A liquid column of height $0.04 \mathrm{~cm}$ balances excess pressure of a soap bubble of certain radius. If density of liquid is $8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and surface tension of soap solution is $0.28 \mathrm{~Nm}^{-1}$, then diameter of the soap bubble is __________ $\mathrm{cm}$. (if $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$ )

<p>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.</p> <p>The excess pressure inside a soap bubble can be calculated using the formula:</p> <p>$\Delta P = \frac{4T}{r}$</p> <p>where:</p> <ul> <li>$\Delta P$ is the excess pressure.</li> <li>$T$ is the surface tension of the soap solution.</li> <li>$r$ is the radius of the soap bubble.</li> </ul> <p>The column of liquid balances this excess pressure, and the pressure exerted by the liquid column is given by:</p> <p>$P = \rho g h$</p> <p>where:</p> <ul> <li>$\rho$ is the density of the liquid.</li> <li>$g$ is the acceleration due to gravity.</li> <li>$h$ is the height of the liquid column.</li> </ul> <p>Thus, we have:</p> <p>$\rho g h = \frac{4T}{r}$</p> <p>Given values:</p> <ul> <li>$\rho = 8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$</li> <li>$g = 10 \mathrm{~m} \mathrm{~s}^{-2}$</li> <li>$h = 0.04 \mathrm{~cm} = 0.04 \times 10^{-2} \mathrm{~m}$</li> <li>$T = 0.28 \mathrm{~Nm}^{-1}$</li> </ul> <p>Substituting the values into the pressure balance equation:</p> <p>$8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = \frac{4 \times 0.28}{r}$</p> <p>Simplifying, we get:</p> <p>$8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = 3200 \times 10^{-4} = 0.32$</p> <p>So,</p> <p>$0.32 = \frac{1.12}{r}$</p> <p>Solving for $r$, we get:</p> <p>$r = \frac{1.12}{0.32}$</p> <p>$r \approx 3.5 \mathrm{~cm}$</p> <p>We need the diameter, which is twice the radius:</p> <p>$\text{Diameter} = 2r = 2 \times 3.5 = 7 \mathrm{~cm}$</p> <p>Therefore, the diameter of the soap bubble is $7 \mathrm{~cm}$.</p>