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Step-by-Step Solution
Step 1: Identify the Relevant Pressure Equations
The pressure inside a soap bubble is greater than the outside pressure by an amount given by
$ \Delta P = \frac{4T}{R} $
(where $T$ is the surface tension and $R$ is the radius of the bubble). The pressure at some depth $Z_{0}$ in a fluid (water, in this case) is given by
$ P = P_{0} + \rho \, g \, Z_{0} $
(where $\rho$ is the density of water, and $g$ is the acceleration due to gravity).
Step 2: Set the Bubble Pressure Equal to Hydrostatic Pressure
We want the pressure inside the bubble to match the hydrostatic pressure at depth $Z_{0}$. Thus, we set:
$ P_{0} + \frac{4T}{R} = P_{0} + \rho \, g \, Z_{0} $
The initial atmospheric pressure $P_{0}$ on both sides cancels out, leaving:
$ \rho \, g \, Z_{0} = \frac{4T}{R} $
Step 3: Solve for the Depth $Z_{0}$
Rearranging for $Z_{0}$:
$ Z_{0} = \frac{4T}{\rho \, g \, R} $
Now substitute the given values:
$T$ (Surface tension)$2.5 \times 10^{-2}\,\text{N/m}$
$\rho$ (Density of water)$10^{3}\,\text{kg/m}^{3}$
$g$ (Acceleration due to gravity)$10\,\text{m/s}^{2}$
$R$ (Radius of bubble)$1 \times 10^{-3}\,\text{m}$
Substitute these into the formula:
$ Z_{0} = \frac{4 \times \left(2.5 \times 10^{-2}\right)}{\left(10^{3}\right) \times 10 \times \left(10^{-3}\right)} $
Compute step by step:
Numerator: $4 \times 2.5 \times 10^{-2} = 10 \times 10^{-2} = 10^{-1}$
Denominator: $10^{3} \times 10 \times 10^{-3} = 10^{3} \times 10^{-3} \times 10 = 10
So,
$ Z_{0} = \frac{10^{-1}}{10} = 10^{-2}\,\text{m} = 0.01\,\text{m} = 1\,\text{cm} $
Final Answer
The depth $Z_{0}$ is 1 cm.
