Can a very thin sheet of any material float inside a liquid? When a body is immersed in a liquid,buoyancy is the net force of all the forces acting on it.
Now the forces are equivalent to those which will act on the same volume of liquid.Rightly so,but considering the chaotic motion of the molecules of the liquid,for every force acting on that particular volume of liquid,there will be another force  in the opposite direction but of different magnitude.The magnitude differs because of the difference in height.
Now let's consider a very very thin sheet immersed in the liquid.The opposite forces now will be of the same magnitude as difference in height is negligible and hence will cancel each other!so the only force acting will be its weight.
So the thin sheet will never be balanced!So does that imply that no thin sheet will ever float inside a liquid??
 A: You seem to be considering an (idealised) infinitely thin sheet. You're right that such an object would have no volume and thus displace no fluid --- but it would also have no mass, since its mass is equal to its density times its volume, which is zero. I guess you could conclude that it should have neutral buoyancy, because both its weight and the buoyancy force are zero.
However, one should be careful about this conclusion, because, any real sheet of material will be made of atoms. It will have a finite mass and a finite volume, and its buoyancy will depend on the ratio of the two, as it does for any material.
A: Let's assume your sheet is a disk of height $h$ and radius $r$, and we'll consider what happens when we take the height down towards zero.

The pressure at some depth $d$ is given by:
$$ P = \rho g d $$
where $\rho$ is the liquid density and $g$ is the acceleration due to gravity. So the pressures on the upper and kower faces of our disc are:
$$ P_2 = \rho g d $$
$$ P_1 = \rho g (d + h) $$
The total force on the faces is just pressure times area:
$$ F_2 = \rho g d \pi r^2$$
$$ F_1 = \rho g (d + h) \pi r^2$$
So the net force (we'll take upwards to be positive) is just:
$$ \Delta F = P_1 - P_2 = \rho g h \pi r^2 $$
But $h \pi r^2$ is just the volume of our disk so:
$$ \Delta F = \rho g V $$
which you should recognise as Archimedes' principle.
The point of all this is that our expression for the upwards force does not include the height of the disk. As long as we keep the volume $V$ constant we can flatten our disk into as thin a sheet as we want, and the magnitude of the upwards force will be unaffected. So thin sheets behave just like lumps of matter of any shape.
A: If the density of the material of your sheet is higher than the density of the liquid, then, the situation will be unstable, and it will collapse immediately by gravitational forces.
The process you are describing is basically just diffusion. If your material is also a fluid, and basically dissolved in the fluid, then the sheet, however thin, will always diffuse away.
If your material is a separate liquid, then surface tension will probably immediately break up the sheet. (If there is no surface tension, then it is dissolve and diffusion will take over).
If your material is a solid, then force by the random motion of water molecules (I don't want to call this diffusion, but it is probably somehow related) will tend to deform the solid sheet. You will then come into the field of solid mechanics. I would think that elasticity and bulk parameters come into play, which for infinitely thin sheets can not withstand the forces by water. 
