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

Track the change in total surface area using the ratio of radii.

Radius of merged drop. Equal volumes add, so \[R^{3}=2r^{3},\qquad \frac{R}{r}=2^{1/3}\approx1.26.\] Surface energy formula. For a sphere the surface energy is $E=S\cdot 4\pi(\text{radius})^{2}$.

Before. Two drops: \[E_i=2\times S\,4\pi r^{2}=8\pi r^{2}S.\] After. One drop: \[E_f=S\,4\pi R^{2}=4\pi r^{2}S\left(\frac{R}{r}\right)^{2}=4\pi r^{2}S\,(2^{1/3})^{2}=4\pi r^{2}S\times1.587.\] So $E_f=6.35\,\pi r^{2}S$.

Released energy. Since the merged drop has less surface area, energy is released: \[\Delta E=E_i-E_f=(8-6.35)\pi r^{2}S=1.65\,\pi r^{2}S.\] The surface shrinks on merging, so this energy comes out as heat. \[\boxed{\Delta E=1.65\,\pi r^{2}S}\]