1. Relation between contact angle and meniscus

   In a capillary tube:

   • If $\theta < 90^\circ$, then $\cos\theta > 0$ and the liquid wets the surface, so the meniscus is concave.
   • If $\theta > 90^\circ$, then $\cos\theta < 0$ and the liquid does not wet the surface, so the meniscus is convex.
   • If $\theta = 90^\circ$, then $\cos\theta = 0$.

2. Given quantity

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

   We now examine the sign of $K$.

3. If $K$ is negative

   For the ratio $\dfrac{\cos\theta_A}{\cos\theta_B}$ to be negative, the numerator and denominator must have opposite signs.

   So one of $\cos\theta_A, \cos\theta_B$ is positive and the other is negative.

   That means:

   • one liquid has $\theta<90^\circ$ $\Rightarrow$ concave meniscus,
   • the other has $\theta>90^\circ$ $\Rightarrow$ convex meniscus.

   Thus, both cannot be concave and both cannot be convex.

4. Check options

   • A: If $K$ is negative, both convex meniscus. ❌ Impossible.
   • B: If $K$ is negative, both concave meniscus. ❌ Impossible.
   • C: If $K=0$, then $\cos\theta_A=0$ (assuming denominator nonzero), so $\theta_A=90^\circ$. This does not imply convex meniscus for $A$. ❌
   • D: If $K$ is negative, one is concave and the other is convex. This matches the sign argument. ✅

5. About the order in option D

   Option D states: liquid $A$ is concave and liquid $B$ is convex.

   This corresponds to: $\cos\theta_A>0,\quad \cos\theta_B<0$ so that $K=\frac{(+)}{(-)}<0$

   Hence option D is consistent.

Final Answer

$\boxed{\text{D}}$