A) \[\text{24}0\text{ N}/{{\text{m}}^{\text{2}}}\]
B) \[0.\text{48 N}/{{\text{m}}^{\text{2}}}\]
C) \[\text{48}0\text{ N}/{{\text{m}}^{\text{2}}}\]
D) \[0.\text{24 N}/{{\text{m}}^{\text{2}}}\]
Correct Answer: C
Solution :
Excess pressure \[p=\frac{4T}{r}\] where T is surface tension and r the radius of drop. Given, \[r=\frac{d}{2}=\frac{1}{2}mm=\frac{1}{2}\times {{10}^{-3}}m\], \[T=60\times {{10}^{-3}}N/m\] \[\therefore \] \[P=\frac{4\times 60\times {{10}^{-3}}}{\frac{1}{2}\times {{10}^{-3}}}=480\,N/{{m}^{2}}\]
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