A glass capillary tube of inner diameter 0.28 mm is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as the vessel in `(N)/(m^(2))` is (surface tension of water `=0.07(N)/(m)` atmospheric pressure `=10^(5)(N)/(m^(2))`

Options

A. `10^(3)`
B. `99xx10^(3)`
C. `100xx10^(3)`
D. `101xx10^(3)`

Explanation

To solve the problem, we need to determine the pressure that must be applied to the water in a capillary tube so that the water level in the tube is the same as the water level in the vessel. We will use the concept of capillary action and the formula for excess pressure due to surface tension.

### 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, \( \gamma = 0.07 \, \text{N/m} \)
- Atmospheric pressure, \( P_0 = 10^5 \, \text{N/m}^2 \)

2. **Calculate Excess Pressure**:
The excess pressure (\( P_{\text{excess}} \)) in the capillary tube due to surface tension is given by the formula:
\[ P_{\text{excess}} = \frac{2\gamma}{r} \]
Substituting the values:
\[ P_{\text{excess}} = \frac{2 \times 0.07}{0.14 \times 10^{-3}} = \frac{0.14}{0.14 \times 10^{-3}} = 10^3 \, \text{N/m}^2 \]

3. **Calculate Total Pressure**:
The total pressure (\( P \)) that needs to be applied to keep the water level in the capillary tube the same as in the vessel is given by:
\[ P = P_0 + P_{\text{excess}} \]
Substituting the values:
\[ P = 10^5 + 10^3 = 10^5 + 0.01 \times 10^5 = 1.01 \times 10^5 \, \text{N/m}^2 \]

4. **Final Answer**:
Therefore, the pressure that must be applied on the water in the capillary tube is:
\[ P = 1.01 \times 10^5 \, \text{N/m}^2 \]