In the diagram shown, the difference in the levels of the liquid in the two tubes of the manometer is 5 cm. The manometer tubes rise from the two points A and B of a horizontal tube carrying water, and the cross-sections of the tube at A and B are 6 mm² and 10 mm² respectively. The rate at which wa…

Question

In the diagram shown, the difference in the levels of the liquid in the two tubes of the manometer is 5 cm. The manometer tubes rise from the two points A and B of a horizontal tube carrying water, and the cross-sections of the tube at A and B are 6 mm² and 10 mm² respectively. The rate at which water flows through the tube is (g=10 m s⁻²)

Accepted Answer

(1) 7.5 cm³ s⁻¹. Given: A_A=6 mm²=6×10⁻⁶ m², A_B=10 mm²=10⁻⁵ m², h=5 cm=0.05 m, g=10 m s⁻². The manometer reads the static-pressure difference, Δ p=ρ gh, and the tube is horizontal, so ρ gh=1/2ρ(v_A²-v_B²) with A the narrower, faster section. v_A²-v_B²=2gh=2(10)(0.05)=1 m² s⁻². Write both speeds through the common rate Q: v_A=Q/(A_A) and v_B=Q/(A_B), so Q²(1/(36×10⁻¹²)-1/(100×10⁻¹²))=1. Q²(2.778×10¹⁰-10¹⁰)=1 gives Q²=5.625×10⁻¹¹, so Q=7.5×10⁻⁶ m³ s⁻¹=7.5 cm³ s⁻¹.

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