Question

Work done in splitting a drop of water of $1mm$ radius into ${10}^6$ droplets is (surface tension of water $=72\times {10}^{-3}J/m^2$ )

• A. $89.5\times {10}^{-5}J$
• B. 0.895 J
• C. $895\times {10}^{-5}J$
• D. None of these

Answer 1

Answer: (B)

The volume of big drop $={10}^6\times$ volume of small drop $\Rightarrow$ $\frac{4}{3}\pi R^3={10}^6\times \frac{4}{3}\pi r^3$ $\Rightarrow$ $R=100r$ Therefore, work done $W=T.\Delta A$ $=T[n.4\pi r^2-4\pi R^2]$ $=4\pi T[{10}^6{r}^2-{10}^4{r}^2]$ $=4\pi \times 72\times {10}^{-3}\times 990000\times {10}^{-6}$ $=0.895J$