Water flows in a streamlined manner through a capillary of radius $a$,the pressure difference being $p$ and the rate of flow $Q$. If the radius is reduced to $a/2$ and the pressure is increased to $4p$,the rate of flow becomes:

Water is conveyed through a uniform tube of $8 \text{ cm}$ in diameter and $3140 \text{ m}$ in length at the rate of $2 \times 10^{-3} \text{ m}^3/\text{s}$. The pressure required to maintain the flow is (Viscosity of water $= 10^{-3} \text{ SI units}$):

The rate of steady volume flow of water through a capillary tube of length $l$ and radius $r$,under a pressure difference of $p$ is $V$. This tube is connected with another tube of the same length but half the radius,in series. Then,the rate of steady volume flow through them is (The pressure difference across the combination is $p$.)

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

$A$ fluid of density $\rho$ is flowing through a uniform tube of diameter $d$. The coefficient of viscosity of the fluid is $\eta$. The critical velocity of the fluid is: