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):
Show 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.
Upon reaching terminal velocity, the ball experiences zero net force. This condition implies that the viscous drag force \( F_d \) is equal in magnitude to the ball's weight:
\[
F_d = Mg
\]
For an object moving in a viscous fluid, Stokes' law describes the drag force:
\[
F_d = 6 \pi \eta r v
\]
Here, \( \eta \) represents the medium's viscosity, \( r \) is the ball's radius, and \( v \) is its velocity. Terminal velocity is achieved when the gravitational force pulling the object downwards is counteracted by the opposing viscous force. Given that the glycerine's density is half that of the ball, the drag force \( F_d \) at terminal velocity is equivalent to the ball's weight, \( Mg \).
Therefore, the viscous force on the ball is \( Mg \). The correct answer is \( \boxed{Mg} \).
