The contact angle, $$\theta,$$ of a liquid in a capillary tube determines its meniscus shape:

If $$\theta < 90^\circ,$$ then $$\cos \theta > 0$$ and the liquid wets the tube, giving a concave meniscus.

If $$\theta > 90^\circ,$$ then $$\cos \theta < 0$$ and the liquid is non-wetting, resulting in a convex meniscus.

If $$\theta = 90^\circ,$$ then $$\cos \theta = 0$$ and the meniscus is essentially flat.

We are given the ratio:

$$K = \frac{\cos \theta_A}{\cos \theta_B}.$$

For $$K$$ to be negative:

The numerator and denominator must have opposite signs.

This means one of the liquids has $$\cos \theta > 0$$ (concave meniscus) and the other has $$\cos \theta < 0$$ (convex meniscus).

Consider Option D:

It states that if $$K$$ is negative, then liquid A has a concave meniscus (so $$\theta_A < 90^\circ$$, $$\cos \theta_A > 0$$) and liquid B has a convex meniscus (so $$\theta_B > 90^\circ$$, $$\cos \theta_B < 0$$).

Hence, the ratio becomes negative:

$$K = \frac{(+)}{(-)} < 0.$$

Therefore, the correct statement is:

Option D