Question

if the work done in blowing a soap bubble of volume 'v' is W, then the work is done in blowing a soap bubble of volume '2v'is​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Work\:Done=\sqrt{2}\:W}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Work \: done \: in \: blowing \: bubble \: of \: volume (v) = W \\ \\\red{\underline \bold{To \: Find :}} \\ \tt: \implies Work \: done \: in \: blowing \: bubble \: of \: volume(2v) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Volume \: of \: bubble = v \\ \\ \tt: \implies \frac{4}{3} \pi {r_{1} }^{3} = v - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Volume \: of \: buble = 2v \\ \\ \tt: \implies \frac{4}{3} \pi { r_{2}}^{2} = 2v \\ \\ \tt: \implies \frac{2}{3} \pi { r_{2} }^{2} = v - - - - - (2) \\ \\ \text{From \: (1) \: and \: (2)} \\ \\ \tt: \implies \frac{4}{3}\pi r_{1}^{3} = \frac{2}{3}\pi { r_{2}}^{3} \\ \\ \tt: \implies \bigg( \frac{ r_{2} }{ r_{1} } \bigg) = \sqrt[3]{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: done \: of \: blowing \: v \: volume \: bubble = surface \: area \times surface \: tension \\ \\ \tt: \implies W= 4\pi r_{1}^{2} \times T- - - - - (3)$

$\bold{Similarly : } \\ \tt: \implies Work \: done \: in \: blowing \: 2v \: volume \: bubble = surface \: area \times surface \: tension \\ \\ \tt: \implies W_{1} = 4\pi { r_{2}}^{2} \times T - - - - - (4)\\ \\ \text{Dividing \: (3) \: and \: (4)} \\ \\ \tt: \implies \frac{ W}{ W_{1} } = \frac{4\pi { r_{1}}^{2} \times t }{ 4\pi r_{2} \times t } \\ \\ \tt: \implies \frac{W}{ W_{1}} = \bigg(\frac{ r_{1}}{ r_{2} } \bigg)^{2} \\ \\ \tt: \implies \frac{ W_{1}}{ W} = \bigg( \frac{ r_{2}}{ r_{1}} \bigg ) ^{2} \\ \\ \tt: \implies \frac{W_{1}}{W} = (\sqrt[3]{2} )^{2} \\ \\ \green{\tt: \implies W_{1} = \sqrt{2} W} \\ \\ \green{\tt \therefore Work \: done \:in \: blowing \: volume \: of \: bubble \: 2v \: is \: \sqrt{2} W}$