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Question #42120

Difficulty: 8/10 | 🟢 Free

The Question

A spherical region of radius $R$ contains a viscous gas of viscosity $\eta$ and a fixed spherically symmetric volume charge density $\rho(r) = \rho_0 \frac{r}{R}$ , where $r$ is the radial distance from the center and $\rho_0$ is a positive constant. A small oil droplet of charge $q$ and radius $a$ is released from $r = R/2$ and moves radially outward. Assuming the mass of the droplet is negligible so that it instantaneously reaches its local terminal velocity, what is the time taken by the droplet to reach the boundary of the region ( $r = R$ )? (Ignore gravity)

Options

A.
$\frac{24 \pi \epsilon_0 \eta a}{q \rho_0}$
B.
$\frac{18 \pi \epsilon_0 \eta a}{q \rho_0} \ln 2$
C.
$\frac{12 \pi \epsilon_0 \eta a}{q \rho_0}$
D.
$\frac{18 \pi \epsilon_0 \eta a}{q \rho_0}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Viscosity and Stokes' Law|Continuous Charge Distribution|Electrostatic Force

Why This Question Is Tricky

Common Mistakes

This problem tests the synthesis of electrostatics (continuous charge distributions) and fluid dynamics (Stokes' drag).

The primary trap is assuming a constant electric field or uniform charge density, leading to incorrect algebraic kinematics ( $t = d/v$ ).

Students must recognize that negligible mass implies the droplet is always in a state of local equilibrium ( $F_{net} = 0$ ), transitioning the problem into a first-order separable differential equation $\frac{dr}{dt} \propto r^2$ .

Failing to integrate the non-uniform velocity profile correctly is a classic pitfall.

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