Question

The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of travelling it at the rate of 30 paise per square meter.

Answer 1

Hint: As we have cost of travelling per square meter. So we will calculate the area to find the total cost for leveling the playground. As a roller in the shape of a cylinder so in one revolution it will cover the area equal to its curved surface area.
Formula for curved surface area(S) of cylinder is
$S=2\pi rh$
Where r is radius and h is height of cylinder.

Complete step-by-step answer:
In the given question the diameter of the cylinder is 120 cm.
So radius(r) of cylinder is $r=\dfrac{120cm}{2}=60cm$
Height of the cylinder is 84 cm.
Hence curved surface area(S) of cylinder is
$\Rightarrow S=2\pi rh$
$\Rightarrow S=2\times \dfrac{22}{7}\times 60\times 84$
$\Rightarrow S=2\times 22\times 60\times 12$
$\Rightarrow S=31680\,c{{m}^{2}}$
The cost of travelling per square meter is 30 paise.
So we need to convert cm2 to m2.
In general $1\,c{{m}^{2}}=\dfrac{1}{10000}\,{{m}^{2}}$
So $31680\,c{{m}^{2}}=\dfrac{31680}{10000}\,{{m}^{2}}=3.1680\,{{m}^{2}}$
So the area covered in one revolution is 3.1680 m2.
So area covered in 500 revolution is $500\times 3.1680\,{{m}^{2}}=1584\,{{m}^{2}}$
So cost(C) for 500 revolution is = area covered in 500 revolution times cost per square meter.
 $\Rightarrow C=1584\times 30=47520\,paise$
To convert paise to rupees we need to divide by 100 because in one rupee there is 100 paise.
$\Rightarrow C=\dfrac{47520}{100}=Rs.\,475.20$

Note: In this we should take care that in one revolution the roller will cover only the area equal to its curved surface area. So we will not include the area by the circular part of the cylinder.
As we used $1\,c{{m}^{2}}=\dfrac{1}{10000}\,{{m}^{2}}$. We got this formula as given below
$\because 1\,m=100\,cm$
So we can write
Hence $1\,c{{m}^{2}}\,=\dfrac{1}{10000}\,{{m}^{2}}$ .