JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन Question Bank Fluid Flow

  • question_answer
    Two capillary of length L and 2L and of radius 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, \[X=\pi P{{R}^{4}}/8\eta L)\]                                              [DCE 2005]

    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                    \[{{R}_{e}}={{R}_{1}}+{{R}_{2}}=\frac{8\eta L}{\pi {{r}^{4}}}+\frac{8\eta \times 2L}{\pi {{(2R)}^{4}}}=\left( \frac{8\eta L}{\pi {{r}^{4}}} \right)\times \frac{9}{8}\]            Equivalent resistance becomes \[\frac{9}{8}\]times so rate of flow will be \[\frac{8}{9}X\]


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