The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2 , then the viscous force acting on the ball will be :

The correct option is Mg/2.

Let us analyze the forces acting on the ball of mass $M$ and density $d$ when it is dropped into the container filled with glycerine.

When the ball falls through the glycerine, three forces act on it:
1. Gravitational Force ($F_g$): Acting vertically downwards.
$F_g=Mg$
2. Buoyant Force ($F_b$): Acting vertically upwards due to the displaced glycerine.
3. Viscous Force ($F_v$): Acting vertically upwards, opposing the downward motion of the ball.

The mass of the ball $M$ can be expressed in terms of its volume $V$ and density $d$:
$M=V·d$
Which gives the volume of the ball as:
$V=\frac{M}{d}$

The buoyant force $F_b$ is equal to the weight of the glycerine displaced by the ball:
$F_b=V·\sigma ·g$
where $\sigma$ is the density of glycerine. Given that $\sigma =\frac{d}{2}$, we substitute this and $V$ into the buoyant force formula:
$F_b=\left(\frac{M}{d}\right)·\left(\frac{d}{2}\right)·g=\frac{Mg}{2}$

According to the problem, the velocity of the ball becomes constant after some time. This constant velocity is known as the terminal velocity. When the ball moves with constant velocity, the net force acting on it is zero (equilibrium condition):
$F_g=F_b+F_v$

We can now solve for the viscous force $F_v$:
$F_v=F_g-F_b$
Substituting the expressions for $F_g$ and $F_b$:
$F_v=Mg-\frac{Mg}{2}=\frac{Mg}{2}$

Therefore, the viscous force acting on the ball when it reaches terminal velocity is $\frac{Mg}{2}$.