Question

The circular end of a road roller has a diameter
0.7 m and its width is 4 m. The least number of
revolutions that the roller must make in order to level
a playground measuring 264 m by 160 m is
(1) 4300 (2) 4100
(3) 4800 (4) 4700

Answer 1

The CSA of the Cylinder is ....
h=0.7m
r=4/2=2m

So CSA=2πrh=2×(22/7)×0.7×2
=88×0.7/7
=88×0.1
=8.8m²

Now Surface area of The Playground
=264×160=42240m²

So the total No: of revolutions=42240/8.8
=4800.
{[Area of Playground÷CSA of Cylinder]}

So the correct answer is (3)→[4800]

Hope it Helps...
Regards,
Leukonov.