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.

Step-by-step Explanation:

When a ball is dropped into a container filled with a viscous fluid like glycerine, it experiences three forces acting on it:

1. Gravitational Force (Weight, F_g): Acting vertically downwards.
The mass of the ball is M. Therefore, the downward gravitational force is:

$F_g=Mg$

2. Buoyant Force (Upthrust, F_b): Acting vertically upwards due to the displaced fluid.
The buoyant force is given by the formula:

$F_b=V·\sigma ·g$

where V is the volume of the ball and σ is the density of the glycerine.
Since the ball has mass M and density d, its volume V is:

$V=\frac{M}{d}$

Given that the density of glycerine is σ = d/2, we can substitute these values into the buoyant force formula:

$F_b=\left(\frac{M}{d}\right)·\left(\frac{d}{2}\right)·g=\frac{Mg}{2}$

3. Viscous Force (F_v): Acting vertically upwards, opposing the motion of the ball.

According to the problem, the velocity of the ball becomes constant after some time (attaining terminal velocity). When the velocity becomes constant, the net force acting on the ball must be zero. Therefore, the upward forces must balance the downward forces:

$F_g=F_b+F_v$

Now, substitute the values of F_g and F_b into this equilibrium equation:

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

Solving for the viscous force F_v:

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

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

Thus, the viscous force acting on the ball is Mg/2.