What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L (min)$$^{-1}$$ to 0.48 L (min)$$^{-1}$$? 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$$^{-3}$$)

The nature of flow is determined by the Reynolds number $$Re = \frac{\rho v D}{\eta}$$, where $$\rho$$ is the density, $$v$$ is the flow velocity, $$D$$ is the diameter, and $$\eta$$ is the viscosity.

The radius of the tap is $$r = 0.5$$ cm $$= 5 \times 10^{-3}$$ m, so the diameter $$D = 0.01$$ m and the cross-sectional area is $$A = \pi r^2 = \pi \times 25 \times 10^{-6}$$ m$$^2$$.

For the first flow rate $$Q_1 = 0.18$$ L/min $$= \frac{0.18 \times 10^{-3}}{60} = 3 \times 10^{-6}$$ m$$^3$$/s, the velocity is $$v_1 = \frac{Q_1}{A} = \frac{3 \times 10^{-6}}{\pi \times 25 \times 10^{-6}} = \frac{3}{25\pi} \approx 0.0382$$ m/s. The Reynolds number is $$Re_1 = \frac{10^3 \times 0.0382 \times 0.01}{10^{-3}} \approx 382$$.

For the second flow rate $$Q_2 = 0.48$$ L/min $$= \frac{0.48 \times 10^{-3}}{60} = 8 \times 10^{-6}$$ m$$^3$$/s, the velocity is $$v_2 = \frac{Q_2}{A} = \frac{8}{25\pi} \approx 0.1019$$ m/s. The Reynolds number is $$Re_2 = \frac{10^3 \times 0.1019 \times 0.01}{10^{-3}} \approx 1019$$.

Since $$Re_1 \approx 382$$ is well below the critical value (around 1000 for flow from a tap), the initial flow is steady (laminar). However, $$Re_2 \approx 1019$$ exceeds this threshold, indicating a transition to unsteady flow.

Therefore, the flow changes from steady flow to unsteady flow.