Why does torque produce a force on the axis of rotation? If a door is rotated about its fixed axis in (outer) space, a force parallel to the door on the hinges will arise due to centripetal force on the centre of mass and conservation of momentum (Newton's third law). 
But any torque on the door will create a force on the hinges which is equal to $t/r$ or torque divided by radius. I'm looking both an intuitive and mathematically based explanation for this fact. I can sort of 'see' why, but my understanding is vague and uncertain. 
 A: My guess at what you're asking
I think I understand your question but I'm not 100% sure. Let me draw a picture.
The forces on a bar moving in a 2D plane about a center of rotation (black dot) look something like this:

And that 2D bar is a model for a 3D door whose hinges are on a fixed axis.
Now the force of the door on the hinge is just the equal-and-opposite force demanded by Newton's third law to this "constraint force" in blue. So you are asking, why does this constraint force have as one of its components this "skinny red component" arrow that corresponds to the torquing force? And why is it pointing in this perhaps-surprising direction downward, and when does it instead point upward? These are great questions.
Note that if the door had no hinge, and you wanted to make it rotate around its center rather than its edge, you would pull "up" on the left hand side and push "down" on the right hand side, in the diagram above. So this little red component would actually point upwards to meet that constraint. But now when it's on the edge we have this problem that the center-of-mass can move, so there's nothing super-wrong with having the two forces both point in the same direction or opposite. It turns out that this will depend a lot on $r$, the radial length at which the torquing force is applied. Still speaking very crudely, if $r$ is very very far (perhaps even further than the door length $L$ if it has a massless prong inserted into it!) then you will find that the torque "rotates the door too much" to keep the hinge at the right place, and therefore the constraint force points in the same direction; but if $r$ is small then the torque "doesn't rotate the door enough" and the constraint force has to step in to help.
Start with the constrained motion.
Well the key in doing this precisely is in the name I gave it, it is a "constraint force" and needs to be whatever is necessary to keep the hinge point from moving anywhere else in the plane. So what is necessary?
Well if the hinge is truly fixed then the door can be described in 2D purely by one angle, $\theta$, which the door makes at the hinge with the horizontal direction; the above diagram happens to show $\theta = 0.$ Now the center-of-mass of the door is at the position $[x, y] = \frac L2 [\cos\theta, \sin\theta].$ We sometimes invent two new perpendicular unit vectors, one called $\hat r = [\cos\theta, \sin\theta],$ and one called $\hat \theta = [-\sin\theta, \cos\theta].$ They are a little confusing! The normal unit vectors $\hat x = [1,0]$ and $\hat y = [0,1]$ are the same everywhere, these depend on what $\theta$ is! But they are very useful for just saying "the center of mass is at position $\hat r ~L/2$."
Anyway this expression gives us just what we need when we start taking time derivatives. The first time derivative is $$[v_x, v_y] = \frac L2 [-\sin\theta, \cos\theta] \frac{d\theta}{dt} = \frac L2 ~\omega~ \hat \theta,$$ where $\omega = d\theta/dt$ is the instantaneous angular velocity. So this is obvious, "the thing only moves in the $\theta$-direction perpendicular to the $r$-direction." The next time-derivative however needs to be done with the "product rule" and gives us,$$\begin{array}{rl}[a_x, a_y] =& \frac L2 [-\cos\theta, -\sin\theta] \left(\frac{d\theta}{dt}\right)^2 + \frac L2 [-\sin\theta, \cos\theta] \frac{d^2\theta}{dt^2}\\
=& -\frac L2 ~\omega^2~ \hat r + \frac L2 ~\alpha~\hat \theta\end{array}$$ where $\alpha = d\omega/dt$ is the angular acceleration. So these are the only ways the thing can accelerate, if the hinge is fixed. The first term is the centripetal force term that we know and love; the second term is due to the torques on the thing.
Now derive the constraint force from that motion.
We now know the only way the thing is allowed by the "hinge has to stay at one place" constraint to accelerate, and we can see some interesting results.
So first off, as a no-brainer: if the torquing force $\vec T$ is not perfectly perpendicular to the door, we have to extract its normal component $T_\theta = \vec T \cdot \hat \theta$ to influence $\alpha$ and anything else is eaten by the hinges. If the force is provided at a distance $r$ from the hinge, its torque is therefore $\tau = T_\theta ~r$ and this produces an angular acceleration $\alpha = \tau/I = T_\theta~r/I,$ which I'll be plugging into the above expression in a moment.
The constraint force $\vec C$ of course comes to fill this gap between "what motions are possible?" and "what are the forces?". For example if you're standing on the ground, then the constraint says "You are not accelerating in the up/down direction" and therefore the normal force provides whatever force it needs to, to make these balance out: this is why the normal force equals your weight. (It is not because of Newton's third law; it is because you are in a state of up-down "equilibrium" which can only happen if the forces on you balance out.) Something similar is happening with this hinge.
So because we know from Newton's second law, the sum of these constraint and torquing forces $\vec C + \vec T$ must be equal to the mass times the acceleration, but I just told you the only possible acceleration:
$$ \vec C + \vec T = m \vec a = -m ~\frac L2 ~\omega^2~ \hat r + m~\frac L2 ~\frac{\tau}I~\hat \theta,$$ because we know the only way that this thing can accelerate, as per the above. Since $\vec T = T_r ~\hat r + T_\theta~ \hat \theta$ we can combine these $\hat \theta$ parts into saying,$$\vec C = -\left(m  ~\frac L2 ~\omega^2 + T_r\right)~ \hat r + T_\theta ~ \left(\frac {mLr}{2I} - 1\right)~\hat\theta.$$
So the component of the constraint force "parallel to the door," as you say, is equal to $-T_r$ plus the centripetal force due to its existing rotation. The component "perpendicular to the door" is more complicated; for this you need to know that $I = \frac13 m L^2$ and hence this is $(\frac32 r/L - 1)~T_\theta.$ That means that if you place the torquing force 2/3rds of the way out, it happens to rotate the door exactly as much as it pushes it in the direction it wants to go, leading to no constraint-force component; any further and it tries to "twist the door too much." So we don't even need the "massless prong" above; just pushing on the last third of the door will force the hinge to push in the same direction that we do.
A: It is only due to the hinges that that when the door is pushed ,that it doesn't translate along the line of action of force applied and the hinges changes the velocity of the door and thus a force is necessary for that.This force is variable , changing its direction every instant
A: An object can be in rotational non-equilibrium while simultaneously being in translational equilibrium depending upon the arrangement of the forces. In other words, an object can undergo an accelerating spin without translating. You can add rotational energy to an object without adding translational energy.
What I wrote above was meant as a hopefully helpful hint. I'll clarify below.
This is a very simple structural engineering problem. A hinge is a structural component that, by definition, can only have a resultant x-direction force and a resultant y-direction force.
You apply a force to the outer region of the door, setting up a torque acceleration of the door. You can mathematically slide the torque force to the hinge. That force now acts on the door at the hinge location. You've also mathematically noted the torque about the hinge as a circular arrow in the proper direction. The one reaction component at the door hinge exactly counters your applied force so that translation in that direction is impossible because the forces are balanced. The OTHER reaction at the hinge directly balances the centripetal acceleration of the rotationally accelerating door. It's much like whirling a ball on a string.
A: Pure torque does not produce any forces. So it is not true that "any torque on the door will create a force on the hinges".
A: If you view the torque as Force * Radius * Sin(angle between force and door) then you can see that t/r is the component of force perpendicular to the door if that force were applied at radius r from the hinge. Now imagine the forces on the hinge if you wanted to generate the same torque at various radii.  The closer I push to the hinge, the more the hinge has to push back to get the door to rotate and the further out I push, the smaller the force, thus the 1/r relationship between the force on the hinge and the radius of the applied force.
A: "Any" torque will not create a force. Take the simple example of a force $F$ (and the only force in this problem) perpendicular to the door at a distance $r$ from the hinge. The torque's magnitude is $|\tau|=|rF|$ and there is no force on the hinge by the problem statement.
A: I think you are asking for a situation where a pinned object has a pure torque applied (no net force) that causes rotation and reaction forces on the hinges. 
The intuition behind this scenario is that the net force on a body goes towards moving the center of mass. In this case, although no applied force exists, the net force is not zero because the center of mass orbits about the joint.
So the force is a result of the joint constraint and that the body is forced to rotate about an axis away from the center of mass.
Imagine the body is stationary to begin with. If the distance between the center of mass and the joint is $c$ then the reaction force $F_R$ causes the center of mass to move by $$ F_R = m a_{CM} = m c \ddot{\theta} $$
The applied torque $\tau$ as well as the torque due to the reaction causes the body to rotate $$ \tau - c F_R = I \ddot{\theta} $$
These two equations are solved for $$\begin{align} F_R & = \frac{m\, c \,\tau}{I+m c^2} \\ \ddot{\theta} &= \frac{\tau}{I+m c^2} \end{align} $$
So to answer your question it is the motion of the center of mass that causes the forces to arise at the hinges. The first equation above indicates the magnitude of this force depending not only on the applied torque $\tau$ but also the inertial properties of the body. Thus indicating this is a dynamic force.
A: 
But any torque on the door will create a force on the hinges which is equal to t/r or torque divided by radius. 

