The work done in increasing the diameter of a | vedclass

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(C) soap bubble has two surfaces (inner and outer), so the change in surface area is $\Delta A = 2 	imes (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1

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$264 	imes 10^{-6} \ J$

Vedclass · Answer

(C) soap bubble has two surfaces (inner and outer), so the change in surface 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: Surface tension $T = 3.5 	imes 10^{-2} \ N/m$, initial radius $r_1 = 1 \ cm = 1 	imes 10^{-2} \ m$, final radius $r_2 = 2 \ cm = 2 	imes 10^{-2} \ m$.Work done $W = T 	imes \Delta A = T 	imes 8\pi(r_2^2 - r_1^2)$.$W = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14 	imes [(2 	imes 10^{-2})^2 - (1 	imes 10^{-2})^2]$.$W = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14 	imes (4 	imes 10^{-4} - 1 	imes 10^{-4})$.$W = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14 	imes 3 	imes 10^{-4}$.$W = 263.76 	imes 10^{-6} \ J \approx 264 	imes 10^{-6} \ J$.

The work done in increasing the diameter of a soap bubble from $2 \ cm$ to $4 \ cm$ is (Surface tension of soap solution $= 3.5 \times 10^{-2} \ N/m$)

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