Question

One drop of soap bubble of diameter$D$breaks into 27 drops having surface tension $\sigma .$The change in surface energy is?
(A) $2\pi \sigma D^2$
(B) $4\pi \sigma D^2$
(C) $8\pi \sigma D^2$
(D) $\pi \sigma D^2$

Answer 1

Hint: As one drop breaks into number of drops then by volume conservation. We can conclude that, volume of one drop of bubble will be equal to the sum of volumes of 27 drops of soap bubbles.
Formula used:
 $\mathrm{\Delta }E=T.\mathrm{\Delta }A$
Here, $\mathrm{\Delta }E$ is the change in energy
$T$ is the surface tension
$A$ is the cross-sectional area
Complete step by step answer:
Soap bubble is of spherical shape and the volume of sphere is ${\displaystyle \frac{4}{3}}\pi r^3$
Since the volume of larger drop is equal to the sum of volumes of the smaller drops
${\displaystyle \frac{4}{3}}\pi R^3=n{\displaystyle \frac{4}{3}}\pi r^3$
Where, $R$ is radius of large drop
$r$ is radius of small drops
$n=27$is the number of drops
Put these values in the above equation
$\Rightarrow {\displaystyle \frac{4}{3}}\pi R^3=$$27\times {\displaystyle \frac{4}{3}}\pi r^3$
Common factors will cancel each other
$\Rightarrow R=27{}^{{\displaystyle \frac{1}{3}}}\times r$
Simplifying and rearranging the equation, we get
$r={\displaystyle \frac{R}{3}}$ . . . (1)
We know that the surface tension is equal to the surface energy per unit area.
$\therefore$Surface energy $=T.A$
Final surface area of Bubble droplets $(A_2)=n.4\pi r^2$
Initial surface area of Bubble drop$(A_1)=4\pi R^2$
Change in energy$\mathrm{\Delta }E=T.\mathrm{\Delta }A$
Where $\mathrm{\Delta }A$is changed in the area.
$\Rightarrow \mathrm{\Delta }E=T(A_2-A_1)$
$=T[n.4\pi r^2-4\pi R^2]$
Take common terms out
$\therefore \mathrm{\Delta }E=T.4\pi [n{r}^2-R^2]$
Substitute the value of $r$ from equation (1) in the above equation
$\Rightarrow \mathrm{\Delta }E=T.4\pi \left[27{\left({\displaystyle \frac{R}{3}}\right)}^2-R^2\right]$
Take $R^2$common
$\Rightarrow \mathrm{\Delta }E=T.4\pi R^2\left[{\displaystyle \frac{27}{9}}-1\right]$
On simplifying we get
$\mathrm{\Delta }E=T.4\pi R^2[3-1]$
$\Rightarrow \mathrm{\Delta }E=T.4\pi R^2[2]$
$=8\pi R^{2}T$
Since, it is given that the surface tension is $\sigma ,$we get
$\mathrm{\Delta }E=2.4\pi R^2.\sigma$
$\Rightarrow \mathrm{\Delta }E=2\pi D^2\sigma$
Hence,option (A) is the correct one.

Note:Surface tension is the tendency of liquid surface to shrink into the minimum surface area possible. We should know that surface energy decreases with increase in temperature and it will also change when larger drop breaks into smaller droplets.