A spherical ball of density `sigma` is freely falling in a viscous liquid of density `p (p lt sigma).` The initial acceleration of the ball will be

Options

A. `((p-sigma)/(p))g`
B. `((sigma-p)/(sigma))g`
C. `((p-sigma)/(sigma))g`
D. `((p+sigma)/(sigma))g`

Explanation

To find the initial acceleration of a spherical ball of density \( \sigma \) falling freely in a viscous liquid of density \( \rho \) (where \( \sigma > \rho \)), we can follow these steps:

### Step 1: Identify the Forces Acting on the Ball
When the ball is falling, there are three main forces acting on it:
1. **Weight (W)**: The downward force due to gravity, given by:
\[ W = \text{mass} \times g = \sigma V g \]
where \( V \) is the volume of the ball and \( g \) is the acceleration due to gravity.

2. **Buoyant Force (F_B)**: The upward force exerted by the liquid, given by Archimedes' principle:
\[ F_B = \rho V g \]
where \( \rho \) is the density of the liquid.

3. **Viscous Force (F_v)**: The upward force due to the viscosity of the liquid. Initially, since the ball is at rest, the velocity \( v \) is zero, which means:
\[ F_v = 0 \]

### Step 2: Write the Equation of Motion
According to Newton's second law, the net force acting on the ball is equal to the mass of the ball multiplied by its acceleration (\( a \)):
\[ \text{Net Force} = W - F_B - F_v = m a \]
Substituting the expressions for the forces, we have:
\[ \sigma V g - \rho V g - 0 = (\sigma V) a \]

### Step 3: Simplify the Equation
We can simplify the equation:
\[ (\sigma - \rho) V g = (\sigma V) a \]
Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ (\sigma - \rho) g = \sigma a \]

### Step 4: Solve for Acceleration (a)
Rearranging the equation to solve for \( a \):
\[ a = \frac{(\sigma - \rho) g}{\sigma} \]

### Final Result
Thus, the initial acceleration of the ball is given by:
\[ a = \left(1 - \frac{\rho}{\sigma}\right) g \]