For Q = 2.5 m^3/s and B = 2 m, with g = 9.81, the critical depth y_c is approximately:

• A. 0.30 m
• B. 0.54 m
• C. 0.80 m
• D. 1.20 m

Answer: (B)

Critical depth is the depth at which the flow is exactly at the transition between subcritical and supercritical, which means the Froude number equals 1. For a rectangular channel, the velocity is V = Q/(B y). At the critical state, V = sqrt(g y). Setting Q/(B y) = sqrt(g y) leads to Q = B sqrt(g) y^(3/2). Solving gives y_c = [Q / (B sqrt(g))]^(2/3).

Plugging in the numbers: sqrt(g) ≈ 3.132, so B sqrt(g) ≈ 6.264. Q/(B sqrt(g)) ≈ 2.5 / 6.264 ≈ 0.399. Therefore y_c ≈ 0.399^(2/3). Since 0.399^(1/3) ≈ 0.736, squaring gives y_c ≈ 0.542 m, i.e., about 0.54 m.

This depth is the critical depth for the given discharge in the 2 m wide channel.