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JCECE Medical JCECE Medical Solved Paper-2005

  • question_answer
    Typical silt (hard mud) particle of radius \[20\,\mu m\]is on the top of lake water, its density is \[2000\,kg/{{m}^{3}}\]and the viscosity of lake water is\[1.0\,m\,Pa,\], density is\[1000\,kg/{{m}^{3}}.\] If the lake is still (has no internal fluid motion), the terminal speed with which the particle hits the bottom of the lake in mm/s is:

    A)  0.67           

    B)  0.77

    C)  0.87           

    D)  0.97

    Correct Answer: C

    Solution :

     Terminal velocity \[({{v}_{T}})\] is given by \[{{v}_{T}}=\frac{2}{9}.\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] where, \[\eta \]is coefficient of viscosity,\[\rho \] is density of silt,\[\sigma \]is density of lake .water and g is acceleration due to gravity. Given, \[r=20\mu m=20\times {{10}^{-6}}m,\] \[\rho =2000\,kg/{{m}^{3}},\] \[\sigma =1000\,kg/{{m}^{3}},g=9.8\,m/{{s}^{2}},\,\eta =1\times {{10}^{-3}}Pa\] \[\therefore \] \[{{v}_{T}}=\frac{2}{9}\frac{{{(20\times {{10}^{-6}})}^{2}}(2000-1000)\times 9.8}{1\times {{10}^{-3}}}\] \[\Rightarrow \] \[{{v}_{T}}=8.7\times {{10}^{-4}}m/s\] \[\Rightarrow \] \[{{v}_{T}}=0.87\,mm/s\]


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