Question

A falling ball viscometer is used to determine the dynamic viscosity of sunflower oil. The viscometer has a tube length of 10 cm with the ball diameter of 0.68 mm. The densities of oil and the ball are 921 kg m$^{-3$ and 2420 kg m$^{-3}$, respectively. If the ball takes 44.5 s to fall from top of the tube, the viscosity of the oil in Pa s is _____.} \textit{[Round off to three decimal places]}

Hint:

In a falling ball viscometer, the dynamic viscosity is determined based on the radius of the ball, densities of the ball and fluid, and the time taken for the ball to fall through the fluid.

Answer 1

Correct Answer: 0.165

The viscosity \( \mu \) can be calculated using the falling ball viscometer formula: \[ \mu = \frac{2 r^2 (\rho_b - \rho_o) g t}{9 v} \] where:
- \( r = 0.34 \, \text{mm} = 0.34 \times 10^{-3} \, \text{m} \) is the radius of the ball,
- \( \rho_b = 2420 \, \text{kg/m}^3 \) is the density of the ball,
- \( \rho_o = 921 \, \text{kg/m}^3 \) is the density of the oil,
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t = 44.5 \, \text{s} \) is the time taken by the ball to fall,
- \( v = 10 \, \text{cm} = 0.1 \, \text{m} \) is the length of the tube.
Substitute the values into the formula: \[ \mu = \frac{2 \times (0.34 \times 10^{-3})^2 \times (2420 - 921) \times 9.81 \times 44.5}{9 \times 0.1} \] \[ \mu \approx \frac{2 \times 0.1156 \times 1499 \times 9.81 \times 44.5}{0.9} \approx 0.169 \, \text{Pa s}. \] Thus, the viscosity of the oil is approximately \( \boxed{0.172} \, \text{Pa s} \) (rounded to three decimal places).