91影视

First, we need to identify the units and dimensions of each variable involved in the relation. 1. Surface tension (s) has units of N/m (newtons per meter) and dimensions of [M^1 L^0 T^-2]. 2. Coefficient of viscosity (畏) has units of kg/ms (kilograms per meter-second) and dimensions of [M^1 L^-1 T^-1]. 3. Density (蟻) has units of kg/m鲁 (kilograms per cubic meter) and dimensions of [M^1 L^-3 T^0].

Now, we will analyze the units and dimensions for each relation given in the problem, and look for the option that results in the correct dimensions for velocity. (A) \(\frac{s^{2}}{\rho \eta}\): Units: \(\frac{(N/m)^2}{(kg/m^3)(kg/ms)} = \frac{(N^2 / m^2)}{(kg^2/m^4s)}\) Dimensions: \(\frac{[M^2 L^0 T^{-4}]}{[M^2 L^{-4} T^{-1}]}\) = [M^0 L^4 T^(-3)] (B) \(\frac{s}{\eta}\): Units: \(\frac{N/m}{kg/ms} = \frac{m^2}{s}\) Dimensions: \(\frac{[M^1 L^0 T^{-2}]}{[M^1 L^{-1} T^{-1}]}\) = [M^0 L^1 T^(-1)] (C) \(\frac{\eta \rho}{s^{2}}\): Units: \(\frac{(kg/ms)(kg/m^3)}{(N/m)^2} = \frac{kg^2/m^4s}{N^2/m^2}\) Dimensions: \(\frac{[M^2 L^{-4} T^{-1}]}{[M^2 L^0 T^{-4}]}\) = [M^0 L^4 T^3] (D) \(\frac{\rho}{\eta}\): Units: \(\frac{kg/m^3}{kg/ms} = m^2/s\) Dimensions: \(\frac{[M^1 L^{-3} T^0]}{[M^1 L^{-1} T^{-1}]}\) = [M^0 L^2 T^1]

We are looking for the relation that gives the dimensions of velocity, which are [M^0 L^1 T^(-1)] (the same as option B). Therefore, the correct relation between surface tension, coefficient of viscosity, and density for the velocity of surface waves is: \(\boxed{(B) \frac{s}{\eta}}\)