Question

A water drop of radius ${10^{ - 2}}\,m$ is broken into $1000$ equal droplets.Calculate the gain in surface energy. Surface tension of water is $0.075\,N\,{m^{ - 1}}$.

Answer 1

In order to answer this question, to find the gain in energy or the work done, first we will find the radius of 1000 droplets by equating the formulas of volume of spherical droplets. And then we will apply the formula for work done.

Formula used:
Work done or gain in energy,
$W = T\Delta A$

Complete step by step answer:
Radius of water droplets $= {10^{ - 2}}m$. As we know that- water droplets are spherical shaped. So, here we will apply the formula of volume of sphere-
$\text{Volume of 1 droplet = Volume of 1000 droplets} \\\ \Rightarrow \dfrac{4}{3}\pi {R^3} = 1000 \times \dfrac{4}{3}\pi {r^3} \\\$
where, $R$ is the radius of 1 droplet.
And we have to find the value of $r$ -
$\Rightarrow {({10^{ - 2}})^3} = 1000{r^3} \\\ \Rightarrow r = {10^{ - 3}}m \\\$
Now, as we have the value of tension.
Surface tension is given, $T = 0.075\,N\,{m^{ - 1}}$
So, apply work-done formula here-
$\because W = T.\Delta A \\\ \Rightarrow W = 0.075[n \times 4\pi {r^2} - 4\pi {R^2}] \\\ \Rightarrow W = 0.075[1000 \times {({10^{ - 3}})^2} - ({10^{ - 2}})] \times 4\pi \\\ \Rightarrow W = 0.075 \times [{10^{ - 3}} - {10^{ - 4}}] \times 4\pi \\\ \therefore W = 8.5 \times {10^{ - 4}}J \\\$
Hence, the gain in energy is the work done, i.e. $8.5 \times {10^{ - 4}}J$.

Note: No energy is "lost" with a conservative force; it is all stored. As a result, work done by conservative forces and potential energy can be considered equivalent and equal. However, keep them separate in your mind as distinct terms and actual quantities.