Question

The surface energy of a liquid drop is $u$. It is sprayed into $1000$ equal droplets. Then find out its surface energy.

Answer 1

Hint
According to the question there is a liquid drop who’s surface energy is u and it is sprayed in 1000 equal droplets. If we increase the free surface area of a liquid then work has to be done against the force of surface tension.

Complete step by step answer
according to the surface energy formula
The initial surface energy
S.E.= surface tension * surface area
$S.{E_i} = \sigma \times 4\pi {R^2}$ ................[ R is equal to radius of the big drop]
Surface energy of the big drop is u
So, $u = \sigma \times 4\pi {R^2}$ ............equation 1
Now, the final surface energy is,
$S.{E_f} = \sigma \times 4\pi {r^2}$ .........[r=radius of the small drops]........equation 2
So , for the 1000 equal droplets the equation is,
$S.{E_f} = \sigma \times 4\pi {r^2} \times 1000$ ...............equation3
Now, the volume of the big drop and 1000 small droplets are equal
$\dfrac{4}{3}\pi {R^3} = 1000 \times \pi {r^3}$
${R^3} = 1000 \times {r^3}$
$r = {(\dfrac{{{R^3}}}{{1000}})^{\dfrac{1}{3}}}$
$\Rightarrow r = \dfrac{R}{{10}}$
Now put the value on equation 3
$S.{E_f} = \sigma \times 4\pi {(\dfrac{R}{{10}})^2} \times 1000$
$\Rightarrow S.{E_f} = \sigma \times 4\pi {R^2} \times \dfrac{{1000}}{{100}}$
$\Rightarrow S.{E_f} = u \times 10$
So final surface energy is $10u$.

Note
The surface energy may therefore be defined as the excess energy at the surface of a material compared to the bulk, or it is the work required to build an area of a particular surface. Another way to view the surface energy is to. relate it to the work required to cut a bulk sample, creating two surfaces.