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Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionEasy2 minPYQ_2020

PhysicsEasysingle choice

A water drop is divided into eight equal droplets. The pressure difference between the inner and outer side of the big drop will be

Options:

Answer:

Solution:

When break into eight drop radius of small drop is $r = \frac{R}{n^{\frac{1}{3}}} \Rightarrow r = \frac{R}{8^{\frac{1}{3}}} = \frac{R}{2}$

For large drop $Δ P_{1} = \frac{2 T}{R}$

For small drop $Δ P_{2} = \frac{2 T}{r} = \frac{2 T}{(\frac{R}{2})} = 2 (\frac{2 T}{R})$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2020

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