The radius of a roller is 35 cm and its length is 3 m. If it takes 30 revolutions to level a playground, find: a) the are of the playground b) the cost of levelling the ground at rs 5 per m2.
Asked by Vru23 | 3rd Oct, 2020, 10:59: PM
Expert Answer:
Radius of the roller = 35 cm
Length of the roller = 3 m = 300 cm
Curved Surface Area of the Roller (Cylindrical shape) = 
Therefore the area covered by the roller in one revolution = 66000 cm2
Therefore the area covered by the roller in 30 revolutions = 66000 × 30 = 1980000 cm2
Therefore the area of the play ground = 1980000 cm2 = 198 m2
Cost of levelling the ground = 5 x 198 = Rs. 990
Length of the roller = 120 cm
Curved Surface Area of the Roller (Cylindrical shape) = 
Therefore the area covered by the roller in one revolution = 31680 cm2
Therefore the area covered by the roller in 1000 revolutions = 31680 × 1000 = 31680000 cm2
Therefore the area of the play ground = 31680000 cm2 = 3168 m2
→ l × b = 3168
→ l × 32 = 3168
→ l = 99 m
Length of the roller = 3 m = 300 cm
Curved Surface Area of the Roller (Cylindrical shape) =
Therefore the area covered by the roller in one revolution = 66000 cm2
Therefore the area covered by the roller in 30 revolutions = 66000 × 30 = 1980000 cm2
Therefore the area of the play ground = 1980000 cm2 = 198 m2
Cost of levelling the ground = 5 x 198 = Rs. 990
Length of the roller = 120 cm
Curved Surface Area of the Roller (Cylindrical shape) =
Therefore the area covered by the roller in one revolution = 31680 cm2
Therefore the area covered by the roller in 1000 revolutions = 31680 × 1000 = 31680000 cm2
Therefore the area of the play ground = 31680000 cm2 = 3168 m2
→ l × b = 3168
→ l × 32 = 3168
→ l = 99 m
Answered by Renu Varma | 5th Oct, 2020, 12:08: PM
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