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. What is the viscous force on a glass sphere of radius `r=1mm` falling through water `(eta=1xx10^(-3)Pa-s)` when the sphere has speed of 3m/s?

Options

A. `2.7xx10^(-2)N`
B. `2.5xx10^(-5)N`
C. `3.7xx10^(-3)N`
D. `5.6xx10^(-5)N`

Answer

To find the viscous force on a glass sphere falling through water using Stokes' law, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the sphere, \( r = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Coefficient of viscosity, \( \eta = 1 \times 10^{-3} \text{ Pa·s} \)
- Speed of the sphere, \( v = 3 \text{ m/s} \)

2. **Use Stokes' Law Formula:**
Stokes' law states that the viscous force \( F_w \) acting on a sphere moving through a viscous fluid is given by:
\[ F_w = 6 \pi \eta r v \]

3. **Substitute the Values into the Formula:**
Plugging in the values we have:
\[ F_w = 6 \pi (1 \times 10^{-3}) (1 \times 10^{-3}) (3) \]

4. **Calculate the Value:**
- First, calculate \( 6 \pi \):
\[ 6 \pi \approx 6 \times 3.14 \approx 18.84 \]
- Now, calculate the entire expression:
\[ F_w = 18.84 \times (1 \times 10^{-3}) \times (3) \times (1 \times 10^{-3}) = 18.84 \times 3 \times 10^{-6} \]
\[ F_w = 56.52 \times 10^{-6} \text{ N} = 5.652 \times 10^{-5} \text{ N} \]

5. **Final Result:**
Thus, the viscous force on the glass sphere is:
\[ F_w \approx 5.65 \times 10^{-5} \text{ N} \]