A ball of mass m and density d1 is dropped into a tall container of glycerin of density d2. After some time its velocity becomes constant (terminal velocity). At that instant the viscous (drag) force acting on the ball is

m*g*(d1/d2)
m*g*(1 - d2/d1)
m*g*((d1 + d2)/d1)
m*g*((d1 + d2)/d2)

Correct answer: m*g*(1 - d2/d1)

Solution

At terminal velocity acceleration is zero, so the viscous drag balances weight minus buoyancy. The buoyant force is the weight of displaced glycerin, equal to (d2/d1) times the ball's weight. Subtracting it from the weight gives the viscous force.

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