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 _______________.
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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Correct Answer: 4
Approach Solution - 1
Approach Solution -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: $\displaystyle \boxed{\,4\ \text{cm}\,}$
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