Two spherical soap bubbles of diameters 10 cm and 6 cm are formed, one at each end of a narrow horizontal glass tube. If the surface tension of the soap solution is `0.03 Nm^(-1)`, then the pressure difference in pascal between the two ends of the tube is

Options

A. 16
B. 1.6
C. 0.016
D. 0.08

Explanation

To solve the problem of finding the pressure difference between two soap bubbles formed at each end of a narrow horizontal glass tube, we can follow these steps:

### Step 1: Understand the Problem
We have two soap bubbles with diameters of 10 cm and 6 cm. We need to find the pressure difference between the two bubbles due to their different sizes. The surface tension of the soap solution is given as \( \gamma = 0.03 \, \text{N/m} \).

### Step 2: Convert Diameters to Radii
First, we need to convert the diameters of the bubbles into radii:
- For the larger bubble (diameter = 10 cm):
\[ R_1 = \frac{10 \, \text{cm}}{2} = 5 \, \text{cm} = 0.05 \, \text{m} \]
- For the smaller bubble (diameter = 6 cm):
\[ R_2 = \frac{6 \, \text{cm}}{2} = 3 \, \text{cm} = 0.03 \, \text{m} \]

### Step 3: Calculate Pressure Inside Each Bubble
The pressure inside a soap bubble is given by the formula:
\[ P = P_0 + \frac{4\gamma}{R} \]
Where \( P_0 \) is the atmospheric pressure (which will cancel out later), \( \gamma \) is the surface tension, and \( R \) is the radius of the bubble.

- For the larger bubble:
\[ P_1 = P_0 + \frac{4 \times 0.03}{0.05} \]
- For the smaller bubble:
\[ P_2 = P_0 + \frac{4 \times 0.03}{0.03} \]

### Step 4: Calculate the Pressure Difference
Now, we find the pressure difference \( \Delta P \) between the two bubbles:
\[ \Delta P = P_2 - P_1 \]
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ \Delta P = \left( P_0 + \frac{4 \times 0.03}{0.03} \right) - \left( P_0 + \frac{4 \times 0.03}{0.05} \right) \]
This simplifies to:
\[ \Delta P = \frac{4 \times 0.03}{0.03} - \frac{4 \times 0.03}{0.05} \]
\[ \Delta P = 4 \times 0.03 \left( \frac{1}{0.03} - \frac{1}{0.05} \right) \]

### Step 5: Simplify the Expression
Calculating the fractions:
\[ \frac{1}{0.03} = \frac{100}{3}, \quad \frac{1}{0.05} = 20 \]
Thus,
\[ \Delta P = 4 \times 0.03 \left( \frac{100}{3} - 20 \right) \]
Finding a common denominator:
\[ \Delta P = 4 \times 0.03 \left( \frac{100 - 60}{3} \right) = 4 \times 0.03 \left( \frac{40}{3} \right) \]
\[ \Delta P = \frac{4 \times 0.03 \times 40}{3} = \frac{4.8}{3} = 1.6 \, \text{Pa} \]

### Final Answer
The pressure difference between the two ends of the tube is:
\[ \Delta P = 1.6 \, \text{Pa} \]

---