What will be the nature of flow of water from a circular tap,
Question:

What will be the nature of flow of water from a circular tap, when its flow rate increased from $0.18 \mathrm{~L} / \mathrm{min}$ to $0.48 \mathrm{~L} / \mathrm{min}$ ? The radius of the tap and viscosity of water are $0.5 \mathrm{~cm}$ and $10^{-3} \mathrm{~Pa}$ s, respectively.

(Density of water: $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )

  1. (1) Unsteady to steady flow

  2. (2) Remains steady flow

  3. (3) Remains turbulent flow

  4. (4) Steady flow to unsteady flow


Correct Option: , 4

Solution:

(4)

The nature of flow is determined by Reynolds Number.

$\mathrm{R}_{\mathrm{e}}=\frac{\rho \mathrm{vD}}{\eta}$

$\rho \rightarrow$ density of fluid $; \eta \rightarrow$ coefficient of

$\mathrm{v} \rightarrow$ velocity of flow

$\mathrm{D} \rightarrow$ Diameter of pipe

If $\mathrm{R}_{\mathrm{e}}<1000 \rightarrow$ flow is steady

$1000<\mathrm{R}_{\mathrm{e}}<2000 \rightarrow$ flow becomes unsteady

$\mathrm{R}_{\mathrm{e}}>2000 \rightarrow$ flow is turbulent

$\mathrm{R}_{\text {einitial }}=10^{3} \times \frac{0.18 \times 10^{-3}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2} \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$         $=382.16$

$\mathrm{R}_{\mathrm{e} \text { final }}=10^{3} \times \frac{0.48 \times 10^{-3}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2} \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$           $=1019.09$

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