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MHT CET · Physics · Thermodynamics

Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

  1. A \(2^{1 / 2}: 1\)
  2. B \(1: 1\)
  3. C \(2^{2 / 3}: 1\)
  4. D None of these
Verified Solution

Answer & Solution

Correct Answer

(C) \(2^{2 / 3}: 1\)

Step-by-step Solution

Detailed explanation

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
\(
\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1
\)