Balancing the forces on drop
Buoyant force + surface tension =mg
σ2Vg+2πRT=ρVg
2πRT=2(2ρ−σ)⋅34πR3g;[As V=34πR3]
⇒R3=(2ρ−σ)g3T⇒R=(2ρ−σ)×103×7.5×10−2N−m−1
R=20(2ρ−σ)3m=2ρ−σ15cm
A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5×10−4Ncm−1. The radius of drop in cm will be : (Take : g=10ms−2)
Held on 25 Jul 2022 · Verified 6 Jul 2026.
2ρ−σ15
ρ−σ15
2ρ−σ3
202ρ−σ3
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