A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let `P_(2)` be the pressure inside the inner bubble and `P_(0)`, the pressure outside the outer bubble. Radius of another bubble with pressure difference `P_(2)-P_(0)` between its inside and outside would be :

Options

A. 6 cm
B. 12 cm
C. 4.8 cm
D. 2.4 cm

Answer

To solve the problem, we need to apply the concept of pressure difference in soap bubbles. The pressure inside a soap bubble is given by the formula:

\[ P = P_0 + \frac{4\sigma}{r} \]

where:
- \( P \) is the pressure inside the bubble,
- \( P_0 \) is the external pressure,
- \( \sigma \) is the surface tension of the soap solution,
- \( r \) is the radius of the bubble.

### Step 1: Calculate the pressure inside the inner bubble \( P_2 \)

For the inner bubble with radius \( r_1 = 4 \) cm:

\[ P_2 = P_0 + \frac{4\sigma}{r_1} \]

Substituting \( r_1 = 4 \) cm:

\[ P_2 = P_0 + \frac{4\sigma}{4} = P_0 + \sigma \]

### Step 2: Calculate the pressure inside the outer bubble \( P_1 \)

For the outer bubble with radius \( r_2 = 6 \) cm:

\[ P_1 = P_0 + \frac{4\sigma}{r_2} \]

Substituting \( r_2 = 6 \) cm:

\[ P_1 = P_0 + \frac{4\sigma}{6} = P_0 + \frac{2\sigma}{3} \]

### Step 3: Find the pressure difference between the inner bubble and the outside pressure

The pressure difference between the inner bubble and the outside pressure is:

\[ P_2 - P_0 = \sigma \]

### Step 4: Find the pressure difference between the inner and outer bubbles

The pressure difference between the inner bubble and the outer bubble is:

\[ P_2 - P_1 = \sigma - \left(P_0 + \frac{2\sigma}{3}\right) \]

Calculating this gives:

\[ P_2 - P_1 = \sigma - P_0 - \frac{2\sigma}{3} = \frac{3\sigma}{3} - \frac{2\sigma}{3} - P_0 = \frac{\sigma}{3} - P_0 \]

### Step 5: Determine the radius of another bubble with pressure difference \( P_2 - P_0 \)

Let the radius of the new bubble be \( r \). The pressure difference between the inside and outside of this new bubble can be expressed as:

\[ P_{inside} - P_{outside} = \frac{4\sigma}{r} \]

Setting this equal to the pressure difference \( P_2 - P_0 \):

\[ \frac{4\sigma}{r} = \sigma \]

### Step 6: Solve for \( r \)

Rearranging gives:

\[ r = \frac{4\sigma}{\sigma} = 4 \text{ cm} \]

### Final Answer

The radius of the new bubble with pressure difference \( P_2 - P_0 \) is 4 cm.

---