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(A) Given: Radius of big drop $R = 1 \,cm = 10^{-2} \,m$, Number of small drops $n = 10^6$, Surface tension $T = 460 	imes 10^{-3} \,N/m$.Since the t

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$0.057$

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(A) Given: Radius of big drop $R = 1 \,cm = 10^{-2} \,m$, Number of small drops $n = 10^6$, Surface tension $T = 460 	imes 10^{-3} \,N/m$.Since the total volume remains constant:$n 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$10^6 	imes r^3 = R^3 \implies r = \frac{R}{10^2} = 10^{-4} \,m$.The energy expended is equal to the increase in surface area multiplied by surface tension:$W = \Delta A 	imes T = (n 	imes 4 \pi r^2 - 4 \pi R^2) 	imes T$$W = 4 \pi (n r^2 - R^2) T$Substitute $r = R/100$:$W = 4 \pi R^2 (n 	imes \frac{1}{10^4} - 1) T$$W = 4 	imes 3.14 	imes (10^{-2})^2 	imes (10^6 	imes 10^{-4} - 1) 	imes 460 	imes 10^{-3}$$W = 4 	imes 3.14 	imes 10^{-4} 	imes (100 - 1) 	imes 0.46$$W = 12.56 	imes 10^{-4} 	imes 99 	imes 0.46 \approx 0.057 \,J$ (Wait, re-calculating: $4 	imes 3.14 	imes 10^{-4} 	imes 99 	imes 0.46 = 0.057 \,J$).

$A$ mercury drop of radius $1 \,cm$ is sprayed into $10^6$ drops of equal size. The energy expended in joules is (Surface tension of mercury is $460 \times 10^{-3} \,N/m$)

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