Q9 of 32 Page 123

The length of a road roller is 2.1m and its diameter is 1.4m. For levelling ground 500 rotations of the road roller were required. How much area of ground was leveled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m.

The road roller is cylindrical in shape as described.

Given: length of cylindrical road roller (h) = 2.1 m


Diameter of the road roller = 1.4 m


Radius of the cylindrical road roller (r) = 1.4 /2 = 0.7 m


The figure is given below:



When the cylindrical road roller will roll on the ground, the only portion of the cylinder that will touch the ground is the curved surface of cylindrical road roller.


Thus, if we find curved surface area of cylinder, we can know area of ground levelled.


Curved surface area of cylinder is given by,


CSA = 2πrh


Where h = height of cylinder when placed vertical or length of the cylinder when placed horizontal


Substituting values in above equation, we get



CSA = 9.24


This implies that, area of ground leveled in 1 rotation = 9.24 m2


Then, area of ground leveled in 500 rotations = 9.24 × 500 = 4620 m2


If cost of levelling 7 m2 ground = Rs. 7


Then cost of levelling 4620 m2 ground = 4620 × 7 = Rs. 32340


Thus, area of the ground that was levelled by the road roller is 4620 m2 and cost of levelling it is Rs. 32340.


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