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
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- 2.18
- 2.38
- 1.68
- 1.99
The Correct Option is D
Approach Solution - 1
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.
Approach Solution -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.
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