What will be the nature 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 :${10}^3 kg/m^3$)

Speed ,  $V=\frac{Q}{A}$

$$\begin{gathered} V_1=\frac{0.18\times {10}^{-3}}{\pi {(0.5\times {10}^{-2})}^2\times 60}=0.03819 m/s \\ {V}_2=\frac{0.48\times {10}^{-3}}{\pi {(0.5\times {10}^{-2})}^2\times 60}=0.1019 m/s \\ Reynolds number R_e=\frac{\rho VD}{\eta } \\ For Q_1 , {\mathrm{R}}_e=\frac{{10}^3\times 0.03819\times {10}^{-2}}{{10}^{-3}}=381.9<1000\text{\hspace{0.17em}\hspace{0.17em}}\left(steadyflow\right) \\ For Q_{2 , }{\mathrm{R}}_e=\frac{{10}^3\times 0.1019\times {10}^{-2}}{{10}^{-3}}=1019>1000\text{\hspace{0.17em}\hspace{0.17em}}(unsteady flow) \end{gathered}$$