Timescale for unstable stratification to resolve itself The ocean is typically vertically stratified, that is the deeper you go the higher the density should be. This can be because of increased salinity or decreased temperature.
If that is not the case then we have an unstable stratification and this instability is called Rayleigh-Taylor instability.
My question is what is the timescale for this stratification to resolve itself? If say at 7m depth the density is 0.1% higher than at 8m, then how much time it takes to get resolved by convection? Is it in minutes, hours, days?
 A: In real oceans this is complicated by currents, eddies, wind driven surface flows, thermohaline gradients, …, but in an idealized two-layer system with a sharp boundary and a density difference of about 0.1%, convective mixing over a 1 m distance should take less than a minute.
For an upper and lower layer with densities $\rho_u$, $\rho_l$ and dynamic viscosities $\mu_1$ and $\mu_2$, Rayleigh–Taylor turbulence has characteristic, length, time, and velocity scales:
$$l_c=\left(\frac{\nu^2}{g A}\right)^{1/3},\qquad
t_c=\left(\frac{\nu}{g^2 A^2}\right)^{1/3},\qquad u_c=\frac{l_c}{t_c}=\left(\nu g A\right)^{1/3}$$
where $g$ is the local acceleration due to gravity, $\nu$ is the average kinematic viscosity, and $A$ is the Atwood number:
$$\nu=\frac{\mu_u+\mu_l}{\rho_u+\rho_l},\qquad A=\frac{\rho_u-\rho_l}{\rho_u+\rho_l}$$
For warm seawater, $\mu\approx 0.001$ Pa s and $\rho \approx 1.02\,\textrm{lg/m}^3$ are reasonable values. A $0.1$% density difference gives $A=5\times 10^{-4}$ and
$$l_c\approx 1\,\textrm{mm,}\qquad t_c\approx 0.4\,\textrm{s,}\qquad u_c\approx 2\,\textrm{mm/s}$$
So it should take the order of a second or so for the Rayleigh-Taylor turbulence to start.  Once started, the buoyancy acceleration of water from the lower layer rising in the upper layer is
$a_B = 2 A g = 0.0098 \textrm{m/s}^2$.
In the absence of viscous drag and turbulence, the time needed rise a distance $h\approx 1$m will be
$$t_L = \sqrt{\frac{2h}{a_B}} = \sqrt{\frac{h}{A g}} \approx 14\,\textrm{s}$$
and its final vertical velocity would be
$$v_f = \sqrt{2 a_B h } = \sqrt{4 A g h} \approx 0.14  \,\textrm{m/s}$$
It turns out ignoring viscous drag is not a terrible approximation in this case.  The viscous de-acceleration (force per unit mass) of turbulent eddies of size $l_t$ rising with velocity $u$ is (Lawrie Eq. 4.31)
$$a_v \sim \frac{\nu u}{l_t^2}$$
The wavelength of the maximally unstable Rayleigh-Taylor mode is $\lambda_{mx} = 4 \pi l_c$, and the viscous de-acceleration for $u\sim v_f$ and $l_t \sim \lambda_{mx}$ is $0.002\,\textrm{m/s}^2$. Not completely negligible, but still significantly less than the buoyancy acceleration $\sim 0.010\,\textrm{m/s}^2$, so our rough estimate for the mixing time should still be about right.
Whenever turbulence is involved, however, we should worry about any simple theoretical analysis, so it is nice that our rough estimates are consistent with this video from Stuart Dalziel's Cambridge DAMTP Lab showing the mixing in real time in a $0.5$ m tall, $0.4$ m wide, $0.2$ m thick tank with $A=7\times 10^{-4}$.
Figure 5.12 (p. 69) of Lawrie shows a nice sequence of timed photos in the same tank.
