The diameter of a rain-drop is 0.02 , cm. If | vedclass

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(A) The excess pressure $\Delta P$ inside a liquid drop is given by the formula $\Delta P = \frac{2T}{r}$.Given: Diameter $d = 0.02 \, cm = 0.02 	ime

Vedclass · Accepted Answer

$1.44 	imes 10^4 \, dyne/cm^2$

Vedclass · Answer

(A) The excess pressure $\Delta P$ inside a liquid drop is given by the formula $\Delta P = \frac{2T}{r}$.Given: Diameter $d = 0.02 \, cm = 0.02 	imes 10^{-2} \, m = 2 	imes 10^{-4} \, m$.Radius $r = \frac{d}{2} = 0.01 \, cm = 10^{-4} \, m$.Surface tension $T = 72 	imes 10^{-3} \, N/m$.Substituting the values: $\Delta P = \frac{2 	imes 72 	imes 10^{-3}}{10^{-4}} = 144 	imes 10^1 = 1440 \, N/m^2$.To convert $N/m^2$ to $dyne/cm^2$: $1 \, N/m^2 = 10 \, dyne/cm^2$.Therefore,$\Delta P = 1440 	imes 10 = 14400 \, dyne/cm^2 = 1.44 	imes 10^4 \, dyne/cm^2$.

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:

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