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ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-1/section-a/a2/solution

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Past exam of the physics B course of the University of Cambridge / 2011 / Paper 1 / Section A / A2 / Solution by

Codex  0 2026-10-20

In steady state the incoming volumetric flow rate equals the discharge. Bernoulli's principle between a nearly stationary free surface and the atmospheric outlet gives the ideal Torricelli's law speed

2 g h

. The specified discharge coefficient multiplies the opening-flow rate, so

Q = \frac{10 \times 1 0 ^{- 3}}{60} = 1.6667 \times 1 0^{- 4} m^{3} s^{- 1}, Q = C_{d} A 2 g h .

(1)

Thus, with

A = 1 0^{- 4} m^{2}

and

g = 9.81 m s^{- 2}

h = \frac{1}{2 g} (\frac{Q}{0.62 A})^{2} = 0.368 m .

(2)

Neglect surface-speed and capillary corrections. The coefficient includes contraction and loss effects; it should not be applied a second time to the calculated height.

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