Two capillary tubes of same radius $r$ but of lengths $l_1$ and $l_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the length of a single tube of the same radius $r$ that can replace the two tubes so that the rate of flow is same as before?

When a liquid flows through a tube,the Reynolds number is $900$. The flow of the liquid is:

What is the value of Reynolds number for streamline flow?

$A$ highly viscous liquid of viscosity coefficient $\eta$ flows through a fixed horizontal cylindrical tube (fixed from the outer surface) of internal radius $r$,thickness $t$ $(t \ll r)$,and length $l$. The volume of liquid flowing per second is $Q$ and the pressure difference across the tube is $P$. If the modulus of rigidity of the material of the tube is $\beta$,the shear strain in the tube will be:

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 $p$.)

Consider an increase of $1 \%$ in each of the radius of the artery, the viscosity of blood, and the density of blood, respectively. The percentage change in the flow rate of blood in the artery is (in $\%$)