Question

A road roller has a diameter 0.7 M and its width is 1.2 M Find the least number of revolution that the roller must take
in order to level a playground of size 120 M and 44 metre

Answer 1

Heya Mate !!!

Here's Your Answer ; 2,000

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Step by Step Explanation ;-
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=> Area of playground =120*44 m^2

In cylinder ;

=> Height ( h ) =1.2 m
=> Diameter ( d ) =0.7 m
=> Radius ( r ) = 0.35

=> c. s. a =2πrh

=> 2*22/7*0.35*1.2

=> 2.64

No. of revolution =area of playground/c.s.a

=> 120*44/2.64

=> 2000

Answer 2

Solutions :-

Given :
Dimension of playground = 120 m × 44 m

Find the area of playground :-

Area of rectangle = (Length × Breadth) sq. unit
= (120 × 44) m²
= (5280) m²


Given :
Diameter of wheel of road roller = 0.7 m
Radius = r = 0.7/2 = 0.35 m
Width = h = 1.2 m


Find the curved surface area of wheel of road roller :-

Curved surface area of cylinder = 2π️rh
️️️️️️= 2 × 22/7 × 0.35 × 1.2
= 2.64


Find the number of revolution taken by the road roller :-

2.64 m² = 1 revolution
5280 m² = (1 × 5280)/2.64 m² = 2000


Hence,
Number of revolution taken by the road roller = 2000

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✯ @shivamsinghamrajput ✯