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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

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Solution

## 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. $\mathrm{Excess}\mathrm{pressure}\mathrm{inside}\mathrm{the}2\mathrm{nd}\mathrm{soap}\mathrm{bubble}:\phantom{\rule{0ex}{0ex}}\mathrm{P}=\frac{4\mathrm{S}}{\mathrm{R}}...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Excess}\mathrm{pressure}\mathrm{inside}\mathrm{the}1\mathrm{st}\mathrm{soap}\mathrm{bubble}:\phantom{\rule{0ex}{0ex}}2\mathrm{P}=\frac{4\mathrm{S}}{\mathrm{x}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2\left(\frac{4\mathrm{S}}{\mathrm{R}}\right)=\frac{4\mathrm{S}}{\mathrm{x}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{x}=\frac{\mathrm{R}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{bubble}=\frac{4}{3}{\mathrm{\pi x}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{bubble}=\frac{4}{3}{\mathrm{\pi R}}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{3}{\mathrm{\pi x}}^{3}=\mathrm{n}\frac{4}{3}{\mathrm{\pi R}}^{3}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{x}}^{3}={\mathrm{nR}}^{3}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{\mathrm{R}}{2}\right)}^{3}={\mathrm{nR}}^{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{n}=\frac{1}{8}=0.125\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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