Answer:
[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine ([tex] \sf \eta [/tex])
Explanation:
[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]
Substituting values of r, ρ, σ, v & g in the equation:
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]
[tex]\sf \implies \eta = 2 \times 7.191[/tex]
[tex]\sf \implies \eta = 14.382 \: poise[/tex]
