Given:
Bed slope: \( s = 0.0001 \)
Manning’s coefficient: \( n = 0.02 \)
Depth of flow: \( y = 1 \, \text{m} \)
Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)
For a very wide rectangular channel (\( B \gg y \)):
Step 1: Hydraulic radius
The hydraulic radius for a very wide rectangular channel is given by:
\[
R = \frac{A}{P} = \frac{By}{B+2y} \approx y \quad \text{(since \( B \gg y \))}
\]
Thus, the hydraulic radius \( R \) is approximately equal to the flow depth \( y \). Therefore:
\[
R = 1 \, \text{m}
\]
Step 2: Critical depth of flow
The critical depth \( y_c \) is given by the formula:
\[
y_c = \left( \frac{q^2}{g} \right)^{1/3}
\]
where \( q \) is the discharge per unit width. Now we calculate \( q \) using Manning's equation for discharge:
\[
q = A \times v = B \times \frac{R^{2/3}}{n} \times s^{1/2}
\]
Simplifying this, we get:
\[
q = \frac{1}{n} \times y^{5/3} \times s^{1/2}
\]
Substituting the values:
\[
q = \frac{1}{0.02} \times (1)^{5/3} \times (0.0001)^{1/2} = 0.5 \, \text{m}^3/\text{s/m}
\]
Step 3: Calculation of critical depth
Now, using the formula for critical depth:
\[
y_c = \left( \frac{(0.5)^2}{9.81} \right)^{1/3}
\]
Solving this:
\[
y_c = 0.294 \, \text{m} \approx 0.3 \, \text{m}
\]
Thus, the critical depth is:
\[
\boxed{0.3} \, \text{m}
\]