If water flows with a velocity of 20 ,cm ,s^{ | vedclass

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(A) To determine the nature of the flow, we calculate the Reynolds number $(R_e)$.The formula for the Reynolds number is $R_e = \frac{\rho v D}{\eta}$

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turbulent

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(A) To determine the nature of the flow, we calculate the Reynolds number $(R_e)$.The formula for the Reynolds number is $R_e = \frac{ho v D}{\eta}$, where $ho$ is the density, $v$ is the velocity, $D$ is the diameter of the pipe, and $\eta$ is the coefficient of viscosity.Given:Density $ho = 10^3 \,kg \,m^{-3}$Velocity $v = 20 \,cm \,s^{-1} = 0.2 \,m \,s^{-1}$Radius $r = 2 \,cm = 0.02 \,m$, so diameter $D = 2r = 0.04 \,m$Viscosity $\eta = 10^{-3} \,kg \,m^{-1} \,s^{-1}$Substituting these values:$R_e = \frac{10^3 	imes 0.2 	imes 0.04}{10^{-3}}$$R_e = \frac{10^3 	imes 0.008}{10^{-3}} = 8 	imes 10^3 = 8000$Since the Reynolds number $R_e > 2000$, the flow is turbulent.

If water flows with a velocity of $20 \,cm \,s^{-1}$ in a pipe of radius $2 \,cm$, then the flow is (The coefficient of viscosity of water is $10^{-3} \,kg \,m^{-1} \,s^{-1}$ and density of water is $10^3 \,kg \,m^{-3}$).

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