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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
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The correct answer is:
$(4)^{\frac{1}{3}} \mathrm{~W}$
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 $\mathrm{A}=4 \pi \mathrm{r}^2$
$\therefore \quad \mathrm{A} \propto \mathrm{V}^{\frac{2}{3}}$
$\therefore \quad \mathrm{W} \propto \mathrm{V}^{\frac{2}{3}}$
$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{\mathrm{V}^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}}$
$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{(2 \mathrm{~V})^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}} \quad \ldots .(\because \mathrm{V}=2 \mathrm{~V})$
$\therefore \quad \frac{\mathrm{W}^{\prime}}{\mathrm{W}}=2^{\frac{2}{3}}=4^{\frac{1}{3}}$
$\therefore \quad \mathrm{W}^{\prime}=4^{\frac{1}{3}} \mathrm{~W}$
Volume of sphere is $\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3$
Area of sphere is $\mathrm{A}=4 \pi \mathrm{r}^2$
$\therefore \quad \mathrm{A} \propto \mathrm{V}^{\frac{2}{3}}$
$\therefore \quad \mathrm{W} \propto \mathrm{V}^{\frac{2}{3}}$
$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{\mathrm{V}^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}}$
$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{(2 \mathrm{~V})^{\frac{2}{3}}}{\mathrm{~V}^{\frac{2}{3}}} \quad \ldots .(\because \mathrm{V}=2 \mathrm{~V})$
$\therefore \quad \frac{\mathrm{W}^{\prime}}{\mathrm{W}}=2^{\frac{2}{3}}=4^{\frac{1}{3}}$
$\therefore \quad \mathrm{W}^{\prime}=4^{\frac{1}{3}} \mathrm{~W}$
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