A liquid drop having surface energy ' E ' is spread into 512 droplets of same size. The final surface energy of the droplets is

Step-by-step solution

Surface area of drop, A ₁=4 R ^2 Surface area of 512 droplets, A ₂=512 (4 r ^2 ) volume of drop = n (volume of droplet) 4 3 R ^3=512 4 3 r ^3 R =8 r A ₂= 512 (4 R ^2 ) 64 A ₂=8 (4 R ^2 ) Surface energy Area E₂ E₁ = A₂ A₁ = 8 (4 R^2 ) 4 R^2 E ₂=8 E ₁=8 E . ( E₁=E )