The excess pressure inside a spherical soap bubble of radius $1 \mathrm{~cm}$ is balanced by a column of oil (Specific gravity $=0.8$ ), $2 \mathrm{~mm}$ high , the surface tension of the bubble is

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Aicharya · Accepted Answer

The excess pressure of soap bubble $p=\frac{4 T}{R}$ $h ho g=\frac{4 T}{R}$ $	herefore T=\frac{R h ho g}{4}$ $=\frac{1 	imes 10^{-2} 	imes 2 	imes 10^{-3} 	imes 0.8 	imes 10^3 	imes 9.8}{4}$ $=3.92 	imes 10^{-2} \mathrm{~N} / \mathrm{m}$ $=0.0392 \mathrm{~N} / \mathrm{m}$

The excess pressure inside a spherical soap bubble of radius $1 \mathrm{~cm}$ is balanced by a column of oil…

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