Question

A solid steel ball of diameter 3.6 mm acquired terminal velocity \( 2.45 \times 10^{-2} \) m/s while falling under gravity through an oil of density \( 925 \, \text{kg m}^{-3} \). Take density of steel as \( 7825 \, \text{kg m}^{-3} \) and \( g \) as \( 9.8 \, \text{m/s}^2 \). The viscosity of the oil in SI unit is

• A. 2.18
• B. 2.38
• C. 1.68
• D. 1.99

Hint:

Use Stokes' Law for terminal velocity of a sphere falling through a viscous fluid: \( v_T = \frac{2 r^2 (\rho_s - \rho_l) g}{9 \eta} \). Ensure all units are in SI. Calculate the radius from the diameter. Rearrange the formula to solve for the viscosity \( \eta \). Substitute the given values and perform the calculation carefully.

Answer 1

The Correct Option is D

Stokes' Law is utilized to determine the viscosity of the oil. This law establishes a relationship between the drag force acting on a sphere moving through a viscous fluid, the sphere's radius, its speed, and the fluid's viscosity. The equation for the terminal velocity of a sphere in a viscous medium under gravitational influence is:

\[v_t = \frac{2}{9} \frac{r^2 ( \rho_s - \rho_f ) g}{\eta}\]

The variables are defined as follows:

• \(v_t\) represents the terminal velocity.
• \(r\) is the sphere's radius.
• \(\rho_s\) denotes the density of the sphere's material (steel).
• \(\rho_f\) indicates the density of the fluid (oil).
• \(g\) is the acceleration due to gravity.
• \(\eta\) is the viscosity of the fluid.

The following values are provided:

• Sphere diameter: \(= 3.6 \, \text{mm} = 3.6 \times 10^{-3} \, \text{m}\)
• Terminal velocity: \(v_t = 2.45 \times 10^{-2} \, \text{m/s}\)
• Density of steel: \(\rho_s = 7825 \, \text{kg/m}^3\)
• Density of oil: \(\rho_f = 925 \, \text{kg/m}^3\)
• Acceleration due to gravity: \(g = 9.8 \, \text{m/s}^2\)

The sphere's radius is calculated first:

\(r = \frac{3.6 \times 10^{-3}}{2} \, \text{m} = 1.8 \times 10^{-3} \, \text{m}\)

These values are then substituted into the terminal velocity equation to solve for the viscosity (\(\eta\)):

\[2.45 \times 10^{-2} = \frac{2}{9} \times \frac{(1.8 \times 10^{-3})^2 \times (7825 - 925) \times 9.8}{\eta}\]

The equation is rearranged to isolate \(\eta\):

\[\eta = \frac{2}{9} \times \frac{(1.8 \times 10^{-3})^2 \times (7825 - 925) \times 9.8}{2.45 \times 10^{-2}}\]

Intermediate calculations are as follows:

• Density difference: \(7825 - 925 = 6900 \, \text{kg/m}^3\)
• Radius squared: \((1.8 \times 10^{-3})^2 = 3.24 \times 10^{-6} \, \text{m}^2\)

These values are inserted into the viscosity equation:

\[\eta = \frac{2 \times 3.24 \times 10^{-6} \times 6900 \times 9.8}{9 \times 2.45 \times 10^{-2}}\]

The calculated viscosity of the oil (\(\eta\)) is approximately:

\(1.99 \, \text{Pa s}\)

The final answer is 1.99.