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  4. A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5 × 10-4 Ncm -1. The radius of drop in cm will be: (. Take: .g =10 m / s 2)

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Q. A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5 \times 10^{-4} Ncm ^{-1}$. The radius of drop in $cm$ will be : $\left(\right.$ Take : $\left.g =10 \,m / s ^2\right)$

JEE MainJEE Main 2022Mechanical Properties of Fluids

Solution:

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Buoyant force $+$ surface tension $= mg$
$ \sigma \frac{ V }{2} g +2 \pi R T=\rho V g $
$2 \pi R T=\frac{(2 \rho-\sigma)}{2} \cdot \frac{4}{3} \pi R ^3 g ; \left[ V =\frac{4}{3} \pi R ^3\right]$
$R^3=\frac{3 T}{(2 \rho-\sigma) g} \Rightarrow R=\sqrt{\frac{3 \times 7.5 \times 10^{-2} N - m ^{-1}}{(2 \rho-\sigma) \times 10}}$
$R =\frac{3}{20 \sqrt{(2 \rho-\sigma)}} m =\frac{15}{\sqrt{2 \rho-\sigma}} cm$