find the number of revolutions required to level a playground of area 22cm^2 by a road roller whose diameter is 14 cm and length is 100 cm
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area covered by road roller( cylinder)= 2πrh= 2×22/7×14×100=8800cm^2
Therefore , no. of revolution required to level a playground of area 22cm^2= 22/8800=1/400.
Actually area of playground to be level is very small.
I hope you will get it
Therefore , no. of revolution required to level a playground of area 22cm^2= 22/8800=1/400.
Actually area of playground to be level is very small.
I hope you will get it
Santosh1729:
I have done a little mistake,
At the place of r=7 I have written 14
then revolution = 1/200
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