Question

A liquid drop of diameter D breaks into 27 tiny drops. The resultant change in energy is :-

• A. $2\pi T{D}^2$
• B. $4\pi T{D}^2$
• C. $\pi T{D}^2$
• D. none of these

Answer 1

The correct option is A $2\pi T{D}^2$

Since, the volume of water in both the large drop and small droplets would remain the same, the radius of each small drop is,

$1.33\pi R^3=27\times 1.33\pi r^3$

Or, r = $\frac{R}{3}$

Now, the change in energy is defined as the surface tension times the change in surface area. Or,

$\delta U=T\delta A$

$\delta A=27\times 4\pi r^2-4\pi R^2$

Substituting r = R/3

$\delta A=8\pi R^2$

Since, surface tension is only a function of the surrounding temperatures, therefore it will remain constant during this change.

Hence, $\delta U=2\pi D^2\times T$

Or, $\delta U=2\pi T{D}^2$