Question

We have two (narrow) capillary tubes $T_{1}$ and $T_{2}$ . Their lengths are $l_{1}$ and $l_{2}$ and radii of the cross-section are $r_{1}$ and $r_{2}$ respectively. The rate of water under a pressure difference $P$ through tube $T_{1}$ is $ \, 8 \, cm^{3} \, s^{- 1}$ . 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 the pressure difference across the combination is same as before(= $P$ )

• A. $4\,cm^{3}s^{- 1}$
• B. $\frac{16}{3} \, cm^{3} \, s^{- 1}$
• C. $\frac{8}{17} \, cm^{3} \, s^{- 1}$
• D. $8 \, cm^{3} \, s^{- 1}$

Answer 1

Answer: (B)

$V=\frac{\pi P r^{4}}{8 \eta l}=8\,cm^{3}s^{- 1}$
For composite tube
$V_{1}=\frac{P \pi r^{4}}{8 \eta \left(\right. l + \frac{l}{2} \left.\right)}=\frac{2}{3}\frac{\pi P r^{4}}{8 \eta l}=\frac{2}{3}\times 8=\frac{1 6}{3}\left(cm\right)^{3}s^{- 1}\left[\therefore l_{1} = l = 2 l_{2} \text{or} l_{2} = \frac{l}{2}\right]$