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 )

Question

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 ), 15 2 ρ - σ, 15 ρ - σ, 3 2 ρ - σ, 3 20 2 ρ - σ

MARKS App · Accepted Answer

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

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