A) W
B) 2 W
C) \[\sqrt{2}\]W
D) \[{{4}^{1/3}}W\]
Correct Answer: D
Solution :
From the definition of surface tension (T), the surface tension of a liquid is equal to the work (W) required to increase the surface area (A) of the liquid film by unity at constant temperature. \[\therefore \] \[W=T\times \Delta \,A\] Since, surface area of a sphere is \[4\pi {{R}^{2}}\] and there are two free surfaces, we have \[W=T\times 8\pi {{R}^{2}}\] ?(i) and volume of sphere \[=\frac{4}{3}\pi {{R}^{3}}\] i.e., \[V=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[R={{\left( \frac{3V}{4\pi } \right)}^{1/3}}\] ?(ii) From Eqs. (i) and (ii), we get \[W=T\times 8\pi \times {{\left( \frac{3V}{4\pi } \right)}^{2/3}}\] \[\Rightarrow \] \[W\propto {{V}^{2/3}}\] \[\therefore \] \[{{W}_{1}}\propto {{V}_{1}}^{2/3}\]and \[{{W}_{2}}\propto {{V}_{2}}^{2/3}\] \[\therefore \] \[\frac{{{W}_{2}}}{{{W}_{1}}}={{\left( \frac{2{{V}_{1}}}{{{V}_{1}}} \right)}^{2/3}}\] \[\Rightarrow \] \[{{W}_{2}}={{2}^{2/3}}{{W}_{1}}-{{4}^{1/3}}W\]
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