A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of another single soap bubble of radius $$r$$ cm. The value of $$r$$ is _____

Consider a spherical soap bubble of radius 3 cm inside another spherical soap bubble of radius 6 cm. We need to find the radius $$r$$ of a single soap bubble whose internal pressure equals the internal pressure of the smaller bubble (radius 3 cm) in this system.

For any soap bubble, there are two surfaces (inner and outer), and the excess pressure inside is given by $$\Delta P = \frac{4S}{R}$$, where $$S$$ is the surface tension and $$R$$ is the radius. Let $$P_0$$ be the atmospheric pressure outside the outer bubble.

Define the pressures:

• $$P_1$$: Pressure inside the outer bubble but outside the inner bubble (in the space between them).
• $$P_2$$: Pressure inside the inner bubble (radius 3 cm).

For the outer bubble (radius $$R_1 = 6$$ cm), the pressure difference is:

$$P_1 - P_0 = \frac{4S}{R_1}$$ $$P_1 = P_0 + \frac{4S}{6}$$

For the inner bubble (radius $$R_2 = 3$$ cm), the pressure difference is:

$$P_2 - P_1 = \frac{4S}{R_2}$$ $$P_2 = P_1 + \frac{4S}{3}$$

Substitute $$P_1$$ from the outer bubble into this equation:

$$P_2 = \left(P_0 + \frac{4S}{6}\right) + \frac{4S}{3}$$ $$P_2 = P_0 + 4S \left(\frac{1}{6} + \frac{1}{3}\right)$$

Simplify the expression inside the parentheses:

$$\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$ So, $$P_2 = P_0 + 4S \times \frac{1}{2} = P_0 + \frac{4S}{2} = P_0 + 2 \times \frac{4S}{4}$$ But we can write it as: $$P_2 = P_0 + \frac{4S}{2}$$ Note that $$\frac{4S}{2} = 2 \times (2S)$$, but we'll keep it as is for comparison.

Now, for a single soap bubble of radius $$r$$, the internal pressure $$P_r$$ is:

$$P_r = P_0 + \frac{4S}{r}$$

The problem states that $$P_2 = P_r$$, so:

$$P_0 + \frac{4S}{2} = P_0 + \frac{4S}{r}$$

Subtract $$P_0$$ from both sides:

$$\frac{4S}{2} = \frac{4S}{r}$$

Divide both sides by $$4S$$ (assuming $$S \neq 0$$):

$$\frac{1}{2} = \frac{1}{r}$$

Therefore, $$r = 2$$ cm.

Hence, the correct answer is 2.