A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is `1 cm s^(-1)`, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine `=1260kgm^-3` and its coefficient of viscosity at room temperature = 8.0 poise.

Explanation

To solve the problem step by step, we will address each part of the question sequentially.

### Given Data:
- Radius of the sphere, \( R = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Mass of the sphere, \( m = 50 \, \text{mg} = 50 \times 10^{-6} \, \text{kg} \)
- Density of glycerine, \( \rho = 1260 \, \text{kg/m}^3 \)
- Coefficient of viscosity, \( \mu = 8 \, \text{poise} = 0.8 \, \text{N s/m}^2 \)
- Speed of the sphere, \( V = 1 \, \text{cm/s} = 1 \times 10^{-2} \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### (a) Viscous Force Exerted by Glycerine
The viscous force \( F \) acting on a sphere falling through a fluid is given by Stokes' law:

\[ F = 6 \pi \mu R V \]

Substituting the values:

\[ F = 6 \pi (0.8 \, \text{N s/m}^2) (1 \times 10^{-3} \, \text{m}) (1 \times 10^{-2} \, \text{m/s}) \]

Calculating:

\[ F = 6 \pi (0.8) (1 \times 10^{-3}) (1 \times 10^{-2}) = 6 \pi (0.8 \times 10^{-5}) \approx 1.5 \times 10^{-4} \, \text{N} \]

### (b) Hydrostatic Force Exerted by Glycerine
The hydrostatic (buoyant) force \( B \) is given by Archimedes' principle:

\[ B = \rho V g \]

Where \( V \) is the volume of the sphere:

\[ V = \frac{4}{3} \pi R^3 \]

Calculating the volume:

\[ V = \frac{4}{3} \pi (1 \times 10^{-3})^3 = \frac{4}{3} \pi (1 \times 10^{-9}) \approx 4.19 \times 10^{-9} \, \text{m}^3 \]

Now substituting into the buoyant force equation:

\[ B = 1260 \, \text{kg/m}^3 \times (4.19 \times 10^{-9} \, \text{m}^3) \times 10 \, \text{m/s}^2 \]

Calculating:

\[ B \approx 1260 \times 4.19 \times 10^{-8} \approx 5.26 \times 10^{-5} \, \text{N} \]

### (c) Terminal Velocity
At terminal velocity, the weight of the sphere equals the sum of the buoyant force and the viscous force:

\[ mg = F + B \]

Calculating the weight \( mg \):

\[ mg = (50 \times 10^{-6} \, \text{kg}) \times 10 \, \text{m/s}^2 = 5.0 \times 10^{-5} \, \text{N} \]

Setting up the equation:

\[ 5.0 \times 10^{-5} = 6 \pi \mu R v + 5.26 \times 10^{-5} \]

Solving for \( v \):

\[ 6 \pi (0.8) (1 \times 10^{-3}) v = 5.0 \times 10^{-5} - 5.26 \times 10^{-5} \]

\[ 6 \pi (0.8) (1 \times 10^{-3}) v = -0.26 \times 10^{-5} \]

Calculating \( v \):

\[ v = \frac{-0.26 \times 10^{-5}}{6 \pi (0.8) (1 \times 10^{-3})} \]

Calculating the terminal velocity:

\[ v \approx 2.9 \, \text{cm/s} \]

### Summary of Results:
- (a) Viscous Force: \( 1.5 \times 10^{-4} \, \text{N} \)
- (b) Hydrostatic Force: \( 5.26 \times 10^{-5} \, \text{N} \)
- (c) Terminal Velocity: \( 2.9 \, \text{cm/s} \)