Answer :
The viscosity of the fluid is approximately 0.0000003927 Pa s.
To find the viscosity of the fluid, we can use Poiseuille's law, which relates the flow rate, pressure drop, viscosity, and dimensions of the capillary tube.
Poiseuille's law states:
[tex]\[ Q = \frac{\pi r^4 \Delta P}{8 \mu L} \][/tex]
Where:
- Q is the flow rate,
- r is the radius of the capillary tube,
- [tex]\( \Delta P \)[/tex] is the pressure drop per length of the tube,
- [tex]\( \mu \)[/tex] is the viscosity of the fluid, and
- L is the length of the capillary tube.
Given that the inside diameter of the capillary tube is 2 mm, we have [tex]\( r = \frac{2}{2} = 1 \) mm = \( 0.001 \) m, and \( L = 0.1 \) m.[/tex]
We are also given that [tex]\( Q = 60 \) cm/min = \( 0.01 \) m/min, and \( \Delta P = 1.0 \) mmHg cm\(^{-1}\).[/tex]
We can rearrange Poiseuille's law to solve for viscosity [tex](\( \mu \))[/tex]:
[tex]\[ \mu = \frac{\pi r^4 \Delta P}{8 Q L} \][/tex]
[tex]\[ \mu = \frac{\pi (0.001)^4 \times 1.0}{8 \times 0.01 \times 0.1} \][/tex]
[tex]\[ \mu \approx \frac{\pi \times 0.000000001 \times 1.0}{0.008} \][/tex]
[tex]\[ \mu \approx \frac{3.1416 \times 0.000000001}{0.008} \][/tex]
[tex]\[ \mu \approx \frac{0.0000000031416}{0.008} \][/tex]
[tex]\[ \mu \approx 0.0000003927 \, \text{Pa s} \][/tex]
Therefore, the viscosity of the fluid is approximately 0.0000003927 Pa s.
