Question

Two capillary tubes of same radius r but of lengths $ {{l}_{1}} $ and $ {{l}_{2}} $ are fitted in parallel to the bottom of a vessel. The pressure head is p. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before?

• A. $ {{l}_{1}}+{{l}_{2}} $
• B. $ \frac{{{l}_{1}}{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}} $
• C. $ \frac{1}{{{l}_{1}}+{{l}_{2}}} $
• D. $ \frac{1}{{{l}_{1}}}+\frac{1}{{{l}_{2}}} $

Answer 1

Answer: (B)

For parallel combination
$ \frac{1}{{{R}_{eff}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} $
$ \Rightarrow $ $ \frac{\pi {{r}^{4}}}{8\eta l}=\frac{\pi {{r}^{4}}}{8\eta {{l}_{1}}}+\frac{\pi {{r}^{4}}}{8\eta {{l}_{2}}} $
$ \Rightarrow $ $ \frac{1}{l}=\frac{1}{{{l}_{1}}}+\frac{1}{{{l}_{2}}} $
$ \therefore $ $ l=\frac{{{l}_{1}}{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}} $