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. \(F=Mg(1+\frac{\rho_0}{\rho})\)
• B. \(F=Mg(1+\frac{\rho}{\rho_0})\)
• C. \(F=Mg(1-\frac{\rho_0}{\rho})\)
• D. \(F=Mg(1±\rho\rho_0)\)

Answer 1

The Correct Option is C

To determine the viscous force acting on the ball, we need to consider the forces acting on the ball when it falls through the viscous liquid at a constant velocity. This constant velocity indicates that the net force on the ball is zero, as per Newton's first law of motion.

1. Gravitational Force: The gravitational force \( F_g \) acting downward on the ball is given by: \( F_g = Mg \).
2. Buoyant Force: According to Archimedes' principle, the buoyant force \( F_b \) on the ball acts upward and is equal to the weight of the liquid displaced by the ball. It is calculated as: \( F_b = V \rho_0 g \), where \( V \) is the volume of the ball, and \( \rho_0 \) is the density of the liquid.
3. Volume of the Ball: The volume \( V \) of the ball can be expressed in terms of its mass and density as: \( V = \frac{M}{\rho} \).
4. Substituting the value of \( V \) into the expression for buoyant force gives: \( F_b = \frac{M}{\rho} \rho_0 g \) or \( F_b = M \frac{\rho_0}{\rho} g \).
5. Viscous Force: Since the ball falls with a constant velocity, the upward forces (buoyant and viscous) balance the downward gravitational force: \( F_g = F_b + F_v \), where \( F_v \) is the viscous force.
6. Using the expressions from above, we get: \( Mg = M \frac{\rho_0}{\rho} g + F_v \).
7. Rearranging for \( F_v \) gives: \( F_v = Mg - M \frac{\rho_0}{\rho} g \) or \( F_v = Mg \left(1 - \frac{\rho_0}{\rho}\right) \).

Thus, the viscous force acting on the ball is \( F = Mg \left(1 - \frac{\rho_0}{\rho}\right) \), which corresponds to the correct answer.