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(D) The nature of flow is determined by the Reynolds number $(R_e)$: $R_e = \frac{\rho v D}{\eta}$.Given: $\rho = 10^3\, kg/m^3$,$\eta = 10^{-3}\, Pa\

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Steady flow to unsteady flow

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(D) The nature of flow is determined by the Reynolds number $(R_e)$: $R_e = \frac{ho v D}{\eta}$.Given: $ho = 10^3\, kg/m^3$,$\eta = 10^{-3}\, Pa\cdot s$,$r = 0.5\, cm = 0.005\, m$,$D = 2r = 0.01\, m$.Flow rate $Q = A \cdot v = \pi r^2 v$,so $v = \frac{Q}{\pi r^2}$.Substituting $v$ in $R_e$: $R_e = \frac{ho Q D}{\eta \pi r^2} = \frac{ho Q (2r)}{\eta \pi r^2} = \frac{2 ho Q}{\eta \pi r}$.For $Q_1 = 0.18\, L/min = \frac{0.18 	imes 10^{-3}}{60}\, m^3/s = 3 	imes 10^{-6}\, m^3/s$:$R_{e1} = \frac{2 	imes 10^3 	imes 3 	imes 10^{-6}}{10^{-3} 	imes \pi 	imes 0.005} \approx 382.16$.Since $R_{e1} For $Q_2 = 0.48\, L/min = \frac{0.48 	imes 10^{-3}}{60}\, m^3/s = 8 	imes 10^{-6}\, m^3/s$:$R_{e2} = \frac{2 	imes 10^3 	imes 8 	imes 10^{-6}}{10^{-3} 	imes \pi 	imes 0.005} \approx 1018.59$.Since $1000 Thus,the flow changes from steady to unsteady.

What will be the nature of flow of water from a circular tap,when its flow rate increases from $0.18\, L/min$ to $0.48\, L/min$? The radius of the tap is $0.5\, cm$ and the viscosity of water is $10^{-3}\, Pa\cdot s$. (Density of water: $10^{3}\, kg/m^{3}$)

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