
From the equation of continuity, v2=2v1.
Applying the Bernoulli's Theorem,
P1+2ρv12=P2+2ρ4(v1)2 ⇒P1−P2=23ρv12
⇒4500=23×750×v12⇒v1=2ms−1
Now volume flow rate,Q=Av1=1.2×10−2×2=24×10−3m3s−1
JEE Main 2022 — Physics Mechanics
A liquid of density 750kgm−3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1=1.2×10−2m2 to A2=2A1. The pressure difference between the wide and narrow sections of the pipe is 4500Pa. The rate of flow of liquid is _____ ×10−3m3s−1.
Held on 28 Jun 2022 · Verified 6 Jul 2026.
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