Q. The rate of steady volume flow of water through a capillary tube of length $l$ and radius $r$, under a pressure difference of $p\ si\ V$. 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)
A) $\frac{V}{16}$
B) $\frac{V}{17}$
C) $\frac{16V}{17}$
D) $\frac{17V}{16}$
Explanation
(b) The rate of flow of water inside a capillary
$V = \frac{\pi\ p\ r^{4}}{8\ n\ l},$
$Pressure\ difference\ \ \ \ \ \ \ p = \frac{V(8\ n\ l)}{\pi\ r^{4}}$
In series combination$\ \ \ p = p_{1} + p_{2}$
$where\ p_{1\ }$and$\ p_{2}\ $are the pressure difference in the two tubes$.$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{V(8\ n\ l)}{\pi\ r^{4}} = \ \frac{V'(8\ n\ l)}{\pi\ r^{4}} + \ \frac{V'(8\ n\ l)}{\pi\ (\pi/2)^{4}}$
In series combination, rate of flow of water$\ \left( V^{'} \right)$will be same in both the tubes.
$\frac{V}{r^{4}} = \frac{V^{'}}{r^{4}} + \frac{V^{'} \times 16}{r^{4}}$
$V = V^{'} + 16V^{'}$
$\ \ V^{'} = \frac{V}{17}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.