A soap bubble of surface tension 0.04 N/m is | vedclass

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(A) The work done in blowing a soap bubble is given by the change in surface energy: $W = \Delta U = S 	imes \Delta A$.Since a soap bubble has two su

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$11304$

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(A) The work done in blowing a soap bubble is given by the change in surface energy: $W = \Delta U = S 	imes \Delta A$.Since a soap bubble has two surfaces, the change in area is $\Delta A = 2 	imes (4 \pi r_2^2 - 4 \pi r_1^2) = 8 \pi (r_2^2 - r_1^2)$.Given: $S = 0.04 \ N/m$, $r_1 = 3.5 \ cm = 0.035 \ m$, $r_2 = 7 \ cm = 0.07 \ m$.$W = 0.04 	imes 8 	imes \frac{22}{7} 	imes [(0.07)^2 - (0.035)^2]$.$W = 0.32 	imes \frac{22}{7} 	imes [0.0049 - 0.001225] = 0.32 	imes \frac{22}{7} 	imes 0.003675$.$W = 0.32 	imes 22 	imes 0.000525 = 0.003696 \ J = 3696 \ \mu J$.Equating to the given expression: $15000 - x = 3696$.$x = 15000 - 3696 = 11304$.

$A$ soap bubble of surface tension $0.04 \ N/m$ is blown to a diameter of $7 \ cm$. If $(15000 - x) \ \mu J$ of work is done in blowing it further to make its diameter $14 \ cm$, then the value of $x$ is . . . . . . . (Take $\pi = 22/7$)

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