Q. The rate of flow of liquid through a capillary tube of radius $r$ is $V$ when the pressure difference across the two ends of the capillary is $p$. If pressure is increased by 3 $p$ and radius is reduced to $r/2$, then the rate of flow becomes
A) $V/9$
B) $3V/8$
C) $V/4$
D) $V/3$
Explanation
(c) $V = \frac{\pi\ p\ r^{4}}{8\eta l}$ and $V^{'} = \frac{\pi(3p + p)(r/2)^{4}}{8\eta l}$
$\therefore\frac{V^{'}}{V} = 4 \times (1/2)^{4} = \frac{1}{4}$ or $V^{'} = \frac{V}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.