Question

The velocity of a small ball of mass '\( M \)' and density '\( d_1 \)' when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is '\( d_2 \)', the viscous force acting on the ball is ( \( g = \) acceleration due to gravity )

• A. \(Mg \frac{d_1}{d_2}\)
• B. \(Mg d_1 d_2\)
• C. \(Mg (d_1 - d_2)\)
• D. \(Mg \left( 1 - \frac{d_2}{d_1} \right)\)

Hint:

At terminal velocity, net force zero: viscous drag = weight – buoyancy. Buoyancy = (mass/density of ball) × density of fluid × g.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Ball of mass \(M\), density \(d_1\), falls in glycerine of density \(d_2\). It attains terminal velocity (constant velocity). At terminal velocity, net force = 0: viscous force + buoyant force = weight.

Step 2: Key Formula or Approach:
Weight \(W = Mg\). Buoyant force = weight of displaced fluid = \(V d_2 g\), where \(V = M/d_1\) (since density \(d_1 = M/V\)). So buoyant force = \(\frac{M}{d_1} d_2 g = Mg \frac{d_2}{d_1}\).
At terminal velocity: viscous force \(F_v = W - \text{buoyant} = Mg - Mg \frac{d_2}{d_1} = Mg \left(1 - \frac{d_2}{d_1}\right)\).

Step 3: Detailed Explanation:
Thus viscous force equals apparent weight.

Step 4: Final Answer:
Option (D) \(Mg \left( 1 - \frac{d_2}{d_1} \right)\).