Question

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

• A. K is negative, then liquid A and liquid B have convex meniscus.
• B. K is negative, then liquid A and liquid B have concave meniscus.
• C. K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.
• D. K is zero, then liquid A has convex meniscus and liquid B has concave meniscus.

Hint:

The sign of $K$ indicates the nature of the meniscus of the liquids.

Answer 1

The Correct Option is C

We are given two liquids A and B having contact angles \( \theta_A \) and \( \theta_B \) in a capillary tube. The ratio is defined as:

\[ K = \frac{\cos \theta_A}{\cos \theta_B} \]

We need to determine which statement about the nature of their meniscus (concave or convex) is correct when \( K \) is negative or zero.

Concept Used:

The shape of a meniscus depends on the contact angle \( \theta \):

• If \( \theta < 90^\circ \), then \( \cos \theta > 0 \) → concave meniscus (liquid wets the surface).
• If \( \theta > 90^\circ \), then \( \cos \theta < 0 \) → convex meniscus (liquid does not wet the surface).

Step-by-Step Solution:

Step 1: Analyze the expression for \( K \).

\[ K = \frac{\cos \theta_A}{\cos \theta_B} \]

The sign of \( K \) depends on the signs of \( \cos \theta_A \) and \( \cos \theta_B \).

Step 2: Consider the case when \( K \) is negative.

If \( K \) is negative, then one cosine term must be positive and the other negative.

• So, one liquid has \( \theta < 90^\circ \) (concave meniscus), and
• the other liquid has \( \theta > 90^\circ \) (convex meniscus).

Step 3: Determine which one corresponds to which case.

For \( K \) to be negative, \( \cos \theta_A \) and \( \cos \theta_B \) have opposite signs. Hence:

• If \( \cos \theta_A > 0 \) → Liquid A has concave meniscus.
• If \( \cos \theta_B < 0 \) → Liquid B has convex meniscus.

Step 4: Now, consider the case when \( K = 0 \).

\[ K = 0 \implies \cos \theta_A = 0 \]

This means \( \theta_A = 90^\circ \). At this angle, the liquid does not rise or fall, and the meniscus is flat. Thus, Liquid A neither wets nor repels the surface, while Liquid B’s nature depends on its own contact angle \( \theta_B \).

Final Computation & Result:

Hence, the correct interpretation is:

\[ \boxed{\text{If } K \text{ is negative, then liquid A has concave meniscus and liquid B has convex meniscus.}} \]

Final Answer: The correct statement is — If \( K \) is negative, then liquid A has concave meniscus and liquid B has convex meniscus.