A steel ball of diameter `d=3.0mm` starts sinking with zero initial velocity in olive oil whose viscosity is `eta=0.90P`. How soon after the beginning of motion will the velocity of the ball differ from the steady-state velocity by `n=1.0%`?

Explanation

`m(dv)/(dt)=mg-6pietarv`
or `(dv)/(dt)+(6pietar)/(m)v=g`
or `(dv)/(dt)+kv=g, k=(6pietar)/(m)`
or `e^(kt)(dv)/(dt)+ke^(kt)v="ge"^(kt)` or `(d)/(dt)e^(kt)v="ge"^(kt)`
or `ve^(kt)=g/ke^(kt)+C` or `v=g/k+Ce^(-kt)` (where C is const.)
Since `v=0` for `t=0`, `0=g/k+C`
So `C=-g/k`
Thus `v=g/k(1-e^(-kt))`
The steady state velocity is `g/k`.
`v` differs from `g/k` by n where `e^(-kt)=n`
or `t=1/k1n n`
Thus `1/k=-((4pi)/(3)r^3rho)/(6pietar)=-(4r^2rho)/(18eta)=-(d^2rho)/(18eta)`
We have neglected buoyancy in olive oil.