If 1000 droplets of water of surface tension 0 . 07 N m - 1 . having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is- Take π = 22 7

Step-by-step solution

Let the radius of bigger droplet is R , then by volume conservation: 4 3 π R 3 = 1000 4 3 π 1 3 ⇒ R = 10   mm Final potential energy: U f = 1000 4 π R 2 T = 1000 × 4 π 10 × 10 - 3 2 T Initial potential energy: U i = 4 π × r 2 T = 4 π × 10 - 3 2 T Therefore, change in potential energy, ∆ E = T × 1000 × 4 π 10 - 3 2 - T × 4 π 10 × 10 - 3 2 ∆ E = 4 × π × 7 × 10 - 2 1000 - 100 × 10 - 6 ∆