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

Explanation

To solve the problem, we need to analyze the forces acting on the ball as it rises through the liquid. Let's break it down step by step:

### Step 1: Identify the forces acting on the ball
When the ball is rising in the liquid, it experiences three main forces:
1. **Weight of the ball (W)**: This acts downward and is given by \( W = mg \), where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.
2. **Buoyant force (B)**: This acts upward and is equal to the weight of the liquid displaced by the ball. It can be expressed as \( B = \rho_L \cdot V \cdot g \), where \( \rho_L \) is the density of the liquid, \( V \) is the volume of the ball, and \( g \) is the acceleration due to gravity.
3. **Viscous force (F_v)**: This opposes the motion of the ball and acts downward.

### Step 2: Establish the relationship between forces
Since the ball rises with a constant velocity, the net force acting on it is zero. Therefore, we can set up the equation:
\[ B - W - F_v = 0 \]
This can be rearranged to:
\[ B = W + F_v \]

### Step 3: Substitute the expressions for weight and buoyant force
We know:
- The weight of the ball: \( W = m \cdot g \)
- The buoyant force: \( B = \rho_L \cdot V \cdot g \)

Substituting these into the equation gives:
\[ \rho_L \cdot V \cdot g = mg + F_v \]

### Step 4: Use the relationship between the densities
According to the problem, the density of the liquid (\( \rho_L \)) is four times the density of the ball (\( \rho_B \)):
\[ \rho_L = 4\rho_B \]

### Step 5: Substitute the density relationship into the buoyant force equation
Now substituting \( \rho_L \) into the buoyant force equation:
\[ 4\rho_B \cdot V \cdot g = mg + F_v \]

### Step 6: Express the mass of the ball in terms of its density
The mass of the ball can be expressed as:
\[ m = \rho_B \cdot V \]

Substituting this into the equation gives:
\[ 4\rho_B \cdot V \cdot g = \rho_B \cdot V \cdot g + F_v \]

### Step 7: Rearranging the equation to find the viscous force
Now, we can simplify this equation:
\[ 4\rho_B \cdot V \cdot g - \rho_B \cdot V \cdot g = F_v \]
\[ (4\rho_B - \rho_B) \cdot V \cdot g = F_v \]
\[ 3\rho_B \cdot V \cdot g = F_v \]

### Step 8: Find the ratio of the viscous force to the weight of the ball
Now we need to find the ratio of the viscous force \( F_v \) to the weight of the ball \( W \):
\[ \frac{F_v}{W} = \frac{3\rho_B \cdot V \cdot g}{\rho_B \cdot V \cdot g} \]

### Step 9: Simplify the ratio
The \( \rho_B \), \( V \), and \( g \) terms cancel out:
\[ \frac{F_v}{W} = 3 \]

### Conclusion
The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of **3**.

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