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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    Typical silt (hard mud) particle of radius 20 um is on the top of lake water, its density is \[2000\text{ }kg/{{m}^{3}}\]and the viscosity of lake water is 1.0 mPa, 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 is ... mm/s.

    A)  0.67                      

    B)         0.77                      

    C)  0.87                      

    D)         0.97

    E)  1.07

    Correct Answer: C

    Solution :

    \[v=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] \[=\frac{2}{9}\times \frac{{{(20\times {{10}^{-6}})}^{2}}(2000-1000)\times 9.8}{1.0\times {{10}^{-3}}}\] \[=8.7\times {{10}^{-4}}m/s\] \[=0.87\text{ }mm/s\]


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