Under standard conditions helium fills up the space between two long coaxial cylinders. The mean radius of the cylinders is equal to `R`, the gap between them is equal to `Delta R`, with `Delta R lt lt R`. The outer cylinder rotates with a fairly low angular velocity `omega` about the stationary inner cylinder. Down to what magnitude should the helium pressure be lowered (keeping the temperature constant) to decrease the sought moment of friction forces `n = 10` times if `Delta R = 6 mm` ?

Explanation

In this case
`N_1 (r_2^2 - r_1^2)/(r_1^2 r_2^2) = 4 pi eta omega`
or `N_1 (2 R Delta R)/(R^4) ~~ 4 pi eta omega` or `N_1 = (2 pi eta omega r^3)/(Delta R)`
To decrease `N_1, n` times `eta` must be decreased `n` times. Now `eta` does not depend on pressure until the pressure is so low that the mean free path equals, say, `(1)/(2) Delta R`. Then the mean free path is fixed and `eta` decreases with pressure. The mean free path equals `(1)/(2) Delta R` when
`(1)/(sqrt(2) pi d^2 n_0) = Delta R(n_0 = concentration)`
Corresponding pressure is `p_0 = (sqrt(2) kT)/(pi d^2 Delta R)`
The sought pressure is `n` times less
`p = (sqrt(2) k T)/(pi d^2 n Delta R) = 70.7 xx (10^-23)/(10^-18 xx 10^-3) ~~ 0.71 Pa`
the answer is qualitative and depends on the choice `(1)/(2) Delta R` for the mean free path.