Q. 94.8( 14 Votes )

# 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.

Answer :

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 m^{2}

Then, area of ground leveled in 500 rotations = 9.24 × 500 = 4620 m^{2}

If cost of levelling 7 m^{2} ground = Rs. 7

Then cost of levelling 4620 m^{2} ground = 4620 × 7 = Rs. 32340

Thus, area of the ground that was levelled by the road roller is 4620 m^{2} and cost of levelling it is Rs. 32340.

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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?

MHB - Math Part-II