QJEEM24Ph82a7: JEE Main 2024 Physics - Class 11 JEEM24Ph82a7 Solution: A liquid column of height balances excess pressure of a soap…

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 $7 \mathrm{~cm}$ .