A) \[2\times {{10}^{-8}}N{{m}^{-1}}\]
B) \[2\times {{10}^{-6}}N{{m}^{-1}}\]
C) \[2\times {{10}^{-4}}N{{m}^{-1}}\]
D) \[2\times {{10}^{-2}}N{{m}^{-1}}\]
Correct Answer: D
Solution :
The surface tension (T), 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. \[\therefore \] \[T=\frac{W}{\Delta \Alpha }\] Initial size of soap film is \[10\,cm\,\times 6\,cm\,\,={{A}_{1}}\] Final size of soap bubble is \[10\,cm\,\,\times \,\,11\,cm\,\,=\,{{A}_{2}}\] \[\Delta A={{A}_{2}}-{{A}_{1}}\] \[=10\times 11-10\times 6\] \[=50\times {{10}^{-4}}{{m}^{2}}\] Since, film consists of two surfaces, therefore, total increase in surface area is \[=2\times 50\times {{10}^{-4}}=100\times {{10}^{-4}}\,{{m}^{2}}\] \[\therefore \] \[T=\frac{2\times {{10}^{-4}}}{100\times {{10}^{-4}}}\,=2\times {{10}^{-2}}\,N{{m}^{-1}}\]
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