There is a circular hole of diameter `d =140 mu m ` at the bottom of a vessel containing mercury . The minimum height of mercury layer so that the mercury will not flow out of this hole is
( Surface tension, `sigma =490 xx 10^(-3) Nm^(-1))`

Options

A. 1.03 mm
B. 1.53mm
C. 1.03 mm
D. 1.53cm

Answer

To determine the minimum height of the mercury layer required to prevent mercury from flowing out of a circular hole at the bottom of a vessel, we can use the concept of excess pressure due to surface tension.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Diameter of the hole, \( d = 140 \, \mu m = 140 \times 10^{-6} \, m \)
- Surface tension, \( \sigma = 490 \times 10^{-3} \, N/m \)
- Density of mercury, \( \rho = 13.6 \, g/cm^3 = 13600 \, kg/m^3 \) (conversion from grams per cubic centimeter to kilograms per cubic meter)
- Acceleration due to gravity, \( g = 10 \, m/s^2 \)

2. **Calculate the Radius of the Hole**:
\[ R = \frac{d}{2} = \frac{140 \times 10^{-6}}{2} = 70 \times 10^{-6} \, m \]

3. **Use the Formula for Excess Pressure**:
The excess pressure \( \Delta P \) due to surface tension can be expressed as:
\[ \Delta P = \frac{4\sigma}{d} \]
Substituting the values:
\[ \Delta P = \frac{4 \times (490 \times 10^{-3})}{140 \times 10^{-6}} \]

4. **Calculate the Excess Pressure**:
\[ \Delta P = \frac{1960 \times 10^{-3}}{140 \times 10^{-6}} = \frac{1960}{140} \times 10^3 = 14 \times 10^3 \, Pa = 14000 \, Pa \]

5. **Relate Excess Pressure to Hydrostatic Pressure**:
The hydrostatic pressure due to the height of the mercury column is given by:
\[ P = H \rho g \]
Setting the excess pressure equal to the hydrostatic pressure:
\[ H \rho g = \Delta P \]
Rearranging for height \( H \):
\[ H = \frac{\Delta P}{\rho g} \]

6. **Substituting Values**:
\[ H = \frac{14000}{13600 \times 10} = \frac{14000}{136000} = 0.1029 \, m \]

7. **Final Result**:
The minimum height of the mercury layer so that the mercury does not flow out of the hole is:
\[ H \approx 0.103 \, m \]