Two liquids A and B have θA and θB as contact angles in a capillary tube. If K = cos θA / cos θB, 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 zero, then liquid A has convex meniscus and liquid B has concave meniscus. (d) K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.

Answer: (d). Concave meniscus: θ 0. Convex meniscus: θ > 90°, cos θ < 0. K = (cos θA)/(cos θB) < 0 means cos θA and cos θB have opposite signs, so one meniscus is concave and the other convex. K = 0 would need θA = 90° (flat surface), so (c) is wrong; (a) and (b) are wrong since both menisci cannot be the same type. So the answer is (d) A concave and B convex.

Two liquids A and B have θA and θB as contact angles in a capillary… | JEE Main 2025 (4 Apr, Shift 1)

Q21JEE Main 2025 · 4 APR · SHIFT 1Medium

Angle of Contact & Capillarity

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:

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Answer: (d)

Concave meniscus: $\theta < 90^\circ$ , $\cos\theta > 0$ . Convex meniscus: $\theta > 90^\circ$ , $\cos\theta < 0$ .
$K = \frac{\cos\theta_A}{\cos\theta_B} < 0$ means $\cos\theta_A$ and $\cos\theta_B$ have opposite signs, so one meniscus is concave and the other convex.
$K = 0$ would need $\theta_A = 90^\circ$ (flat surface), so (c) is wrong; (a) and (b) are wrong since both menisci cannot be the same type.
So the answer is (d) A concave and B convex.

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