Question

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

Answer 1

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.

$\huge \pink{~} \red {τ} \green {α} \blue {v} \orange {v} \pink {u} \red {♡}$

Answer 2

Answer:

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.

Step-by-step explanation:

Hope it helps you

Please mark me as the brainliest and drop some thanks