Question

A sphere liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$, then the work done in this process will be:

• A. $2\pi R^{2}T$
• B. $3\pi R^{2}T$
• C. $4\pi R^{2}T$
• D. $2\pi {RT}^2$

Answer 1

Answer: (C)

$4\pi R^{2}T$

Radius of bigger drop is $R$

Let the radius of smaller drops be $r$

Mass of bigger drop must be equal of total mass of smaller drops.

$\therefore$ $M=8m$

$\rho \times \frac{4}{3}\pi R^3=8\times \rho \times \frac{4}{3}\pi r^3$ where , $\rho$ is the density of liquid

OR $R^3=8r^3$ $⟹R=2r$ ........(1)

Surface area of bigger drop $A_1=4\pi R^2$

Total surface area of smaller drops $A_2=8\times 4\pi r^2=8\times 4\pi \times (\frac{R}{2}{)}^2=8\pi R^2$

Change in surface area $\Delta A=A_2-A_1=4\pi R^2$

$\therefore$ Work done $W=T\Delta A=T\times 4\pi R^2=4\pi R^{2}T$