Q. 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 that can replace the two tubes so that the rate of flow is same as before
A) $l_{1} + l_{2}$
B) $\frac{1}{l_{1}} + \frac{1}{l_{2}}$
C) $\frac{l_{1}l_{2}}{l_{1} + l_{2}}$
D) $\frac{1}{l_{1} + l_{2}}$
Explanation
(c) For parallel combination $\frac{1}{R_{eff}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\Rightarrow \frac{\pi r^{4}}{8\eta l} = \frac{\pi r^{4}}{8\eta l_{1}} + \frac{\pi r^{4}}{8\eta l_{2}} \Rightarrow \frac{1}{l} = \frac{1}{l_{1}} + \frac{1}{l_{2}}\ \therefore l = \frac{l_{1}l_{2}}{l_{1} + l_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.