A road 1600m long and 4.4 m wide.A cylindrical road roller of length 2m and diameter 1.4 m.Find the number of revolutions required to level the road using the road roller
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area of road=l*b
=1600*4.4
=7040
CSA of roller=2πrh
=2*π*0.7*2
=8.8
no. of revolutions=7040/8.8=800 revolutions
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Answer:
1091.3482 (approx.)
Step-by-step explanation:
Diameter of roller (let it be D) is 1.4 m
therefore circumference (C) = π×D [2×π×r]
⇒C= 1.4×π ≅ 4.398 m
this is the distance covered in one revolution of the roller
length of road (L) = 1600 m
number of revolutions needed to cover the whole length is
L/C
=1600/(1.4×π)
=363.783 (approx)
we aint done yet.
width of road is 4.4 m but our roller is only 2 m long
so it has to come back and do the whole thing at least thrice to over the whole width
total number of revolutions is
=363.783 × 3
=1091.3482 (approx.)
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