A mercury drop of radius I cm is broken into 106 droplets of equal size. The work done is (T = 35 ×10-2 N/m)

Initial radius of drop, R = 1 cm = 1×10-2mSuppose, r is the radius of droplets.Since, volume remain constant.   43πR3 = 106×43πr3[Since, number of droplets = 106]∴  r = 1×10-2102=10-4      [∵ R = 1 ×10-2 m]Increase in surface area,∆S = 106×4πr2-4πR2      = 106×4×π×10-8-4π×10-4       = 4×9.9×π×10-3∴ Work done = Increase in surface area × Surface tension      = 4 × 9.9 ×π×10-3×35×10-2       = 4.35×10-2 J