A wide unlined channel carries sediment-free water. The depth of water is 1 m. The specific weight of water is 10 kN/m3. To prevent scouring, the maximum permissible tractive stress on bed is 10 N/m2. The maximum slope of the channel bed to prevent scouring is 1 in n. The value of n is_________ (in integer).

The correct option is 1000.

To understand why this is the correct answer, let us break down the problem step-by-step using the principles of open channel hydraulics and sediment transport.

For a wide rectangular or unlined channel, the hydraulic radius (R) is approximately equal to the depth of flow (y) because the width (B) is much greater than the depth (y).
Therefore, we have:
$R\approx y=1\text{ m}$

The average boundary shear stress (also known as tractive stress, τ) on the channel bed is given by the formula:
$\tau =\gamma \times R\times S$
where:
- γ is the specific weight of water = 10 kN/m³ = 10,000 N/m³
- R is the hydraulic radius ≈ depth of water (y) = 1 m
- S is the bed slope of the channel = 1 / n

To prevent scouring, the maximum tractive stress on the bed must not exceed the maximum permissible tractive stress. Setting the shear stress equal to the maximum permissible value of 10 N/m², we get:
${\tau }_{\text{max}}$

Substituting the given values into the equation:
$10=10000\times 1\times \left(\frac{1}{n}\right)$

Solving for n:
$10=\frac{10000}{n}$
$n=\frac{10000}{10}=1000$

Thus, the maximum slope of the channel bed to prevent scouring is 1 in 1000, confirming that the value of n is indeed 1000.