Question

Find the number of revolution required to level a playground of area 22m^2 by a road roller whose diameter is 14cm and length is 100 cm

Answer 1

Answer:

50 revolutions

Step-by-step explanation:

area of playground = 22m^2 = 220000cm^2

diameter of the road roller = 14cm

therefore radius = 7cm

no.of revolutions = $\frac{area of playground}{area of road roller(csa)}$

                    =$\frac{220000}{2*22/7*7*100}$

                =50 revolutions

hope u understand