- When a drop of water is broken into smaller drops, the total amount of water (and thus its total mass and volume) remains conserved, assuming no evaporation or loss during the process. If a large drop of volume V breaks into two smaller drops with volumes V1 and V2, then V = V1 + V2. - Surface area is related to volume, but it is not conserved. Breaking a drop into smaller drops increases the total surface area. For example, if a spherical drop of radius R breaks into two equal spherical drops of radius r, the total volume is conserved (V_large = 2 V_small), which implies R³ = 2r³. The surface area of the large drop is A_large = 4πR², and the total surface area of the two small drops is 2 A_small = 2 4πr² = 8πr². Since R³ = 2r³, R = r (2)^(1/3). Then R² = r² (2)^(2/3). So A_large = 4π r² (2)^(2/3). The total surface area of the two small drops is 8πr². Since (2)^(2/3) ≈ 1.587 and 8πr² = 2 4πr², it is clear that 8πr² > 4πR². - Surface energy is directly proportional to surface area (Surface Energy = Surface Tension Surface Area). Since breaking the drop increases the total surface area, the total surface energy of the two small drops is greater than the surface energy of the single large drop. This increase in surface energy is why work must be done to break a drop. - Radius is not conserved; the radius of the smaller drops is less than the radius of the original drop (r