An oil drop of density ρ is floating half immersed inside a liquid of density σ. If the surface tension of the liquid be T, then the radius R of the oil drop is:
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a
3⊤g(2ρ+σ)
b
3⊤2g(ρ−σ)
c
3Tg(2ρ−σ)
d
none of these
answer is C.
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Detailed Solution
In floating condition weight of drop = upthrust +force due to surface tension.43πR3ρg=1243πR3σg+2πRTR=3Tg(2ρ−σ)