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Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionMedium2 minQB

PhysicsMediumsingle choice

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

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Answer:

Solution:

$\frac{4}{3} \times π R^{3} = 8 \times \frac{4}{3} π (r)^{3}$ $r = R /2$ Change in surface area $= 8 \times 4 π [\frac{R}{2}]^{2} - 4 π R^{2} = 4 π R^{2}$ $W = a .4 π R^{2}$ .

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: QB

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