Drop of liquid of density \\rho is floating half immersed in a liquid of density \\rho_{0} . If surface tension of liquid is s, the radius of drop is

Correct Answer is : √3sg(2ρ−ρ0)

\\sqrt{{3 s}/{g(\\rho-\\rho_{0})}}

\\sqrt{{3 s}/{g(2\\rho-\\rho_{0})}}

\\sqrt{{3 s}/{g(3\\rho-\\rho_{0})}}

\\sqrt{{3 s}/{g(4\\rho-\\rho_{0})}}

CGPET 2019 Solved Paper

Section: Physics

Question : 39 of 150

Marks: +1, -0

Drop of liquid of density ρ is floating half immersed in a liquid of density ρ0. If surface tension of liquid is s, the radius of drop is

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