Q. Two capillaries of same length and radii in the ratio 1:2 are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is $1\ m$ of water, the pressure difference across first capillary of
A) $9.4\ m$
B) $4.9\ m$
C) $0.49\ m$
D) $0.94\ m$
Explanation
(d) Given, $l_{1} = l_{2} = 1$, and $\frac{r_{1}}{r_{2}} = \frac{1}{2}$
$V = \frac{\pi P_{1}r_{1}^{4}}{8\eta l} = \frac{\pi P_{2}r_{2}^{4}}{8\eta l} \Rightarrow \frac{P_{1}}{P_{2}} = \left( \frac{r_{2}}{r_{1}} \right)^{4} = 16 \Rightarrow P_{1} = 16P_{2}$
Since both tubes are connected in series, hence pressure difference across combination
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.