Physics - Fluid Question with Solution | TestHub

PhysicsFluidMedium2 minQB

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If $r = 5 \times 1 0^{- 4} m, ρ = 1 0^{3} kgm^{- 3}, g = 10 ms^{- 2}, T = 0.11 Nm^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately

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Drop detaches from dropper When :

$F_{V} = mg \frac{2 π r ^{2} T}{R} = ρ \cdot \frac{4}{3} π R^{3} g R = (\frac{3 r ^{2} T}{2 g ρ})^{1/4} = [\frac{3 \times ( 5 \times 1 0 ^{- 4} ) ^{2} \times ( 0.11 )}{2 \times 10 \times 1000}]^{1/4} \approx 1.4 \times 10 - 3 m$

Stream:JEESubject:PhysicsTopic:Fluid

⏱ 2mℹ️ Source: QB

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Physics - Fluid Question with Solution | TestHub

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