Question

the circumference of the road roller is 30.8 m and its length is 7 m . find the number of revolution made by it to level an area of 11858 m square​

Answer 1

Given:-

circumference of roller=30.8m

length of roller=7m

so,

CSA of road roller = 2πrH

=>30.8*7=215.6m²

Total area given=11858m²

therefore,

215.6 * no.of revolution= 11858

no.of revolution=> 11858/215.6=55

Hence, total no.of revolution =55

good luck

Answer 2

Answer:

110 revolution

Step-by-step explanation:

circle circumference =2×22/7×r

which is equal to 30.8

so, 30.8=2×22/7×r=30.8

r=30.8×7/22×2

r=215.6/44

r=4.9

2×22/7×4.9×7

215.6

11858=v×215.6

11858/215.6

55

therefore 55 revolution