A vessel whose bottom has round holes with diameter of 1 mm is filled with water Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is (given surface tension of water is `75xx10^(-3)N//m)` and `g=10m//s^(2)`

Answer

To solve the problem of determining the maximum height to which water can be filled in a vessel with holes at the bottom, we will use the concept of surface tension and hydrostatic pressure.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of the hole, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Radius of the hole, \( r = \frac{d}{2} = \frac{1 \times 10^{-3}}{2} = 5 \times 10^{-4} \, \text{m} \)
- Surface tension of water, \( T = 75 \times 10^{-3} \, \text{N/m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

2. **Calculate the Force Due to Surface Tension:**
The force due to surface tension acting at the hole can be calculated as:
\[ F_T = 2 \pi r T \]
Substituting the values:
\[ F_T = 2 \pi (5 \times 10^{-4}) (75 \times 10^{-3}) = 2 \pi \times 5 \times 10^{-4} \times 75 \times 10^{-3} \]

3. **Calculate the Weight of the Water Column:**
The weight of the water column above the hole can be expressed as:
\[ W = mg = \rho V g = \rho (A h) g \]
Where \( A \) is the area of the hole, \( A = \pi r^2 \):
\[ W = \rho (\pi r^2 h) g \]

4. **Set the Forces Equal:**
For the water to remain in the vessel without leaking, the upward force due to surface tension must equal the downward weight of the water column:
\[ F_T = W \]
Thus,
\[ 2 \pi r T = \rho (\pi r^2 h) g \]

5. **Simplify the Equation:**
Canceling \( \pi \) from both sides and rearranging gives:
\[ 2rT = \rho r^2 h g \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ 2T = \rho r h g \]

6. **Solve for Maximum Height \( h \):**
Rearranging for \( h \):
\[ h = \frac{2T}{\rho r g} \]

7. **Substituting the Values:**
Substitute the known values into the equation:
\[ h = \frac{2 \times (75 \times 10^{-3})}{(1000) \times (5 \times 10^{-4}) \times (10)} \]
Simplifying this gives:
\[ h = \frac{150 \times 10^{-3}}{1000 \times 5 \times 10^{-4} \times 10} = \frac{150 \times 10^{-3}}{5 \times 10^{-1}} = \frac{150 \times 10^{-3}}{0.5} = 0.03 \, \text{m} \]

8. **Convert to Centimeters:**
Converting meters to centimeters:
\[ h = 0.03 \, \text{m} = 3 \, \text{cm} \]

### Final Answer:
The maximum height to which the water can be filled in the vessel without leakage is **3 cm**.