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 is rising through a liquid layer at a constant terminal velocity, which means it is in a state of dynamic equilibrium and the net force acting on it must be zero. We need to find the ratio of the downward viscous force to the downward gravitational weight of the ball.

Step 2: Key Formula or Approach:
Let the volume of the ball be $V$, the density of the ball material be $\rho_b$, and the density of the liquid be $\rho_l = 4\rho_b$. The forces acting on the rising ball are: 1. Weight ($W$) acting vertically downwards: $W = V\rho_b g$ 2. Upward Buoyant force ($F_B$) due to displaced liquid: $F_B = V\rho_l g = V(4\rho_b)g = 4W$ 3. Viscous drag force ($F_v$) acting downwards (opposing the upward motion) Equating upward and downward forces: $$F_B = W + F_v \implies F_v = F_B - W$$

Step 3: Detailed Explanation:
Substitute the weight relationship into the force balance equation: $$F_v = 4W - W$$ $$F_v = 3W$$ To find the factor by which the viscous force is greater than the weight of the ball, we take the ratio: $$\frac{F_v}{W} = 3$$

Step 4: Final Answer:
The viscous force is greater than the weight of the ball by a factor of 3, which corresponds to option (B).