If work done in blowing a soap bubble of volume ' \(\mathrm{V}\) ' is \(\mathrm{W}\) then the work done in blowing the bubble of volume \(2 \mathrm{~V}\) from same soap solution is

The work done is given as \(\mathrm{W}=\mathrm{T} \Delta \mathrm{A}\)
Volume of sphere is \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3\)
Area of sphere is \(\mathrm{A}=4 \pi \mathrm{r}^2\)
\(\therefore \quad \mathrm{A} \propto \mathrm{V}^{\frac{2}{3}}\)
\(\therefore \quad \mathrm{W} \propto \mathrm{V}^{\frac{2}{3}}\)
\(\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{\mathrm{V}^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}}\)
\(\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{(2 \mathrm{~V})^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}} \quad \ldots .(\because \mathrm{V}=2 \mathrm{~V})\)
\(\therefore \quad \frac{\mathrm{W}^{\prime}}{\mathrm{W}}=2^{\frac{2}{3}}=4^{\frac{1}{3}}\)
\(\therefore \quad \mathrm{W}^{\prime}=4^{\frac{1}{3}} \mathrm{~W}\)