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Question #42109

Difficulty: 8/10 | 🟢 Free

The Question

In a theoretical model of a fluid, the molecular cohesive force field near the liquid-gas interface is anisotropic. The interface is parallel to the $xy$ -plane at $z = 12L$ . For a molecule at position $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ (where $z=0$ represents the deep bulk), the cohesive force is directed entirely perpendicular to the interface and is given by $\vec{F}_c = - \frac{F_0 z^2}{L^2} \hat{k}$ An external agent quasi-statically moves a molecule from the origin $(0,0,0)$ to a point $(3L, 4L, 12L)$ on the liquid surface. What is the work done by the external agent?

Options

A.
576 F_0 L
B.
$\frac{2197}{3} F_0 L$
C.
676 F_0 L
D.
1872 F_0 L

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Molecular Forces of Cohesion|Potential Gradient Direction|Conservative Force Field|Dot Product Geometry

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of work done in a non-uniform, anisotropic conservative force field.

A common trap is assuming molecular cohesion acts as a central force ( $\vec{F} \propto r^2 \hat{r}$ ), which would lead students to integrate over the radial distance $r = \sqrt{(3L)^2 + (4L)^2 + (12L)^2} = 13L$ to incorrectly obtain $\frac{2197}{3} F_0 L$ .

By defining the cohesive gradient strictly perpendicular to the interface, the problem forces the application of the dot product $\vec{F} \cdot d\vec{r}$ , demonstrating that displacement parallel to the surface requires no work against cohesion.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Molecular Forces of Cohesion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free