A liquid of density 600 kg·m⁻³ flows steadily through a horizontal tube of varying cross-section. At point A, the cross-sectional area is 1.0 cm² and the liquid speed is 10 cm·s⁻¹. At point B, the cross-sectional area is 20 mm². What is the pressure difference P(A)-P(B)?
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Question

A liquid of density 600 kg·m⁻³ flows steadily through a horizontal tube of varying cross-section. At point A, the cross-sectional area is 1.0 cm² and the liquid speed is 10 cm·s⁻¹. At point B, the cross-sectional area is 20 mm². What is the pressure difference P(A)-P(B)?

Options

G1 Physics · Accepted Answer

First convert the area at B: 20 mm² = 0.20 cm². Using continuity, A(A)v(A) = A(B)v(B). Therefore, 1.0×10 = 0.20×v(B), so v(B) = 50 cm·s⁻¹. Convert speeds to SI units: v(A) = 0.10 m·s⁻¹, v(B) = 0.50 m·s⁻¹. The two points are at the same height, so Bernoulli's equation gives P(A)+(1 ÷ 2)ρv(A)² = P(B)+(1 ÷ 2)ρv(B)². Hence, P(A)-P(B) = (1 ÷ 2)ρ[v(B)²-v(A)²]. Substitute: P(A)-P(B) = (1 ÷ 2)(600)[0.50²-0.10²] = 300(0.25-0.01) = 300×0.24 = 72 Pa. Correct option: D.