Two blocks are placed side by side on a surface (friction is present). What is the frictional force distribution between two blocks? 
Two blocks are placed as shown in the figure above.
Maximum possible friction between 5kg block and surface is 5 * 10 * 0.5 = 25 N (10 m/s^2 is the acceleration due to gravity).
Maximum possible friction between 10kg block and surface is 10 * 10 * 0.5 = 50 N.
The blocks are not going to move as maximum friction is 75N. But what are the frictional forces on each block.
Friction on 5 kg block is enough to oppose the 20N force. So, which of the following is correct?

*

*friction on 5 kg = 20 N, friction on 10 kg = 0 N (5 kg block completely opposes the 20 N force)

*friction on 5 kg = 20/3 N, friction on 10 kg = 40/3 N (distributed according to mass)

*not possible to find the frictional force distribution.
Also, what is the friction distribution if 40 N is applied on the 5 Kg block instead of 20 N?
edit: Assume the simplistic situation, where there are no deformations and the friction is uniform on the surface.
Thanks.
 A: Option 3 is correct. Without more information the distribution of friction force is indeterminate. It depends how this situation has been reached, and the exact model of static friction (which could be different for each block). Any combination of friction forces is possible (at least in theory) subject to the static limits given.
Whether the applied force is 20N or 40N the overall static limit is not reached, so there is no motion and the distribution of friction force is indeterminate. It is only when the overall static limit of 75N has been reached that the distribution of friction force is determinate (25N and 50N respectively).

One model is that static friction is generated from a microscopic elastic displacement, similar to Hooke's Law. (See The Dahl Model on p 6 of A Study of Friction Models and Friction Compensation.) Using this model, suppose that the 5kg block has an elastic displacement of 25um before it starts moving when the applied force is 25N. ie It has a spring constant of 1N/um. And that the 10kg block also has a displacement of 25um before it starts moving when the applied force is 50N. ie It has a spring constant of 2N/um.
First place the blocks initially separated by 20um. Then the 20N applied force will move the 5kg block 20um bringing it into contact with the 10kg block. We are left with the 5kg block providing all of the friction and the 10kg block providing none. (Option 1.)
As a second scenario suppose the blocks are initially placed in contact. Friction is generated such that any displacements are the same for both blocks - ie the blocks behave like springs in parallel. The 20N friction force is distributed 6.7N from the 5kg block and 13.3N from the 10kg block. (Option 2.)
In another scenario a 20N pull to the right is applied on the 10kg block, sustained by a 10um elastic displacement to the right. Then the 5kg block is placed in contact with it on the left. It is not displaced so there is no friction force on it. Now a pushing force on the 5kg block is applied from the left, building up from zero to 20N, exactly as the pulling force is reduced to zero on the right. The 5kg block remains undisplaced bearing no friction force, while the 10kg block retains its 10um displacement and 20N friction force. (The converse of Option 1.)
More complicated scenarios could be imagined. It is even possible to arrange for the friction force on the 5kg block to point to the right, in the same direction as the applied force.
For example, start with a 50N pull to the right on the 10kg block. This creates 50N of static friction to the left. Then set the 5kg block in contact on the left. Gradually reduce the 50N pulling force to 20N. This generates a friction force of 10N to the right on the 5kg block while the friction on the 10kg block reduces to 30N to the left. Finally gradually reduce the pulling force on the right to zero while increasing a compensating pushing force to the right on the 5kg from zero up to 20N. We are left with friction forces 30N to the left on the 10kg block and 10N to the right on the 10kg block.
Other models of static friction also exist, such as mixed elastic-plastic deformations and 'creep' (ie relaxation, time-dependence). (See link above.)
A: First option is correct i.e

friction on 5 kg = 20 N, friction on 10 kg = 0 N (5 kg block completely opposes the 20 N force

As long as the external force overcome the frictional force on the smaller block the contact force between the block will be 0.
In the second case , the blocks will not move as limiting value of static friction is greater than external force.
Hint

*

*To find the frictional force i suggest you to draw the free body of diagram of both blocks and write equation of motion for each block , and then set up acceleration =0.


*The normal reaction/ the contact force between the block is the frictional force on the bigger block. It will be very easy to see if you draw the free body diagram.
Edit

Assume block of 5 kg is split in a stack of 5 blocks of 1 kg. According to this answer, what will do first of these 1kg blocks ? Its maximum static friction 5N ? Move ?

I believed that in newtonian mechanics,  every object is treated as point object rather than the mass is distributed along the surface .
A: Assuming a simplistic model (no deformation or microscopic movements, etc).
If the applied force is insufficient to overcome the force of friction of the first block, there is no applied force left over with which the first block can transfer to the next block. Therefore, the second block does not contribute to the static friction.

If the force applied is enough to overcome the static friction of the first block, but not the total static friction of two blocks then the first block "uses up" all of its friction force to oppose the applied force, at which point it is forced against the second block where the force of friction of the second block opposes the remainder of the applied force.
The force between the blocks is treated as an internal stress of the two-block system and is in equal and opposite directions.

A: Could be some day of some millennium teachers start to draw surfaces not as perfect skating surfaces when talking about friction:

Assuming that the "past" history of the experiment has been:

*

*place the two blocks over the surface and in contact between them.

*apply the push force.

Then each micro-element (bump) of the horizontal surface produces a friction force against the push force, proportional to the pressure in this contact element.
At macro level, that means body of 5 kg will produce a friction $f_5=5a$ and body of 10 kg a friction $f_{10}=10a$, being the addition of both frictions equal to the push force of 20N.
$$f_5+f_{10}=5a+10a\underset{\cdot}{=} 20 N \rightarrow a = 4/3 N$$
$$f_5=5a=20/3 N$$
$$f_{10}=10a=40/3 N$$
Conclusion: b)
In case push force is 40N instead of 20N, friction forces are double.
Curiosity: this answer is valid even in other planets, $g \ne 9.8$, with the only condition maximal static friction is not surpassed.
Note1: Sammy Gerbil answer is very good and detailed, but it covers cases (except second one) that probably are out of the question. You can answer c), but then prepare to visit the teacher's office to defend your answer.
Note2: At atomic level we have not "bumps" but electromagnetic interactions, the "bump" similitude is only a teaching resource.
