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JCECE Medical JCECE Medical Solved Paper-2012

  • question_answer
    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)  \[\frac{1}{{{l}_{1}}}+\frac{1}{{{l}_{2}}}\]

    Correct Answer: B

    Solution :

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


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