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PhysicsFluidSurface TensionMedium2 minQB

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A mercury droplet of radius R and surface tension $α$ is broken into 8 smaller droplets of equal size. The work done by the external agency is :

Answer:

Solution:

$\frac{4}{3} \times π R^{3} = 8 \times \frac{4}{3} π (r)^{3} r = R /2$

Change in surface area $= 8 \times 4 π [\frac{R}{2}]^{2} - 4 π R^{2} = 4 π R^{2}$

$W = α .4 π R^{2} .$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: QB

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