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Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?
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Verified Answer
The correct answer is:
$2^{2 / 3}: 1$
Volume remains constant after coalescing. Thus,
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3}
$$
where $R$ is radius of bigger drop and $r$ is radius of each smaller drop.
$\therefore$
$$
R=2^{1 / 3} r
$$
Now, surface energy per unit surface area is the surface tension. So,
Surface energy, $W=T \Delta A$
Or
$$
W=4 \pi R^{2} T
$$
Therefore, surface energy of bigger drop
$$
\begin{aligned}
W_{1} &=4 \pi\left(2^{1 / 3} r\right)^{2} T \\
&=\left(2^{2 / 3}\right) 4 \pi r^{2} T
\end{aligned}
$$
Surface energy of smaller drop
$$
W_{2}=4 \pi r^{2} T
$$
Hence, required ratio
$$
\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1
$$
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3}
$$
where $R$ is radius of bigger drop and $r$ is radius of each smaller drop.
$\therefore$
$$
R=2^{1 / 3} r
$$
Now, surface energy per unit surface area is the surface tension. So,
Surface energy, $W=T \Delta A$
Or
$$
W=4 \pi R^{2} T
$$
Therefore, surface energy of bigger drop
$$
\begin{aligned}
W_{1} &=4 \pi\left(2^{1 / 3} r\right)^{2} T \\
&=\left(2^{2 / 3}\right) 4 \pi r^{2} T
\end{aligned}
$$
Surface energy of smaller drop
$$
W_{2}=4 \pi r^{2} T
$$
Hence, required ratio
$$
\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1
$$
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