1.

a road roller has a diameter 0.7m and its width is 1.2 m.Find the least number of revolutions that the roller must take in order to level a playground of size 120m×44m.

Answer»

Area of PLAYGROUND = 120 x 44m^2
in cylinder,
height = 1.2m
diameter = 0.7m
radius = 0.35m
c.s.a. = 2πrh
=2 x 22/7 x 0.35 x 1.2
=2.64m
no.. of revolution = area of playground/ c.s.a.
=120 x 44/ 2.64
= 2000
I HOPE it's HELP U :-)


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