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Water from a tap emerges vertically downwards with an initial speed of 1.0 m s⁻¹. The cross-sectional area of the tap is 10⁻⁴ m². Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: (Take g = 10 m s⁻²)
- 5 × 10⁻⁴ m²
- 2 × 10⁻⁵ m²
- 5 × 10⁻⁵ m²
- 1 × 10⁻⁵ m²
Correct answer: 5 × 10⁻⁵ m²
Solution
Speed after falling 0.15 m: v = sqrt(1^2 + 2*10*0.15) = 2 m/s. By continuity A0*v0 = A*v, so A = 1e-4 * 1/2 = 5e-5 m^2.
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