Question

Two mercury drops, each with same radius \( r \), merged to form a bigger drop. If \( T \) is the surface tension of mercury, then the surface energy of the bigger drop is given by

• A. \( 2\pi r^2 T \)
• B. \( \frac{5}{3} \pi r^2 T \)
• C. \( 2\pi r^2 T^2 \)
• D. \( \frac{8}{3} \pi r^2 T \)

Hint:

When merging drops, the total volume is conserved, but the surface area changes. The resulting surface area will generally be less than the sum of the surface areas of the individual smaller drops.

Answer 1

The Correct Option is D

The surface area of a spherical drop is given by \( 4\pi r^2 \). When two drops of radius \( r \) combine, the volume of the new drop is the sum of the volumes of the two smaller drops: \[ V = \frac{4}{3}\pi r^3 + \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3 \] The radius \( R \) of the new drop is found by equating its volume to that of a sphere: \[ \frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3 \quad \Rightarrow \quad R = r\sqrt[3]{2} \] The surface area of the new drop is: \[ 4\pi R^2 = 4\pi (r\sqrt[3]{2})^2 = \frac{8}{3}\pi r^2 \] Thus, the surface energy \( E \) of the bigger drop is: \[ E = {Surface Area} \times T = \frac{8}{3}\pi r^2 T \]