Question

The velocity of a small ball of mass '\( M \)' and density '\( d_1 \)' when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is '\( d_2 \)', the viscous force acting on the ball is ( \( g = \) acceleration due to gravity )

• A. \(Mg \frac{d_1}{d_2}\)
• B. \(Mg d_1 d_2\)
• C. \(Mg (d_1 - d_2)\)
• D. \(Mg \left( 1 - \frac{d_2}{d_1} \right)\)

Hint:

At terminal velocity, net force zero: viscous drag = weight – buoyancy. Buoyancy = (mass/density of ball) × density of fluid × g.

Answer 1

The Correct Option is D

Step 1: Balance the forces at terminal speed.
Once the speed is steady, the net force is zero, so the viscous force plus the buoyant force equals the weight.

Step 2: Write the buoyant force.
The ball's volume is $V = \tfrac{M}{d_1}$, so the upthrust is $V d_2 g = Mg\tfrac{d_2}{d_1}$.

Step 3: Solve for the viscous force.
\[ F_v = Mg - Mg\frac{d_2}{d_1} = Mg\left(1 - \frac{d_2}{d_1}\right). \]

Step 4: State it.
\[ \boxed{Mg\left(1 - \tfrac{d_2}{d_1}\right)} \]