Question

A road roller is cylindrical in shape. Its circular end has a diameter 200 cm and its length is 1.4 m. Find the least number of revolutions that the roller must make in order to level a playground of dimensions 125m \[ \times \] 20m.

Answer 1

Hint: In this problem a road roller is rolled on the playground. So, only the curved surface of the road roller is with the contact of ground. So, to find the curved surface area of the cylinder we use the formula \[2\pi rh\].

Complete step-by-step answer:

As we know that road roller is of cylindrical shape.
And it is obvious that when a road roller rolls on the playground then it's only curved surface area is with the contact of ground.
Curved surface area of the cylinder is the total area of the cylinder excluding the area of its circular face.
Now the area covered in one revolution will be equal to the curved surface area of the road roller.
So, the number of revolutions will be equal to the total area of the ground divided by the area covered in one revolution because the road roller has to level the complete playground.
Now as we know that the curved surface area of the cylinder of the cylinder of length h and radius r is \[2\pi rh\].
And the radius of the road roller is half of the diameter.
So, radius = 100 cm = 1m (because 1m = 100cm)
The length of the road roller is 1.4m.
So, curved surface area of the road roller will be \[2\pi \times 1 \times 1.4 = 2.8\pi = 2.8 \times \dfrac{{22}}{7} = 8.8{\text{ }}{{\text{m}}^2}\]
And, the area covered by the road roller in one complete revolution will be \[8.8{\text{ }}{{\text{m}}^2}\]
And the area of the playground will be \[125 \times 20 = 2500{\text{ }}{{\text{m}}^2}\].
So, the least number of revolutions made by the road roller will be = area of the playground divided by the area covered in one revolution.
So, the number of revolutions = \[\dfrac{{2500}}{{8.8}} = 284.09\] revolutions.
Hence, the least number of revolutions the road roller had to make to level the complete playground will be equal to 284.09 revolutions.

Note: Whenever we come up with this type of problem then first, we have to find the curved surface area of the cylinder using formula \[2\pi rh\] (where r is the radius and h is the length) and this will be equal to the area covered in one revolution. And then find the area of the playground. After that divide the total area of the playground with the area covered by the road roller in one complete revolution to get the least number of revolutions because the least number of revolutions will be only when the road roller moves at each place only once.