$A$ glass tube of uniform cross-section is connected to a tap with a rubber tube. The tap is opened slowly. Initially, the flow of water in the tube is streamline. The speed of flow of water to convert it into a turbulent flow is (radius of tube $= 1 \,cm$, $\eta = 1 \times 10^{-3} \,Ns/m^2$, $R_{n} = 2500$, and density of water $\rho = 10^3 \,kg/m^3$) (in $m/s$)

The rate of flow of liquid in a tube of radius $r$ and length $l$,whose ends are maintained at a pressure difference $P$,is given by $V = \frac{\pi Q P r^4}{\eta l}$,where $\eta$ is the coefficient of viscosity. What is the value of $Q$?

What is the value of Reynolds number for turbulent flow?

Under a constant pressure head,the rate of flow of liquid through a capillary tube is $V$. If the length of the capillary is doubled and the diameter of the bore is halved,the rate of flow would become

Consider a steady flow of oil in a pipeline. The cross-sectional radius of the pipeline decreases gradually as $r = r_0 e^{-\alpha x}$, where $\alpha = \frac{1}{3} \text{ m}^{-1}$ and $x$ is the distance from the pipeline inlet. If $R_1$ is the Reynolds number for a certain pipeline cross-section at a distance $x_1$ metre from the inlet and $R_2$ is for distance $(x_1 + 3)$ metre, then the ratio $\frac{R_1}{R_2}$ is

Deduce the formula of Reynolds number in the form of inertial force and viscous force.