Explanation

(c) Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles A and B be r.
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Excess of pressure as compared to atmosphere inside A is
$p_{1} = \frac{4T}{r_{1}} = \frac{4T}{0.03}$
Excess of pressure inside B is
$p_{2} = \frac{4T}{r_{2}} = \frac{4T}{0.04}$
In the double bubble the pressure difference between A and B on either side of the common surface is
$\frac{4T}{0.03} - \frac{4T}{0.04} = \frac{4T}{r}$
$\Longrightarrow \ \ \frac{1}{0.03} - \frac{1}{0.04} = \frac{1}{r}$
$\Longrightarrow \ r = \frac{0.03 \times 0.04}{0.01} = 0.12m$