Question

Narrow capillary tubes A and B are of length l and l/2 and of the same radius r. The rate of flow of water through the tube A under a constant pressure head 'P' is 3 cm³/sec. If A and B are connected in series and the same pressure difference 'P' is maintained between the ends of the composite tube, then the rate of flow of water is

• (A) 1.5 cm³/s
• (B) 3 cm³/s
• (C) 2 cm3/s
• (D) None of these

Answer 1

The Correct Option is C

Explanation:
The volumetric flow is given by:$$\frac{V}{t}\left(\frac{\pi r^4}{8\eta l}\right)P$$The resistance of the pipe is $R=\frac{8\eta l}{\pi r^4}$For the first pipe, length is 1 and for the second pipe, length is $\frac{t}{2}$The resistance of the combination in series is:$$\begin{array}{r}{R}_s=R_1+R_2=\frac{8\eta l}{\pi r^4}+\frac{8\eta }{\pi r^4}\left(\frac{l}{2}\right)\\ R_s=\frac{3}{2}\times \frac{8\eta L}{\pi r^4}\Rightarrow R_s=\frac{3}{2}R\end{array}$$The new volumetric flow is $\mathrm{V}$$$\frac{V^{\prime }}{V}=\frac{R}{R_s}=\frac{2}{3}$$$$V^{\prime }=\frac{2}{3}V=\frac{2}{3}\times 3=2{\mathrm{cm}}^3/\mathrm{s}$$