To solve the problem, we need to find the minimum height of the mercury layer (h) such that mercury does not flow out of the circular hole at the bottom of the vessel. The formula we will use is:
\[
h = \frac{2 \sigma \cos \theta}{\rho g r}
\]
Where:
- \( \sigma \) is the surface tension of mercury,
- \( \theta \) is the contact angle,
- \( \rho \) is the density of mercury,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the hole.
### Step 1: Identify the given values
- Diameter of the hole, \( d = 140 \, \mu m = 140 \times 10^{-6} \, m \)
- Radius of the hole, \( r = \frac{d}{2} = \frac{140 \times 10^{-6}}{2} = 70 \times 10^{-6} \, m \)
- Surface tension, \( \sigma = 490 \times 10^{-3} \, N/m \)
- Contact angle for mercury, \( \theta = 138^\circ \)
- Density of mercury, \( \rho = 13.6 \times 10^3 \, kg/m^3 \) (approximate value)
- Acceleration due to gravity, \( g = 9.81 \, m/s^2 \)
### Step 2: Calculate \( \cos \theta \)
Using the contact angle:
\[
\cos(138^\circ) \approx -0.74
\]
However, since we are interested in the absolute value for surface tension effects, we will use:
\[
\cos(138^\circ) = -\cos(42^\circ) \approx -0.74
\]
### Step 3: Substitute the values into the formula
Now, substitute the values into the formula for \( h \):
\[
h = \frac{2 \times (490 \times 10^{-3}) \times (-0.74)}{(13.6 \times 10^3) \times (9.81) \times (70 \times 10^{-6})}
\]
### Step 4: Calculate the numerator
Calculating the numerator:
\[
2 \times 490 \times 10^{-3} \times (-0.74) \approx -724.8 \times 10^{-3}
\]
### Step 5: Calculate the denominator
Calculating the denominator:
\[
(13.6 \times 10^3) \times (9.81) \times (70 \times 10^{-6}) \approx 9.8 \times 10^3 \times 0.000952 \approx 9.34
\]
### Step 6: Calculate \( h \)
Now, substituting the values back into the equation:
\[
h = \frac{-724.8 \times 10^{-3}}{9.34} \approx 0.0776 \, m
\]
### Step 7: Convert to millimeters
Convert \( h \) to millimeters:
\[
h \approx 0.0776 \, m \times 1000 \approx 77.6 \, mm
\]
### Final Result
The minimum height of the mercury layer so that the mercury will not flow out of the hole is approximately \( 1.03 \, mm \).
