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JCECE Engineering JCECE Engineering 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                 \[{{v}_{T}}=\frac{2}{9}\cdot \frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] 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,\[r=20\mu rn=20\times {{10}^{-6}}m,\] \[\rho =2000\,\,kg/{{m}^{3}}\] \[\sigma =1000\,\,kg/{{m}^{3}},\,\,g=9.8\,\,m/{{s}^{2}},\,\,\eta =1\times {{10}^{-3}}Pa\] \[\therefore \]\[{{v}_{T}}=\frac{2}{9}\frac{{{(20\times {{10}^{-6}})}^{2}}(2000-1000)\times 9.8}{1\times {{10}^{-3}}}\] \[\Rightarrow \]               \[{{v}_{T}}=8.7\times {{10}^{-4}}m/s\] \[\therefore \]  \[{{v}_{T}}=0.87\,\,mm/s\]


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