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 :

When the ball falls through glycerine it eventually reaches a constant (terminal) velocity, which means the net force on it becomes zero.

At terminal velocity the downward forces (weight) are balanced by the upward forces (buoyant force due to displaced fluid and the viscous drag). Hence

$Mg\; =\; F\_\{\backslash text\{buoy\}\}\; +\; F\_\{\backslash text\{viscous\}\}$

We need to find the magnitude of the viscous force $F\_\{\backslash text\{viscous\}\}$.

First compute the buoyant force. The ball’s volume $V$ can be expressed in terms of its mass $M$ and its density $d$:

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

The density of glycerine is given as half the ball’s density, i.e., $\backslash rho\_\{\backslash text\{glycerine\}\}\; =\; \backslash frac\{d\}\{2\}$. The buoyant force equals the weight of the displaced fluid:

$F\_\{\backslash text\{buoy\}\}\; =\; \backslash rho\_\{\backslash text\{glycerine\}\}\; \backslash ,\; V\; \backslash ,\; g\; =\; \backslash frac\{d\}\{2\}\backslash ,\backslash frac\{M\}\{d\}\backslash ,g\; =\; \backslash frac\{Mg\}\{2\}$

Now apply the force balance at terminal velocity:

$Mg\; =\; \backslash frac\{Mg\}\{2\}\; +\; F\_\{\backslash text\{viscous\}\}$

Solving for the viscous force gives

$F\_\{\backslash text\{viscous\}\}\; =\; Mg\; -\; \backslash frac\{Mg\}\{2\}\; =\; \backslash frac\{Mg\}\{2\}$

Thus the magnitude of the viscous force acting on the ball is

$\backslash frac\{Mg\}\{2\}$