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Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionMedium2 minPYQ_2021

PhysicsMediumsingle choice

Let ' $R_{1}$ ' and ' $R_{2}$ ' are radii of two mercury drops. A big mercury drop is formed from then under isothermal conditions. The radius of the resultant drop is

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Solution:

The volume of the bigger drop will be equal to the sum of the volumes of the smaller drops $\frac{4}{3} π R^{3} = \frac{4}{3} π R_{1}^{3} + \frac{4}{3} π R_{2}^{3} ∴ R = (R_{1}^{3} + R_{2}^{3})^{1/3}$

Stream:MHT_CETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2021

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