Direction :
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):}`
Which object has greatest terminal speed in the liquid ?

Options

A. `I`
B. `II`
C. `III`
D. `IV`

Explanation

To determine which object has the greatest terminal speed in the liquid, we will use the formula for terminal velocity derived from the forces acting on the objects.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( v_T \) of an object moving through a viscous medium is given by the formula:
\[ v_T = \frac{mg}{b} \]
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( b \) is the resistive constant.

2. **Given Data**:
We have four objects with the following properties:
- Object I: Mass = 1 kg, \( b = 3.7 \, \text{N-s/m} \)
- Object II: Mass = 2 kg, \( b = 1.4 \, \text{N-s/m} \)
- Object III: Mass = 3 kg, \( b = 1.4 \, \text{N-s/m} \)
- Object IV: Mass = 4 kg, \( b = 2.8 \, \text{N-s/m} \)

3. **Calculating Terminal Velocity for Each Object**:
- For Object I:
\[ v_{T1} = \frac{1 \times 10}{3.7} \approx 2.70 \, \text{m/s} \]
- For Object II:
\[ v_{T2} = \frac{2 \times 10}{1.4} \approx 14.29 \, \text{m/s} \]
- For Object III:
\[ v_{T3} = \frac{3 \times 10}{1.4} \approx 21.43 \, \text{m/s} \]
- For Object IV:
\[ v_{T4} = \frac{4 \times 10}{2.8} \approx 14.29 \, \text{m/s} \]

4. **Comparing Terminal Velocities**:
Now we compare the calculated terminal velocities:
- \( v_{T1} \approx 2.70 \, \text{m/s} \)
- \( v_{T2} \approx 14.29 \, \text{m/s} \)
- \( v_{T3} \approx 21.43 \, \text{m/s} \)
- \( v_{T4} \approx 14.29 \, \text{m/s} \)

From the calculations, we can see that \( v_{T3} \) is the largest.

5. **Conclusion**:
Therefore, the object with the greatest terminal speed in the liquid is **Object III**.