A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7 . 5 × 10 - 4 N cm - 1 . The radius of drop in cm will be : (Take : g = 10 ms - 2 )

Step-by-step solution

Balancing the forces on drop Buoyant force + surface tension = mg σ V 2 g + 2 π RT = ρ V g 2 π RT = 2 ρ - σ 2 · 4 3 π R 3   g ;             As  V = 4 3 π R 3 ⇒ R 3 = 3 T 2 ρ - σ g ⇒ R = 3 × 7 . 5 × 10 - 2   N - m - 1 2 ρ - σ × 10 R = 3 20 2 ρ - σ   m = 15 2 ρ - σ   cm