There is an air bubble of radius 1 . 0mm in a liquid of surface...

The pressure inside the bubble is given by

$\DeltaP = P - {P}_{0} = ρgh + \dfrac{2T}{r}$

The data given is

$r = {10}^{- 3}mT = 0 . 075N{m}^{- 1}ρ = 1000kg{m}^{- 3}h = 10cm$

Therefore, the pressure in the bubble is

$\DeltaP = \left(1000 \times 10 \times 10 \times {10}^{- 2}\right) + \dfrac{2 \times 0 . 075}{{10}^{- 3}} = 1000 + 150 = 1150Pa$