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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are\[\rho \]and\[\sigma \]respectively, and the viscosity. of the liquid is\[\eta \]):

    A)  \[\frac{{{r}^{2}}g}{9\eta }(\rho -2\sigma )\]

    B)         \[\frac{{{r}^{2}}g}{9\eta }(2\rho -\sigma )\]

    C)  \[\frac{{{r}^{2}}g}{9\eta }(\rho -\sigma )\]          

    D)         \[\frac{2{{r}^{2}}g}{9\eta }(\rho -\sigma )\]

    E)  \[\frac{{{r}^{2}}g}{18\eta }(\rho -2\sigma )\]

    Correct Answer: A

    Solution :

    Net force on the ball        = downward force - upward force\[=\frac{mg}{2}\] \[\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )-6\pi \eta rv=\frac{mg}{2}\] \[\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g-6\pi \eta rv=\frac{1}{2}\left( \frac{4}{3}\pi {{r}^{3}}\rho  \right)g\] \[{{r}^{2}}(\rho -\sigma )g-\frac{9}{2}\eta v=\frac{1}{2}{{r}^{2}}\rho g\] \[\frac{9}{2}v\eta ={{r}^{2}}(\rho -\sigma )g-\frac{1}{2}{{r}^{2}}\rho g\] \[=\frac{1}{2}{{r}^{2}}\rho g-{{r}^{2}}\sigma g\] \[\frac{9}{2}v\eta =\frac{1}{2}{{r}^{2}}g(\rho -2\sigma )\] \[v=\frac{{{r}^{2}}g}{9\eta }(\rho -2\sigma )\]


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