If the excess pressure inside a soap bubble of radius \(3 \mathrm{~mm}\) is equal to the pressure of a water column of height \(0.8 \mathrm{~cm}\), then the surface tension of the soap solution is ( \(\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^3, g=9.8 \mathrm{~m} / \mathrm{s}^2\) )

The excess pressure inside the bubble is given as, \(P=\frac{4 T}{r}\) as the soap bubble has two liquid-gas interfaces.
The pressure to height column of water is given as, \(P=\rho h g\)
Since the excess pressure inside the bubble is balanced by the height of the column and it can be written as,
\(\frac{4 T}{r}=\rho h g\)
\(\Rightarrow T=\frac{r \rho h g}{4}\)
On plugging the given values:
\(T=\frac{1}{4}\left(3 \times 10^{-3} \mathrm{~m} \times \frac{10^3 \mathrm{~kg}}{\mathrm{~m}^3} \times 8 \times 10^{-3} \mathrm{~m} \times 9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\right)\) \(=58.8 \times 10^{-3} \mathrm{~N} / \mathrm{m}\)