Q. A capillary tube of radius $R$ and length $L$ is connected in series with another tube of radius $R/2$ and length$L/4$. If the pressure difference across the two tubes taken together is $p$, then the ratio of pressure difference across the first tube to that across the second tube is
A) 1 : 4
B) 1 : 1
C) 4 : 1
D) 2 : 1
Explanation
(a) Volume of liquid flowing per second through each of the two tubes in series will be the same. So
$V = \frac{\pi\ p_{1}\ R^{4}}{8\ \eta\ L} = \frac{\pi p_{2}(R/2)^{4}}{8\eta(L/2)}$ or $\frac{p_{1}}{p_{2}} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.