Is there a physically realizable inductive turing machine that can solve Hilbert's $10$th problem and can it overcome Church-Turing Hypothesis? There is a claim on https://en.wikipedia.org/wiki/Super-recursive_algorithm#Inductive_Turing_machines that 'Simple inductive Turing machines are equivalent to other models of computation such as general Turing machines of Schmidhuber, trial and error predicates of Hilary Putnam, limiting partial recursive functions of Gold, and trial-and-error machines of Hintikka and Mutanen.[1] More advanced inductive Turing machines are much more powerful. There are hierarchies of inductive Turing machines that can decide membership in arbitrary sets of the arithmetical hierarchy(Burgin 2005). In comparison with other equivalent models of computation, simple inductive Turing machines and general Turing machines give direct constructions of computing automata that are thoroughly grounded in physical machines'. 
Wiki also says these are different from infinite time Turing machines.
Are inductive turing machines different from turing machines with oracles?
Are inductive turing machines physically realizable (at least in the same sense of realizaility of Turing machines as Intel processors with bounded RAM and one that degrades over time)?
Can inductive turing machines solve the halting problem (essentially Hilbert's $10$th as well)?
Please refer here for overcoming Church-Turing Hypothesis with Inductive Turing machines https://en.wikipedia.org/wiki/Super-recursive_algorithm#Relation_to_the_Church.E2.80.93Turing_thesis.
Here is another article (published in communications of the ACM and well cited) http://www.columbia.edu/itc/hs/medinfo/g6080/misc/p82-burgin.pdf.
 A: Consider the following paper, written by A. Steven Younger, Emmett Redd, Hava Siegelmann, and Conrad Bell:

"A Physical Machine Based on a Super-Turing Computational Model" [found under title on the Web].

I quote the Abstract verbatim:

We present evidence that the Turing Machine is too restrictive a model to sufficiently describe the computation of our analog computer and, therefore a more conprehensive model is needed.  We report on the construction of a prototype, the Optical Analog Recurrent Neural Network (OpticARRN), and experimental results showing that it performs computations which are beyond those of computers based on the Turing machine.  we conclude that the behavior of OpticARRN is better described by the super-Turing computational model proposed by Siegelmann.  To the best of our knowledge, this is the first application of analog recurrent neural networks realized in a physical computer based on this model.

(Suffice it to say, I leave the judgement as to the truth or falsity of the claim(s) made in the Abstract and the paper to the Reader.)
What I find personally (for what that's worth) interesting in this paper is this particular claim (found in Section 3, "Testing for Computation Beyond the Turing Limit"):

In order to test super-Turing computation, a suitable problem must be found.  In this case, the answer came from the area of chaotic systems.  The dynamics of chaos are both aperiodic and defined on a continuous phase space.  As such, they cannot be mimicked by a Turing machine [Siegelmann, H. (1998). Neural Networks and Analog Computation Beyond the Turing Limit.  Boston: Birkhauser(p. 155)].

It is the validity of this claim that (seemingly) makes or breaks the experiment. Also, the reader should pay particular attention to their methodology for the interpretation of the experimental data.
Note that the experimental setup is shown in figure 1 on pg. 4 of the paper (at least on the copy of the paper I found on the Web). 
A: Let me try to answer the actual question that was asked. The Wikipedia
page
defines inductive Turing machines as follows:

An inductive Turing machine is a definite list of well-defined
  instructions for completing a task which, when given an initial
  state, will proceed through a well-defined series of successive
  states, eventually giving the final result. The difference between
  an inductive Turing machine and an ordinary Turing machine is that
  an ordinary Turing machine must stop when it has obtained its
  result, while in some cases an inductive Turing machine can
  continue to compute after obtaining the result, without stopping.

Two remarks. 


*

*I assume that when the description says "eventually giving
the final result," what is meant is that there is a stage after
which the computation is always displaying that result as output.
This makes the concept identical to what has also been known by the term computability-in-the-limit, as well as other terminology. One naturally extends the concept to partial functions, by insisting that for  inputs not in the domain, what we want is for the outputs not to converge or stabilize. This is evidently the simple model of inductive machine; the wikipedia page makes 
references to a hierarchy of more powerful machines. 

*Although the Wikipedia page makes numerous references to Mark
Burgin — his name appears 24 times in the linked article — to my understanding of the history of the subject, the particular concept of computability-in-the-limit has been well understood and analyzed by computability theorists much earlier than Burgin's writings.
To my way of thinking, the main thing to say about this notion of computability is the following, which is commonly given as an exercise in computability theory courses.
Theorem. For any function $f$, the following are equivalent.


*

*$f$ is computable by an inductive Turing machine; that is, $f$ is computable in the limit.

*$f$ is computable (in the usual sense) by a Turing machine
equipped with an oracle for the halting problem.

