Answer :
To find the radius of a small oil drop falling with a terminal velocity of \(4.0 \, \text{m/s}\) through air, given the viscosity of air is \(1.8 \times 10^{-5} \, \text{Ns/m}^2\), the density of the oil is \(900 \, \text{kg/m}^3\), and the acceleration due to gravity \(g\) is \(10 \, \text{m/s}^2\), we proceed as follows:
1. Identify the terminal velocity formula for a falling sphere:
The terminal velocity \(v_t\) of a sphere falling through a fluid can be given by:
[tex]\[ v_t = \frac{2 r^2 (\rho_{\text{oil}} - \rho_{\text{air}}) g}{9 \eta_{\text{air}}} \][/tex]
where:
- \(v_t\) is the terminal velocity,
- \(r\) is the radius of the sphere,
- \(\rho_{\text{oil}}\) is the density of the oil,
- \(\rho_{\text{air}}\) is the density of air,
- \(g\) is the acceleration due to gravity,
- \(\eta_{\text{air}}\) is the viscosity of air.
2. Assumptions and given values:
[tex]\[ v_t = 4.0 \, \text{m/s} \][/tex]
[tex]\[ \eta_{\text{air}} = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \][/tex]
[tex]\[ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 \][/tex]
Since the density of air is negligible compared to the density of oil, we assume \(\rho_{\text{air}} \approx 0\):
[tex]\[ g = 10 \, \text{m/s}^2 \][/tex]
3. Rearrange the formula to solve for \(r\):
[tex]\[ v_t = \frac{2 r^2 \rho_{\text{oil}} g}{9 \eta_{\text{air}}} \][/tex]
Solving for \(r^2\):
[tex]\[ r^2 = \frac{9 \eta_{\text{air}} v_t}{2 g \rho_{\text{oil}}} \][/tex]
4. Substitute the given values:
[tex]\[ r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times 4.0}{2 \times 10 \times 900} \][/tex]
5. Calculate \(r^2\):
[tex]\[ r^2 = \frac{9 \times 1.8 \times 4.0 \times 10^{-5}}{2 \times 10 \times 900} \][/tex]
[tex]\[ r^2 = \frac{64.8 \times 10^{-5}}{18000} \][/tex]
[tex]\[ r^2 = 3.6 \times 10^{-8} \, \text{m}^2 \][/tex]
6. Calculate the radius \(r\):
[tex]\[ r = \sqrt{3.6 \times 10^{-8}} \][/tex]
[tex]\[ r \approx 1.897 \times 10^{-4} \, \text{m} \][/tex]
or
[tex]\[ r \approx 0.0001897 \, \text{m} \][/tex]
Therefore, the radius of the oil drop is approximately [tex]\(0.0001897\)[/tex] meters (or [tex]\(189.7 \, \mu\text{m}\)[/tex]).
1. Identify the terminal velocity formula for a falling sphere:
The terminal velocity \(v_t\) of a sphere falling through a fluid can be given by:
[tex]\[ v_t = \frac{2 r^2 (\rho_{\text{oil}} - \rho_{\text{air}}) g}{9 \eta_{\text{air}}} \][/tex]
where:
- \(v_t\) is the terminal velocity,
- \(r\) is the radius of the sphere,
- \(\rho_{\text{oil}}\) is the density of the oil,
- \(\rho_{\text{air}}\) is the density of air,
- \(g\) is the acceleration due to gravity,
- \(\eta_{\text{air}}\) is the viscosity of air.
2. Assumptions and given values:
[tex]\[ v_t = 4.0 \, \text{m/s} \][/tex]
[tex]\[ \eta_{\text{air}} = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \][/tex]
[tex]\[ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 \][/tex]
Since the density of air is negligible compared to the density of oil, we assume \(\rho_{\text{air}} \approx 0\):
[tex]\[ g = 10 \, \text{m/s}^2 \][/tex]
3. Rearrange the formula to solve for \(r\):
[tex]\[ v_t = \frac{2 r^2 \rho_{\text{oil}} g}{9 \eta_{\text{air}}} \][/tex]
Solving for \(r^2\):
[tex]\[ r^2 = \frac{9 \eta_{\text{air}} v_t}{2 g \rho_{\text{oil}}} \][/tex]
4. Substitute the given values:
[tex]\[ r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times 4.0}{2 \times 10 \times 900} \][/tex]
5. Calculate \(r^2\):
[tex]\[ r^2 = \frac{9 \times 1.8 \times 4.0 \times 10^{-5}}{2 \times 10 \times 900} \][/tex]
[tex]\[ r^2 = \frac{64.8 \times 10^{-5}}{18000} \][/tex]
[tex]\[ r^2 = 3.6 \times 10^{-8} \, \text{m}^2 \][/tex]
6. Calculate the radius \(r\):
[tex]\[ r = \sqrt{3.6 \times 10^{-8}} \][/tex]
[tex]\[ r \approx 1.897 \times 10^{-4} \, \text{m} \][/tex]
or
[tex]\[ r \approx 0.0001897 \, \text{m} \][/tex]
Therefore, the radius of the oil drop is approximately [tex]\(0.0001897\)[/tex] meters (or [tex]\(189.7 \, \mu\text{m}\)[/tex]).
