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Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionMedium2 minPYQ_2022

PhysicsMediumsingle choice

The radii of two mercury drops are $R_{1}$ and $R_{2}$ . Under isothermal conditions, a single drop of radius $R$ is formed from them. The relations between $R, R_{1}$ and $R_{2}$ is

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Answer:

Solution:

Total volume remains same, $\frac{4}{3} π R^{3} = \frac{4}{3} π R_{1}^{3} + \frac{4}{3} π R_{2}^{3} R^{3} = R_{1}^{3} + R_{2}^{3}$

Stream:MHT_CETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2022

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