A) \[3.9\text{ }N/m\]
B) \[3.9\times {{10}^{-1}}N/m\]
C) \[3.9\times {{10}^{-2}}N/m\]
D) \[3.9\text{ }dyne/m\]
Correct Answer: C
Solution :
Excess pressure inside a soap bubble of radius R is \[=\frac{4T}{R}\] ??. (i) where T is surface tension of liquid film. Pressure due to oil column \[=h\rho g\] ?? (ii) where h is height of column, p the density and g the gravity. From Eqs. (i) and (ii), we get \[\frac{4T}{R}=h\rho g\] \[\Rightarrow \] \[T=\frac{4\rho gR}{4}\] Given, \[h=2\text{ }mm=0.2\text{ }cm,\text{ }g=980\text{ }cm/{{s}^{2}}\] \[\rho =0.8g/cc,\text{ }R=1\,cm\] \[\therefore \] \[T=\frac{0.2\times 0.8\times 980}{4}\] \[=3.92\times 10\,dyne/cm\] In N/m \[T=3.9\times 10\times \frac{{{10}^{-5}}}{{{10}^{-2}}}\] \[=3.9\times {{10}^{-2}}N/m\]
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