If 8 identical droplets are formed from a sin | vedclass

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(A) Let $R$ be the radius of the large drop and $r$ be the radius of the small droplets. Given $R = 2\,mm = 2 	imes 10^{-3}\,m$ and $n = 8$.Since the

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$23.42$

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(A) Let $R$ be the radius of the large drop and $r$ be the radius of the small droplets. Given $R = 2\,mm = 2 	imes 10^{-3}\,m$ and $n = 8$.Since the volume remains constant,$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$,which implies $r = \frac{R}{n^{1/3}} = \frac{R}{8^{1/3}} = \frac{R}{2}$.The change in surface energy $\Delta U$ is given by $\Delta U = T 	imes \Delta A = T(n 	imes 4\pi r^2 - 4\pi R^2)$.Substituting $r = R/2$,we get $\Delta U = T(n 	imes 4\pi (R/2)^2 - 4\pi R^2) = 4\pi R^2 T (n/4 - 1)$.Alternatively,using the formula $\Delta U = 4\pi R^2 T (n^{1/3} - 1)$:$\Delta U = 4 	imes 3.14159 	imes (2 	imes 10^{-3})^2 	imes 0.465 	imes (8^{1/3} - 1)$.$\Delta U = 4 	imes 3.14159 	imes 4 	imes 10^{-6} 	imes 0.465 	imes (2 - 1)$.$\Delta U = 16 	imes 3.14159 	imes 0.465 	imes 10^{-6} \approx 23.376 	imes 10^{-6}\,J$.Rounding to the nearest provided option,the change is $23.4\,\mu J$.

If $8$ identical droplets are formed from a single drop of radius $2\,mm$,what is the change in energy in $\mu J$? (Surface tension $T = 0.465\,J/m^2$)

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