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Question:

Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of the tap is 10⁻⁸ m². Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is (g=10 m/s²)

1.0×10⁻⁹ m²

5.0×10⁻⁸ m²

5.0×10⁻⁹ m²

2.0×10⁻⁹ m²

Solution:

Volume flow rate = 10⁻⁸ m³/sec
ΔP = ρgh = 10³ kg/m³ × 10 m/s² × 0.15 m => ΔP = 1500 Pa
=> ΔP = ½ρ(v₂² - v₁²)
=> 1500 = ½ × 10³ (v₂²)
=> v₂ = √3 m/s ≈ 1.73 m/s
By equation of continuity, A₁v₁ = A₂v₂ => 10⁻⁸ × 1 = A₂ × 1.73 => A₂ = 5.78 × 10⁻⁹ m²
The closest answer is 5.0 × 10⁻⁹ m²