Question

A small metal sphere of mass \( M \) and density \( \rho_1 \), when dropped in a jar filled with liquid moves with terminal velocity after some time. The viscous force acting on the sphere is
\textit{(where \( \rho_2 \) is the density of liquid, \( g \) is the gravitational acceleration)}

• A. \( Mg \left( 1 - \frac{\rho_2}{\rho_1} \right) \)
• B. \( Mg \frac{\rho_2}{\rho_1} \)
• C. \( Mg \left( 1 - \frac{\rho_1}{\rho_2} \right) \)
• D. \( Mg \frac{\rho_1}{\rho_2} \)

Hint:

At terminal velocity, the net force on an object is zero. The buoyant force and viscous drag force balance out the gravitational force.

Answer 1

The Correct Option is A

Step 1: Terminal velocity condition.
At terminal velocity, the net force acting on the sphere is zero. The forces involved are the buoyant force, the gravitational force, and the viscous drag force. The viscous force is given by: \[ F_{\text{viscous}} = Mg \left( 1 - \frac{\rho_2}{\rho_1} \right) \] where \( \rho_1 \) is the density of the sphere, \( \rho_2 \) is the density of the liquid, and \( g \) is the gravitational acceleration.
Step 2: Conclusion.
Thus, the correct answer is (A) \( Mg \left( 1 - \frac{\rho_2}{\rho_1} \right) \).