The surface tension of soap solution is $3.5 \times 10^{- 2} N m^{- 1} .$ The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is ….. $\times 10^{- 4} J$ (take $π = 22 / 7$ )

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answer is 264.

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Detailed Solution

$W = 2 \times 4 π T [{(20 \times 10^{- 2})}^{2} - {(10 \times 10^{- 2})}^{2}]$
$= 8 \times \frac{22}{7} \times 3.5 \times 10^{- 2} \times 300 \times 10^{- 4} = 264 \times 10^{- 4} J$

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