Question

Two soap bubbles in vacuum, of radius $3cm$ and $4cm$ coalesce to form a single bubble under isothermal conditions. Then the radius of bigger bubble is:

• A. $7cm$
• B. $\frac{12}{7}cm$
• C. $12cm$
• D. $5cm$

Answer 1

The correct option is D $5cm$

For isothermal conditions, Boyle's law holds good.
$P_1{V}_1+P_2{V}_2=PV$
where
$P_1=$ excess pressure inside first bubble
$P_2=$ excess pressure inside second bubble
$V_1=$ volume of first bubble
$V_2=$ volume of second bubble
$P=$ excess pressure of new bubble formed
$V=$ volume of new bubble formed

$P_1{V}_1+P_2{V}_2=PV$

$\Rightarrow \frac{4T}{r_1}\times \frac{4}{3}\pi r_1^3+\frac{4T}{r_2}\times \frac{4}{3}\pi r_2^3=\frac{4T}{R}\times \frac{4}{3}\pi R^3$

$\Rightarrow R=\sqrt{r_1^2+r_2^2}=\sqrt{3^2+4^2}=5cm$