Question 227: Properties of Matter

Question

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

Accepted Answer

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:
$$ P = ho g h $$
where:
$$ho$$ is the density of the liquid.
$$g$$ is the acceleration due to gravity.
$$h$$ is the height of the liquid column.
Thus, we have:
$$ ho g h = \frac{4T}{r} $$
Given values:
$$ ho = 8 	imes 10^3 \mathrm{~kg} \mathrm{~m}^{-3} $$
$$ g = 10 \mathrm{~m} \mathrm{~s}^{-2} $$
$$ h = 0.04 \mathrm{~cm} = 0.04 	imes 10^{-2} \mathrm{~m} $$
$$ T = 0.28 \mathrm{~Nm}^{-1} $$
Substituting the values into the pressure balance equation:
$$ 8 	imes 10^3 	imes 10 	imes 0.04 	imes 10^{-2} = \frac{4 	imes 0.28}{r} $$
Simplifying, we get:
$$ 8 	imes 10^3 	imes 10 	imes 0.04 	imes 10^{-2} = 3200 	imes 10^{-4} = 0.32 $$
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:
$$ 	ext{Diameter} = 2r = 2 	imes 3.5 = 7 \mathrm{~cm} $$
Therefore, the diameter of the soap bubble is $$ 7 \mathrm{~cm} $$.

Properties of Matter

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