A soap bubble, having a radius of $1 ext{ mm}$, is blown from a detergent solution having a surface tension

Question

A soap bubble, having a radius of $1 	ext{ mm}$, is blown from a detergent solution having a surface tension of $2.5 	imes 10^{-2} 	ext{ N/m}$. The pressure inside the bubble equals the pressure at a point $Z_0$ below the free surface of the water in a container. Taking $g = 10 	ext{ m/s}^2$ and the density of water $ho = 10^3 	ext{ kg/m}^3$, the value of $Z_0$ is:

Accepted Answer

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 $P_{gauge} = Z_0 ho g$ .
3. **Equating Pressures:** The problem states that the absolute pressure inside the bubble ($P_{atm} + \Delta P$) equals the absolute pressure at depth $Z_0$ ($P_{atm} + Z_0 ho g$). Therefore, the excess pressure equals the gauge pressure due to the water column:
   $$ \frac{4S}{r} = Z_0 ho g $$
4. **Calculation:**
   - Surface Tension, $S = 2.5 	imes 10^{-2} 	ext{ N/m}$
   - Radius, $r = 1 	ext{ mm} = 10^{-3} 	ext{ m}$
   - Density, $ho = 10^3 	ext{ kg/m}^3$
   - Gravity, $g = 10 	ext{ m/s}^2$

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

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