Math, asked by shakshi1997, 9 months ago

a cylindrical garden roller is 75cm.long and its radius is 14cm.how many revolution will it make to level an area of 9900cm2​

Answers

Answered by himanshi94
13

Answer:

Therefore the no. of revolutions are 1.5.....

Hope this will help u...

:)...

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Answered by Anonymous
16

The total number of revolution is 1.5

Given data :

Length of the roller = 75 cm

Radius of the roller = 14 cm

Garden's area = 9900 cm²

So,

Curved surface are of the roller

= 2 × π × r × h

= 2 × 22/7 × 14 × 75

= 2 × 22 × 2 × 75

= 88 × 75

= 6600 cm²

Number of revolutions = (9900/6600) = 1.5 (answer)

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