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Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionEasy2 minPYQ_2026

PhysicsEasysingle choice

Under isothermal conditions, a single drop of radius $R$ is formed from two mercury drops of radii $R_{1}$ and $R_{2}$ . After combination, the relation between $R$ , $R_{1}$ and $R_{2}$ is

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Answer:

Solution:

Since mercury is an incompressible liquid, the total mass and density remain constant under isothermal conditions. Thus, the total volume of the drops is conserved.

Volume of the new drop = Sum of volumes of the two original drops

$\frac{4}{3} π R^{3} = \frac{4}{3} π R_{1}^{3} + \frac{4}{3} π R_{2}^{3}$

$R^{3} = R_{1}^{3} + R_{2}^{3}$

Answer: $R^{3} = R_{1}^{3} + R_{2}^{3}$

Stream:MHT_CETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2026

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