The velocity of water in a river is 18 km h$$^{-1}$$ near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water = $$10^{-2}$$ poise.

The velocity of water near the surface is given as 18 km h$$^{-1}$$, and the river depth is 5 m. To find the shearing stress between horizontal layers of water, we use Newton's law of viscosity, which states that the shearing stress (τ) is equal to the coefficient of viscosity (η) multiplied by the velocity gradient (dv/dy). The formula is τ = η × (dv/dy).

First, we need to convert the velocity from km h$$^{-1}$$ to m s$$^{-1}$$ because SI units are required. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 18 km h$$^{-1}$$ = 18 × (1000 m) / (3600 s) = 18 × (1000 ÷ 3600) m s$$^{-1}$$. Simplifying 1000 ÷ 3600 gives 5/18, so 18 × (5/18) = 5 m s$$^{-1}$$. Therefore, the velocity at the surface is 5 m s$$^{-1}$$.

At the bottom of the river (depth = 5 m), the velocity of water is zero due to the no-slip condition. The velocity gradient dv/dy is the change in velocity per unit depth. Assuming a linear velocity profile from the bottom to the surface, dv/dy = (velocity at surface - velocity at bottom) ÷ depth = (5 m s$$^{-1}$$ - 0 m s$$^{-1}$$) ÷ 5 m = 5 ÷ 5 = 1 s$$^{-1}$$.

The coefficient of viscosity η is given as 10$$^{-2}$$ poise. Poise is a CGS unit, so we convert it to SI units (Pa·s or N s m$$^{-2}$$). We know that 1 poise = 0.1 Pa·s. Therefore, η = 10$$^{-2}$$ poise = 10$$^{-2}$$ × 0.1 Pa·s = 10$$^{-3}$$ Pa·s = 10$$^{-3}$$ N s m$$^{-2}$$.

Now, substituting into the formula for shearing stress: τ = η × (dv/dy) = (10$$^{-3}$$ N s m$$^{-2}$$) × (1 s$$^{-1}$$) = 10$$^{-3}$$ N m$$^{-2}$$.

Comparing with the options, 10$$^{-3}$$ N m$$^{-2}$$ corresponds to option B. Hence, the correct answer is Option B.