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Question #42235

Difficulty: 8/10 | 🟢 Free

The Question

A solid cubical block of side length $L = 2\text{ m}$ and uniform density $\rho_b = 1100\text{ kg/m}^3$ is released from rest such that its lower face just touches the surface of an inviscid liquid. The density of the liquid varies with depth $y$ as $\rho(y) = \rho_0 + ky$ , where $\rho_0 = 1000\text{ kg/m}^3$ and $k = 600\text{ kg/m}^4$ . Assuming the liquid is deep enough and the block does not tumble, what is the maximum depth $H$ reached by the bottom face of the block? (Take $g = 10\text{ m/s}^2$ )

Options

A.
$2.5\text{ m}$
B.
$3.0\text{ m}$
C.
$3.5\text{ m}$
D.
$4.0\text{ m}$

Check Your Answer

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

Archimedes' Buoyancy Dynamics|Mechanical Energy Conservation|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in a stratified fluid where the buoyant force is position-dependent.

The primary trap is assuming the buoyant force takes its fully submerged functional form from the moment the block touches the fluid, completely ignoring the immersion phase where the force is quadratic with respect to depth.

Students often incorrectly write $W_B = \int_0^H F_{B,submerged} dy$ or attempt to use the center of mass of the displaced fluid without accounting for the changing geometry of the displaced volume during entry.

Recognizing the piecewise nature of the non-conservative buoyant force is essential for setting up the correct energy conservation integral.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Archimedes' Buoyancy Dynamics
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free