Question

A road roller takes \[750\] complete revolutions to move once over to level a road. Find the area of road if the diameter of a road roller is \[84cm\] and length is \[1m\].

Answer 1

Hint: In the given question, we have to find the area of the road. In one revolution, the road roller will cover an area equal to its lateral surface. Since the road roller is in the shape of a cylinder, we can find the lateral surface area with the help of formula \[2\pi rh\] and multiply with the number of revolutions to get the area of the road.

Complete step by step solution:
In this given problem,
Road-roller is used to level the ground and its wheels are generally in the shape of a cylinder. It is shown in diagram as follows:

The formula for lateral surface area of cylinder is:
\[ = 2\pi rh\] ---------(1)
We are given that-
\[h = 1m = \] height or length of the cylinder
\[
  r = \dfrac{{Diameter}}{2} = \dfrac{{84}}{2}cm \\
   = 42cm \;
\]
Therefore, Radius of the cylinder, \[r = \dfrac{{42}}{{100}}m\]
Substitute all values in the formula, the lateral surface area of road roller will be:
\[ \Rightarrow 2\pi rh = 2(\dfrac{{22}}{7})(\dfrac{{42}}{{100}})(1)\]
\[ = 2.64{m^2}\]
Now the area of the road will be total revolutions completed by the road-roller multiplied by the LSA of road-roller:
\[ = 750 \times 2.64\]
\[ = 1980{m^2}\]
Thus, the area of the road is \[1980{m^2}\].
So, the correct answer is “\[1980{m^2}\]”.

Note: The key point in this question is to identify the formula and shape of the road roller.
A cylinder is one of the most common three-dimensional shapes, with two parallel circular bases separated by a space.
A curved surface connects the two circular bases at a fixed distance from the middle. The axis of the cylinder is a line segment that connects the centres of two circular bases. The height of the cylinder is the distance between the two circular bases. Example – LPG cylinder.