Question

A small ball of mass \( M \) and density \( \rho \) is dropped in a viscous liquid of density \( \rho_0 \). After some time, the ball falls with a constant velocity. What is the viscous force on the ball?

• A. \( F = Mg \left( 1 -\frac{\rho_0}{\rho} \right) \)
• B. \( F = Mg \left( 1 + \frac{\rho}{\rho_0} \right) \)
• C. \( F = Mg \left( 1 + \frac{\rho_0}{\rho} \right) \)
• D. \( F = Mg \left( 1 \pm \rho \rho_0 \right) \)

Hint:

In problems involving viscous force and terminal velocity, remember that at terminal velocity:
The weight of the object is balanced by the sum of buoyant and viscous forces.

Answer 1

The Correct Option is A

Phase 1: Situational Analysis
A ball achieves constant velocity when the net external force acting upon it is nullified. The forces under consideration are: 
The gravitational force on the ball, denoted as \( Mg \). 
The buoyant force exerted by the liquid, which is equivalent to the weight of the liquid displaced by the ball. 
Phase 2: Force Calculation 
The buoyant force is quantified by the weight of the displaced liquid, expressed as \( \rho_0 V g \), where \( V \) represents the ball's volume. 
The ball's volume can be mathematically represented as \( V = \frac{M}{\rho} \), given that \( \rho = \frac{M}{V} \). 
Phase 3: Force Equilibrium at Constant Velocity 
At the point of terminal velocity, the ball's weight \( Mg \) is counteracted by both the viscous drag force and the buoyant force. Therefore, the equation is: \[ Mg = F_{\text{viscous}} + \rho_0 V g \] Upon substituting \( V = \frac{M}{\rho} \), the equation becomes: \[ Mg = F_{\text{viscous}} + \rho_0 \frac{M}{\rho} g \] The viscous force \( F_{\text{viscous}} \) is determined as follows: \[ F_{\text{viscous}} = Mg \left( 1 -\frac{\rho_0}{\rho} \right) \] Concluding Statement: The viscous force acting on the ball is \( F = Mg \left( 1 -\frac{\rho_0}{\rho} \right) \).