Question

Two water drops each of radius \( r \) coalesce to form a bigger drop. If \( T \) is the surface tension, the surface energy released in this process is:

• A. \( 4 \pi r^2 T \)
• B. \( 8 \pi r^2 T \)
• C. \( 12 \pi r^2 T \)
• D. \( 6 \pi r^2 T \)

Hint:

When two drops coalesce, the volume remains constant, but the surface area decreases, which results in a release of surface energy.

Answer 1

The Correct Option is B

The surface energy of a single drop is calculated as \( 4 \pi r^2 T \), with \( r \) denoting the radius and \( T \) the surface tension. Initially, two drops exist, resulting in a total surface energy of \( 2 \times 4 \pi r^2 T = 8 \pi r^2 T \). Upon coalescing, the new drop's radius increases to \( \sqrt{2}r \). Consequently, the surface energy of this combined drop is \( 4 \pi (\sqrt{2}r)^2 T = 8 \pi r^2 T \). The surface energy released is the disparity between the initial and final surface energies, which is \( 8 \pi r^2 T \). Therefore, the correct value is \( 8 \pi r^2 T \).