Correct Answer - A::B
`r=0.5mm=5xx10^-3m`
`h=5.0mm=5xx10^-2m`
`T=0.075N/m`
the excess pressure at `5cm` before the surface
`P=hrhog=100x(5x10^-2)x9.8`
`=490N/m^2`
Again excess pressure at the surface
`P=(2T)/r=(2xx(0.75))/((5xx10^-4))`
`=300N/m^2`
diffeeence in pressure
`=P=(F/r)`
`=490-300=190N/m^2`