A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four lime 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

Options

A. `2`
B. `3`
C. `4`
D. `6`

Explanation

To solve the problem step by step, we will analyze the forces acting on the ball as it rises through the liquid.

### Step 1: Identify the Forces Acting on the Ball
When the ball is submerged in the liquid, three main forces act on it:
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 (F_b)**: This acts upward and is given by Archimedes' principle. The buoyant force is equal to the weight of the liquid displaced by the ball. Since the density of the liquid (\( \rho_L \)) is four times the density of the ball (\( \rho_B \)), we have:
\[ F_b = V \cdot \rho_L \cdot g = V \cdot (4 \rho_B) \cdot g \]
where \( V \) is the volume of the ball.
3. **Frictional Force (F_f)**: This acts downward, opposing the motion of the ball as it rises through the liquid.

### Step 2: Apply the Condition of Constant Velocity
Since the ball rises with a constant velocity, the net force acting on it is zero. This means that the upward forces must balance the downward forces:
\[ F_b = W + F_f \]

### Step 3: Substitute the Forces into the Equation
From the previous steps, we can express the forces in terms of the volume and densities:
\[ 4V\rho_B g = mg + F_f \]
Substituting \( mg \) with \( V \rho_B g \) (since \( m = V \rho_B \)):
\[ 4V\rho_B g = V\rho_B g + F_f \]

### Step 4: Solve for the Frictional Force
Rearranging the equation to find the frictional force:
\[ F_f = 4V\rho_B g - V\rho_B g \]
\[ F_f = (4 - 1)V\rho_B g = 3V\rho_B g \]

### Step 5: Relate the Frictional Force to the Weight of the Ball
We know that the weight of the ball is:
\[ W = mg = V\rho_B g \]
Now, we can express the frictional force in terms of the weight of the ball:
\[ F_f = 3V\rho_B g = 3W \]

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

### Final Answer
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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