If $\rho$ is the density and $\eta$ is the coefficient of viscosity of a fluid which flows with a speed $v$ in a pipe of diameter $d$,the correct formula for Reynolds number $R_{e}$ is ..............

What is the value of the Reynolds number $(R_e)$ for unsteady flow?

$A$ 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 $m/s$)

Two tubes of radii $r_1$ and $r_2$,and lengths $l_1$ and $l_2$,respectively,are connected in series and a liquid flows through each of them in streamline conditions. $P_1$ and $P_2$ are pressure differences across the two tubes. If $P_2 = 4P_1$ and $l_2 = \frac{l_1}{4}$,then the radius $r_2$ will be equal to:

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

Two capillary tubes of the same length but different radii $r_1$ and $r_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is the same as before?