Q. A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $P$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is
A) $P$
B) $\frac{3P}{4}$
C) $\frac{P}{2}$
D) $\frac{P}{4}$
Explanation
(d) From $V = \frac{P\pi r^{4}}{8\eta l} \Rightarrow P = \frac{V8\eta l}{\pi r^{4}}$
$\Rightarrow \frac{P_{2}}{P_{1}} = \frac{V_{2}}{V_{1}} \times \frac{l_{2}}{l_{1}} \times \left( \frac{r_{1}}{r_{2}} \right)^{4} = 2 \times 2 \times \left( \frac{1}{2} \right)^{4} = \frac{1}{4}$
$\Rightarrow P_{2} = \frac{P_{1}}{4} = \frac{P}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.