A soap bubble of radius

Excess pressure inside the smaller soap bubble

$\Delta \mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{r}_{1}}+\frac{4 \mathrm{~S}}{\mathrm{r}_{2}}$  .........(I)

The excess pressure inside the equivalent soap bubble

$\Delta \mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{R}_{\mathrm{eq}}} \ldots$ (ii)

From (i) \& (ii)

$\frac{4 \mathrm{~S}}{\mathrm{R}_{\mathrm{eq}}}=\frac{4 \mathrm{~S}}{\mathrm{r}_{1}}+\frac{4 \mathrm{~S}}{\mathrm{r}_{2}}$

$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}$

$=\frac{1}{6}+\frac{1}{3}$

$\mathrm{R}_{\mathrm{eq}}=2 \mathrm{~cm}$

Ans. $2.00$