Q. Under a pressure head, the rate of orderly volume flow of liquid through a capillary tube is $Q$. If the length of capillary tube were doubled and the diameter of the bore is halved, the rate of flow would become
A) $\frac{Q}{4}$
B) $16\ Q$
C) $\frac{Q}{8}$
D) $\frac{Q}{32}$
Explanation
(d) $Q = \frac{\pi\ p\ r^{4}}{8\ \eta\ l}$ and $Q_{1} = \frac{\pi\ p(r/2)^{4}}{8\eta(2l)} = \frac{Q}{32}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.