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 : $10^{3}\, kg / m ^{3}$ )
JEE MAIN 2021, Medium
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The nature of flow is determined by Reynolds Number.
$R _{ e }=\frac{\rho vD }{\eta}$
$[\rho \rightarrow$ density of fluid ;
$\eta \rightarrow$ coefficient of ;
$v \rightarrow$ velocity of flow ;
$D \rightarrow$ Diameter of pipe$]$
From NCERT
If $R _{ e }<1000 \quad \rightarrow$ flow is steady
$1000< R _{ e }<2000 \rightarrow$ flow becomes unsteady
$R _{ e }>2000 \rightarrow$ flow is turbulent
$R _{ e \text { initial }}=10^{3} \times \frac{0.18 \times 10^{-3}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2} \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$
$=382.16$
$R _{ efinal }=10^{3} \times \frac{0.48 \times 10^{-3}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2} \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$
$=1019.09$
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