# The diameter of a roller is 42 cm and its length is 100 cm.It takes 400 complete revolutions moving once over to level a play ground.Determine the area of the playground? Also find the cost of leveling the playground at Rs.5 per sq.m

when the roller rolls over the ground once completely, the ground area covered by it will be equal to its curved surface area.

length of the roller h =100cm=1 m.

the curved surface area of the roller = area covered in 400 revolutions = 400*1.32= 528 m square

cost of leveling= Rs. 5 per sq metre

therefore the total cost=528*5=2640/-

• 1

r = 42 / 2 = 21cm

In 1 revolution road covered = 2 * π * r * h

= 2 * 22 / 7 * 21 * 100 = 13200 cm2

In 400 revolutions area covered = 400 * 13200 = 5280000 cm2

In m = 5280000 / 100 = 52800 m2

1 m levelling costs = Rs. 5

52800 m will cost = 5 * 52800 = Rs. 264000

• -1

lateral surface area = 2TTrh

= 2*22/7*100*21

=13200 cm^2 = 1 revolution

400 revolutions = 13200*400=5280000cm^2  ( 1m^2=10000cm^2)

area of the park= 528 m^2

cost of levelling 1 m^2=rs5

cost of levelling 528m^2 = 528 * 5

=Rs 26400

• -1
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