Q. We have two (narrow) capillary tubes $T_{1}$ and $T_{2}$. Their lengths are $l_{1}$ and $l_{2}$ and radii of cross-section are $r_{1}$ and $r_{2}$ respectively. The rate of flow of water under a pressure difference $P$ through tube $T_{1}$ is $8cm^{3}/sec$. If $l_{1} = 2l_{2}$ and $r_{1} = r_{2}$, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before $( = P)$
A) $4\ cm^{3}/sec$
B) $(16/3)cm^{3}/sec$
C) $(8/17)cm^{3}/sec$
D) None of these
Explanation
(b) $V = \frac{\pi Pr^{4}\ }{8\eta l} = \frac{8cm^{3}}{\sec}$
For composite tube
$V_{1} = \frac{P\pi r^{4}}{8\eta\left( l + \frac{l}{2} \right)} = \frac{2}{3}\frac{\pi Pr^{4}}{8\eta l} = \frac{2}{3} \times 8 = \frac{16}{3}\frac{cm^{3}}{\sec}$
$\left\lbrack \because l_{1} = l = 2l_{2}\ or\ l_{2} = \frac{l}{2}\ \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.