a road roller is cylindrical in shape its circular and has a diameter 250 cm and its width is 1 m 40 centimetre Find the least number of revolution that the roller must make in order to level up in a gram 10 metre into 25 metre
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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.
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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