If the excess pressure inside a soap bubble of radius $3 \mathrm{~mm}$ is equal to the pressure of a water…
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 ( $\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^3, g=9.8 \mathrm{~m} / \mathrm{s}^2$ )
$0.588 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
$588 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
$58.8 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
$5.88 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
Solution
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:
$T=\frac{1}{4}\left(3 \times 10^{-3} \mathrm{~m} \times \frac{10^3 \mathrm{~kg}}{\mathrm{~m}^3} \times 8 \times 10^{-3} \mathrm{~m} \times 9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\right)=58.8 \times 10^{-3} \mathrm{~N} / \mathrm{m}$