Question

A road roller takes \[900\] complete revolutions to move once over to level a road. Find the area of the road levelled if the diameter of the road roller is $126\,cm$ and its length is $2.5\,m$.

Answer 1

Hint: When a road roller completes one revolution it will cover the area equal to the circumference of the circle multiplied by the length of the road roller.
Evaluate the area covered in the one revolution by using the circumference of the circle which is $2\pi r$(where $r$ is the radius of the circle) and then multiply it by the length of the road roller.

Complete step-by-step answer:
We are given that a road roller whose diameter is $126\,cm$and length $2.5\,m$takes \[900\] complete revolutions to move once over to level a road.
Let the length $l = 2.5m$and radius is half of the diameter.
Therefore, radius $r = \dfrac{{126}}{2} = 63\,cm$
Convert radius into \[m\].
$
  100\,cm = 1\,m \\
  63\,cm = \,\dfrac{1}{{100}} \times 63 = 0.63\,m \\
 $
$r = 0.63\,m$
We have to find the area of the road levelled in \[900\] complete revolutions.
First, we evaluate the area of the road levelled in \[1\] complete revolution.
Let the area covered in one revolution is $A$.
To evaluate the area, multiply circumference of the circle by the length of the road roller.
The circumference of the circle of radius $r$is $2\pi r$.
Now we evaluate the area covered in one revolution.
Therefore,
$A = 2\pi rl$
Substitute all the values and evaluate the area. Use $\pi = \dfrac{{22}}{7}$
$
  A = 2 \times \dfrac{{22}}{7} \times 0.63 \times 2.5 \\
  A = 9.9\,{m^2} \\
 $
To evaluate the area of road levelled in \[900\] complete revolutions, multiply the area of the road levelled in \[1\] complete revolution by \[900\].
Therefore, the required area is $900 \times 9.9 = 8910{m^2}$
Hence, the area of road levelled in \[900\] complete revolutions is $8910\,{m^2}$.
Note: In this type of questions don’t forget to make the units the same for all the quantities. If units of the quantities are different than the final answer will be wrong.