Question

The diameter of a road roller of length 120cm is 84cm. If it takes 500 complete revolutions to have a playground, then find the cost of levelling it at the cost of 75 paise per square meter. (Take $\pi =\dfrac{22}{7}$ ).

Answer 1

Hint: To solve the question given above, we will first find out the area of the playground to be levelled. For this we will calculate the area of the roller and multiply it with 500 to obtain the levelled area. Then, to find the cost of levelling, we will multiply the levelled area by 75 paise (Rs$0.75/{{m}^{2}}$)

Complete step by step answer:
It is given in the question that the roller is revolved 500 times to level the entire playground. So the total levelled area of the playground will be equal to 500 times the area of the roller. The shape of the roller is a cylinder. So, we have to basically find the curved area of the cylinder which has diameter =84cm and length 120cm. Thus, the radius of cylinder will be:
$R=\dfrac{D}{2}=\dfrac{84}{2}=42cm$.
Let us draw the cylinder to get a better view.

The curved surface of the cylinder with the radius and length given is calculated by the formula shown:
Curved surface area $=2\pi rl$
Where ‘r’ is the radius of base and ‘l’ is the length of the cylinder.
Thus, the curved surface area of cylinder in our case will be:
Curved surface area $=2\times \pi \times 42\times 120c{{m}^{2}}$
$\Rightarrow $ Curved surface area$=2\times \dfrac{22}{7}\times 42\times 120c{{m}^{2}}$
$\Rightarrow $ Curved surface area$=31680c{{m}^{2}}$
Now the area of the playground will be equal to 500 times the curved surface area of the cylinder.
So we have the following equation:
Area of playground$=500\times $ Curved surface area of the roller.
$\Rightarrow $ Area of playground$=500\times 31680c{{m}^{2}}$
$\Rightarrow $ Area of playground$=15840000c{{m}^{2}}$
Now, we have to convert this area to ${{m}^{2}}$ from $c{{m}^{2}}$ because the cost given is 75 paise$/{{m}^{2}}$. We have the following conversion:
$1{{m}^{2}}=1m\times 1m=100cm\times 100cm$
$\Rightarrow 1{{m}^{2}}=10000c{{m}^{2}}$
Thus, the area of the playground$=15840000c{{m}^{2}}$
Area of the playground$=1584\times 10000c{{m}^{2}}$
$\Rightarrow $ Area of the playground$=1584{{m}^{2}}$.
Now it is given that cost of levelling is 75 paise per square meter. Thus, the cost of 1 meter square is 75 paise.
Cost of $1{{m}^{2}}=75$ paise.
Cost of $1584{{m}^{2}}=1584\times 75$ paise.
Cost of $1584{{m}^{2}}=118800$ paise.
Now, we know that, 1 Rupee = 100 paise. Thus, we have:
Cost of $1584{{m}^{2}}=118800\times Rs1$
Cost of $1584{{m}^{2}}=Rs1188$

Hence, the total cost of levelling is Rs 1188.

Note: We have assumed that the roller is cylindrical in shape i.e. the radius of the roller is constant everywhere and thus we have applied the area of the cylinder. If the radius is not constant, then the roller would be friction. Also, we have assumed that the cylinder is the right cylinder i.e. the axis of the cylinder is perpendicular to circular base.