A drop of liquid of radius R=10^{-2} ,m havin | vedclass

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(B) The initial surface area of the large drop is $A_i = 4 \pi R^2$. The final surface area of $K$ small drops of radius $r$ is $A_f = K 	imes 4 \pi

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$6$

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(B) The initial surface area of the large drop is $A_i = 4 \pi R^2$. The final surface area of $K$ small drops of radius $r$ is $A_f = K 	imes 4 \pi r^2$.Since the volume is conserved, $\frac{4}{3} \pi R^3 = K 	imes \frac{4}{3} \pi r^3$, which implies $r = R K^{-1/3}$.Substituting $r$ into the final area: $A_f = K 	imes 4 \pi (R K^{-1/3})^2 = 4 \pi R^2 K^{1/3}$.The change in surface energy is $\Delta U = S(A_f - A_i) = S 	imes 4 \pi R^2 (K^{1/3} - 1)$.Given $\Delta U = 10^{-3} \,J$, $R = 10^{-2} \,m$, and $S = \frac{0.1}{4 \pi} \,Nm^{-1}$:$10^{-3} = \left(\frac{0.1}{4 \pi}ight) 	imes 4 \pi 	imes (10^{-2})^2 	imes (K^{1/3} - 1)$.$10^{-3} = 0.1 	imes 10^{-4} 	imes (K^{1/3} - 1) = 10^{-5} 	imes (K^{1/3} - 1)$.$K^{1/3} - 1 = \frac{10^{-3}}{10^{-5}} = 10^2 = 100$.$K^{1/3} = 101 \approx 100 = 10^2$.$K = (10^2)^3 = 10^6$.Since $K = 10^\alpha$, we have $\alpha = 6$.

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

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