Sure, here is the step-by-step solution to the problem where we need to calculate the radius of the oil drop in Millikan's experiment:
Given Data:
1. Fall distance of the oil drop, [tex]\( d = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)[/tex]
2. Fall time of the oil drop, [tex]\( t = 71 \, \text{s} \)[/tex]
3. Density of oil, [tex]\( \rho_{\text{oil}} = 880 \, \text{kg/m}^3 \)[/tex]
4. Density of air, [tex]\( \rho_{\text{air}} = 1.29 \, \text{kg/m}^3 \)[/tex]
5. Viscosity of air, [tex]\( \eta_{\text{air}} = 1.80 \times 10^{-5} \, \text{Ns/m}^2 \)[/tex]
6. Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
Step-by-Step Solution:
1. Calculate the Velocity of the Falling Drop:
The velocity [tex]\( v \)[/tex] is given by the distance divided by time:
[tex]\[
v = \frac{d}{t} = \frac{4 \times 10^{-3} \, \text{m}}{71 \, \text{s}}
\][/tex]
2. Use Stokes’ Law to Relate the Radius of the Drop:
Stokes' law helps us understand the force of viscosity acting on a spherical object moving through a fluid. For a sphere falling under gravity in a viscous medium, the terminal velocity is given by:
[tex]\[
v = \frac{2 r^2 g (\rho_{\text{oil}} - \rho_{\text{air}})}{9 \eta_{\text{air}}}
\][/tex]
Where:
[tex]\[
v = \text{velocity of the drop}, \\
r = \text{radius of the drop}, \\
\rho_{\text{oil}} = \text{density of oil}, \\
\rho_{\text{air}} = \text{density of air}, \\
\eta_{\text{air}} = \text{viscosity of air}, \\
g = \text{acceleration due to gravity}
\][/tex]
3. Rearrange to Solve for the Radius [tex]\( r \)[/tex]:
Solving for [tex]\( r \)[/tex]:
[tex]\[
r^2 = \frac{9 \eta_{\text{air}} v}{2 g (\rho_{\text{oil}} - \rho_{\text{air}})}
\][/tex]
Taking the square root of both sides:
[tex]\[
r = \sqrt{\frac{9 \eta_{\text{air}} v}{2 g (\rho_{\text{oil}} - \rho_{\text{air}})}}
\][/tex]
4. Substitute the Known Values:
Plugging in the values to find [tex]\( r \)[/tex]:
[tex]\[
r = \sqrt{\frac{9 \times 1.80 \times 10^{-5} \, \text{Ns/m}^2 \times \frac{4 \times 10^{-3} \, \text{m}}{71 \, \text{s}}}{2 \times 9.8 \, \text{m/s}^2 \times (880 \, \text{kg/m}^3 - 1.29 \, \text{kg/m}^3)}}
\][/tex]
5. Perform the Computation:
After performing the necessary arithmetic operations:
[tex]\[
r \approx 7.3 \times 10^{-7} \, \text{m}
\][/tex]
Thus, the radius of the drop is approximately [tex]\( 7.3 \times 10^{-7} \)[/tex] meters, which matches the given answer.
