Question

A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The viscous force of the liquid on the rising ball is greater than the weight of the ball by a factor of

• A. 4
• B. 3
• C. 2
• D. 5

Hint:

For any object rising or sinking at constant velocity, if the fluid's density is $n$ times the object's density, the buoyant force is $n W$. The drag force is always $|n - 1|W$. Since $n = 4$, the factor is simply $4 - 1 = 3$ instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A ball rises at constant terminal velocity through a liquid of density 4ρ_b; find the ratio of downward viscous force to the ball's weight.

Step 2: Key Formula or Approach:
At terminal velocity, net force is zero: upward buoyant force = downward weight + viscous drag.

Step 3: Detailed Explanation:
Weight W = Vρ_b g. Buoyant force F_B = V(4ρ_b)g = 4W. Force balance: F_B = W + F_v → F_v = 3W. Ratio F_v/W = 3.

Step 4: Final Answer:
Viscous force is greater than weight by a factor of 3, matching option (B).