The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground

Radius of the roller (r) = 84/ 2 cm = 42 cm 

Length of the roller (h) = 120 cm 

Curved surface area of the roller = 2πrh 

= 2 × 22/ 7 × 42 × 120 cm² = 44 × 720 cm² = 31680 cm² 

∴ area covered by the roller in 1 revolution = 31680 cm² 

∴ area covered by the roller in 500 revolutions = 31680 × 500 cm² = 15840000 cm

Hence, area of the playground = 15840000/ 100× 100 m² = 1584 m² Ans.

It can be observed that a roller is cylindrical.

Height (h) of cylindrical roller = Length of roller = 120 cm

Radius (r) of the circular end of roller = (84/2)cm = 42cm

CSA of roller = 2πrh

= (2 x 22/7 x 42 x 120)cm²

= 31680cm²

Area of field = 500 × CSA of roller

= (500 × 31680) cm²

= 15840000 cm²

= 1584 m²