The pressure inside two soap bubbles, (A) is 1.01 and that of (B) is 1.02 atmosphere respectively. The ratio…
The pressure inside two soap bubbles, (A) is 1.01 and that of (B) is 1.02 atmosphere respectively. The ratio of their respective radii (A to B) is (outside pressure $=1 \mathrm{~atm}$.)
$2: 1$
$4: 1$
$6: 1$
$8: 1$
Solution
Outside pressure $=1 \mathrm{~atm}$
Pressure inside first bubble $=1.01 \mathrm{~atm}$
Pressure inside second bubble $=1.02 \mathrm{~atm}$
$\therefore \quad$ Excess pressures will be $\Delta \mathrm{P}_{\mathrm{A}}=1.01-1=0.01 \mathrm{~atm}$ and $\Delta \mathrm{P}_{\mathrm{B}}=1.02-1=0.02 \mathrm{~atm}$ Now, $\Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{r} \propto \frac{1}{\Delta \mathrm{P}}$ $\therefore \quad \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{A}}}=\frac{0.02}{0.01}=\frac{2}{1}$