A) \[2\pi T{{D}^{2}}\]
B) \[4\pi T{{D}^{2}}\]
C) \[\pi T{{D}^{2}}\]
D) \[8\pi T{{D}^{2}}\]
Correct Answer: A
Solution :
Surface tension\[(T)\]of a liquid is equal to the work\[(W)\]required to increase the surface area\[(\Delta A)\]of the liquid film by unity at constant temperature. \[W=T\Delta \,\,A=\]surface energy Also, volume of big drop\[=27\times \]volume of small drop \[ie\], \[V=27V\] where\[V\]is volume of big drop of diameter\[D\]and\[V\]the volume of small drop of diameter\[d.\] \[\therefore \] \[\frac{4}{3}\pi {{\left( \frac{D}{2} \right)}^{3}}=27\times \frac{4}{3}\pi {{\left( \frac{d}{2} \right)}^{3}}\] \[\Rightarrow \] \[\frac{D}{2}=3\times \frac{d}{2}\] \[\Rightarrow \] \[d=\frac{D}{3}\] Radius of small drop,\[r=\frac{d}{2}=\frac{D}{6}\]. \[\therefore \] Change in surface energy\[=T({{A}_{2}}-{{A}_{1}})\] \[=T[27\cdot 4\pi {{r}^{2}}-4\pi {{R}^{2}}]\] \[=T\,\,4\pi \left[ 27{{\left( \frac{D}{6} \right)}^{2}}-{{\left( \frac{D}{2} \right)}^{2}} \right]\] \[=4\pi \,\,T\left[ \frac{3{{D}^{2}}}{4}-\frac{{{D}^{2}}}{4} \right]=2\pi \,\,{{D}^{2}}T\]
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