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Resistive force proportional to object velocity
At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as
`R = -bv`
Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity.
Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here.
Thus `mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v`
Solving the equation
`v = (mg)/(b) (1- e^(-bt//m))`
where e=2.71 is the base of the natural logarithm
The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed `v_(1)` and then on it continues to move with zero acceleration
`mg - b_(T) =0`
`rArr m_(T) = (mg)/(b)`
Hence `v = v_(T) (1-e^((vt)/(m)))`
In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table.
`{:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):}`
At the start of motion when object is released in the liquid, its acceleration is :

Options

A. 0
B. 0.63 g
C. g
D. Depends on the mass of the object

Explanation

To solve the problem, we need to analyze the forces acting on the sphere when it is released in the liquid. Here are the steps to find the acceleration at the start of the motion:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere**:
- The weight of the sphere acting downwards: \( F_g = mg \)
- The resistive force acting upwards, which is proportional to the velocity: \( R = -bv \)

2. **Apply Newton's Second Law**:
- According to Newton's second law, the net force acting on the object is equal to the mass of the object multiplied by its acceleration:
\[ F_{\text{net}} = ma \]

3. **Set Up the Equation**:
- The net force acting on the sphere can be expressed as:
\[ F_{\text{net}} = mg - R \]
- Substituting the expression for the resistive force:
\[ F_{\text{net}} = mg - (-bv) = mg + bv \]
- Therefore, we can write:
\[ ma = mg - bv \]

4. **Consider the Initial Conditions**:
- At the start of the motion, the sphere is released from rest, so the initial velocity \( v = 0 \).
- This means the resistive force \( R = -b(0) = 0 \).

5. **Simplify the Equation**:
- Substituting \( v = 0 \) into the equation gives:
\[ ma = mg - 0 \]
- Thus, we have:
\[ ma = mg \]

6. **Solve for Acceleration**:
- Dividing both sides by \( m \):
\[ a = g \]

### Conclusion:
The acceleration of the sphere at the start of the motion when it is released in the liquid is equal to \( g \).