Question

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

Answer 1

Given that $r=42cm$ and $h=120cm$
Area covered by the roller in one revolution=Curved Surface Area of the road roller.
$=2\pi rh$
$=2\times \frac{22}{7}\times 42\times 120$
$31680{cm}^2$
Area covered by the roller in $500$ revolutions $=31680\times 500$
$=15840000{cm}^2$
$=\frac{15840000}{10000}=1584m^2[10000{cm}^2=1sq.m]$
Cost of levelling per $1sq.m$ $=$ Rs.$\frac{75}{100}$
Thus, cost of levelling the play ground $=\frac{1584\times 75}{100}=$ Rs.$1188$.