When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius `r` moving is given by stoke's law, `F_(w)=6pietarv`. in this formula `eta` in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by `F=etaAv//z`. For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference `P_(2)-P_(1)`. The flow rate of expressed by the formula `Q=(piR^(4)(P_(2)-P_(1)))/(8etaL)` poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula `R_(e)=(2overline(v)rhoR)/(eta)` In which `overline(v)` is the average speed `rho` is the density `eta` is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be `rho=1000kg//m^(3)`
Q. If the sphere in previous question has mass of `1xx10^(-5)kg` what is its terminal velocity when falling through water? `(eta=1xx10^(-3)Pa-s)`

A. 1.3m/s
B. 3.4m/s
C. 5.2m/s
D. 6.5m/s

Options

A. 1.3m/s
B. 3.4m/s
C. 5.2m/s
D. 6.5m/s

Answer

To find the terminal velocity of a sphere falling through water, we can use Stokes' law, which relates the viscous force acting on the sphere to its velocity. The formula given by Stokes' law is:

\[ F_w = 6 \pi \eta r v \]

Where:
- \( F_w \) is the viscous force,
- \( \eta \) is the coefficient of viscosity of the fluid,
- \( r \) is the radius of the sphere,
- \( v \) is the velocity of the sphere.

At terminal velocity, the viscous force equals the weight of the sphere. Therefore, we can set the two forces equal:

\[ F_w = mg \]

Where:
- \( m \) is the mass of the sphere,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Setting these equal gives us:

\[ 6 \pi \eta r v_t = mg \]

Now, we can solve for the terminal velocity \( v_t \):

\[ v_t = \frac{mg}{6 \pi \eta r} \]

### Step 1: Identify the given values
- Mass of the sphere, \( m = 1 \times 10^{-5} \, \text{kg} \)
- Coefficient of viscosity, \( \eta = 1 \times 10^{-3} \, \text{Pa.s} \)
- Radius of the sphere, \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Substitute the values into the terminal velocity formula
Now we substitute the known values into the formula:

\[ v_t = \frac{(1 \times 10^{-5} \, \text{kg})(10 \, \text{m/s}^2)}{6 \pi (1 \times 10^{-3} \, \text{Pa.s})(1 \times 10^{-3} \, \text{m})} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ mg = (1 \times 10^{-5})(10) = 1 \times 10^{-4} \, \text{N} \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ 6 \pi \eta r = 6 \pi (1 \times 10^{-3})(1 \times 10^{-3}) = 6 \pi \times 10^{-6} \approx 1.884 \times 10^{-5} \, \text{N.s/m} \]

### Step 5: Calculate terminal velocity
Now substituting the values back into the equation for \( v_t \):

\[ v_t = \frac{1 \times 10^{-4}}{1.884 \times 10^{-5}} \approx 5.3 \, \text{m/s} \]

### Conclusion
The terminal velocity of the sphere falling through water is approximately \( 5.2 \, \text{m/s} \).

### Final Answer
Thus, the correct answer is **C. 5.2 m/s**.