Question

The diameter of a roller 1.5m long is 84cm .if it take 100 revolutions to level a ground. find the cost of leveling this ground at the rate of 50 paise pe r square meter​

Answer 1

Answer:

Answer

r=

2

1.5

=0.75m

h=84cm=0.84m

∴ Curved surface area=2πrh=2×

7

22

×0.75×0.84=3.96m

2

∴ Area of the ground levelled in 1 revolution=3.96m

2

∴ Area of the ground levelled in 100 revolution=3.96m

2

×100=396m

2

∴ Cost of levelling=Rs.396×

100

50

=Rs.198

Answer 2

Given that, Diameter of the roller = 84 cm = 0.84 m

Length of the roller = 1.5 m

Radius of the roller = D/2 = 0.84/2 = 0.42 Area covered by the roller on one revolution = covered surface area of roller Curved surface of roller = 2πrh = 2 x 22/7 x 0.42 x 1.5 = 0.12 x 22 x 1.5 m2

Area of the playground = 100 x Area covered by roller in one revolution = (100 x 0.12 x 22 x 1.5) m2 = 396 m2

Now,

now multiply the cost per square meter with the answer.

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