Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of soap bubble from 3.5 cm to 7cm will be:
Take $[π = \frac{22}{7}]$

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$9.24 \times 10^{- 4} J$

$5.76 \times 10^{- 4} J$

$18.48 \times 10^{- 4} J$

$0.72 \times 10^{- 4} J$

answer is A.

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Detailed Solution

Work done
$\begin{array}{l} = S \times ΔA \\ = S \times 2 \times 4 π (r_{2}^{2} - r_{1}^{2}) \\ = 16 π \times 10^{- 2} (7^{2} - 3 . 5^{2}) \times 10^{- 4} \\ = 1847 \times 10^{- 6} J \\ = 18.47 \times 10^{- 4} J \end{array}$

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