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