You are correct : the diagram is unrealistic. The entire problem leaves out a lot of important information. It is a very disappointing effort from such a well-known teaching site.
Mass $m_1$ does not appear to be moving on a frictionless horizontal table - although that would be a reasonable assumption. Gravity must be assumed to act on mass $m_2$ otherwise there could be no centripetal force. And the preceding paragraph refers to $m_2$ as a counterweight. Gravity must act on $m_1$ also. Without a table to rest on, mass $m_1$ would not rotate in a perfectly horizontal circle however fast it moved. The string would always be inclined downwards by some small angle (see Conical Pendulum). For this reason alone the centripetal force would indeed be less than the counter-weight.
The tube has a rounded opening or bell which, like a pulley, allows the string to change direction smoothly. The implication is that the tube is frictionless, and that the tension is the same in both sections of the string. If static friction were included the tension in the nearly-horizontal section of the string could be greater or less than that in the vertical section, because friction could act in either direction.
The bell opens out horizontally. Perhaps the person who devised the question intended that this would ensure that the string remained horizontal when it left contact with the bell, so that mass $m_1$ is guaranteed to travel in a horizontal circle. If so, this is a naive assumption which is incorrect. The string will change direction abruptly at the rim where it is no longer supported by the bell.
Lastly, it is not stated that $m_2$ hangs freely, but that seems to be implied. Like $m_1$, counterweight $m_2$ could also be resting on an invisible horizontal surface, so that part of its weight is balanced by the normal contact force.
The length of the string is fixed, and the problem states that $m_1$ moves in a circle, therefore $m_2$ does not move vertically. It is held in equilibrium by the upward tension in the string and the downward force of gravity. It is not necessary to assume that $m_2$ rests on a horizontal surface. The fact that $m_2$ is not moving vertically (ie that it is not falling) does not mean that it is resting on a horizontal surface which stops it from falling.
I think I got the solution. I chose (C) at the poll, but the correct answer is (B).
Break the cable in $N$ chunks, each of mass $m$. Call $1$ the chunk at the bottom of the rope, and $N$ the one of the top. Draw a free body diagram for chunk 1. There if a vertical force of $mg$ due to gravity, a horizontal force $D$ due to the drag, and a force $\vec T_1$ exerted by the chunk $2$ of rope. Call $\theta_1$ the angle of $\vec T_1$ away from the vertical. Since the chunk is moving at constant velocity by assumption, we have the equations
$$
\left\{
\begin{align}
mg &= T_1 \cos\theta_1 \\
D &= T_1 \sin\theta_1
\end{align}
\right.
$$
Let's move to the next chunk of rope. Again, we have a vertical force $mg$, a horizontal force $D$, the force $\vec T_2$ exerted by the chunk of rope $3$ and, this time, by Newton's third law, we also have $-\vec T_1$ exerted by chunk 1. Then by Newton's second:
$$
\left\{
\begin{align}
mg + T_1\cos\theta_1 &= T_2 \cos\theta_2 \\
D + T_1 \sin\theta_1 &= T_2 \sin\theta_2
\end{align}
\right.
$$
It shouldn't be hard to convince yourself that the signs are right. Note also we are not assuming $\theta_2=\theta_1$ for now, as that will be our conclusion.
Note that we have already found $\vec T_1$, so let's replace it in here:
$$
\left\{
\begin{align}
2mg &= T_2 \cos\theta_2 \\
2D &= T_2 \sin\theta_2
\end{align}
\right.
$$
Pretty neat, eh? We have an expression for $\vec T_2$ that does not depend on the details of $\vec T_1$. Actually, you can generalise this formula by induction, or simply inspection, to get $\vec T_k$ for all $k=1,...,N$:
$$
\left\{
\begin{align}
kmg &= T_k \cos\theta_k \\
kD &= T_k \sin\theta_k
\end{align}
\right.
$$
Ok, now were are almost done. Notice how the magnitude $T_k$ of $\vec T_k$ depends on $k$:
$$T_k = \sqrt{(kmg)^2 + (kD)^2} = k \sqrt{(mg)^2+D^2} = kT_1.$$
The tension grows linearly in $k$. Cute. Here is the kicker though:
$$\tan\theta_k = \frac{\sin\theta_k}{\cos\theta_k}=\frac{kD}{kmg} = \frac{D}{mg}.$$
So the angles $\theta_k$ do not depend on $k$. Since the angle $\theta_k$ how much the line connecting the chunk $k$ and the chunk $k+1$ deviates from vertical, this tells you that the rope is a straight line. $\square$
You can sanity check: if you ignore air resistance, set $D=0$ then $\tan \theta = 0$ and $\theta = 0$: rope is vertical. As $D$ becomes larger and larger, $\tan\theta$ grows more and more, and $\theta$ approaches $\pi/2$, horizontal rope.
