Q107 / 120 · VITEEE 2018 · Physics

A large number of liquid drops each of radius $\mathrm{r}$ coalesce to from a single drop of radius $\mathrm{R}$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given, surface tension of liquid $\mathrm{T}$, density $\rho$ )

  • A
    $\sqrt{\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}$
  • B
    $\sqrt{\frac{2 \mathrm{~T}}{\rho}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)}$
  • C
    $\sqrt{\frac{4 \mathrm{~T}}{\rho}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)}$
  • Correct answer
    D
    $\sqrt{\frac{6 \mathrm{~T}}{\rho}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)}$

Solution

When drops combine to form a single drop of radius $\mathrm{R}$. Then energy released, $$ \mathrm{E}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] $$ If this energy is converted into kinetic energy then $$ \begin{array}{l} \frac{1}{2} \mathrm{mv}^{2}=4 \pi \mathrm{R}^{3} \mathrm{~T}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \frac{1}{2} \times\left[\frac{4}{3} \pi \mathrm{R}^{3} \rho\right] \mathrm{v}^{2}=4 \pi \mathrm{R}^{3} \mathrm{~T}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \mathrm{v}=\sqrt{\frac{6 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]} \end{array} $$

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

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