Given data

• Radius of air bubble: $r = 1.0\,\text{mm} = 1.0 \times 10^{-3}\,\text{m}$
• Surface tension: $T = 0.075\,\text{N m}^{-1}$
• Density of liquid: $\rho = 1000\,\text{kg m}^{-3}$
• Depth below free surface: $h = 10\,\text{cm} = 0.10\,\text{m}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

Pressure outside the bubble at depth $h$

At depth $h$, the liquid pressure is greater than atmospheric pressure by $\rho g h$ So, $\rho g h = 1000 \times 10 \times 0.10 = 1000\,\text{Pa}$

Excess pressure inside an air bubble due to surface tension

For an air bubble in a liquid, there is only one liquid-air interface, so excess pressure is $\Delta P_{\text{surface}} = \frac{2T}{r}$

Substituting values: $\frac{2T}{r} = \frac{2 \times 0.075}{1.0 \times 10^{-3}} = \frac{0.15}{10^{-3}} = 150\,\text{Pa}$

Total pressure inside bubble above atmospheric pressure

Pressure inside bubble exceeds atmospheric pressure by: $\rho g h + \frac{2T}{r}$

Therefore, $1000 + 150 = 1150\,\text{Pa}$

Final answer $\boxed{1150}$

Comparison with stored correct answer

Stored correct answer = $1150$

This matches the derived answer exactly.