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

A

3Mg/23Mg/2

B

2Mg2Mg

C

Mg/2Mg/2

D

MgMg

Step-by-Step Solution

When the ball's velocity becomes constant, it has reached its terminal velocity, a state where the net force acting on the object is zero , . In this equilibrium state, the downward force of gravity (weight, WW) is perfectly balanced by the sum of the upward forces: the buoyant force (BB) and the viscous force (FvF_v) .

  1. Weight (WW): Given as W=MgW = Mg.
  2. Buoyant Force (BB): This force is equal to the weight of the displaced fluid. Since the ball has mass MM and density dd, its volume is V=M/dV = M/d . The density of glycerine is given as d/2d/2. Therefore, B=V×(d/2)×g=(M/d)×(d/2)×g=Mg/2B = V \times (d/2) \times g = (M/d) \times (d/2) \times g = Mg/2 .
  3. Equilibrium Equation: At terminal velocity, Fv+B=WF_v + B = W. Substituting the values, we get Fv+Mg/2=MgF_v + Mg/2 = Mg.
  4. Result: Solving for the viscous force, Fv=MgMg/2=Mg/2F_v = Mg - Mg/2 = Mg/2.
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