Question

A metal sphere of mass \( m \) and density \( \sigma_1 \) falls with terminal velocity through a container containing liquid. The density of liquid is \( \sigma_2 \). The viscous force acting on the sphere is

• A. \( mg \left( 1 - \frac{\sigma_1}{\sigma_2} \right) \)
• B. \( mg \left( 1 - \frac{\sigma_2}{\sigma_1} \right) \)
• C. \( mg \left( 1 + \frac{\sigma_1}{\sigma_2} \right) \)
• D. \( mg \left( 1 + \frac{\sigma_2}{\sigma_1} \right) \)

Hint:

In problems involving terminal velocity, remember to account for both the gravitational force and the buoyant force. The net force determines the viscous force at terminal velocity.

Answer 1

The Correct Option is B

Step 1: Understanding terminal velocity and viscous force.
The terminal velocity occurs when the force due to gravity is balanced by the buoyant force and the viscous drag force. The force due to gravity is \( mg \), and the buoyant force is proportional to the density of the liquid. The viscous drag force is proportional to the density difference between the sphere and the liquid. Step 2: Applying the density relation.
At terminal velocity, the net force is zero, so the viscous force \( F_v \) is: \[ F_v = mg \left( 1 - \frac{\sigma_2}{\sigma_1} \right) \] Thus, the correct answer is (B) \( mg \left( 1 - \frac{\sigma_2}{\sigma_1} \right) \).