$A$ capillary tube is connected to the bottom of a container. If its radius is increased by $10\%$,what is the percentage change in the rate of flow of the liquid (in $\%$)?

$A$ capillary tube of length $l$ and radius $r$ is connected to the bottom of a container. When the pressure difference across it is $P$,the volume of water flowing out per second is $V$. If another capillary tube of the same length but half the radius is connected in series with the first one,what will be the new volume of liquid flowing out per second? (The total pressure difference across the system is $P$.)

The flow rate of water from a tap of diameter $1.25 \text{ cm}$ is $3 \text{ litres per min}$. If the coefficient of viscosity of water is $10^{-3} \text{ Pa-s}$,the nature of the flow is:

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

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.

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