# The surface tension of a soap solution is $30×{10}^{-3}{\mathrm{Nm}}^{-1}$. The work done in stretching a bubble of this solution of surface area $5\mathrm{cm}×5\mathrm{cm}$, to an area of $10\mathrm{cm}×10\mathrm{cm}$, is

1. A

$4.5×{10}^{-4}\mathrm{J}$

2. B

$6.0×{10}^{-4}\mathrm{J}$

3. C

$4.5×{10}^{-5}\mathrm{J}$

4. D

$7.5×{10}^{-4}\mathrm{J}$

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### Solution:

Work done

$\begin{array}{l}=2\left({\mathrm{A}}_{2}-{\mathrm{A}}_{1}\right)×\mathrm{S}\\ =2\left(10×10-5×5\right)×{10}^{-4}{\mathrm{m}}^{2}×30×{10}^{-3}{\mathrm{Nm}}^{-1}\\ =4.5×{10}^{-5}\mathrm{J}.\end{array}$

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