Well, actually it is in fact not that torque that causes the force. The force on the hinges comes not because of the torque but because of the motion!
You almost explain it yourself here:

If a door is rotated about its fixed axis in (outer) space, a force parallel to the door on the hinges will arise due to centripetal force on the centre of mass and conservation of momentum (Newton's third law).

What causes the centripetal force? Not some external force, but merely the fact that the particles of the door must start turning. When they are in motion (having a speed) they wish to stay on the straight track, but the hinge holds on to them and pulls them inwards to make them turn. This "turning" means acceleration, and Newton's 2nd law says that such acceleration is the result of a force - that force is at the hinge.
A: 
But any torque on the door will create a force on the hinges

This is not true.
There is no torque in principals of motion. In a nutshell, first law of Newton defines force and second law describes relation between force (defined by first law) and position vector of the particle. Third law is about action and reaction forces.
When two forces equal in magnitude $F$ but opposite in direction that act in a distance $r$ with respect to each other, we define their torque by $t=Fr$.
We should go in the right direction from principals (or laws) to definitions not vice versa. If we want to do inverse path, we should take into account the definitions or we must change our accepted principals. So, when there is a torque, certainly there have been two forces that have produced the torque according to its definition.
Torque doesn’t create any force. There is no torque at all. What exists according to the principals accepted by us, is force and if something should create another thing, it is force that create torque by getting help from another force.
