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

At terminal velocity, the net force on the ball is zero. This implies the viscous drag force \( F_d \) equals the ball's weight: \[ F_d = Mg \] For an object in a viscous medium, drag force is described by Stokes' law: \[ F_d = 6 \pi \eta r v \] where \( \eta \) is the medium's viscosity, \( r \) is the ball's radius, and \( v \) is the velocity. Velocity stabilizes when gravitational force is counteracted by viscous force. Given the glycerine's density is half the ball's, the drag force \( F_d \) at terminal velocity matches the ball's weight, \( Mg \). Consequently, the viscous force on the ball is \( Mg \), leading to the correct answer \( \boxed{Mg} \).