Question

A big drop is formed by coalescing \[1000\] small drops of water. What will be the change in surface energy? What will be the total surface energy of the droplets and the surface energy of the big drop?

Answer 1

Hint: We know that when a big drop is split in small n droplets, the total volume of all the n small droplets after splitting will be equal to the volume of the bigger drop. When a bigger droplet splits into smaller droplets, Surface energy reduces and as a result energy is released. Similarly, when multiple smaller droplets are joined together to form one single bigger drop, Surface energy increases and as a result energy is absorbed.

Complete answer:
As we know, the ratio of the energy of the big drops to the energy of the small drops is the formula that can be used to find the ratio of the initial surface energy of the droplets to the final surface energy of the single droplet. The one thousand similar drops are connected to make a big drop. Now this big water drop is formed by the combination of n small equal radius water drops. Let the radius of the small water drop be r. Now as we know that the water drops are always in spherical shape, so the volume of the n small water drops is equal to the volume of the big water drops. As we know, the volume of a big drop is the volume of \[1000\] small droplets.
$\dfrac{4}{3}\times \pi {{R}^{3}}=100\times \dfrac{4}{3}\pi {{r}^{3}}\Rightarrow R=10r$
Thus, $r=\dfrac{R}{10}$ let T be the surface tension of water, then surface tension energy of $1000$ droplets is $1000\left( T\times 4\pi {{r}^{2}} \right)=1000\left[ T\times 4\pi {{\left( \dfrac{R}{10} \right)}^{2}} \right]=10\left( T\times 4\pi {{R}^{2}} \right).$ Here the surface tension energy of big drop is $T\times 4\pi {{R}^{2}}.$ Thus, the surface energy will decrease.
$\therefore \dfrac{Total\text{ }surface\text{ }energy\text{ }of\text{ }1000\text{ }droplets}{Surface\text{ }energy\text{ }of\text{ }the\text{ }big\text{ }drop}=\dfrac{10\left( T\times 4\pi {{R}^{2}} \right)}{T\times 4\pi {{R}^{2}}}=\dfrac{10}{1}.$
Therefore, it will decrease to \[{{\dfrac{1}{10}}^{th}}\] or \[1/{{10}^{th}}\] of its previous value.

Note:
Remember that the sign has to be omitted. Another thing to note is that in the question, we have neglected the viscosity of the drops since no energy is lost in that and we assumed all the energy loss is because of the change in surface tension of the system.