A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm . If

Question

A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm . If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm , then the value of $x$ is $\_\_\_\_$ .
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(\pi=22 / 7)
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Professor PC Thomas Classes · Answer

A soap bubble has two free surfaces. The work done in expanding it equates to the change in total surface energy: $W = T 	imes \Delta A = T 	imes 2 	imes 4\pi (r_2^2 - r_1^2)$. The initial radius is $r_1 = 3.5 	ext{ cm} = 0.035 	ext{ m}$ and the final radius is $r_2 = 7 	ext{ cm} = 0.07 	ext{ m}$. The change in the square of radii is $r_2^2 - r_1^2 = (0.07)^2 - (0.035)^2 = 0.0049 - 0.001225 = 0.003675 	ext{ m}^2$. Substituting the values: $W = 8 	imes \frac{22}{7} 	imes 0.04 	imes 0.003675 = \frac{176}{7} 	imes 0.000147 = 176 	imes 0.000021 = 0.003696 	ext{ J}$. This equates to $3696 \mu	ext{J}$. We are given $W = 15000 - x$. Thus, $x = 15000 - 3696 = 11304$.