If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm , then the surface tension of soap solution will be (r = 1 cm and density `d = 0.8 g "cc"^(-1)) , g = 10 m s ^(-2)`

Options

A. `4 Nm^(-1)`
B. `4xx10 ^(-1)Nm^(-1)`
C. `4xx10 ^(-2)Nm^(-1)`
D. `4xx10 ^(-3)Nm^(-1)`

Answer

To solve the problem of finding the surface tension of the soap solution given the excess pressure inside a soap bubble and the height of an oil column, we will follow these steps:

### Step 1: Understand the relationship between excess pressure and surface tension
The excess pressure (\( \Delta P \)) inside a soap bubble is given by the formula:
\[ \Delta P = \frac{4T}{R} \]
where \( T \) is the surface tension and \( R \) is the radius of the bubble.

### Step 2: Calculate the excess pressure due to the oil column
The excess pressure can also be expressed in terms of the height of the oil column (\( h \)), the density of the oil (\( \rho \)), and the acceleration due to gravity (\( g \)):
\[ \Delta P = \rho g h \]

Given:
- Height of the oil column, \( h = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \)
- Density of the oil, \( d = 0.8 \text{ g/cm}^3 = 0.8 \times 10^3 \text{ kg/m}^3 = 800 \text{ kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)

Now, substituting these values into the equation for excess pressure:
\[ \Delta P = \rho g h = (800 \text{ kg/m}^3)(10 \text{ m/s}^2)(2 \times 10^{-3} \text{ m}) \]

### Step 3: Calculate the excess pressure
Calculating the above expression:
\[ \Delta P = 800 \times 10 \times 2 \times 10^{-3} = 8000 \times 10^{-3} = 8 \text{ Pa} \]

### Step 4: Set the two expressions for excess pressure equal to each other
Now, we can set the two expressions for excess pressure equal to each other:
\[ \frac{4T}{R} = 8 \]
Given that the radius \( R = 1 \text{ cm} = 0.01 \text{ m} \), we can substitute this into the equation:
\[ \frac{4T}{0.01} = 8 \]

### Step 5: Solve for surface tension \( T \)
Rearranging the equation to solve for \( T \):
\[ 4T = 8 \times 0.01 \]
\[ 4T = 0.08 \]
\[ T = \frac{0.08}{4} = 0.02 \text{ N/m} \]

### Final Answer
The surface tension of the soap solution is:
\[ T = 0.02 \text{ N/m} \]