Two capillaries of same length and radii in the ratio $1:2$ are connected in series. $A$ liquid flows through them in a streamlined condition. If the pressure across the two extreme ends of the combination is $1 \ m$ of water,the pressure difference across the first capillary is...... $m$.

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 $= 10^{-3} \text{ SI units}$):

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

In deriving Bernoulli's equation,we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
$(a)$ What is the largest average velocity of blood flow in an artery of diameter $2 \times 10^{-3} \;m$ if the flow must remain laminar?
$(b)$ Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

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

$A$ fluid of density $\rho$ and viscosity $\eta$ is flowing through a pipe of diameter $d$, with a velocity $v$. The Reynolds number is: