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?

Two different liquids are flowing in two tubes of equal radius. The ratio of coefficients of viscosity of liquids is $52:49$ and the ratio of their densities is $13:1$,then the ratio of their critical velocities will be

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

$A$ liquid of density $10^3 \, kg/m^3$ and coefficient of viscosity $8 \times 10^{-2} \, \text{decapoise}$ is flowing in a tube of radius $2 \, cm$ with speed $2 \, m/s$. The Reynold's number is ..........

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

What is inertial force in the context of fluid dynamics?