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?
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 = 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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