Question

The diameter of a roller \[1.5\] m long is 84 cm. If it takes 100 revolutions to level this playground, find the cost of levelling this ground at the rate of 50 paise per square meter.

Answer 1

Hint: First, we will convert the diameter of the roller is meters by using that \[1{\text{ cm}} = \dfrac{1}{{100}}{\text{ m}}\] And then find the radius. Then we will use the formula of the curved surface area of the cylinder is \[2\pi rh\], where \[r\] is the radius and \[h\] is the height of the cylinder. Then we will calculate the cost of levelling this ground by the roller at 50 paise per square meter.

Complete step by step solution: We are given that the diameter \[d\] of a roller is 84 cm and the length of the roller is \[1.5\] m.

First, we will convert the diameter of the roller to meters by using that \[1{\text{ cm}} = \dfrac{1}{{100}}{\text{ m}}\].

\[
   \Rightarrow 84{\text{ cm}} \\
   \Rightarrow \dfrac{{84}}{{100}}{\text{ m}} \\
   \Rightarrow 0.84{\text{ m}} \\
 \]

We will now find the radius \[r\] of the given roller from the above value of the diameter of a roller.

\[
   \Rightarrow r = \dfrac{{0.84}}{2}{\text{ m}} \\
   \Rightarrow r = 0.42{\text{ m}} \\
 \]

We will use the formula of the curved surface area of the cylinder is \[2\pi rh\], where \[r\] is the radius and \[h\] is the height of the cylinder.

Substituting the values of \[r\] and \[h\] in the above formula of the curved surface area of a roller, we get

\[
   \Rightarrow {\text{Curved Surface Area of Roller = }}2 \times \dfrac{{22}}{7} \times 0.42 \times 1.5 \\
   \Rightarrow {\text{Curved Surface Area of Roller}} = 2 \times 22 \times 0.6 \times 1.5 \\
   \Rightarrow {\text{Curved Surface Area of Roller}} = 44 \times 0.6 \times 1.5 \\
   \Rightarrow {\text{Curved Surface Area of Roller}} = 39.6{\text{ }}{{\text{m}}^2} \\
 \]

Since we know that the area covered by a roller in one revolution is the same as the covered surface area of the roller.

Thus, the area of the ground leveled in 1 revolution is \[3.96{\text{ }}{{\text{m}}^2}\].

We will now find the area of the ground levelled in 100 revolutions from the above value.

\[3.96{\text{ }}{{\text{m}}^2} \times 100 = 396{\text{ }}{{\text{m}}^2}\]

Calculating the cost of levelling this ground by the roller at 50 paise per square meter.

\[
  {\text{Cost of leveling}} = 396 \times \dfrac{{50}}{{100}} \\
   = {\text{Rs. }}198 \\
 \]

Therefore, the cost of levelling this ground is Rs. 198.

Note:
In solving these types of questions, you should be familiar with the formula of the curved surface area of a cylinder. Some students use the formula of the curved surface area instead of the total surface area of the cylinder, which is wrong. In general the total surface area of the cylinder is \[2\pi rh + 2\pi {r^2}\], where \[r\] is radius and \[h\] is height. But in this question we only need to consider curved surface area, that is, \[2\pi rh\]. So we need to take care of calculating the area for one revolution.