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A certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, which one of the following statements is true for the energy $(E)$ in the process?
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Verified Answer
The correct answer is:
$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is released.
Change in surface energy is given by
$E=T(\Delta A) \quad---(1)$
The initial area is given by,
$\mathrm{A}=\left(4 \pi r^2\right) n$
The final area is given by,
$\mathrm{a}=4 \pi R^2$
Therefore, the change in area is given by,
$\begin{aligned}
& \Delta A=a-A \\
& \Rightarrow \Delta A=4 \pi\left(n r^2-R^2\right)---(2)
\end{aligned}$
Now, using volume conservation: $\left(\frac{4}{3} \pi r^2\right) n=\frac{4}{3} R^3$
$\therefore n=\frac{R^3}{r^3} \quad---(3)$
$\Delta A=4 \pi\left[\frac{R^3}{r^3} \cdot r^2-R^2\right]=4 \pi\left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4}{3} \pi R^3\right) 3\left[\frac{1}{r}-\frac{1}{R}\right]=3 V\left[\frac{1}{r}-\frac{1}{R}\right]$
Introducing above value in equation (1)
$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$
$E=T(\Delta A) \quad---(1)$
The initial area is given by,
$\mathrm{A}=\left(4 \pi r^2\right) n$
The final area is given by,
$\mathrm{a}=4 \pi R^2$
Therefore, the change in area is given by,
$\begin{aligned}
& \Delta A=a-A \\
& \Rightarrow \Delta A=4 \pi\left(n r^2-R^2\right)---(2)
\end{aligned}$
Now, using volume conservation: $\left(\frac{4}{3} \pi r^2\right) n=\frac{4}{3} R^3$
$\therefore n=\frac{R^3}{r^3} \quad---(3)$
$\Delta A=4 \pi\left[\frac{R^3}{r^3} \cdot r^2-R^2\right]=4 \pi\left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4}{3} \pi R^3\right) 3\left[\frac{1}{r}-\frac{1}{R}\right]=3 V\left[\frac{1}{r}-\frac{1}{R}\right]$
Introducing above value in equation (1)
$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$
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