The excess pressure inside a soap bubble
`/_P=4S/R=(4xx0.03Nm^-1)/(7.5xx10^-3m)=15Nm^-2`
The pressure due to 0.02 cm of the liquid column ils
`/_P=hrhog`
`=(0.02xx10^-2m)rho(9.8ms^2)`
Thus, `16 Nm^-2=(0.02xx10^-2m)rho(9.8ms^-2)`
`or rho=8.2x10^3kgm^-3`