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)

When the small rigid spherical ball is dropped in glycerine, it initially accelerates due to gravity. However, due to the resistive viscous force exerted by the fluid, the ball eventually reaches terminal velocity, where the net force on it becomes zero (i.e., no further acceleration).

Forces Acting on the Ball at Terminal Velocity

At terminal velocity, the forces acting on the ball are:

1. Gravitational Force ($F_g$): $$F_g = Mg$$
2. Buoyant Force ($F_b$) (due to displaced glycerine): $$F_b = \rho_{\text{glycerine}} V g$$ Since the density of glycerine is half the density of the ball ($\rho_{\text{glycerine}} = \frac{1}{2} \rho_{\text{ball}}$), we can write: $$F_b = \frac{1}{2} \rho_{\text{ball}} V g$$ Since the volume of the sphere is $V = \frac{M}{\rho_{\text{ball}}}$, we get: $$F_b = \frac{1}{2} Mg$$
3. Viscous Force ($F_v$) (opposes motion):
   • According to Stokes' law, at terminal velocity, the viscous force balances the excess force after accounting for buoyancy.

Since the ball moves with terminal velocity, the net force is zero:

$$F_g = F_b + F_v$$

Substituting values:

$$Mg = \frac{1}{2} Mg + F_v$$

Solving for $F_v$:

$$F_v = Mg - \frac{1}{2} Mg = \frac{1}{2} Mg$$