A spherical solid ball of volume V is made of a material of density `rho_1`. It is falling through a liquid of density `rho_2(rho_2ltrho_1)`. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., `F_(viscous)=-kv^2(kgt0)`. The terminal speed of the ball is

Options

A. (a) `sqrt((Vg(rho_1-rho_2))/(k))`
B. (b) `(Vgrho_1)/(k)`
C. (c) `sqrt((Vgrho_1)/(k))`
D. (d) `(Vg(rho_1-rho_2))/(k)`

Explanation

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

### Step 1: Identify the forces acting on the ball
When the ball is falling through the liquid, three main forces act on it:
1. Weight of the ball (downward)
2. Buoyant force (upward)
3. Viscous force (upward)

### Step 2: Write the expressions for these forces
- **Weight of the ball (W)**: The weight can be calculated using the formula:
\[ W = V \cdot \rho_1 \cdot g \]
where \( V \) is the volume of the ball, \( \rho_1 \) is the density of the ball, and \( g \) is the acceleration due to gravity.

- **Buoyant force (B)**: The buoyant force can be expressed as:
\[ B = V \cdot \rho_2 \cdot g \]
where \( \rho_2 \) is the density of the liquid.

- **Viscous force (F_viscous)**: The viscous force acting on the ball is given by:
\[ F_{\text{viscous}} = -k v^2 \]
where \( k \) is a constant and \( v \) is the speed of the ball.

### Step 3: Set up the equation for terminal velocity
At terminal velocity, the net force acting on the ball is zero. Therefore, we can set up the equation:
\[ W = B + F_{\text{viscous}} \]
Substituting the expressions for weight, buoyant force, and viscous force, we get:
\[ V \cdot \rho_1 \cdot g = V \cdot \rho_2 \cdot g + k v_t^2 \]
where \( v_t \) is the terminal velocity.

### Step 4: Rearrange the equation to solve for terminal velocity
Rearranging the equation gives:
\[ V \cdot \rho_1 \cdot g - V \cdot \rho_2 \cdot g = k v_t^2 \]
Factoring out \( V \cdot g \) from the left side:
\[ V \cdot g (\rho_1 - \rho_2) = k v_t^2 \]

### Step 5: Solve for terminal velocity
Now, we can isolate \( v_t^2 \):
\[ v_t^2 = \frac{V \cdot g (\rho_1 - \rho_2)}{k} \]
Taking the square root gives us the terminal velocity:
\[ v_t = \sqrt{\frac{V \cdot g (\rho_1 - \rho_2)}{k}} \]

### Final Answer
Thus, the terminal speed of the ball is:
\[ v_t = \sqrt{\frac{V \cdot g (\rho_1 - \rho_2)}{k}} \]

---