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 _______________.
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The radius of curvature of the common surface of two soap bubbles in contact is calculated using the formula \( r = \frac{r_1 \cdot r_2}{r_1 - r_2} \).
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 \( \boxed{4} \).
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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 \[ \frac{1}{r}=\frac{1}{R_1}-\frac{1}{R_2}. \]
