JEE Main 16 March 2021 Shift 2 Solved Paper

As we know that the nature of flow of water is determined by Reynold's number,
Re=‌

ρvD

η

Here, ρ=103kgm−3,r=0.5cm=0.5×10−2m
D=2r=1×10−2m
η=10−3Pa−s
Q1=0.18L/min=‌

0.18×10−3

60

m3/s
Q2=0.48L/‌min=‌

0.48×10−3

60

m3/s
Also, v=‌

Q

πr2

∴v1=‌

Q1

πr2

Also, v=‌

Q

πr2

∴v1=‌

Q1

πr2

∴Re‌initial ‌=‌

ρv1D

η

=‌

ρQ1D

πr2η

=‌

103×0.18×10−3

π×(0.5×10−2)2×60

×‌

1×10−2

10−3

=381.97∼eq382
and Re‌final ‌=‌

ρv2D

η

=‌

ρQ2D

πr2η

=‌

103×0.48×10−3×1×10−2

π×(0.5×10−2)2×60×10−3

=1019.09
We know that,
When Re<1000, the flow of water is considered to be steady.
When 1000<Re<2000, the flow of water becomes unsteady.
When Re>2000, the flow of water becomes turbulent.
As the Re‌final ‌ value lies in between 1000 and 2000 and
Re‌initial ‌ value is less than 1000 , so the nature of flow of water will be steady flow to the unsteady one.