A road roller is cylindrical in shape. Its circular end has a diameter $250 c m$ and its width is $1 m 40 c m$ . Find the least number of revolution that the roller must make in order to level a playground $110 m \times 25 m$ .

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Solution

Given:-
diameter of roller = 250 cm = 2.5m
width of roller = 1.4m
radius of the roller = 1.25m

we know that
curved surface area = 2×22/7×1.25×1.40
= 44 × 1.25 × 0.20
=11m^2
area of ground = length × breath
= 25 × 10
= 250 m^2
number of revolutions to cover the field= Area of field/ CSA of roller.
= 250/11
= 22.727 revolutions approx.

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A road roller is cylindrical in shape. Its circular end has a diameter 250 cm and its width is 1 m 40 cm. Find the least number of revolution that the roller must make in order to level a playground 110 m ×25 m.

A road roller is cylindrical in shape. Its circular end has a diameter $250 c m$ and its width is $1 m 40 c m$ . Find the least number of revolution that the roller must make in order to level a playground $110 m \times 25 m$ .