Question 21: Properties of Matter

Question

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

Accepted Answer

The contact angle, $$ heta,$$ of a liquid in a capillary tube determines its meniscus shape: If $$ heta 0$$ and the liquid wets the tube, giving a concave meniscus . If $$ heta > 90^\circ,$$ then $$\cos heta 0$$ (concave meniscus) and the other has $$\cos heta 0$$) and liquid B has a convex meniscus (so $$ heta_B > 90^\circ$$, $$\cos heta_B < 0$$). Hence, the ratio becomes negative: $$K = \frac{(+)}{(-)} < 0.$$ Therefore, the correct statement is: Option D

Answer

K is negative, then liquid A and liquid B have convex meniscus.

Answer

$K$ is negative, then liquid $A$ and liquid $B$ have concave meniscus.

Answer

K is zero, then liquid A has convex meniscus and liquid B has concave meniscus.

Answer

K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.

Properties of Matter

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:

Choose the correct answer: