PhysicsMedium148×since 2002Q7993A drop of liquid of density ρ\rhoρ is floating half immersed in a liquid of density σ{\sigma}σ and surface tension 7.5×10−47.5 \times 10^{-4}7.5×10−4 Ncm^−-−1. The radius of drop in cm\mathrm{cm}cm will be : (g = 10 ms^−-−2)A15(2ρ−σ)\frac{15}{\sqrt{(2 \rho-\sigma)}}(2ρ−σ)15B15(ρ−σ)\frac{15}{\sqrt{(\rho-\sigma)}}(ρ−σ)15C32(ρ−σ)\frac{3}{2 \sqrt{(\rho-\sigma)}}2(ρ−σ)3D320(2ρ−σ)\frac{3}{20 \sqrt{(2 \rho-\sigma)}}20(2ρ−σ)3Check answerSkip