JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    Two capillary of lengths \[L\] and \[2L\] and of radii \[R\] and \[2R\] are connected in series. The net rate of flow of fluid through them will be (given rate of the flow through single capillary,\[X=\frac{\pi p{{R}^{4}}}{8\eta L})\]

    A) \[\frac{8}{9}X\]                                

    B) \[\frac{9}{8}X\]

    C)  \[\frac{5}{7}X\]                                               

    D)  \[\frac{7}{5}X\]

    Correct Answer: A

    Solution :

    Fluid resistance is given by\[R=\frac{8\eta L}{\pi {{r}^{4}}}\] When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is                 \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}=\frac{8\eta L}{\pi {{R}^{4}}}+\frac{8\eta \times 2L}{\pi {{(2R)}^{4}}}\]                       \[=\left[ \frac{8\eta L}{\pi {{R}^{4}}}\times \frac{9}{8} \right]\] Equivalent resistance becomes \[\frac{9}{8}\] times so, rate of flow will be\[\frac{8}{9}X\].


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