Question

The diameter of a roller $1.5$m long is $84$cm. If it takes $100$ revolutions to level this play ground, find the cost of levelling this ground at the rate of $50$ paise per square meter.

Answer 1

$r=\frac{1.5}{2}=0.75$m

$h=84$cm$=0.84$m

$\therefore$ Curved surface area=$2\pi rh=2\times \frac{22}{7}\times 0.75\times 0.84=3.96m^2$

$\therefore$ Area of the ground levelled in $1$ revolution$=3.96m^2$

$\therefore$ Area of the ground levelled in $100$ revolution$=3.96m^2\times 100=396m^2$

$\therefore$ Cost of levelling$=$Rs.$396\times \frac{50}{100}=$Rs.$198$