A soap bubble is blown to a diameter of 7 cm | vedclass

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

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$7$

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(B) The work done in blowing a soap bubble is given by the change in surface energy: $W = S 	imes \Delta A$.Since a soap bubble has two surfaces,the change in area is $\Delta A = 2 	imes 4\pi (r_2^2 - r_1^2) = 8\pi (r_2^2 - r_1^2)$.Given: $S = 40 \ dyne/cm$,$W = 36960 \ erg$,and initial diameter $d_1 = 7 \ cm$,so initial radius $r_1 = 3.5 \ cm = \frac{7}{2} \ cm$.Substituting the values: $36960 = 40 	imes 8 	imes \frac{22}{7} 	imes (r_2^2 - (3.5)^2)$.$36960 = 320 	imes \frac{22}{7} 	imes (r_2^2 - 12.25)$.$36960 = \frac{7040}{7} 	imes (r_2^2 - 12.25)$.$r_2^2 - 12.25 = \frac{36960 	imes 7}{7040} = 36.75$.$r_2^2 = 36.75 + 12.25 = 49$.$r_2 = 7 \ cm$.

$A$ soap bubble is blown to a diameter of $7 \ cm$. $36960 \ erg$ of work is done in blowing it further. If the surface tension of the soap solution is $40 \ dyne/cm$,then the new radius is . . . . . . $cm$. Take $\pi = \frac{22}{7}$.

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