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((V g(rho_(1)-rho_(2)))/(k))`
B. `(Vg rho_(1))/(k)`
C. `sqrt((Vg rho_(1))/(k))`
D. `(Vg(rho_(1)-rho_(1)))/(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 spherical solid
When the spherical solid is falling through the liquid, there are three main forces acting on it:
1. Gravitational force (weight) acting downward.
2. Buoyant force acting upward.
3. Viscous force acting upward, which is proportional to the square of the velocity.

### Step 2: Write down the expressions for the forces
1. The gravitational force (weight) \( F_g \) is given by:
\[ F_g = \text{mass} \times g = V \cdot \rho_1 \cdot g \]
where \( V \) is the volume of the sphere, \( \rho_1 \) is the density of the sphere, and \( g \) is the acceleration due to gravity.

2. The buoyant force \( F_b \) is given by Archimedes' principle:
\[ F_b = V \cdot \rho_2 \cdot g \]
where \( \rho_2 \) is the density of the liquid.

3. The viscous force \( F_{viscous} \) is given by:
\[ F_{viscous} = -k v^2 \]
where \( k \) is a constant of proportionality and \( v \) is the speed of the sphere.

### 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, we get:
\[ V \cdot \rho_1 \cdot g - V \cdot \rho_2 \cdot g - k v^2 = 0 \]

### Step 4: Simplify the equation
Factoring out \( V \cdot g \) from the first two terms gives:
\[ V \cdot g (\rho_1 - \rho_2) - k v^2 = 0 \]
Rearranging this equation, we have:
\[ k v^2 = V \cdot g (\rho_1 - \rho_2) \]

### Step 5: Solve for terminal velocity \( v \)
Dividing both sides by \( k \) gives:
\[ v^2 = \frac{V \cdot g (\rho_1 - \rho_2)}{k} \]
Taking the square root of both sides, we find the terminal speed \( v \):
\[ v = \sqrt{\frac{V \cdot g (\rho_1 - \rho_2)}{k}} \]

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

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