AFMC AFMC Solved Paper-2007

  • question_answer
    Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced what will be the new velocity?

    A) \[{{\left( 2 \right)}^{\frac{1}{3}}}v\]                                        

    B) \[{{\left( 2 \right)}^{\frac{3}{2}}}v\]

    C) \[{{\left( 2 \right)}^{\frac{2}{3}}}v\]                                        

    D) \[{{\left( 2 \right)}^{\frac{1}{4}}}v\]

    Correct Answer: C

    Solution :

    Key Idea: Rain drops attain a constant velocity, called as terminal velocity. Let r be the radius of the each drop. The terminal velocity of drop will be given by \[v=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\]    ?(i) where\[\rho \]is density of drop and a is density of viscous medium of coefficient of viscosity \[\eta .\]When two drops each of radius r coalesce to form a new drop, then the radius of coalesced drop will be \[R={{(2)}^{\frac{1}{3}}}r\] Hence, new terminal velocity of coalesced drop will be \[v'=\frac{2}{9}\left[ \frac{{{({{2}^{1/3}}r)}^{2}}(\rho -\sigma )g}{\eta } \right]\]                ?(ii) From Eqs. (i) and (ii), we get \[\frac{v'}{v}={{(2)}^{\frac{2}{3}}}\]or     \[v'={{(2)}^{\frac{2}{3}}}v\]


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