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

Step 1: Understanding the situation 
The ball reaches a constant velocity when the net force acting on it becomes zero. The forces involved are: 
The weight of the ball, \( Mg \). 
The buoyant force due to the liquid, which is equal to the weight of the liquid displaced by the ball. 
Step 2: Finding the force on the ball 
The buoyant force is given by the weight of the liquid displaced, which is \( \rho_0 V g \), where \( V \) is the volume of the ball. 
The volume of the ball can be expressed as \( V = \frac{M}{\rho} \) since \( \rho = \frac{M}{V} \). 
Step 3: Equating the forces at constant velocity 
At terminal velocity, the weight of the ball \( Mg \) is balanced by the viscous force and the buoyant force. Thus: \[ Mg = F_{\text{viscous}} + \rho_0 V g \] Substituting \( V = \frac{M}{\rho} \): \[ Mg = F_{\text{viscous}} + \rho_0 \frac{M}{\rho} g \] The viscous force \( F_{\text{viscous}} \) is: \[ F_{\text{viscous}} = Mg \left( 1 -\frac{\rho_0}{\rho} \right) \] Final Answer: The viscous force on the ball is \( F = Mg \left( 1 -\frac{\rho_0}{\rho} \right) \).