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JCECE Medical JCECE Medical Solved Paper-2009

  • question_answer
    An air bubble of radius 1 cm rises from the bottom portion through a liquid of density \[1.5~g/cc\]at a constant speed of \[0.25\,cm\,{{s}^{-1}}.\]If the density of air is neglected, the coefficient of viscosity of the liquid is approximately (in Pa)

    A)  13000           

    B)  1300

    C)  130              

    D)  13

    Correct Answer: C

    Solution :

     \[v=\frac{2}{9}\frac{{{r}^{2}}\rho g}{\eta }\] \[\Rightarrow \] \[\eta =\frac{2}{9}.\frac{{{r}^{2}}\rho g}{v}\] \[=\frac{2}{9}\frac{{{(1\times {{10}^{-2}})}^{2}}\times (1.5\times {{10}^{3}})\times 9.8}{0.25\times {{10}^{-2}}}\] \[=130Pa-s\]


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