For this device the liquid to be tested is drawn into the tube to a level above the top etched line. the time is then obtained for the liquid to drain to the bottom etched line. the kinematic viscosity, ν, in m2/s is then obtained from the equation ν = kr4t where k is a constant, r is the radius of the capillary tube in mm, and t is the drain time in seconds. when glycerin at 20 ˚c (υ=1.19·10-3 m2/s) is used as a calibration fluid in a particular viscometer the drain time is 1430 s. when a liquid having a density of 970 kg/m3 is tested in the same viscometer the drain time is 900 s. what is the dynamic viscosity of this liquid?

The dynamic viscosity of the liquid is 0.727 kg/m*s

Explanation:

In the equation for that viscosimeter, ν = KR⁴t, the terms K and R are not dependent on the liquid that is being tested, unlike ν and t.

Using that equation and the data given in the problem, we can calculate the product of K and R⁴.

1.19*10⁻³m²/s = (KR⁴)* 1430 s

KR⁴=8,32*10⁻⁷m²/s²

We can now calculate the kinematic viscosity of the unknown liquid.

ν=8,32*10⁻⁷m²/s²*900s

ν=7.49*10⁻⁴m²/s

The relationship between the kinematic viscosity and the dynamic viscosity is given by the equation μ=ν * ρ, where μ is the dynamic viscosity and ρ is the density. Thus:

μ=7.49*10⁻⁴m²/s * 970 kg/m³

μ=0.727 kg/m*s

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

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