An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density `10^(3)" kgm"^(-3).` The pressure inside the bubble is 1100 Nm greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? `(g = 9.8 ms^(-2))`

Options

A. 0.1m
B. 0.15m
C. 0.20m
D. 0.25m

Explanation

To find the depth at which the air bubble is located below the surface of the liquid, we can use the following steps:

### Step 1: Understand the Pressure Difference
The pressure inside the bubble (P_i) is given to be 1100 N/m² greater than the atmospheric pressure (P_a). Therefore, we can express this as:
\[ P_i = P_a + 1100 \, \text{N/m}^2 \]

### Step 2: Apply the Hydrostatic Pressure Formula
The pressure difference between the inside of the bubble and the atmospheric pressure can be related to the hydrostatic pressure due to the liquid column above the bubble and the effect of surface tension. The equation can be written as:
\[ \Delta P = \rho g h + \frac{2 \gamma}{r} \]
Where:
- \(\Delta P = 1100 \, \text{N/m}^2\)
- \(\rho = 10^3 \, \text{kg/m}^3\) (density of the liquid)
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)
- \(h\) = depth (what we want to find)
- \(\gamma = 0.06 \, \text{N/m}\) (surface tension)
- \(r = 0.1 \, \text{cm} = 0.001 \, \text{m}\) (radius of the bubble)

### Step 3: Substitute Known Values into the Equation
We can substitute the known values into the equation:
\[ 1100 = (10^3)(9.8)h + \frac{2(0.06)}{0.001} \]

### Step 4: Calculate the Surface Tension Contribution
First, calculate the term involving surface tension:
\[ \frac{2(0.06)}{0.001} = 120 \, \text{N/m}^2 \]

### Step 5: Rewrite the Equation
Now we can rewrite the equation:
\[ 1100 = (9800)h + 120 \]

### Step 6: Solve for h
Rearranging gives:
\[ 9800h = 1100 - 120 \]
\[ 9800h = 980 \]
\[ h = \frac{980}{9800} \]
\[ h = 0.1 \, \text{m} \]

### Conclusion
The depth of the bubble below the surface of the liquid is:
\[ h = 0.1 \, \text{m} \]