Question

A liquid drop of radius R breaks into 27 small drops. What will be the change in surface energy if the surface tension of the liquid is $T$ ?

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

Answer 1

Answer: (A)

: Change in surface energy $=T\times$ change in area. Volume of one big drop = Volume of 27 small drops $\therefore$ $\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$ or $R=3r$ .... (i) Surface area of 27 drops $=27\times 4\pi r^2$ Surface area of one bigger drop $=4\pi R^2$ $\therefore$ change in area $=27\times 4\pi r^2-4\pi R^2$ $=4\pi [27r^2-R^2]=4\pi \left[27\times \frac{R^2}{9}-R^2\right]$ $=4\pi \times 2R^2=8\pi R^2$ . $\therefore$ Change in surface energy $=8\pi R^{2}T$ .