The excess pressure inside the first soap bubble of radius ' $\mathrm{R}_{1}$ ' is two times, that inside the second soap bubble of radius ' $\mathrm{R}_{2}$ '. The ratio of volumes of the first bubble to that of second bubble is

Question

Aicharya · Accepted Answer

Excess pressure $\quad P=\frac{4 T}{R}$ $	herefore \frac{P_{1}}{P_{2}}=\frac{R_{2}}{R_{1}}$ $\frac{P_{1}}{P_{2}}=2$ $\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi R_{1}^{2}}{\frac{4}{3} \pi R_{2}^{3}}=\frac{R_{1}^{3}}{R_{2}^{3}}=\left(\frac{1}{2}ight)^{3}=\frac{1}{8}$

The excess pressure inside the first soap bubble of radius ' $\mathrm{R}_{1}$ ' is two times, that inside…

Solution