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Properties of Matter question

2021 · 16 Mar · Shift 2 · Q46
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  5. /2021 · 16 Mar · Shift 2 · Q46

Properties of Matter question

2021 · 16 Mar · Shift 2 · Q46

JEE MainPhysicsProperties of MatterMCQ+4 / −1
What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10 −-− 3 Pa s, respectively. (Density of water : 103 kg/m3)
  1. A
    Steady flow to unsteady flow
  2. B
    Unsteady to steady flow
  3. C
    Remains turbulent flow
  4. D
    Remains steady flow
View written solutionFree

Correct answer: D

  1. Use Reynolds number to determine the nature of flow

For flow through a pipe/tap,

Re=ρvdηRe = \frac{\rho v d}{\eta}Re=ηρvd​

where:

  • ρ=103 kg m−3\rho = 10^3\,\text{kg m}^{-3}ρ=103kg m−3
  • η=10−3 Pa s\eta = 10^{-3}\,\text{Pa s}η=10−3Pa s
  • radius r=0.5 cm=5×10−3 mr = 0.5\,\text{cm} = 5\times 10^{-3}\,\text{m}r=0.5cm=5×10−3m
  • diameter d=2r=10−2 md = 2r = 10^{-2}\,\text{m}d=2r=10−2m

Cross-sectional area of tap:

A=πr2=π(5×10−3)2=25π×10−6 m2A = \pi r^2 = \pi (5\times 10^{-3})^2 = 25\pi \times 10^{-6}\,\text{m}^2A=πr2=π(5×10−3)2=25π×10−6m2
  1. First flow rate: 0.18 L/min0.18\,\text{L/min}0.18L/min

Convert to SI:

Q1=0.18×10−3 m3/minQ_1 = 0.18\times 10^{-3}\,\text{m}^3/\text{min}Q1​=0.18×10−3m3/min Q1=0.18×10−360=3×10−6 m3/sQ_1 = \frac{0.18\times 10^{-3}}{60} = 3\times 10^{-6}\,\text{m}^3/\text{s}Q1​=600.18×10−3​=3×10−6m3/s

Velocity:

v1=Q1A=3×10−625π×10−6=325π≈0.0382 m/sv_1 = \frac{Q_1}{A} = \frac{3\times 10^{-6}}{25\pi \times 10^{-6}} = \frac{3}{25\pi} \approx 0.0382\,\text{m/s}v1​=AQ1​​=25π×10−63×10−6​=25π3​≈0.0382m/s

Reynolds number:

Re1=(103)(0.0382)(10−2)10−3=382Re_1 = \frac{(10^3)(0.0382)(10^{-2})}{10^{-3}} = 382Re1​=10−3(103)(0.0382)(10−2)​=382

Since Re1<2000Re_1 < 2000Re1​<2000, the flow is streamline/steady (laminar).

  1. Second flow rate: 0.48 L/min0.48\,\text{L/min}0.48L/min

Convert to SI:

Q2=0.48×10−3 m3/minQ_2 = 0.48\times 10^{-3}\,\text{m}^3/\text{min}Q2​=0.48×10−3m3/min Q2=0.48×10−360=8×10−6 m3/sQ_2 = \frac{0.48\times 10^{-3}}{60} = 8\times 10^{-6}\,\text{m}^3/\text{s}Q2​=600.48×10−3​=8×10−6m3/s

Velocity:

v2=Q2A=8×10−625π×10−6=825π≈0.1019 m/sv_2 = \frac{Q_2}{A} = \frac{8\times 10^{-6}}{25\pi \times 10^{-6}} = \frac{8}{25\pi} \approx 0.1019\,\text{m/s}v2​=AQ2​​=25π×10−68×10−6​=25π8​≈0.1019m/s

Reynolds number:

Re2=(103)(0.1019)(10−2)10−3=1019Re_2 = \frac{(10^3)(0.1019)(10^{-2})}{10^{-3}} = 1019Re2​=10−3(103)(0.1019)(10−2)​=1019

Again, Re2<2000Re_2 < 2000Re2​<2000, so the flow is still streamline/steady (laminar).

  1. Conclusion

Initially the flow is steady, and after increasing the flow rate it is still steady.

So the correct option is:

D: Remains steady flow\boxed{\text{D: Remains steady flow}}D: Remains steady flow​
  1. Comparison with stored answer

Stored correct answer is A, but our calculation clearly gives D because both Reynolds numbers are below the critical value for turbulence.

Hence, I disagree with the stored answer.

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