A large containers (with open top) of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area a in its side wall near the bottom. The container contains a liquid de

Question

A large containers (with open top) of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area a in its side wall near the bottom. The container contains a liquid density ρ and mass m₀. If the liquid starts flowing out of the hole at time t = 0, the velocity of eff

Accepted Answer

Answer: √(gm₀/2Ap) — Initially height = h After 75% drained, remaining = 25% ⇒ height = h/4 Velocity: v = √(2gh₁) = √(2g × h/4) = √(gh/2) Now, h = m₀/(Aρ) ⇒ v = √(gm₀/(2Aρ))

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