Q. 14.2( 22 Votes )

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?

Answer :

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.


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