Two liquids $A$ and $B$ have $\theta_A$ and $\theta_B$ as contact angles in a capillary tube. If $K=\cos \theta_A / \cos \theta_B$, then identify the correct statement:

<p>The contact angle, $\theta,$ of a liquid in a capillary tube determines its meniscus shape:</p> <p><p>If $\theta < 90^\circ,$ then $\cos \theta > 0$ and the liquid wets the tube, giving a <strong>concave meniscus</strong>.</p></p> <p><p>If $\theta > 90^\circ,$ then $\cos \theta < 0$ and the liquid is non-wetting, resulting in a <strong>convex meniscus</strong>.</p></p> <p><p>If $\theta = 90^\circ,$ then $\cos \theta = 0$ and the meniscus is essentially flat.</p></p> <p><p>We are given the ratio:</p> <p>$K = \frac{\cos \theta_A}{\cos \theta_B}.$</p></p> <p><p>For $K$ to be negative:</p></p> <p><p>The numerator and denominator must have opposite signs.</p></p> <p><p>This means one of the liquids has $\cos \theta > 0$ (concave meniscus) and the other has $\cos \theta < 0$ (convex meniscus).</p></p> <p>Consider Option D:</p> <p><p>It states that if $K$ is negative, then liquid A has a <strong>concave meniscus</strong> (so $\theta_A < 90^\circ$, $\cos \theta_A > 0$) and liquid B has a <strong>convex meniscus</strong> (so $\theta_B > 90^\circ$, $\cos \theta_B < 0$).</p></p> <p><p>Hence, the ratio becomes negative:</p> <p>$K = \frac{(+)}{(-)} < 0.$</p></p> <p>Therefore, the correct statement is:</p> <p>Option D</p>