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Question

(D) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet. The surface area of each small droplet is $A_s = 4\pi r^2 = 1

Vedclass · Accepted Answer

$336 	imes 10^{-9} \,J$

Vedclass · Answer

(D) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet. The surface area of each small droplet is $A_s = 4\pi r^2 = 10^{-7} \,m^2$.Since the volume is conserved, $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$, which gives $R^3 = 64r^3$, so $R = 4r$.The surface area of the large drop is $A_L = 4\pi R^2 = 4\pi (4r)^2 = 16(4\pi r^2) = 16 	imes 10^{-7} \,m^2$.The total surface area of $64$ droplets is $A_{total} = 64 	imes 10^{-7} \,m^2$.The increase in surface area is $\Delta A = A_{total} - A_L = (64 - 16) 	imes 10^{-7} = 48 	imes 10^{-7} \,m^2$.The increase in surface energy is $\Delta U = T 	imes \Delta A$, where $T = 0.07 \,N/m$.$\Delta U = 0.07 	imes 48 	imes 10^{-7} = 3.36 	imes 10^{-7} \,J = 336 	imes 10^{-9} \,J$.

$A$ water drop breaks into $64$ identical droplets, each with a surface area of $10^{-7} \,m^2$. If the surface tension of water is $0.07 \,N/m$, what is the increase in surface energy during this process?

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