A) 3.92 N/m
B) 0.0392 N/m
C) 0.392 N/m
D) 0.00392 N/m
Correct Answer: B
Solution :
The excess pressure of soap bubble \[p=\frac{4T}{R}\] Or \[h\rho g=\frac{4T}{R}\] \[(\because p=h\rho g)\] \[\therefore \] \[T=\frac{Rh\rho g}{4}\] \[=\frac{1\times {{10}^{-2}}\times 2\times {{10}^{-3}}\times 0.8\times {{10}^{3}}\times 9.8}{4}\] \[=0.0392\text{ }N/m\]
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