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) = begin mathsize 16px style 2 πrh space space equals space 2 space cross times 22 over 7 cross times 35 space cross times space 300 space equals space 66000 space cm squared end style

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) = begin mathsize 16px style 2 πrh space space equals space 2 space cross times 22 over 7 cross times 42 space cross times space 120 space equals space 31680 space cm squared end style

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