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

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.