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: Set the scene.
A ball of mass $M$ and density $d_1$ is dropped into glycerine of density $d_2$. After a while its speed stops changing, so it has reached terminal velocity. We want the viscous force on it then.

Step 2: Use the balance at terminal velocity.
At steady speed the ball does not accelerate, so all forces cancel: viscous force plus buoyant force equals the weight.

Step 3: Write the weight.
The weight pulling down is $W = Mg$.

Step 4: Write the buoyant force.
The buoyant push equals the weight of the glycerine pushed aside. The ball's volume is $V = \frac{M}{d_1}$, so buoyancy is $V d_2 g = \frac{M}{d_1} d_2 g = Mg\frac{d_2}{d_1}$.

Step 5: Solve for the viscous force.
From the balance, viscous force $F_v = W - \text{buoyancy} = Mg - Mg\frac{d_2}{d_1}$.

Step 6: Tidy the expression.
Factor out $Mg$: $F_v = Mg\left(1 - \frac{d_2}{d_1}\right)$.
\[ \boxed{F_v = Mg\left(1 - \frac{d_2}{d_1}\right)} \]