Question

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be

• A. 2Mg
• B. \(\frac{Mg}{2}\)
• C. Mg
• D. \(\frac{3}{2}Mg\)

Answer 1

The Correct Option is B

When the ball is dropped into glycerine, it initially accelerates due to gravity. However, as it moves, it experiences a viscous force from the glycerine, which increases with velocity until it balances the gravitational force, causing the ball to reach a constant terminal velocity. At this point, the net force acting on the ball is zero.

According to the principles of fluid dynamics, the net force acting on the ball can be described by the following forces:

1. Gravitational Force (\( F_g \)): This force acts downward and is given by:

\(F_g = Mg\)

1. Buoyant Force (\( F_b \)): This force acts upward and, according to Archimedes' principle, is equal to the weight of the glycerine displaced by the ball. If the density of the glycerine is \( \frac{d}{2} \), and using the volume \( V \) of the ball, the buoyant force is given by:

\(F_b = V \cdot \left(\frac{d}{2}\right) \cdot g\)

1. Viscous Force (\( F_v \)): This is the force we are trying to determine, and it acts upward.

At terminal velocity, the forces balance, so:

\(F_g = F_b + F_v\)

The volume of the ball \( V \) can be related to its mass \( M \) and density \( d \) as:

\(V = \frac{M}{d}\)

Substituting for the buoyant force:

\(F_b = \frac{M}{d} \cdot \left(\frac{d}{2}\right) \cdot g = \frac{Mg}{2}\)

Insert these into the balance equation:

\(Mg = \frac{Mg}{2} + F_v\)

Solving for the viscous force \( F_v \):

\(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\)

Thus, the viscous force acting on the ball is \(\frac{Mg}{2}\).