Question

find the number of revolutions taken by a road roller whose lsa is 32cm² to level a play ground of area 6400m²​

Answer 1

the number of revolutions that are required to level the playground is 194 revolutions

Explanation:

We know that LSA, that is lateral surface area of a cylinder is the area of a rectangular sheet, which when spread onto the ground, covers the area equal to the LSA of the cylinder.

Hence, area covered by the roller in one revolution = area of LSA of cylinder = 32 cm².

Now, to cover the are of 6200 cm²,

the number of revolutions required = Total area of the playground/ LSA of the cylinder = 6200/32 = 193.75 = 194

Hence the number of revolutions that are required to level the playground is 194.