Q.$1000$ identical small spherical water droplets, each of radius $1.0\times10^{-4}\ \text{m}$, coalesce (join together) to form a single large spherical drop, with no change in the total volume of water. Taking the surface tension of water as $0.072\ \text{N/m}$, calculate

Given: $n = 1000$, $r = 1.0\times10^{-4}\ \text{m}$, $T = 0.072\ \text{N/m}$.

1. Radius of the large drop, from conservation of volume, $n\times\tfrac43\pi r^3 = \tfrac43\pi R^3$, so $R = n^{1/3}r$:

   $R = (1000)^{1/3}\times(1.0\times10^{-4}) = 10\times1.0\times10^{-4} = 1.0\times10^{-3}\ \text{m}$

2. Energy released. Total surface area before coalescence:

   $A_{\text{before}} = n\times4\pi r^2 = 1000\times4\pi\times(1.0\times10^{-4})^2 = 1.2566\times10^{-4}\ \text{m}^2$

   Surface area of the single large drop after coalescence:

   $A_{\text{after}} = 4\pi R^2 = 4\pi\times(1.0\times10^{-3})^2 = 1.2566\times10^{-5}\ \text{m}^2$

   Decrease in total surface area: $\Delta A = A_{\text{before}} - A_{\text{after}} = 1.2566\times10^{-4} - 1.2566\times10^{-5} = 1.1310\times10^{-4}\ \text{m}^2$ …

Use conservation of volume to find the radius of the merged drop ($R = n^{1/3}r$), compute the total surface area before and after coalescence, and multiply the decr …

• Using $R = nr$ instead of $R = n^{1/3}r$ -- volume, not radius, is conserved, and volume scales as the cube of the radius. …