1. Excess Pressure in a Soap Bubble: A soap bubble has two free surfaces (inner and outer). Therefore, the excess pressure ($P_{excess}$) inside the bubble is given by the formula $\Delta P = \frac{4S}{r}$, where $S$ is the surface tension and $r$ is the radius .
2. Hydrostatic Pressure: The gauge pressure at a depth $Z_0$ below the free surface of a liquid is given by $P_{gauge} = Z_0 \rho g$ .
3. Equating Pressures: The problem states that the absolute pressure inside the bubble ($P_{atm} + \Delta P$) equals the absolute pressure at depth $Z_0$ ($P_{atm} + Z_0 \rho g$). Therefore, the excess pressure equals the gauge pressure due to the water column: $\frac{4S}{r} = Z_0 \rho g$
4. Calculation:

• Surface Tension, $S = 2.5 \times 10^{-2} \text{ N/m}$
• Radius, $r = 1 \text{ mm} = 10^{-3} \text{ m}$
• Density, $\rho = 10^3 \text{ kg/m}^3$
• Gravity, $g = 10 \text{ m/s}^2$

Substituting the values: $Z_0 = \frac{4S}{r \rho g}$ $Z_0 = \frac{4 \times (2.5 \times 10^{-2})}{(10^{-3}) \times 10^3 \times 10}$ $Z_0 = \frac{10 \times 10^{-2}}{10} = 10^{-2} \text{ m}$ $Z_0 = 1 \text{ cm}$