A road roller was used for levelling a road of width 2 m. It was observed that, the road roller required 25 complete revolutions to level the entire road. If the radius and the length of the roller is 7 m and 2 m respectively, then length of the road is
[Take π = 3.14]
Given, Radius of Cylinder, r=7 m
length of roller, h=2 m
Width of road, B=2 m
Length of road be ′L′
Number of revolution taken = 25
Roller rolls on its curved surface area.
∴ Surface area of = CSA of cylinder
one revolution
=2πrh=2×227×7×2=88 m2
Total area covered in 25 revolution =25×88=2200 m2
This area will be equal to the area of the road levelled
Area of road = L×B=2200⇒L×2=2200⇒L=1100 m
∴ Length of the level road is 1100 m.