From equation of continuity,
av1=2av2⇒v2=2v1
From Bernoulli's theorem,
P1+21ρv12+ρgh1=P2+21ρv22+ρgh2⇒P1−P2=ρ[g(h2−h1)+21(v22−v12)]⇒4100=800[10(−1)+213v12]⇒8004100+10=23v12⇒v1=4×3121=6363ms−1
Therefore, x=363.
An ideal fluid of density 800 kg m^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to a 2. The…
JEE Main 2022 — Physics Mechanics
2022integermedium
An ideal fluid of density 800kgm−3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to 2a. The pressure difference between the wide and narrow sections of pipe is 4100Pa. At wider section, the velocity of fluid is 6xms−1 for x=_____ . (Given g=10ms−2)
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Held on 26 Jun 2022 · Verified 6 Jul 2026.
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