Surface tension of a soap bubble is 2.0 times | vedclass

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(C) soap bubble has two surfaces (inner and outer),so its total surface area is $A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ in increasing

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$18.48$

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(C) soap bubble has two surfaces (inner and outer),so its total surface area is $A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ in increasing the radius from $R_1$ to $R_2$ is equal to the change in surface energy:$W = T 	imes \Delta A = T 	imes (A_2 - A_1) = T 	imes 8 \pi (R_2^2 - R_1^2)$.Given:$T = 2.0 	imes 10^{-2} \; N m^{-1}$$R_1 = 3.5 \; cm = 3.5 	imes 10^{-2} \; m$$R_2 = 7.0 \; cm = 7.0 	imes 10^{-2} \; m$Substituting the values:$W = 2.0 	imes 10^{-2} 	imes 8 	imes \frac{22}{7} 	imes [(7.0 	imes 10^{-2})^2 - (3.5 	imes 10^{-2})^2]$$W = 16 	imes 10^{-2} 	imes \frac{22}{7} 	imes (49 - 12.25) 	imes 10^{-4}$$W = 16 	imes 10^{-2} 	imes \frac{22}{7} 	imes 36.75 	imes 10^{-4}$$W = 16 	imes 10^{-2} 	imes 22 	imes 5.25 	imes 10^{-4}$$W = 18.48 	imes 10^{-4} \; J$.

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

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