Two separate soap bubbles of radii 3 × 10 ^ - 3 m and 2 × 10 ^ - 3...

Excess pressure inside first bubble of radius ${r}_{1}$

${p}_{1} = {p}_{0} + \dfrac{4T}{r1}$

Excess pressure inside second bubble of radius ${r}_{2}$
${p}_{2} = {p}_{0} + \dfrac{4T}{r2}$

Excess pressure inside double bubble,

$p = {p}_{2} - {p}_{1}$

$p = 4T \left(\dfrac{{r}_{1} - {r}_{2}}{{r}_{1}{r}_{2}}\right)</p><p>If$R$be the radius of double bubble, then</p><p>$p = $\dfrac{4T}{R}</p><p>or 4T\left(\dfrac{{r}_{1} - {r}_{2}}{{r}_{1}{r}_{2}}\right) = \dfrac{4T}{R}</p><p>$⇒ R = $\dfrac{{r}_{1}{r}_{2}}{{r}_{1} - {r}_{2}} = \dfrac{3 \times {10}^{- 3} \times 2 \times {10}^{- 3}}{\left(3 - 2\right) \times {10}^{- 3}} =6 \times {10}^{- 3}$ m$</p><p>$R = 6 $\times {10}^{- 3}$ m$</p>$