A spherical solild of volume V is made of a material of density `rho_(1)`. It is falling through a liquid of density `rho_(2)(rho_(2) lt rho_(1))`. Assume that the liquid applies a viscous froce on the ball that is proportional ti the its speed v, i.e., `F_(viscous)=-kv^(2)(kgt0)`. The terminal speed of the ball is

Options

A. `sqrt((Vg (rho_(1) - rho_(2)))/(k))`
B. `(Vg rho_(1))/(k)`
C. `sqrt((Vg rho_(1))/(k))`
D. `(Vg (rho_(1) - rho_(2)))/(k)`

Answer

To find the terminal speed of a spherical solid falling through a liquid, we can follow these steps:

### Step 1: Identify the forces acting on the sphere
When the sphere is falling, three main forces act on it:
1. The gravitational force (weight) acting downward: \( F_g = mg \)
2. The buoyant force acting upward: \( F_b = V \rho_2 g \)
3. The viscous force acting upward, which is proportional to the speed of the sphere: \( F_{viscous} = -kv^2 \)

### Step 2: Write the expressions for the forces
- The mass \( m \) of the sphere can be expressed as:
\[ m = V \rho_1 \]
- The gravitational force is then:
\[ F_g = V \rho_1 g \]
- The buoyant force is:
\[ F_b = V \rho_2 g \]

### Step 3: Set up the equation for terminal velocity
At terminal velocity, the net force acting on the sphere is zero. Therefore, we can write:
\[ F_g - F_b - F_{viscous} = 0 \]
Substituting the expressions for the forces:
\[ V \rho_1 g - V \rho_2 g - kv^2 = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ V \rho_1 g - V \rho_2 g = kv^2 \]
Factoring out \( Vg \) from the left side:
\[ Vg (\rho_1 - \rho_2) = kv^2 \]

### Step 5: Solve for terminal velocity \( v_t \)
Now, we can solve for \( v \):
\[ kv^2 = Vg (\rho_1 - \rho_2) \]
Dividing both sides by \( k \):
\[ v^2 = \frac{Vg (\rho_1 - \rho_2)}{k} \]
Taking the square root gives us the terminal velocity:
\[ v_t = \sqrt{\frac{Vg (\rho_1 - \rho_2)}{k}} \]

### Final Answer
The terminal speed of the ball is:
\[ v_t = \sqrt{\frac{Vg (\rho_1 - \rho_2)}{k}} \]