Question

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

Answer 1

The roller is a right circular cylinder of height $h=120cm$ and radius of its $base=\frac{d}{2}=42cm.$

Therefore,

Area covered by the roller in one revolution $=$ Curved surface area of the roller

$\Rightarrow 2\times \frac{22}{7}\times 42\times 120=31680c{m}^2$

So, area covered by the roller in 500 revolutions = $(31680\times 500)c{m}^2$

$\Rightarrow \frac{31680\times 500}{100\times 100}=1584m^2\dots (1m^2={10}^{4}c{m}^2)$

Hence, cost of levelling the playground $=Rs.1584\times \frac{30}{100}$

Total cost $=Rs.475.20$