Question

If $T$ is the surface tension of a fluid, then the energy needed to break a liquid drop of radius $R$ into $64$ equal drops is :

• A. $6\pi R^{2}T$
• B. $\pi R^{2}T$
• C. $12\pi R^{2}T$
• D. $8\pi R^{2}T$

Answer 1

Answer: (C)

$12\pi R^{2}T$

Let the radius of the small drop is $r$.

Since, the volume of the bigger drop will be equal to the $64$ smaller drops. therefore,

$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$

$R=4r$

The surface energy of the larger drop is given as,

$E_1=\left(4\pi R^2\right)T$

The surface energy of the small drops is given as,

$E_2=64\times \left(4\pi r^2\right)T$

$=64\times \left(4\pi {\left(\frac{R}{4}\right)}^2\right)T$

$=\left(16\pi R^2\right)T$

The required energy is given as,

$\Delta E=E_2-E_1$

$=\left(16\pi R^2\right)T-\left(4\pi R^2\right)T$

$=\left(12\pi R^2\right)T$

Thus, the required energy to create the $64$ small drops is $\left(12\pi R^2\right)T$.