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PhysicsFluidSurface TensionMedium2 minPYQ_2020

PhysicsMediumsingle choice

small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

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Solution:

$2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} ∴ R = 2^{\frac{1}{3}} r$ Ratio of energies $\frac{E _{2}}{E _{1}} = \frac{4 π R ^{2} T}{2 \times π r ^{2} \times T}$ $\frac{E _{2}}{E _{1}} = \frac{R ^{2}}{2 r ^{2}} = \frac{2 ^{\frac{2}{3}} r ^{2}}{2 r ^{2}} = 2^{\frac{2}{3} - 1} = 2^{- \frac{1}{3}}$

Stream:MHT_CETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2020

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