A small horizontal hole of area 2 cm² is made 5 m below the water surface in a large open tank. Taking ρ = 1000 kg/m³ and g = 10 m/s², what is the horizontal reaction force on the tank due to the emerging water jet?
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Question

A small horizontal hole of area 2 cm² is made 5 m below the water surface in a large open tank. Taking ρ = 1000 kg/m³ and g = 10 m/s², what is the horizontal reaction force on the tank due to the emerging water jet?

Options

G1 Physics · Accepted Answer

The speed of efflux is v = √(2gh). Therefore, v² = 2gh = 2×10×5 = 100 m²/s². The mass-flow rate through the hole is ṁ = ρav. The rate at which horizontal momentum leaves the tank is F = ṁv = ρav². Hole area: a = 2 cm² = 2×10⁻⁴ m². Hence, F = 1000×2×10⁻⁴×100 = 20 N. By Newton's third law, the tank experiences an equal reaction force opposite to the jet direction. Correct option: C.