*The graph of $f$ is $\Sigma_2$-definable.
Proof. ($1\to 3$). If $f$ is computable by an inductive Turing
machine, then $f(a)=b$ if and only if there is some stage of the
inductive computation on input $a$ such that at any later stage,
the output is still $b$. This is a $\Sigma_2$ definition of the
graph of $f$.
($3\to 2$) If the graph of $f$ is $\Sigma_2$-definable, then
$f(a)=b$ just in case $\exists x\forall y\ B(x,y,a,b)$, where $B$
is $\Delta_0$. With an oracle for the halting problem and any
particular $x$, $a$ and $b$, we can ask the oracle if the $\forall
y$ condition holds. In this way, on input $a$, we can search for an
$x$ and $b$ that fulfill the condition. When found, output $b$.
($2\to 1$) If $f$ is computable with respect to an oracle for the
halting problem, then it is computable by an inductive Turing
machine: just compute better and better approximations to the
halting problem, and for each of them, use that approximation as an
oracle for the computation of $f$. This process eventually
stabilizes, because for any given input, the approximation to the
halting problem will be accurate for a long enough time to support
the correct computation of $f$. $\Box$
Note that the argument in the implication ($2\to 1$) exhibits the
feature that is central to some of the commentary about these
machines, namely, that although we can compute better and better
approximations to the halting problem, in a way that will
eventually be correct on any given instance, nevertheless we are
typically not able to recognize computably when our approximation
is correct. Thus, although we may be computing the function $f$
accurately by using that approximation, we have no way of knowing
for sure that we have the final answer.
Corollary. For any set $A$, the following are equivalent.


*

*$A$ is decidable by an inductive Turing machine.

*$A$ is Turing computable from the halting problem.

*$A$ has complexity $\Delta_2$ in the arithmetic hierarchy.
Proof. The characteristic function of $A$ is a total function,
and so its graph is $\Sigma_2$ if and only if $A$ has complexity
$\Delta_2$. $\Box$
In this sense, yes, the so-called inductive Turing machines can compute the halting problem and therefore Hilbert's 10th problem, since that problem is equivalent to the halting problem.
But to be clear, I don't take this to show that the inductive Turing machine model refutes the Church-Turing thesis. 
Unfortunately, it seems that much of the commentary and literature surrounding the claim that it does is of poor quality and in some cases mathematically empty. The discussion seems to have become distracted in the literature and gotten off track in a way; it is a pity.
One of the central achievements of computability theory is the recognition of the subtle distinction between the concept of a set being computably enumerable and it being computably decidable. The recognition that these two aspects of computability are not the same has clarified so many issues in computability. We have known since Turing that the halting problem is computably enumerable but not decidable. Meanwhile, the main arguments for inductive computability violating the Church-Turing thesis seem to my way of understanding things to amount to an attempt to erase this important distinction. After all, the halting problem itself is computable in the limit, since we can say that a program does not halt until we see that it does, and then say from that point on that it does halt. Does this show that the halting problem is computably decidable? No, I don't think so, not in any satisfactory way. And similarly I reject that claim that functions computable-in-the-limit are computable. Since these kinds of simple observations seem to resolve essentially all of the issues on this topic, I cannot recommend following much of the literature surrounding this supposed debate.
A: 
Are inductive turing machines physically realizable (at least in the same sense of realizaility of Turing machines as Intel processors with bounded RAM and one that degrades over time)?
Can inductive turing machines solve the halting problem (essentially Hilbert's 10th as well)?

Others have basically answered this question, but if you're still not sure, the following exercise may help you see what's going on.
Forget about inductive Turing machines for the moment.  I have just invented  a Magic Machine that will solve the halting problem.  The Magic Machine is very simple.  If you give the Magic Machine a Turing machine T, the Magic Machine will simulate T, and will observe T while the simulation is running.  As long as the simulation continues to run, the Magic Machine keeps intoning, "It doesn't halt...it doesn't halt...it doesn't halt..."  However, if and when the simulated T halts, the Magic Machine suddenly changes its tune and says, "It halts!  It halts!  It halts!"
As I said, I claim that my Magic Machine is physically realizable and solves the halting problem.  Do you believe me?  If so, I will sell you one for the low, low price of $1 million.
If you don't think that my Magic Machine is "a physically realizable machine that solves the halting problem," then see if you can articulate clearly why you aren't inclined to buy it.  If you can, then you should be able to articulate equally clearly why no inductive Turing machine is "a physically realizable machine that solves the halting problem."
A: This is not an answer to the OP's question, and is a bit of a tangent. 
But perhaps relevant concerning the physical realizability issue raised by Joel.
I just today heard a talk on a "Fold-and-Cut Machine." 
This leads to a physical model equi-powerful 
to a nondeterministic Turing machine:

"a fold-and-cut machine can decide a 3-SAT instance with $n$ variables and $m$ clauses using $O(nm+m^2)$ operations (...), showing that the machine is at least as powerful as a nondeterministic Turing machine."
An, Byoungkwon, Erik D. Demaine, Martin L. Demaine, and Jason S. Ku. "Computing 3SAT on a Fold-and-Cut Machine."
  Full CCCG Proceedings download,
  p.208ff for the article.

One folds a strip of paper a polynomial number of times, 
snips it to produce holes, 
rearranges the folding, 
and then looks to see if you can see all the way through the
rearranged folding. You can iff the 3-SAT instance is solvable.



