Question

Diameter of a road roller $1.2$m long is $32$ cm. If the cost of leveling the playground at Rs $1.5$ oer $m^2$ is Rs 1425.60, determine number of revolution takes by roller for leveling the ground

Answer 1

$r=\left(\frac{32}{2}\right)\times \frac{1}{100}m$

Total area of playground

$=\frac{1425.6}{1.5}=950.4m^2$

now area covered by roller in one revolution

= area of curved surface

$=L\times \left(2\pi r\right)$

$=1.2\times 2\pi \times \left(\frac{32}{2+100}\right)$

$=1.2057m^2$

$\therefore$ no of revolution $=\frac{950.4}{1.2057}=788.2$

$=789$ (approx)