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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):}`
A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 10.00 cm/s. Time required to achieve speed 6.32 cm/s from start of the motion is (Take `g = 10.00 m//s^(2)`) :

Options

A. `5.00 xx 10^(-3) s`
B. `1.00 xx 10^(-2) s`
C. `2.5 xx 10^(-3) s`
D. `1.00 xx 10^(-3) s`

Explanation

To solve the problem of finding the time required for a small sphere of mass 2.00 g to achieve a speed of 6.32 cm/s in a viscous medium, we will follow these steps:

### Step 1: Convert Units
First, we need to convert the mass of the sphere from grams to kilograms and the speeds from cm/s to m/s.

- Mass of the sphere:
\[ m = 2.00 \, \text{g} = 2.00 \times 10^{-3} \, \text{kg} \]

- Speeds:
\[ v_T = 10.00 \, \text{cm/s} = 10.00 \times 10^{-2} \, \text{m/s} = 0.10 \, \text{m/s} \]
\[ v = 6.32 \, \text{cm/s} = 6.32 \times 10^{-2} \, \text{m/s} = 0.0632 \, \text{m/s} \]

### Step 2: Determine the Constant \( b \)
Using the terminal velocity formula:
\[ v_T = \frac{mg}{b} \]
We can rearrange this to find \( b \):
\[ b = \frac{mg}{v_T} \]
Substituting the known values:
\[ b = \frac{(2.00 \times 10^{-3} \, \text{kg})(10 \, \text{m/s}^2)}{0.10 \, \text{m/s}} = \frac{2.00 \times 10^{-2} \, \text{N}}{0.10 \, \text{m/s}} = 0.20 \, \text{N-s/m} \]

### Step 3: Calculate the Time Constant \( \tau \)
The time constant \( \tau \) is given by:
\[ \tau = \frac{m}{b} \]
Substituting the values we have:
\[ \tau = \frac{2.00 \times 10^{-3} \, \text{kg}}{0.20 \, \text{N-s/m}} = 0.01 \, \text{s} \]

### Step 4: Relate Velocity to Time
The velocity as a function of time is given by:
\[ v = v_T \left(1 - e^{-\frac{t}{\tau}}\right) \]
We want to find the time \( t \) when \( v = 0.0632 \, \text{m/s} \):
\[ 0.0632 = 0.10 \left(1 - e^{-\frac{t}{0.01}}\right) \]

### Step 5: Solve for \( e^{-\frac{t}{\tau}} \)
Rearranging the equation:
\[ \frac{0.0632}{0.10} = 1 - e^{-\frac{t}{0.01}} \]
\[ 0.632 = 1 - e^{-\frac{t}{0.01}} \]
\[ e^{-\frac{t}{0.01}} = 1 - 0.632 = 0.368 \]

### Step 6: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ -\frac{t}{0.01} = \ln(0.368) \]
\[ t = -0.01 \ln(0.368) \]

### Step 7: Calculate \( t \)
Using the value of \( \ln(0.368) \approx -1 \):
\[ t \approx -0.01 \times (-1) = 0.01 \, \text{s} \]

### Final Answer
The time required to achieve a speed of 6.32 cm/s is approximately:
\[ t \approx 0.01 \, \text{s} \text{ or } 10 \, \text{ms} \]