To solve the problem, we need to find the pressure that must be applied to the water in the capillary tube so that the water level in the tube is the same as in the vessel. We will use the concept of capillary rise and the formula for pressure.
### Step-by-Step Solution:
1. **Identify Given Values:**
- Inner diameter of the capillary tube, \( d = 0.28 \, \text{mm} = 0.28 \times 10^{-3} \, \text{m} \)
- Radius of the capillary tube, \( r = \frac{d}{2} = \frac{0.28 \times 10^{-3}}{2} = 0.14 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( \sigma = 0.07 \, \text{N/m} \)
- Density of water, \( \rho \approx 1000 \, \text{kg/m}^3 \) (standard value)
- Acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \)
2. **Calculate Capillary Rise (h):**
The formula for capillary rise is given by:
\[
h = \frac{2\sigma \cos \theta}{\rho g r}
\]
Since the angle of contact \( \theta \) for water in glass is approximately 0 degrees, \( \cos \theta = 1 \). Thus, the formula simplifies to:
\[
h = \frac{2\sigma}{\rho g r}
\]
3. **Substitute the Values into the Formula:**
\[
h = \frac{2 \times 0.07}{1000 \times 9.81 \times (0.14 \times 10^{-3})}
\]
4. **Calculate h:**
\[
h = \frac{0.14}{1000 \times 9.81 \times 0.14 \times 10^{-3}} = \frac{0.14}{0.00137214} \approx 0.102 \, \text{m}
\]
5. **Calculate the Pressure due to the Height (h):**
The pressure due to the height of the water column is given by:
\[
P = \rho g h
\]
Substituting the values:
\[
P = 1000 \times 9.81 \times 0.102
\]
6. **Calculate P:**
\[
P \approx 1000 \times 9.81 \times 0.102 \approx 1000 \times 1.00082 \approx 9810 \, \text{N/m}^2
\]
7. **Calculate the Pressure to be Applied:**
The total pressure that must be applied to maintain the water level in the capillary tube the same as in the vessel is:
\[
P_{\text{applied}} = P_{\text{atmospheric}} - P
\]
Given \( P_{\text{atmospheric}} = 10^5 \, \text{N/m}^2 \):
\[
P_{\text{applied}} = 10^5 - 9810 \approx 90190 \, \text{N/m}^2
\]
### Final Answer:
The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as in the vessel is approximately \( 90190 \, \text{N/m}^2 \).
