Solution:
Volume conservation gives \(R = 5r\). Initial surface area is \(A_i = 125 \times 4\pi r^2 = 500\pi r^2\) and final is \(A_f = 4\pi R^2 = 100\pi r^2\). The decrease in area is \(400\pi r^2\), so loss in surface energy is \(T \times 400\pi r^2\).
Leave a Reply