The correct answer is C. 7RS.
The minimum size of stone that will remain at rest in a channel of longitudinal slope S and hydraulic mean depth R is given by the Shields’ criterion:
$$\tau_c = \frac{1}{2} \rho g d_s^2 S$$
where $\tau_c$ is the critical shear stress, $\rho$ is the density of water, $g$ is the acceleration due to gravity, $d_s$ is the diameter of the stone, and $S$ is the longitudinal slope of the channel.
The Shields’ criterion can be rearranged to give:
$$d_s = \left(\frac{2 \tau_c}{\rho g S}\right)^{1/2}$$
The critical shear stress is the minimum shear stress required to move a particle of a given size. It is a function of the particle’s shape, roughness, and density, as well as the fluid’s viscosity and density.
The longitudinal slope of the channel is the ratio of the change in elevation to the horizontal distance. It is a measure of the steepness of the channel.
The hydraulic mean depth is a measure of the cross-sectional area of the channel divided by the wetted perimeter. It is a measure of the average depth of water in the channel.
The minimum size of stone that will remain at rest in a channel is a function of the channel’s slope, hydraulic mean depth, and the critical shear stress. The critical shear stress is a function of the particle’s shape, roughness, and density, as well as the fluid’s viscosity and density.