A horizontal composite capillary tube has a radius ‘2r’ for a length ‘2L’ and radius ‘r’ for a length ‘L’ …

Q: A horizontal composite capillary tube has a radius ‘2r’ for a length ‘2L’ and radius ‘r’ for a length ‘L’ and is connected to a tank at one end and left free at the other end. The tank contains a liquid of coefficient of viscosity ‘η’. If a constant pressure difference ‘P’ exists across the ends of the capillary tube, the volume flux through the capillary tube is

(a) $ \displaystyle \frac{16}{17} \frac{\pi P r^4}{8 \eta L} $

(b) $ \displaystyle \frac{9}{8} \frac{\pi P r^4}{8 \eta L} $

(c) $ \displaystyle \frac{17}{16} \frac{\pi P r^4}{8 \eta L} $

(d) $ \displaystyle \frac{8}{9} \frac{\pi P r^4}{8 \eta L} $

Ans: (d)

Sol: Rate of flow is

$ \displaystyle Q = \frac{P \pi r^4}{8 \eta l} $

$ \displaystyle Q = \frac{P }{8 \eta l/\pi r^4} = \frac{P}{R}$

Where , R = Fluid Resistance

Now , Equivalent Resistance ,

$ \displaystyle R_{eq} = R_1 + R_2 $

$ \displaystyle = \frac{8 \eta (2L)}{\pi (2r)^4} + \frac{8 \eta L}{\pi r^4} $

$ \displaystyle = \frac{8 \eta L}{\pi r^4} [1/8 + 1] $

$ \displaystyle R_{eq} = \frac{8 \eta L}{\pi r^4} (\frac{9}{8}) $

Hence , Rate of Flow is

$ \displaystyle = \frac{P }{R_{eq}} $

$ \displaystyle = \frac{8}{9} \frac{P \pi r^4}{8 \eta L} $