air inside them are $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ and $\mathrm{P}_{2}=\frac{4 \sigma}{\mathrm{r}_{2}},$ Since
the temperature reamins unchanged, we have from Boyle's law
$\mathrm{P}_{1} \mathrm{V}_{1}+\mathrm{P}_{2} \mathrm{V}_{2}=\mathrm{PV}$
or $\frac{4 \sigma}{\mathrm{r}_{1}} \cdot \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\frac{4 \sigma}{\mathrm{r}_{2}} \cdot \frac{4}{3} \pi \mathrm{r}_{2}^{3}=\frac{4 \sigma}{\mathrm{r}} \cdot \frac{4}{3} \pi \mathrm{r}^{3}$ ......$(i)$
where $r$ is the radius of the single bubble formed.
From $(i),$ we get $r^{2}=r^{2}=r^{2}+r_{2}^{2}$ or $r=\sqrt{r_{1}^{2}+r_{2}^{2}}$
So diameter of bubble is $=5.0 \mathrm{\,mm}$
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