The situation you describe, in which the axis of a partially filled
cylindrical tank is horizontally aligned, is analyzed on this Web
[Partially Filled Horizontal Cylindrical Tank]
Besides the nice illustration (Fig. 1.1) at the top of the page, which
shows clearly the equivalence of the fraction of volume filled in the
tank with the fraction of a circular cross-section "filled" to the
liquid's surface line, there is a very useful graph (Fig. 1.8) at the
bottom of the page.
For practical purposes your Answer may simply be "looked up" on that
graph. In order to fill 10% of the volume of the tank, the vertical
depth of the liquid (at center) should be something more than 0.3
times the radius, or 0.15 times the diameter if you prefer. We make
these figures more precise with the computations below.
A more exact approach requires a bit of mathematical formulation.
From the graceful curve shown in the graph, we recognize that the
relationship between the percentage of tank volume and the ratio of
liquid height (depth) to radius is nonlinear.
To continue with the formulation, let's define our values. Let R be
the radius of the tank. You may assume whatever units of length are
convenient, because we are going to use ratios to eliminate any
dependence on the specific choice of units.
Let H be the vertical height of the liquid (at center), so that H/R is
a dimensionless ratio expressing the height of the liquid in
relationship to the tank's radius. If it were more convenient for
some reason to work with the diameter of the tank, rather than its
radius, note that the diameter is simply 2R.
Now an empty tank will have H/R = 0, while a full tank has H/R = 2.
For that matter when H/R = 1 the tank is precisely half full. The
graph's curve passes smoothly through these three collinear points,
but as noted before, it's not flat.
For convenience let's write h = H/R, so we will have a simple variable
for this dimensionless value, the ratio of liquid height to radius.
Now the fraction v of the tank's volume which is filled is the same as
the fraction of the area of the circle which is "filled", the circle
segment bounded by a chord corresponding to the liquid's surface, as
shown at the right in Figure 1.1 on the Web page above.
Developing a formula for the fraction v which the liquid fills in the
tank is then equivalent to a formula for the fraction of area of the
circle within the circle segment (bottom right of Fig. 1.1).
Some analytical formulas for v in terms of h, or more literally for v
in terms of R and H, are given on that Web page. However I think some
greater simplicity can be preserved if we introduce some intermediate
variable to which both v and h can be functionally related. As you
will see, this will serve our practical needs quite well.
In particular I wish to define an angle A (in radians) which is half
of the central angle subtended by the chord forming the circle
segment. In other words A is the measure of the angle formed by the
vertical centerline of the circle and the ray that extends to the
corner of the circle segment (at the point where the liquid's surface
reaches the inner tank).
We then have the following relations based on the tank's geometry:
v = (A - 0.5 sin(2A))/pi
h = 1 - cos(A)
It is because of the relatively simple relationship of h to A that one
is able to express v in terms of h, i.e. A = arccos(1 - h).
However to answer your question of what value of h gives v = 0.1, it
is still necessary to confront the difficult nonlinear expression
above for v. Essentially we still need to find what angle A will
A - 0.5 sin(2A) = 0.1 pi
Once this angle A is known with the required precision, it is simple
to recover a corresponding value for h = 1 - cos(A).
When only a single solution, e.g. for v = 0.1, is sought, it may be
found with nothing more than a decent scientific calculator and a bit
of trial and error. For example, bearing in mind that we are working
with angles measured in radians here, I found using the Windows
A = 0.81 gives v = (0.3106...)/pi = 0.09887...
While not an especially precise solution, it would be easy to tweak A
upwards to obtain even closer approximations of v to 0.1.
If we consider the corresponding value for h:
h = 1 - cos(A) = 0.31050...
we confirm the earlier observation based on the graph (Fig. 1.8) that
the Answer to your Question is found for the height to radius ratio h
slightly more than 0.3.
If you wish to be assured of more than 10% filled volume, then I would
suggest a target value of h = 0.32. This corresponds to A = 0.8230...
and thus to:
v = (A - 0.5 sin(2A))/pi = 0.103...
Please let me know, by Request for Clarification, if some elaboration
of the formulas or other discussion would be helpful to you.