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Question #42119

Difficulty: 8/10 | 🟢 Free

The Question

A large open container filled with a viscous liquid of density $\rho_L$ and coefficient of viscosity $\eta$ moves horizontally with a constant acceleration $a = \sqrt{3}g$ . A small solid sphere of radius $r$ and density $\rho_S$ (where $\rho_S > \rho_L$ ) is released from rest relative to the liquid. After a long time, the sphere reaches a steady terminal velocity $\vec{v}_t$ relative to the container. What is the magnitude of this terminal velocity?

Options

A.
$\frac{2 r^2 (\rho_S - \rho_L) g}{9\eta}$
B.
$\frac{4 r^2 (\rho_S - \rho_L) g}{9\eta}$
C.
$\frac{2 r^2 g \sqrt{(\rho_S - \rho_L)^2 + 3\rho_S^2}}{9\eta}$
D.
$\frac{2 r^2 (\rho_S - \rho_L) \sqrt{3} g}{9\eta}$

Check Your Answer

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

Viscosity and Stokes' Law|Frame of Reference Inertial vs Non-Inertial|Vector Analysis

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply fluid dynamics and Stokes' Law in a non-inertial reference frame.

A common trap is assuming that the buoyant force remains strictly vertical (opposing only true gravity) while the pseudo-force acts independently on the sphere's mass.

In reality, the accelerating fluid establishes a tilted pressure gradient, meaning the buoyant force aligns anti-parallel to the effective gravity vector $\vec{g}_{\text{eff}} = \vec{g} - \vec{a}$ .

Students who fail to recognize this will incorrectly calculate the vector sum of forces (often leading to Option C), failing to realize that the entire settling dynamic simply rotates to align with $\vec{g}_{\text{eff}}$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Viscosity and Stokes' Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free