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

To find the viscosity of the oil, we can use Stokes' Law, which relates the drag force experienced by a sphere moving through a viscous fluid to its radius, speed, and the fluid's viscosity. The formula for terminal velocity under the influence of gravity for a sphere in a viscous medium is given by:

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

where:

• \(v_t\) is the terminal velocity
• \(r\) is the radius of the sphere
• \(\rho_s\) is the density of the sphere material (steel in this case)
• \(\rho_f\) is the density of the fluid (oil in this case)
• \(g\) is the acceleration due to gravity
• \(\eta\) is the viscosity of the fluid

We are given:

• Diameter of the sphere \(= 3.6 \, \text{mm} = 3.6 \times 10^{-3} \, \text{m}\)
• 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 \(g = 9.8 \, \text{m/s}^2\)

First, calculate the radius of the sphere:

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

Substitute these values into the terminal velocity equation and solve for the viscosity (\(\eta\)):

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

Rearrange to solve for \(\eta\):

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

Calculating the values inside the equation:

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

Substitute these into the equation:

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

After solving, the viscosity of the oil (\(\eta\)) is approximately:

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

Therefore, the correct answer is 1.99.

Answer 2

The terminal velocity \( v_T \) of a sphere falling through a viscous fluid is given by Stokes' Law: \[ v_T = \frac{2 r^2 (\rho_s - \rho_l) g}{9 \eta} \] where \( r \) is the radius of the sphere, \( \rho_s \) is the density of the sphere (steel), \( \rho_l \) is the density of the liquid (oil), \( g \) is the acceleration due to gravity, and \( \eta \) is the viscosity of the liquid. 
Given: 
Diameter of the steel ball \( d = 3.6 \, \text{mm} = 3.6 \times 10^{-3} \, \text{m} \) 
Radius of the steel ball \( r = \frac{d}{2} = \frac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3} \, \text{m} \) 
Terminal velocity \( v_T = 2.45 \times 10^{-2} \, \text{m/s} \) Density of oil \( \rho_l = 925 \, \text{kg m}^{-3} \) 
Density of steel \( \rho_s = 7825 \, \text{kg m}^{-3} \) 
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) 
We need to find the viscosity \( \eta \) of the oil. 
Rearranging the formula for terminal velocity: \[ \eta = \frac{2 r^2 (\rho_s - \rho_l) g}{9 v_T} \] 
Substituting the given values: \[ \eta = \frac{2 (1.8 \times 10^{-3})^2 (7825 - 925) (9.8)}{9 (2.45 \times 10^{-2})} \] \[ \eta = \frac{2 (3.24 \times 10^{-6}) (6900) (9.8)}{9 (2.45 \times 10^{-2})} \] \[ \eta = \frac{2 \times 3.24 \times 10^{-6} \times 6900 \times 9.8}{0.2205} \] \[ \eta = \frac{0.436512}{0.2205} \] \[ \eta \approx 1.98 \, \text{Pa s} \] The viscosity of the oil is approximately 1.98 Pa s, which is close to 1.99.