At room temperature the viscosity of air and water are 1.8 x 10 -5 Pa*s and 1.0

Question

At room temperature the viscosity of air and water are 1.8 x 10-5 Pa*s and 1.0 x 10-3 Pa*s respectively.

What is the ratio of drag forces, Fd(nair)/Fd(nwater), when a diver goes from air to water?

Assume the drag force is given by Stokes' law(i.e.equation 1), and the

only variable that changes in going from air to water is the viscosity.

i.e. this equation (1)   

   Fd = Knrv

   where K is a constant depending on the shape of the object,

      n is the viscosity of the fluid,

   r is an appropriate linear dimension and

   v is the speed.

For a sphere r is the radius, K equals 6, and the drag force is called Stoke's law.

Explanation / Answer

Given force :    Fd   =   Knrv                         where : velocity of the fluid inversely propotional to its refractive index                       thus, Fd ( Kn r/ i ) where i is the refractive index of medium in the given formula, n , i are variables , remaining all constants      thus,    Fd (n air ) / Fd (water) = n_ air *   i _ water /   n_water * i_air                                                        = 1.8 * 10^-5 (1.33) / 1*10^-3 (1)                                                         = 2.394 (appx.)

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