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A soap bubble is blown to a diameter of $7 \mathrm{~cm} .36960 \mathrm{erg}$ of work is done in blowing it further. If surface tension of soap solution is $40 \mathrm{dyne} / \mathrm{cm}$ then the new radius is______ $\mathrm{cm}$ Take $\left(\pi=\frac{22}{7}\right)$
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The correct answer is:
7
$\begin{aligned} & \omega=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A} \\ & 36960 \mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^2-\left(\frac{7}{2}\right)^2\right] \mathrm{cm}^2 \\ & \mathrm{r}=7 \mathrm{~cm}\end{aligned}$
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