Question

The velocity of a small ball of mass ‘m’ and density \(d_1\), when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is \(d_2\), then the viscous force acting on the ball, will be

• A. \(mg\) \(\bigg(1-\frac{d_1}{d_2}\bigg)\)
• B. \(mg\) \(\bigg(1-\frac{d_2}{d_1}\bigg)\)
• C. \(mg\) \(\bigg(\frac{d_1}{d_2}-1\bigg)\)
• D. \(mg\) \(\bigg(\frac{d_2}{d_1}-1\bigg)\)

Answer 1

The Correct Option is B

When a small ball of mass \(m\) and density \(d_1\) is dropped in a fluid like glycerin with density \(d_2\), it experiences three main forces when it reaches terminal velocity:

• Gravitational Force (\(F_g\)): This is the force due to gravity, acting downward. It is given by: F_g = mg
• Buoyant Force (\(F_b\)): As per Archimedes' principle, this force acts upward and is given by: F_b = V \cdot d_2 \cdot g
• Viscous Force (\(F_v\)): This force opposes the motion of the ball and acts upwards when the ball is moving downwards. According to Stoke's Law, it is given by: F_v = 6 \pi \eta r v

At terminal velocity, the ball moves with constant velocity, meaning the net force acting on it is zero. Therefore, the sum of the upward forces (Buoyant force + Viscous force) is equal to the downward gravitational force:

F_g = F_b + F_v
mg = V \cdot d_2 \cdot g + F_v

Solving for the viscous force, we get:

F_v = mg - V \cdot d_2 \cdot g

Since the volume of the ball \(V\) can be related to its density by \(V = \frac{m}{d_1}\), the equation becomes:

F_v = mg - \left(\frac{m}{d_1} \cdot d_2 \cdot g\right)
F_v = mg - \left(\frac{mg \cdot d_2}{d_1}\right)

Simplifying this expression, we find the viscous force:

F_v = mg \left(1 - \frac{d_2}{d_1}\right)

This matches the option given in the question as the correct answer: mg \left(1 - \frac{d_2}{d_1}\right).

Thus, the viscous force acting on the ball when it reaches terminal velocity is:

mg \left(1-\frac{d_2}{d_1}\right).