​ If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? ( π = 22/7 ) ​

Question

If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? ( π = \(\frac{22}{7}\))

Answer 1

Given: For road roller, 

diameter (d) = 0.9 m, length (h) = 1.4 m 

To find: Area of a field pressed in 500 rotations

Solution: 

i. Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller 

∴ Curved surface area of the road roller = 2πrh 

= πdh ,..[∵ d = 2r] 

= \(\frac{22}{7}\) x 0.9 x 1.4 7 

= 22 x 0.9 x 0.2 

= 3.96 sq.m.

ii. Area of land pressed in 1 rotation = 3.96 sq.m. 

∴Area of land pressed in 500 rotations = 500 x 3.96 = 1980 sq.m. 

∴ 1980 sq.m, land will be pressed in 500 rotations of the road roller.