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, let's break it down into three parts as per the question.

### 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, \( \eta = 8 \, \text{poise} = 8 \times 0.1 \, \text{N s/m}^2 = 0.8 \, \text{N s/m}^2 \) (since \( 1 \, \text{poise} = 0.1 \, \text{N s/m}^2 \))
- Speed of the sphere, \( v = 1 \, \text{cm/s} = 0.01 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Part (a): Find the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm/s.

Using Stokes' law, the viscous force \( F_b \) can be calculated using the formula:
\[ F_b = 6 \pi \eta r v \]

Substituting the values:
\[ F_b = 6 \pi (0.8) (1 \times 10^{-3}) (0.01) \]
\[ F_b = 6 \pi (0.8) (1 \times 10^{-3}) (1 \times 10^{-2}) \]
\[ F_b = 6 \pi (0.8) (1 \times 10^{-5}) \]
\[ F_b = 4.8 \pi \times 10^{-5} \, \text{N} \]
Calculating \( 4.8 \pi \):
\[ F_b \approx 4.8 \times 3.14 \times 10^{-5} \approx 1.51 \times 10^{-4} \, \text{N} \]

### Part (b): Find the hydrostatic force exerted by the glycerine on the sphere.

The hydrostatic force (upthrust) can be calculated using the formula:
\[ F_u = \rho V g \]
where \( V \) is the volume of the sphere given by:
\[ V = \frac{4}{3} \pi r^3 \]

Calculating the volume:
\[ V = \frac{4}{3} \pi (1 \times 10^{-3})^3 = \frac{4}{3} \pi (1 \times 10^{-9}) = \frac{4 \pi}{3} \times 10^{-9} \, \text{m}^3 \]

Now substituting into the upthrust formula:
\[ F_u = 1260 \left(\frac{4 \pi}{3} \times 10^{-9}\right) (10) \]
\[ F_u = 1260 \times \frac{4 \pi}{3} \times 10^{-8} \, \text{N} \]
Calculating \( F_u \):
\[ F_u \approx 1260 \times 4.18879 \times 10^{-8} \approx 1.57 \times 10^{-5} \, \text{N} \]

### Part (c): Find the terminal velocity with which the sphere will move down without acceleration.

At terminal velocity, the net force acting on the sphere is zero. Thus, the downward gravitational force equals the sum of the upward viscous force and buoyant force:
\[ mg = F_b + F_u \]

Substituting the values:
\[ (50 \times 10^{-6})(10) = 6 \pi \eta r v_t + F_u \]
\[ 5 \times 10^{-5} = 6 \pi (0.8) (1 \times 10^{-3}) v_t + 1.57 \times 10^{-5} \]

Now, rearranging for \( v_t \):
\[ 5 \times 10^{-5} - 1.57 \times 10^{-5} = 6 \pi (0.8) (1 \times 10^{-3}) v_t \]
\[ 3.43 \times 10^{-5} = 6 \pi (0.8) (1 \times 10^{-3}) v_t \]
Calculating \( 6 \pi (0.8) (1 \times 10^{-3}) \):
\[ 6 \times 3.14 \times 0.8 \times 10^{-3} \approx 15.08 \times 10^{-3} \]
Now substituting back:
\[ 3.43 \times 10^{-5} = 15.08 \times 10^{-3} v_t \]
\[ v_t = \frac{3.43 \times 10^{-5}}{15.08 \times 10^{-3}} \approx 0.00228 \, \text{m/s} \approx 0.228 \, \text{cm/s} \]

### Summary of Results:
(a) Viscous force \( F_b \approx 1.51 \times 10^{-4} \, \text{N} \)
(b) Hydrostatic force \( F_u \approx 1.57 \times 10^{-5} \, \text{N} \)
(c) Terminal velocity \( v_t \approx 0.228 \, \text{cm/s} \)