A vessel, whose bottom has round holes with diameter of 0.1 mm , is filled with water. The maximum height to which the water can be filled without leakage is (S.T. of water =75 dyne/cm , g=1000 cm/s)

Options

A. 100 cm
B. 75 cm
C. 50 cm
D. 30 cm

Answer

To solve the problem, we need to find the maximum height to which water can be filled in a vessel with round holes at the bottom without leaking. The relationship between surface tension, density, and height will help us determine this.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of holes, \( d = 0.1 \, \text{mm} = 0.01 \, \text{cm} \)
- Radius of holes, \( r = \frac{d}{2} = \frac{0.01 \, \text{cm}}{2} = 0.005 \, \text{cm} \)
- Surface tension of water, \( T = 75 \, \text{dyne/cm} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 = 1 \, \text{g/cm}^3 \)
- Acceleration due to gravity, \( g = 1000 \, \text{cm/s}^2 \)

2. **Use the Formula for Maximum Height:**
The maximum height \( h \) to which water can be filled without leaking is given by the formula derived from the balance of surface tension and hydrostatic pressure:
\[ h = \frac{2T}{\rho g r} \]

3. **Substitute the Values:**
- Substitute \( T = 75 \, \text{dyne/cm} \), \( \rho = 1 \, \text{g/cm}^3 \), \( g = 1000 \, \text{cm/s}^2 \), and \( r = 0.005 \, \text{cm} \):
\[ h = \frac{2 \times 75}{1 \times 1000 \times 0.005} \]

4. **Calculate the Height:**
- Calculate the numerator:
\[ 2 \times 75 = 150 \, \text{dyne} \]
- Calculate the denominator:
\[ 1 \times 1000 \times 0.005 = 5 \, \text{g/cm}^2 \]
- Now, calculate \( h \):
\[ h = \frac{150}{5} = 30 \, \text{cm} \]

5. **Conclusion:**
The maximum height to which the water can be filled without leakage is **30 cm**.