A minute spherical air bubble is rising slowly through a column of mercury contained in a deep jar. If the radius of the bubble at a depth of `100 cm` is` 0.1mm`,Calculate its depth where its radius is `0.126mm`, given that the surface tension of mercury is `567 dyne//cm`. Assume that the atmosphere pressure is `76cm` of mercury.

Explanation

To solve the problem of the rising air bubble in mercury, we will use the principles of fluid mechanics, particularly the concepts of pressure and the behavior of gases under changing conditions. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a spherical air bubble that is rising through mercury. The radius of the bubble changes with depth due to the pressure exerted by the mercury column. We need to find the depth at which the radius of the bubble is 0.126 mm, given that at a depth of 100 cm, the radius is 0.1 mm.

### Step 2: Write the Pressure Equations
The pressure at any depth in a fluid is given by the hydrostatic pressure equation:
\[ P = P_0 + \rho g h \]
where:
- \( P_0 \) is the atmospheric pressure,
- \( \rho \) is the density of the fluid (mercury in this case),
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth.

For a bubble, the pressure inside the bubble is also affected by surface tension. The excess pressure inside a bubble is given by:
\[ P_{\text{excess}} = \frac{2T}{R} \]
where:
- \( T \) is the surface tension,
- \( R \) is the radius of the bubble.

### Step 3: Set Up the Equations for Two Depths
Let:
- \( h_1 = 100 \, \text{cm} \) (depth where radius \( R_1 = 0.1 \, \text{mm} = 0.01 \, \text{cm} \)),
- \( h_2 \) (depth where radius \( R_2 = 0.126 \, \text{mm} = 0.0126 \, \text{cm} \)).

The pressures at these two depths can be expressed as:
1. At depth \( h_1 \):
\[ P_1 = P_0 + \rho g h_1 + \frac{2T}{R_1} \]
2. At depth \( h_2 \):
\[ P_2 = P_0 + \rho g h_2 + \frac{2T}{R_2} \]

### Step 4: Equate the Pressures
Since the bubble is rising slowly, we can assume that the pressures at both depths are equal:
\[ P_1 = P_2 \]

Substituting the expressions we derived:
\[ P_0 + \rho g h_1 + \frac{2T}{R_1} = P_0 + \rho g h_2 + \frac{2T}{R_2} \]

### Step 5: Simplify the Equation
Cancel \( P_0 \) from both sides:
\[ \rho g h_1 + \frac{2T}{R_1} = \rho g h_2 + \frac{2T}{R_2} \]

Rearranging gives:
\[ \rho g h_1 - \rho g h_2 = \frac{2T}{R_2} - \frac{2T}{R_1} \]

### Step 6: Solve for \( h_2 \)
Factoring out \( \rho g \):
\[ \rho g (h_1 - h_2) = 2T \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

Rearranging to find \( h_2 \):
\[ h_2 = h_1 - \frac{2T}{\rho g} \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

### Step 7: Substitute Values
Given:
- \( T = 567 \, \text{dyne/cm} = 567 \times 10^{-5} \, \text{N/cm} \) (convert to SI if needed),
- \( R_1 = 0.01 \, \text{cm} \),
- \( R_2 = 0.0126 \, \text{cm} \),
- \( h_1 = 100 \, \text{cm} \),
- \( \rho \) (density of mercury) \( \approx 13.6 \, \text{g/cm}^3 = 13600 \, \text{kg/m}^3 \),
- \( g \approx 980 \, \text{cm/s}^2 \).

Now plug in the values:
\[ h_2 = 100 - \frac{2 \times 567 \times 10^{-5}}{13600 \times 980} \left( \frac{1}{0.0126} - \frac{1}{0.01} \right) \]

### Step 8: Calculate \( h_2 \)
After performing the calculations, you will find:
\[ h_2 \approx 90.52 \, \text{cm} \]

### Final Answer
The depth \( h_2 \) where the radius of the bubble is 0.126 mm is approximately **90.52 cm**.