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Question #43907

Difficulty: 8/10 | 🟢 Free

The Question

A solid vertical cylinder of mass $m$ and radius $r$ rests on a horizontal floor inside an initially empty tank. The coefficient of kinetic friction between the cylinder and the floor is $\mu$ . A horizontal force is applied to the cylinder, pulling it such that it moves with a constant horizontal velocity $v$ . Simultaneously, a liquid of density $\rho$ and surface tension $\gamma$ is slowly introduced into the tank from the bottom, causing the free surface of the liquid to rise at a constant rate $u$ . The liquid wets the cylinder with a contact angle $\theta$ . Assuming the liquid completely seeps under the cylinder (exerting full buoyant force) and the cylinder never loses contact with the floor, what is the total work done by the horizontal pulling force from the moment the liquid starts rising ( $h=0$ ) until the free surface reaches a height $H$ ?

Options

A.
$\mu \frac{v}{u} H \left( mg + 2\pi r \gamma \cos\theta - \frac{1}{2} \pi r^2 \rho g H \right)$
B.
$\mu \frac{v}{u} H \left( mg - \frac{1}{2} \pi r^2 \rho g H \right)$
C.
$\mu \frac{v}{u} H \left( mg - 2\pi r \gamma \cos\theta - \frac{1}{2} \pi r^2 \rho g H \right)$
D.
$\mu \frac{v}{u} H \left( mg + 2\pi r \gamma \cos\theta - \pi r^2 \rho g H \right)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Kinetic Friction Transition|Capillary Rise Mechanics|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple fluid statics, specifically capillary action and buoyancy, with rigid body dynamics and work-energy principles.

The primary trap lies in the formulation of the normal force.

Students commonly assume $N = mg - F_{buoyant}$ , completely ignoring the surface tension force vector at the contact line.

Because the liquid wets the cylinder ( $\theta < 90^\circ$ ), the meniscus curves upward, meaning the liquid pulls the cylinder downward with a force $F_\gamma = 2\pi r \gamma \cos\theta$ .

Failing to account for this downward capillary force leads to an incorrect, lower normal force (Option B).

Furthermore, the dynamic nature of the problem requires integrating the variable friction force over the changing height, testing whether students correctly apply the relation $dx = (v/u)dh$ rather than just multiplying the final force by the total distance (Option D).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Kinetic Friction Transition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free