In an isothermal process,2 water drops of rad | vedclass

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(B) The change in surface energy $\Delta U$ is given by $\Delta U = T \Delta A$,where $\Delta A$ is the change in surface area.Initial surface area $A

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$0.5$

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(B) The change in surface energy $\Delta U$ is given by $\Delta U = T \Delta A$,where $\Delta A$ is the change in surface area.Initial surface area $A_i = 2 	imes (4 \pi r^2) = 8 \pi r^2$.Using volume conservation,$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$,so $R = 2^{1/3} r$.Final surface area $A_f = 4 \pi R^2 = 4 \pi (2^{1/3} r)^2 = 4 \pi (2^{2/3}) r^2$.Change in area $\Delta A = A_f - A_i = 4 \pi r^2 (2^{2/3} - 2)$.Since the surface area decreases,energy is released: $\Delta U = T (A_i - A_f) = T 	imes 4 \pi r^2 (2 - 2^{2/3})$.Given $T = 0.1 \, N/m$ and $r = 10^{-3} \, m$:$\Delta U = 0.1 	imes 4 \pi 	imes (10^{-3})^2 	imes (2 - 1.587) = 0.4 \pi 	imes 10^{-6} 	imes 0.413 \approx 0.519 \, \mu J$.Rounding to the nearest option,the energy change is approximately $0.5 \, \mu J$.

In an isothermal process,$2$ water drops of radius $1 \, mm$ are combined to form a bigger drop. Find the energy change (in $\mu J$) in this process if the surface tension $T = 0.1 \, N/m$.

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