Question

The diameter of a roller 84 cm long is 1.5 m. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of ₹ 0.50 per square metre.

Answer 1

We have d = 1.5 m, $r=\frac{1.5}{2}=\frac{15}{20}m=\frac{3}{4}m$
and $h=84cm=\frac{84}{100}m$
$\therefore$ Area covered in 1 revolution $=2\pi rh=2\times \frac{22}{7}\times \frac{3}{4}\times \frac{84}{100}=\frac{44\times 9}{100}{m}^2$
$\therefore$ Area covered in 100 revolutions $=\frac{100\times 44\times 9}{100}=396m^2$
$\therefore$ Cost of levelling the playground $=₹0.50\times 396=₹198$