JEE Main 2023PhysicsMechanical Properties of FluidsEasyMCQ

JEE Main 2023 — Mechanical Properties of Fluids Question with Solution

JEE Main 2023 (13 Apr Shift 1)

Question

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at A is 1.5 cm2, and B is 25 mm2, if the speed of liquid at B is 60 cm s-1 then (PA–PB) is

(Given PA and PB are liquid pressures at A and B points.

Density ρ=1000 kg m-3

A and B are on the axis of tube)

Choose an option

Show full solutionCorrect option: C
Correct answer
C175 Pa

Step-by-step explanation

Using Bernoulli's theorem,

PA+12ρVA2=PB+12ρVB2   ...(i)

Using equation of continuity,

AAVA=ABVB

So, 1.5×10-4×VA=25×10-6×VB

VA=25×10-61.5×10-4×0.6

=110 m s-1

Substituting in equation (i)

⇒PA-PB=100020.62-0.12=10002×0.7×0.5

=175 Pa

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About this question

This is a previous-year question from JEE Main 2023, covering the Mechanical Properties of Fluids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.