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)`

Explanation

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 through the liquid, there are three main forces acting on it:
1. **Weight of the sphere (W)**: This is the force due to gravity acting downward, given by:
\[ W = \rho_1 \cdot V \cdot g \]
where \( \rho_1 \) is the density of the sphere, \( V \) is its volume, and \( g \) is the acceleration due to gravity.

2. **Buoyant force (F_b)**: This is the upward force exerted by the liquid, given by:
\[ F_b = \rho_2 \cdot V \cdot g \]
where \( \rho_2 \) is the density of the liquid.

3. **Viscous force (F_v)**: This is the force exerted by the liquid on the sphere, which is proportional to the square of its velocity:
\[ F_v = -k v^2 \]
where \( k \) is a constant and \( v \) is the velocity of the sphere.

### Step 2: Set up the equation for terminal velocity
At terminal velocity (\( v_t \)), the net force acting on the sphere is zero. Therefore, the weight of the sphere is balanced by the sum of the buoyant force and the viscous force:
\[ W = F_b + F_v \]
Substituting the expressions for these forces, we get:
\[ \rho_1 \cdot V \cdot g = \rho_2 \cdot V \cdot g + k v_t^2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ \rho_1 \cdot V \cdot g - \rho_2 \cdot V \cdot g = k v_t^2 \]
Factoring out \( V \cdot g \) from the left side:
\[ (\rho_1 - \rho_2) \cdot V \cdot g = k v_t^2 \]

### Step 4: Solve for terminal velocity
Now, we can solve for \( v_t^2 \):
\[ v_t^2 = \frac{(\rho_1 - \rho_2) \cdot V \cdot g}{k} \]
Taking the square root of both sides gives us the terminal velocity:
\[ v_t = \sqrt{\frac{(\rho_1 - \rho_2) \cdot V \cdot g}{k}} \]

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

---