Let's start by understanding the problem. We need to determine the diameter of the soap bubble, given certain properties of the liquid and the soap solution.

The excess pressure inside a soap bubble can be calculated using the formula:

$$ \Delta P = \frac{4T}{r} $$

where:

• $$\Delta P$$ is the excess pressure.
• $$T$$ is the surface tension of the soap solution.
• $$r$$ is the radius of the soap bubble.

The column of liquid balances this excess pressure, and the pressure exerted by the liquid column is given by:

$$ P = \rho g h $$

where:

• $$\rho$$ is the density of the liquid.
• $$g$$ is the acceleration due to gravity.
• $$h$$ is the height of the liquid column.

Thus, we have:

$$ \rho g h = \frac{4T}{r} $$

Given values:

• $$ \rho = 8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} $$
• $$ g = 10 \mathrm{~m} \mathrm{~s}^{-2} $$
• $$ h = 0.04 \mathrm{~cm} = 0.04 \times 10^{-2} \mathrm{~m} $$
• $$ T = 0.28 \mathrm{~Nm}^{-1} $$

Substituting the values into the pressure balance equation:

$$ 8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = \frac{4 \times 0.28}{r} $$

Simplifying, we get:

$$ 8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = 3200 \times 10^{-4} = 0.32 $$

So,

$$ 0.32 = \frac{1.12}{r} $$

Solving for $$r$$, we get:

$$ r = \frac{1.12}{0.32} $$

$$ r \approx 3.5 \mathrm{~cm} $$

We need the diameter, which is twice the radius:

$$ \text{Diameter} = 2r = 2 \times 3.5 = 7 \mathrm{~cm} $$

Therefore, the diameter of the soap bubble is $$ 7 \mathrm{~cm} $$.