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

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$4.5 \times 10^{- 4} J$

$6.0 \times 10^{- 4} J$

$4.5 \times 10^{- 5} J$

$7.5 \times 10^{- 4} J$

answer is C.

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Detailed Solution

Work done

$\begin{array}{l} = 2 (A_{2} - A_{1}) \times S \\ = 2 (10 \times 10 - 5 \times 5) \times 10^{- 4} m^{2} \times 30 \times 10^{- 3} {Nm}^{- 1} \\ = 4.5 \times 10^{- 5} J . \end{array}$

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