Question

Two mercury drops each of radius \(r\) merge to form a bigger drop. If the surface tension of mercury is \(S\), the surface energy released is:

• A. \(1.65\,\pi r^{2} S\)
• B. \(1.33\,\pi r^{2} S\)
• C. \(1.44\,\pi r^{2} S\)
• D. \(1.22\,\pi r^{2} S\)

Hint:

Conserve volume so \(R=2^{1/3}r\), then energy released \(=S(8\pi r^{2}-4\pi R^{2})\).

Answer 1

The Correct Option is A

Step 1: Volume is conserved when the two drops merge. Let \(R\) be the radius of the big drop: \[2\cdot\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}\;\Rightarrow\;R^{3}=2r^{3}\;\Rightarrow\;R=2^{1/3}r.\] Step 2: Surface energy is surface tension times surface area, \(E=S\cdot\text{(area)}\). Initial total area of the two small drops: \[A_i=2\cdot 4\pi r^{2}=8\pi r^{2}.\] Final area of the big drop: \[A_f=4\pi R^{2}=4\pi\left(2^{1/3}r\right)^{2}=4\pi\,2^{2/3}r^{2}.\] Step 3: Energy released is \(S\) times the decrease in area: \[\Delta E=S\left(A_i-A_f\right)=S\,4\pi r^{2}\left(2-2^{2/3}\right).\] Step 4: Compute the number, using \(2^{2/3}=1.587\): \[2-1.587=0.413,\qquad 4\times0.413=1.65.\] Hence \[\Delta E=1.65\,\pi r^{2} S.\] \[\boxed{\Delta E=1.65\,\pi r^{2} S}\]