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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):}`
If an object of mass 2 kg and constant `b = 4 (N-s)/(m)` has terminal speed `v_(T)` in a liquid then time required to reach `0.63 v_(T)` from start of the motion is :

Options

A. 2.0 sec
B. 1.26 sec
C. 0.33 sec
D. 0.5 sec

Explanation

To solve the problem, we need to determine the time required for an object of mass 2 kg and a resistive constant \( b = 4 \, \text{(N-s)/m} \) to reach 63% of its terminal speed \( v_T \).

### Step-by-Step Solution:

1. **Determine the Terminal Velocity \( v_T \)**:
The terminal velocity \( v_T \) can be calculated using the formula:
\[ v_T = \frac{mg}{b} \]
Where:
- \( m = 2 \, \text{kg} \) (mass of the object)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( b = 4 \, \text{(N-s)/m} \)

Plugging in the values:
\[ v_T = \frac{2 \times 9.81}{4} = \frac{19.62}{4} = 4.905 \, \text{m/s} \]

2. **Calculate 63% of Terminal Velocity**:
To find 63% of the terminal velocity:
\[ 0.63 v_T = 0.63 \times 4.905 = 3.09615 \, \text{m/s} \]

3. **Use the Velocity Equation**:
The velocity of the object as a function of time is given by:
\[ v(t) = v_T \left(1 - e^{-\frac{bt}{m}}\right) \]
We need to find the time \( t \) when \( v(t) = 0.63 v_T \):
\[ 3.09615 = 4.905 \left(1 - e^{-\frac{4t}{2}}\right) \]

4. **Rearranging the Equation**:
Dividing both sides by \( 4.905 \):
\[ \frac{3.09615}{4.905} = 1 - e^{-2t} \]
Calculating the left side:
\[ 0.63 = 1 - e^{-2t} \]

5. **Solving for \( e^{-2t} \)**:
Rearranging gives:
\[ e^{-2t} = 1 - 0.63 = 0.37 \]

6. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ -2t = \ln(0.37) \]
Therefore:
\[ t = -\frac{\ln(0.37)}{2} \]

7. **Calculating \( t \)**:
Using a calculator to find \( \ln(0.37) \):
\[ \ln(0.37) \approx -0.9943 \]
Thus:
\[ t = -\frac{-0.9943}{2} \approx 0.49715 \, \text{s} \]

8. **Final Result**:
Rounding to two decimal places, the time required to reach 63% of the terminal speed is approximately:
\[ t \approx 0.50 \, \text{s} \]