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 Rs. 0.50 per square metre. [3 MARKS]

Answer 1

Formula: 1 Mark
Application: 1 Mark
Solution: 1 Mark

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 of levelled 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 levelled in 100 revolutions$=\frac{100\times 44\times 9}{100}=396m^2$
$\therefore$ Cost of levelling the playground $=Rs.0.50\times 396=Rs.198$