Question

The diameter and length of a roller are 70 cm and 100 cm respectively. It moves once over to level a playground of area 4400 sq. m. Find the number of revolutions the roller will take to level the ground.

Answer 1

Answer:

2000 Revolutions

Step-by-step explanation:

The Radius  of the Roller = 35 cm and the Length = 100 cm

Now the roller is a cylindrical object which when revolve covers  an area equal to its side surface area per revolutions.

Now the area of side surface = 2 * Radius * π * Length

Therefore the area here $= 2*35*100*\frac{22}{7} = 22*1000 cm^2$

There area covered per revolution $= 22*1000 cm^2$

Total area to be covered = 4400 sq.m = 4400*100*100 sq.cm

Number of revolution $= \frac{4400*100*100}{22*1000} = 2000$  Revolutions

Total Number of revolutions = 2000