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(B) soap film has two surfaces. The change in surface area $\Delta A$ is given by $2 	imes l 	imes \Delta x$, where $l = 10 \,cm = 0.1 \,m$ and $\De

Vedclass · Accepted Answer

$13.0 	imes 10^{-5} \,J$

Vedclass · Answer

(B) soap film has two surfaces. The change in surface area $\Delta A$ is given by $2 	imes l 	imes \Delta x$, where $l = 10 \,cm = 0.1 \,m$ and $\Delta x = 1 \,mm = 10^{-3} \,m$.$\Delta A = 2 	imes 0.1 \,m 	imes 10^{-3} \,m = 2 	imes 10^{-4} \,m^2$.The work done $W$ is equal to the change in surface energy, given by $W = T 	imes \Delta A$.Given $T = 65 	imes 10^{-3} \,N/m$.$W = (65 	imes 10^{-3} \,N/m) 	imes (2 	imes 10^{-4} \,m^2) = 130 	imes 10^{-7} \,J = 1.3 	imes 10^{-5} \,J$.Wait, re-evaluating the provided surface tension value $65 	imes 10^{-2} \,N/m$ from the prompt: $W = (65 	imes 10^{-2}) 	imes (2 	imes 10^{-4}) = 130 	imes 10^{-6} = 1.3 	imes 10^{-5} \,J$. Given the options, if $T = 65 	imes 10^{-3} \,N/m$ (standard value), $W = 1.3 	imes 10^{-5} \,J$. If $T = 65 	imes 10^{-2} \,N/m$, $W = 1.3 	imes 10^{-4} \,J = 13.0 	imes 10^{-5} \,J$. Thus, option $B$ is correct.

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