Finding vertical spacing between string wound on the cylinder. 
Consider a cylinder of height $h$ cm and radius $r = \dfrac2π$ cm as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of $n$ turns (in other words, the string's length is the minimum length required to wind $n$ turns)

 
$1.)$ what is the vertical spacing in cm between two consecutive turns ?
$a.) \quad \dfrac hn\\
b.)\quad\dfrac {h}{\sqrt{n}}\\
c.)\quad\dfrac{ h}{n^2}\\
d.)\quad \text{cannot be determined}$
answer given is $a.)\quad \dfrac hn $ .
the lateral surface area is  $2\pi rh=2\pi\cdot \dfrac2\pi\cdot h=4h$
i felt that it would be useful to get that , but i can't seem to find the way to the get through the solution
 A: No, surface area has nothing to do with it...
Hint: Suppose that $A$ and $B$ are points at the extremes of the cylinder, lying on the same vertical line (which can't really be inferred from either the statement or the picture): what does it mean that the length of the string is minimal? What if you were to try it with a real cylinder and a real string?
Note: A complete, rigorous answer would likely be too difficult for you to understand, as it involves some calculus. Basically, it boils down to proving that the geodesic curves on a cylinder of radius $r$ are precisely the helixes of radius $r$. It is like saying that the "straight lines" between two points on a cylinder's surface are helixes.

Hypothesis $(*)$: To get an idea of what is going on, suppose you know that the shortest curve on your cylinder between two points $C_1,C_2$ on the segment $\overline{AB}$ is precisely a helix arc with pitch the length of $\overline{C_1C_2}$ and radius $r$.
Call $C_0,\dotsc,C_n$ the points of intersection between the $n$-times wound string $s$ and the segment $A,B$, so that $A = C_0$ and $B = C_n$. By the Hypothesis $(*)$ and the hypothesis of minimality of length of $s$, we can assume that $s$ is the union of $n$ arcs of helix, starting at $C_i$ and ending at $C_{i+1}$ for each $i \in \{0,\dotsc,n-1\}$. If we can prove that all the segments $\overline{C_i,C_{i+1}}$ have the same length, then
$$
\frac{\text{heigth}}{\text{n° of loops}} = \frac hn
$$

To simplify the setting, suppose $n = 2$ and let $c_i$ be the arc of helix between $C_i$ and $C_{i+1}$ for $i \in \{0,1\}$. From the Wikipedia page on helixes, we know that the length of $c_i$ is $\sqrt{r^2 + b_i^2}$ where $b_i$ is $1/2\pi$ times the length of $\overline{C_iC_{i+1}}$. Define
$$
b = \frac{h}{4\pi}
$$
so that $2\pi b$ is exactly half the height of the cylinder. If the the segments $\overline{C_iC_{i+1}}$ have different length, then there is a $c \neq 0$ with $\left|c\right| < b$ such that $b_0 = b - c$ and $b_1 = b + c$. In particular, $s$ will have length
$$
\sqrt{r^2 + (b-c)^2} + \sqrt{r^2 + (b+c)^2}
> 2\sqrt{r^2 + b^2}
$$
Why is this? The RHS is the length of the hypotenuse $H$ of the right triangle $T$ with sides $2r$ and $2b$, while the LHS is the sum of the lengths of the hypotenuses $H_0,H_1$ of the right triangles $T_0,T_1$ with one side $r$ and the other $b-c$ and $b+c$ respectively. The choice $c \neq 0$ guarantees that the triangle with sides $H,H_0,H_1$ is non-degenerate.
A: I am assuming $ AB = h. $
Take the development on a paper sheet. Repeatedly ( n times) wrap it on the cylinder. 
The shortest distance ( called a geodesic) between A and B is a straight line between start A and ending B points, basically climbing along a circular helix like a single start screw -thread.. 
Cylinder perimeter = 4 n, height h, length of wound thread $ hypotenuse =\sqrt{h^2 + 16 n^2} $, $ \tan( Helix Angle) = \frac {h}{4 n}, pitch = \frac{h}{n} $.
Surface Area plays no direct role.In variational calculus it is is the field of excursions of thread to minimize length.
A: Unwrap the $n$-fold covered cylinder so that you obtain a rectangle. The string then appears as a diagonal of this rectangle.
In $n$ full turns the height gain of an ant walking along  the string is $h$. Therefore the height gain per turn is ${h\over n}$. 
