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A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:

1. 12.6 × 10⁻⁶ m³/s
2. 8.9 × 10⁻⁶ m³/s
3. 2.23 × 10⁻⁶ m³/s
4. 6.4 × 10⁻⁶ m³/s

Correct answer: 12.6 × 10⁻⁶ m³/s

Solution

The rate of flow can be calculated using Torricelli's theorem: velocity v = √(2gh). Substituting g = 10 m/s² and h = 2 m, v = √(2 × 10 × 2) = √40 ≈ 6.32 m/s. The flow rate Q = A × v, where A = 2 × 10⁻⁶ m². Thus, Q ≈ 2 × 10⁻⁶ × 6.32 ≈ 12.6 × 10⁻⁶ m³/s.

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