When a liquid of density $\rho$ flows through a tube of diameter $d$ with critical velocity $V$,then the Reynolds number is (where $\eta$ is the coefficient of viscosity of the liquid).

$A$ 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:

$A$ liquid flows through a pipe with a pressure difference of $P$. What should be the pressure difference required to maintain the same flow rate if the radius is doubled and the length is doubled?

We have two narrow capillary tubes $T_1$ and $T_2$. Their lengths are $l_1$ and $l_2$ and radii of cross-section are $r_1$ and $r_2$ respectively. The rate of flow of water under a pressure difference $P$ through tube $T_1$ is $8 \ cm^3/sec$. If $l_1 = 2l_2$ and $r_1 = r_2$,what will be the rate of flow when the two tubes are connected in series and the pressure difference across the combination is the same as before $(= P)$?

If the pressure difference across a pipe of radius $r$ and length $l$ is $p$,then the volume of liquid flowing out per second is $V$. Find the correct expression for $V$.

Which of the following quantities has the dimensions of $\frac{\pi P r^4}{8 Q l}$? (Where $P =$ pressure,$r =$ radius,$Q =$ volume flow rate in $m^3/s$,and $l =$ length.)