Question

the diameter of a roller 120 cm long is 84 cm it takes 500 complete revolutions to level a playground find the cost of levelling it at the rate of rupess 25 sq meter
​

Answer 1

Answer:

Rs475.20

Step-by-step explanation:

h=120 cm

and radius of its base =r=42 cm

∴ area covered by the roller in one revolution = curved surface area of the roller

=2×

7

22

×42×120cm

2

=31680cm

2

so, area covered by the roller is 500 revolution = (31680×500)cm

2

=

100×100

31680×500

m

2

=1584m

2

hence , cost of levelling the playground = 1584×

100

30

=Rs475.20