Question

If the radius of a spherical liquid (of surface tension $S$) drop increases from $r$ to $ r+ \Delta\,r, $ the corresponding increase in the surface energy is

• A. $ 8\pi \,r\,\Delta rS $
• B. $ 4\pi \,r\,\Delta rS $
• C. $ 16\pi \,r\,\Delta rS $
• D. $ 2\pi \,r\,\Delta rS $

Answer 1

Answer: (A)

Work done in blowing a liquid drop or soap bubble
$W=S \times 4 \pi\left[r_{2}^{2}-r_{1}^{2}\right]$
Here, $W=S \times 4 \pi\left[(r +\Delta r)^{2}-(r)^{2}\right]$
$=S \times 4 \pi\left[r^{2}+(\Delta r)^{2}+2 \Delta r \times r-r^{2}\right]$
$=8 \pi \Delta r S r\left[(\Delta r)^{2}\right. $ is very small]