The radius R of the soap bubble is doubled un | vedclass

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(B) The surface energy $E$ of a soap bubble is given by the product of surface tension $T$ and the total surface area. Since a soap bubble has two sur

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$24$

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(B) The surface energy $E$ of a soap bubble is given by the product of surface tension $T$ and the total surface area. Since a soap bubble has two surfaces (inner and outer),the total surface area is $2 	imes 4 \pi R^{2} = 8 \pi R^{2}$.Initial surface energy $E_{i} = T 	imes (2 	imes 4 \pi R^{2}) = 8 \pi R^{2} T$.When the radius is doubled,the new radius becomes $R' = 2R$.The final surface energy $E_{f} = T 	imes (2 	imes 4 \pi (2R)^{2}) = T 	imes (2 	imes 4 \pi 	imes 4R^{2}) = 32 \pi R^{2} T$.The work done $W$ is the change in surface energy:$W = E_{f} - E_{i} = 32 \pi R^{2} T - 8 \pi R^{2} T = 24 \pi R^{2} T$.

The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble,the work done in doing so is given by: (in $pi R^{2} T$)

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