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: $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )
Correct Option: , 4
(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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