If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations?

We have the road roller which is cylindrical,

Now, given that

Length of the road roller, i.e., height of the cylinder (h) = 1.4 m

Diameter of the road roller = 0.9 m

⇒ Radius of base of the cylindrical road roller (r) = Diameter/2

⇒ Radius (r) = 0.9/2 m

Now, since the road roller is rotating, we can deduce that only the curved surface of the cylindrical road roller will be pressed against the field.

⇒ Area of the field pressed in 1 rotation = Curved surface area of the cylindrical road roller

Then, Area of the field pressed in 500 rotations = 500 (Curved surface area of cylindrical road roller) …(i)

So, curved surface area of cylinder is given by

CSA = 2πrh

Substituting values, we get

CSA =

⇒ CSA = 3.96

Now, using equation (i) we get

Area of the field pressed in 500 rotations = 500 × 3.96 = 1980 m²

Thus, area of field pressed by the road roller in 500 rotations is 1980 m².