Question

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125

Answer 1

Correct option: (d)
Let the excess pressure inside the second bubble be P.
​∴ Excess pressure inside the first bubble = 2P
Let the radius of the second bubble be R.
Let the radius of the first bubble be x.

$$\begin{gathered} \mathrm{Excess}\mathrm{pressure}\mathrm{inside}\mathrm{the}2\mathrm{nd}\mathrm{soap}\mathrm{bubble}: \\ \mathrm{P}=\frac{4\mathrm{S}}{\mathrm{R}}...\left(1\right) \\ \mathrm{Excess}\mathrm{pressure}\mathrm{inside}\mathrm{the}1\mathrm{st}\mathrm{soap}\mathrm{bubble}: \\ 2\mathrm{P}=\frac{4\mathrm{S}}{\mathrm{x}} \\ \mathrm{From}\left(1\right),\mathrm{we}\mathrm{get}: \\ 2\left(\frac{4\mathrm{S}}{\mathrm{R}}\right)=\frac{4\mathrm{S}}{\mathrm{x}} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{R}}{2} \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{bubble}=\frac{4}{3}{\mathrm{\pi x}}^3 \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{bubble}=\frac{4}{3}{\mathrm{\pi R}}^3 \\ \Rightarrow \frac{4}{3}{\mathrm{\pi x}}^3=\mathrm{n}\frac{4}{3}{\mathrm{\pi R}}^3 \\ \Rightarrow {\mathrm{x}}^3={\mathrm{nR}}^3 \\ \Rightarrow {\left(\frac{\mathrm{R}}{2}\right)}^3={\mathrm{nR}}^3 \\ \Rightarrow \mathrm{n}=\frac{1}{8}=0.125 \end{gathered}$$