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Question:

A small hole of an 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

12.6×10⁻⁶ m³/s

2.23×10⁻⁶ m³/s

6.4×10⁻⁶ m³/s

8.9×10⁻⁶ m³/s

Solution:

Correct option is A. 12.6×10⁻⁶ m³/s
Rate of flow of liquid Q = av = a√(2gh) = 2×10⁻⁶ m² × √(2 × 10 × 2 m/s) = 2 × √40 × 10⁻⁶ m³/s = 2 × 2√10 × 10⁻⁶ m³/s = 4√10 × 10⁻⁶ m³/s ≈ 12.56 × 10⁻⁶ m³/s ≈ 12.6 × 10⁻⁶ m³/s