If diameter of a road roller is 0.9m and it's length is 1.4m how much area of a field will be pressed in its 500 rotation
Answers
HEY MATE...!!!
HERE IS YOUR ANSWER :
As the part used by the road-roller in pressing the field is its curves surface area of CSA...
Thus, we need to find the CSA of road roller which is cylindrical in shape...
FORMULA TO CALCULATE CSA OF CYLINDER : 2*PI*RADIUS*HEIGHT
HERE, RADIUS = 0.9/2 m AND HEIGHT = 1.4 m
Applying in the formula,
CSA = 2*22/7*0.9/2*1.4
= 3.96 sq. mt.
Now, 3.96 sq. mt. is the area pressed by the roller coaster in one rotation...
TOTAL AREA PRESSED BY THE ROLLER IN 500 ROTATION = 500*3.96
= 1980 sq. mt.
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Diameter = 0.9m ; Radius = 0.9/2 = 0.45
Height = 1.4m
We know that, The shape of rollers wheels is like a cylinder.
Therefore, Curved surface area of roller = Curved surface area of roller
Curved surface area of cylinder = 2πrh
= 2 × 22/7 × 0.45 × 1.4
= 44 × 0.2 × 0.45
= 3.96 sq.m
Therefore, Total area pressed by roller in one rotation = 3.96 sq.m
It is given that, The roller rotates 500 times to press the all ground.
∴ Area pressed in 500 rotations = 500 × 3.96
= 1980 sq.m
Explanation:
In the given question, The diameter of road roller is given as 0.9m. First we converted it into radius i.e. 0.45m. The height of the roller was given as 1.4m. Now, As we have to find the outer area/curved surface area of roller. So, we used the formula of Curved surface area of cylinder. By putting the given values in the formula, we find the curved surface area of roller of 1 sqm. It was given that, to find the Area pressed in 500 rotations. So, We multiplied the Curved surface area of roller for 1sqm with 500. And, we get the final answer as 1980 sqm.