Physics › Kinetic Theory of Gases and Radiation › Pressure and Energy of gas
' $n$ ' number of liquid drops each of radius ' $r$ ' coalesce to form a single drop of radius ' $R$ '. The…
' $n$ ' number of liquid drops each of radius ' $r$ ' coalesce to form a single drop of radius ' $R$ '. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is [T = surface tension of liquid, $\rho=$ density of liquid.]
$\sqrt{\frac{\mathrm{T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$ $\sqrt{\frac{2 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$ $\sqrt{\frac{4 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$ $\sqrt{\frac{6 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$ Solution
$\begin{aligned}
& \frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3... (given) \\
\therefore \quad & R^3=n r^3
\end{aligned}$
Energy released,
$\begin{aligned}
\Delta \mathrm{U} & =\mathrm{T} \times 4 \pi \mathrm{r}^2 \times \mathrm{n}-\mathrm{T} \times 4 \pi \mathrm{R}^2 \\
& =\mathrm{T} \times 4 \pi \frac{\mathrm{R}^3}{\mathrm{r}}-\mathrm{T} \times 4 \pi \mathrm{R}^2
\end{aligned}$
This energy is converted into K.E
$\begin{array}{ll}
\therefore \quad & \frac{1}{2} m v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\
\Rightarrow & \frac{1}{2} \rho \times \frac{4}{3} \pi R^3 \times v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\
& v^2=\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \\
\therefore \quad v & =\sqrt{\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}
\end{array}$
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Asked in: MHT CET 2023 (10 May Shift 2)
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