Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm. if the amount of glycerine collected per second at one end is `4.0 xx 10^(-3) kg s^(-1)`, what is the pressuer difference between the two ends of the tube? (density of glycerine = `1.3xx10^(3)kgm^(-3)` and viscosity of glycerine =`0.83Ns m^(-2)`).

Explanation

To find the pressure difference between the two ends of the tube through which glycerine flows, we can use Poiseuille's law. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Length of the tube, \( L = 1.5 \, \text{m} \)
- Radius of the tube, \( r = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Mass flow rate of glycerine, \( \dot{m} = 4.0 \times 10^{-3} \, \text{kg/s} \)
- Density of glycerine, \( \rho = 1.3 \times 10^{3} \, \text{kg/m}^3 \)
- Viscosity of glycerine, \( \eta = 0.83 \, \text{Ns/m}^2 \)
...