Math, asked by rishabchandrago3879, 10 months ago

The radius of a roller is 49cm and its length is 125 cm how much atea of playground will be levelled in 400 revolutions moving one over the ground

Answers

Answered by bhagyashreechowdhury
16

1540 m² area of the playground will be levelled in 400 revolutions moving one over the ground.

Step-by-step explanation:

Formula Used:

  • CSA of cylinder = 2πrh

The radius of a roller, r = 49 cm

The length of the roller i.e., the height of the cylinder, h = 125 cm

We know that,

[The distance covered by the roller in 1 revolution] = [Curved Surface Area of the cylinder] = 2πrh

The distance covered by the roller in 1 revolution is,

= 2 * 22/7 * 49 * 125

= 38500 cm²

Thus,

The distance covered by the roller in 400 revolutions is given by,

= 400 * 38500 cm²

= 15400000 cm²

Since 1 cm² = 1/10000 m²

= 1540 m²

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Also View:

The radius of a roller is 60 cm and its length is 84cm find the area of the ground pressed by it in 500 revolution?

https://brainly.in/question/7015998

The diameter of a roller is 42 cm and its length is 100 cm. it takes 400 complete revolutions moving once over the level of a playground . determine the area of the playground. also,find the cost of leveling the playground a rs50 per 105sqm?

https://brainly.in/question/2787475

Answered by rajivarya1983
8

Step-by-step explanation:

radius = 49cm

hight = 12cm

2*22/7*49*125 = 38500 cm

revolution =400

400*38500 = 15400000m

1cm=1/ 10000 m

= 15400000/10000 = 1540 m

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