Q135 / 160 · AP EAMCET 2013 · Physics

The terminal velocity of a liquid drop of radius ' $r$ ' falling through air is $v$. If two such drops are combined to form a bigger drop, the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air)

  • A
    $\sqrt{2} v$
  • B
    $2 v$
  • Correct answer
    C
    $\sqrt[3]{4} v$
  • D
    $\sqrt[3]{2} v$

Solution

Terminal velocity $v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}$ When, the two drops of same radius $r$ coalesce then radius of new drop is $R$. $$ \begin{array}{llrl} & \therefore & \frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\ & \Rightarrow & R & =2^{1 / 3} \cdot r \\ \text { Critical velocity } & \propto r^2 \\ & \therefore & \frac{v}{v_1} & =\frac{r^2}{2^{2 / 3} \cdot r^2} \\ & \Rightarrow & V_1 & =\sqrt[3]{4} \cdot v \end{array} $$

Hence, the correct answer is option: 'C'.

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