Question

Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is _______________.

Hint:

The radius of curvature of the common surface of two soap bubbles in contact is calculated using the formula \( r = \frac{r_1 \cdot r_2}{r_1 - r_2} \).

Answer 1

Correct Answer: 4

For two soap bubbles in contact, the radius of curvature \( r \) of the common surface is given by: \[ r = \frac{r_1 \cdot r_2}{r_1 - r_2} \] where \( r_1 = 4 \, \text{cm} \) and \( r_2 = 2 \, \text{cm} \): \[ r = \frac{2 \times 4}{4 - 2} = \frac{8}{2} = 4 \, \text{cm} \] Thus, the answer is \( \boxed{4} \).

Answer 2

Step 2 — Pressures in the two bubbles: 
Let $p_1$ and $p_2$ be the pressures inside the bubbles of radii $R_1$ and $R_2$ respectively. Taking outer atmospheric pressure as $p_0$, \[ p_1 = p_0 + \frac{4\gamma}{R_1},\qquad p_2 = p_0 + \frac{4\gamma}{R_2}. \] Hence the pressure difference between the two bubbles is \[ p_1 - p_2 = 4\gamma\!\left(\frac{1}{R_1}-\frac{1}{R_2}\right). \]

Step 3 — Pressure difference across the common soap film:
The common surface (soap film) separating the two bubbles has radius of curvature $r$. Because the film has two surfaces, the pressure jump across it is \[ p_1 - p_2 = \frac{4\gamma}{r}. \]

Step 4 — Equate the two expressions for the pressure difference:
\[ 4\gamma\!\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{4\gamma}{r}. \] Cancel $4\gamma$ (nonzero) to get \[ \frac{1}{r}=\frac{1}{R_1}-\frac{1}{R_2}. \]

Step 5 — Substitute numerical values:
\[ \frac{1}{r}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\quad\Rightarrow\quad r=4\ \text{cm}. \]

Final Answer: $\displaystyle \boxed{\,4\ \text{cm}\,}$