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JCECE Engineering JCECE Engineering 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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