Two spherical soap bubbles formed in vacuum h | vedclass

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Question

Since the bubbles are in vacuum, the pressure of

air inside them are $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ and $\mathrm{P}_{2}=\frac{4 \

Vedclass · Accepted Answer

$5.0$

Vedclass · Answer

Since the bubbles are in vacuum, the pressure of

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 $=5.0 \mathrm{\,mm}$

Two spherical soap bubbles formed in vacuum has diameter $3.0\,mm$ and $4.0\,mm$ . They coalesce to form a single spherical bubble. If the temperature remains unchanged, find the diameter of the bubble so formed ....... $mm$

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