The base diameter and length of a cylindrical road roller are 3.5 m and 5 m respectively. What area can be flattened by the roller in 100 revolutions?
Answers
Answered by
0
Step-by-step explanation:
h=1.4m
r=100cm=1m
CSA=2πrh=2×
7
22
×1×1.4=8.8m
2
Total area =125×2−=2500m
2
No. of revolutions =
8.8
2500
=284.09.
Answered by
1
Answer:
diameter=3.5m
then,radius will be d÷2
=3.5÷2
=1.75m
Height=5m
A.T.Q
c.s.a of cylinder=2πrh
=2×22÷7×1.75×5
=55m²
area can be flattened in 100 revolutions will be
=100×55
=5500m²
Similar questions