BCECE Medical BCECE Medical Solved Papers-2003

  • question_answer
    A rain drop of radius 0.3 mm has a terminal velocity of 1 m/s and the viscosity of 1 m/s and the viscosity of air is \[18\times {{10}^{-5}}\] poise. The viscous force on the drop is :

    A)  \[16.95\times {{10}^{-9}}N\]

    B)  \[1.695\times {{10}^{-9}}N\]

    C)  \[10.17\times {{10}^{-9}}N\]

    D)  \[101.74\times {{10}^{-9}}N\]

    Correct Answer: D

    Solution :

    For a drop of radius r and terminal velocity v, the viscous force is given by \[F=6\pi \,\eta rv\] where \[\eta \] is coefficient of viscosity. Putting the numerical values from the question, we have \[\eta =18\times {{10}^{-5}}\] poise \[=18\times {{10}^{-6}}kg/m-s\] \[r=0.3\,mm=0.3\times {{10}^{-3}}m,\,\,v=1\,\,m/s\] \[=101.74\times {{10}^{-9}}N\]


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