Question

A small rigid spherical ball of mass \( M \) is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider \( g \) as acceleration due to gravity):

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

Hint:

When the terminal velocity is reached, the drag force equals the weight of the object. For spherical objects in a viscous medium, use Stokes' law to calculate drag force and balance it with gravitational force to find terminal velocity.

Answer 1

The Correct Option is B

When the ball reaches its terminal velocity, the net force acting on the ball becomes zero. This means the viscous drag force \( F_d \) equals the weight of the ball: \[ F_d = Mg \] For an object moving through a viscous medium, the drag force is given by Stokes' law: \[ F_d = 6 \pi \eta r v \] where \( \eta \) is the viscosity of the medium, \( r \) is the radius of the ball, and \( v \) is the velocity. The velocity becomes constant when the downward force due to gravity is balanced by the upward viscous force. Since the density of the glycerine is half that of the ball, the drag force \( F_d \) at terminal velocity is equal to the weight of the ball, which is \( Mg \). Thus, the viscous force acting on the ball is \( Mg \), so the correct answer is \( \boxed{Mg} \).

Answer 2

Step 1: Forces acting on the ball. 
When the ball falls through the viscous liquid, three forces act on it:

1. Weight of the ball, \( W = Mg \), downward.
2. Buoyant force, \( F_b = V\rho_f g \), upward (where \( \rho_f \) is the density of the fluid).
3. Viscous force, \( F_v \), upward.

 

Step 2: Condition at terminal velocity.
At terminal velocity, the ball moves with constant speed. Hence, net force = 0: \[ Mg = F_b + F_v. \]

Step 3: Express buoyant force.
Let the density of the ball be \( \rho_b \). Then: \[ F_b = V\rho_f g. \] The mass of the ball is \( M = V\rho_b \). Substitute into the equilibrium equation: \[ V\rho_b g = V\rho_f g + F_v. \] \[ F_v = Vg(\rho_b - \rho_f). \]

Step 4: Use the given condition.
Given: \( \rho_f = \dfrac{\rho_b}{2}. \) \[ F_v = Vg\left(\rho_b - \dfrac{\rho_b}{2}\right) = Vg\left(\dfrac{\rho_b}{2}\right) = \dfrac{1}{2}V\rho_b g. \] Since \( M = V\rho_b \): \[ F_v = \dfrac{1}{2}Mg. \]

Step 5: Verify direction and balance.
At terminal velocity: \[ Mg = F_b + F_v = \dfrac{Mg}{2} + \dfrac{Mg}{2} = Mg. \] Thus, equilibrium holds true