EDIT: what happens if we do not assume that drag is constant?
In the above, we assumed that each chunk of rope experiences the same amount of drag $D$. This simplifies matters a lot. In reality, the drag will be proportional to the areas of the projection of the chunk of rope along the plane perpendicular to the movement. In formulas,
$$D_k = \gamma(v) \cos\theta_k,$$
where $\gamma(v)$ has dimensions of acceleration, and is a monotone increasing function of the velocity, with $\gamma(0)=0$. Assuming that $D=D_k$ for all $k$ is compatible with taking a small angle approximation.
At first sight, not taking this approximation might seem to allow for a different shape to the rope. I will now argue that this is not the case, using a purely physical argument.
To simplify matters, let's assume that the rope is hanging from a pole in a wind tunnel. Thus the rope is not moving in our frame, and instead we a have a wind at constant velocity $v$. When $v=0$, the rope is vertical. Now, let a little bit of wind in. The small angle approximation is expected to hold, and we have the whole rope at a small angle $\theta$. Each chunk of rope is experiencing exactly the same amount of drag.
Now, what happens as we increase $v$ by an amount $\delta v$? Each single chunk of rope experiences a little more drag. But, crucially, they all experience the same increase in drag! So again the rope is in a straight line, with a slightly larger angle. So, no matter how strong the wind is, there is a steady configuration where the rope is straight.
Let me put it in a different way. The force balance for the $k$th chunk of rope is
$$
\left\{
\begin{align}
kmg &= T_k \cos\theta_k \\
k\gamma(v)\cos\theta_k &= T_k \sin\theta_k
\end{align}
\right.
$$
Now, let's solve all $2N$ of these equations simultaneously by assuming $\theta_k=\theta$ for all $k$. The equations for a chunk $k$ become:
$$
\left\{
\begin{align}
kmg &= T_k \cos\theta \\
k\gamma(v)\cos\theta &= T_k \sin\theta
\end{align}
\right.
$$
Then take the ratio of these two formulas to get:
$$\tan\theta = \frac{\sin\theta}{\cos\theta}=\frac{k\gamma(v)\cos\theta}{kmg} = \frac{\gamma(v)}{mg}\cos\theta.$$
Note that $k$ dropped out of the formula!
We can rearrange this to
$$\sin\theta = \frac{\sin\theta}{\cos\theta}=\frac{k\gamma(v)\cos\theta}{kmg} = \frac{\gamma(v)}{mg}(1-sin^2\theta)$$
(thanks @MichaelSeifert!), which has exactly one positive solution:
$$\sin\theta = \frac{mg}{2\gamma}\left(\sqrt{1+\left(\frac{2\gamma}{mg}\right)^2}-1\right).$$
If $\theta_k=\theta$, each we solve all the equations. $\square$
"Wait!", you might say, "you assumed that the rope was in a straight line, isn't that what you wanted to prove?". Well, yes, I did assume that the rope was in a straight line. But what I proved, is that there a straight line configuration is a static configuration, an equilibrium. In other words, I proved that if the rope is in a straight line at that angle, it will feel no net forces.
To really know that this is the configuration the rope will take, one would need to show that this is a stable equilibrium, so that small deviations will decay and lead there. I'm happy to take that as an assumption.
Sanity check. Let $\alpha = \frac{2\gamma}{mg}$, this is a dimensionless quantity that quantifies how strong is the wind. We can rewrite the angle as a function of $\alpha$:
$$\sin\theta = \frac{\sqrt{1+\alpha^2}-1}{\alpha}.$$
With weak winds, $\alpha\ll 1$ and
$$\sin\theta \approx \frac\alpha2 = \frac\gamma{mg},$$
which is compatible with the computation before. This is neat, as there we assumed that the drag was independent of the angle, which is a valid approximation only for small angles, i.e. small winds. (By the way, I realise now that I should have put $\gamma=D$)
If instead we have huge winds, $\alpha\gg1$, we have
$$\sin\theta \approx 1 - \frac{mg}{2\gamma},$$
This model predicts a maximum angle of $\theta = \pi/2$ for the rope.
Best Answer
I think the author has forgotten to mention that the part of the rope on the table is stretched out in a straight line, otherwise, I would say the way it's piled does make a difference, and the problem is not solvable as stated.
In that case, the equation of motion is simply
$$M \ddot x = m_{hang} g$$
$$M \ddot x = M \frac x l g$$
$$\ddot x = \frac g l x$$
This indeed gives us a general solution of
$$x(t) = A e^{\lambda t} + B e^{-\lambda t}$$
where $\lambda = \sqrt{\frac x l}$