What is the length (in cm) of chord PQ in a circle with a radius of 7 cm, where a diameter AB and non-diameter chord PQ intersect perpendicularly at point C, and the ratio of AC to BC is 4 : 3?

The correct option is 8√3.

We are given the following information:
- The radius of the circle, $r=7\text{cm}$.
- $\mathrm{AB}$ is a diameter of the circle. Therefore, the length of $\mathrm{AB}$ is:
$AB=2r=2\times 7=14\text{cm}$

Point $C$ lies on the diameter $\mathrm{AB}$ such that the ratio of the segments is:
$AC:BC=4:3$

Let $AC=4k$ and $BC=3k$ for some constant $k$.
Since $AC+BC=AB$, we can write:
$4k+3k=14$
$7k=14$
$k=2$

Substituting $k=2$ back into our segment lengths, we get:
$AC=4\times 2=8\text{cm}$
$BC=3\times 2=6\text{cm}$

The chord $\mathrm{PQ}$ intersects the diameter $\mathrm{AB}$ perpendicularly at point $C$. A diameter that is perpendicular to a chord bisects it. Therefore, $C$ is the midpoint of $\mathrm{PQ}$, which means:
$PC=CQ$

By the intersecting chords theorem, when two chords intersect inside a circle, the products of their segments are equal:
$AC\times BC=PC\times CQ$

Using $PC=CQ$ and substituting the values of $AC$ and $BC$:
$8\times 6=P{C}^2$
$P{C}^2=48$
$PC=\sqrt{48}=4\sqrt{3}\text{cm}$

Since $\mathrm{PQ}$ is twice the length of $\mathrm{PC}$:
$PQ=2\times PC=2\times 4\sqrt{3}=8\sqrt{3}\text{cm}$