Easy
The diameter of a rain-drop is $0.02 \, cm$. If the surface tension of water is $72 \times 10^{-3} \, N/m$,then the pressure difference between the internal and external surfaces of the drop will be:
- A$1.44 \times 10^4 \, dyne/cm^2$
- B$1.44 \times 10^4 \, N/m^2$
- C$1.44 \times 10^3 \, dyne/cm^2$
- D$1.44 \times 10^5 \, N/m^2$
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