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 problem asks for the magnitude of the viscous (drag) force acting on a small ball when it reaches its constant (terminal) velocity while moving through glycerine.

At terminal velocity the net force on the ball is zero, so the upward forces balance the downward weight.

Write the force balance:

$F\_\{\backslash text\{viscous\}\}\; +\; F\_\{\backslash text\{buoyant\}\}\; =\; F\_\{\backslash text\{weight\}\}$

Re‑arrange to solve for the viscous force:

$F\_\{\backslash text\{viscous\}\}\; =\; F\_\{\backslash text\{weight\}\}\; -\; F\_\{\backslash text\{buoyant\}\}$

Now express each term.

Weight of the ball:

$F\_\{\backslash text\{weight\}\}\; =\; M\; g$

Buoyant force equals the weight of the displaced fluid:

$F\_\{\backslash text\{buoyant\}\}\; =\; \backslash rho\_\{\backslash text\{fluid\}\}\; V\; g$

where \(\rho_{\text{fluid}}\) is the density of glycerine and \(V\) is the volume of the ball.

Given data:

• Density of the ball = \(d\)
• Density of glycerine = \(\dfrac{d}{2}\)

The volume of the ball can be written using its mass and density:

$V\; =\; \backslash dfrac\{M\}\{\backslash text\{density\; of\; ball\}\}\; =\; \backslash dfrac\{M\}\{d\}$

Insert the fluid density and the volume into the buoyant force:

$F\_\{\backslash text\{buoyant\}\}\; =\; \backslash left(\backslash dfrac\{d\}\{2\}\backslash right)\backslash !\backslash left(\backslash dfrac\{M\}\{d\}\backslash right)\; g\; =\; \backslash dfrac\{M\}\{2\}\; g$

Finally compute the viscous force:

$F\_\{\backslash text\{viscous\}\}\; =\; M\; g\; -\; \backslash dfrac\{M\}\{2\}\; g\; =\; \backslash dfrac\{M\}\{2\}\; g$

Thus the magnitude of the viscous force acting on the ball at terminal velocity is:

$F\_\{\backslash text\{viscous\}\}\; =\; \backslash dfrac\{M\; g\}\{2\}$

This matches the option “Mg/2”.