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MHT CET 2026 (12 May, Shift 2) · Velocity of Efflux (Torricelli)

Velocity of Efflux (Torricelli) question from MHT CET 2026 (12 May, Shift 2)

Q1MHT CET 2026 · 12 MAY · SHIFT 2Medium

Velocity of Efflux (Torricelli)

A water tank of height ’h’\text{'}h\text{'} has a hole at the bottom. It is completely filled with water. The total pressure at the bottom is 4H4H and atmospheric pressure is HH. The velocity of water flowing out of the hole is (ρ\rho = density of water)

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Answer: (a)

  1. Gauge pressure at the bottom: ρgh=4H−H=3H\rho g h = 4H - H = 3H
  2. Torricelli: 12ρv2=ρgh=3H\frac{1}{2}\rho v^2 = \rho g h = 3H
  3. v=6Hρv = \sqrt{\frac{6H}{\rho}}
  4. So the answer is (a) (6Hρ)1/2\left(\frac{6H}{\rho}\right)^{1/2}.
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