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← JEE Main 2022 · 25 Jul · Shift 2
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JEE Main 2022 · 25 Jul · Shift 2, Physics question 33

A drop of liquid of density ρ\rho is floating half immersed in a liquid of density σ\sigma and surface tension 7.5×10−47.5\times10^{-4} Ncm−1^{-1}. The radius of drop in cm will be : (g=10g=10 ms−2^{-2})

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Answer and worked solution

Official answer: A

Balance. Weight equals buoyancy on half the volume plus surface tension along the waterline: 43πr3ρg=23πr3σg+2πrT\frac43\pi r^3\rho g=\frac23\pi r^3\sigma g+2\pi rT, so r2=3Tg(2ρ−σ)r^2=\dfrac{3T}{g(2\rho-\sigma)}.

Numbers. T=0.075T=0.075 N m−1^{-1}: r=0.152ρ−σr=\dfrac{0.15}{\sqrt{2\rho-\sigma}} m =152ρ−σ=\dfrac{15}{\sqrt{2\rho-\sigma}} cm.

(A)

Topics:Surface tensionBuoyancy and Archimedes' principle
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