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Physics - Fluid Question with Solution | TestHub

PhysicsFluidBernoulli's theorem and applicationsMedium2 minQB

PhysicsMediumsingle choice

Passage / Comprehension

The inner cross-sectional area of the fixed horizontal tube is A; its right-hand section has area KA, where K < 1. The two sections join smoothly without sharp turns. A liquid of density ρ and negligible viscosity is pushed by a piston on the left. The water leaving the tube hits a vertical wall and spreads into a tapering film.

Shared passage diagram

Value of horizontal force which is exerted by tube on support

Options:

Answer:

Solution:

FBD of water in tube:

FBD of water in tube: applied force F to the right and horizontal support force F_HS to the left Let V be the piston speed. Continuity gives exit speed: $u = \frac{V}{K}$ From Question 11: $ρ A V^{2} = \frac{2 F K ^{2}}{1 - K ^{2}}$ Horizontal momentum balance on water in the tube: $F - F_{hs} = ρ A V^{2} (\frac{1}{K} - 1)$ Substitute V and simplify: $F_{hs} = F - \frac{2 F K}{1 + K} = F \frac{1 - K}{1 + K}$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Bernoulli's theorem and applications

⏱ 2mℹ️ Source: QB

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