A hollow sphere has a small hole in it. On lowering the sphere in a tank of water, it is observed that water enters into the hollow sphere at a depth of `40 cm` below the surface. Surface tension of water is `7 xx 10^(-2) N//m`. The diameter of the hole is

Options

A. `1/28mm`
B. `1/21mm`
C. `1/14mm`
D. `1/7mm`

Explanation

To solve the problem, we need to determine the diameter of the hole in the hollow sphere based on the given conditions. The key concept here is the relationship between the excess pressure created by surface tension and the depth of water.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a hollow sphere with a small hole.
- Water enters the sphere at a depth of 40 cm below the surface.
- The surface tension of water is given as \( \sigma = 7 \times 10^{-2} \, \text{N/m} \).

2. **Excess Pressure Calculation**:
- The excess pressure (\( \Delta P \)) due to surface tension for a hole can be calculated using the formula:
\[ \Delta P = \frac{4\sigma}{r} \]
where \( r \) is the radius of the hole.

3. **Hydrostatic Pressure at Depth**:
- The hydrostatic pressure (\( P \)) at a depth \( h \) in a fluid is given by:
\[ P = \rho g h \]
where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)), \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the depth (40 cm or 0.4 m).
- Thus, the hydrostatic pressure at 40 cm depth is:
\[ P = 1000 \times 9.81 \times 0.4 = 3924 \, \text{Pa} \]

4. **Setting Up the Equation**:
- At the depth of 40 cm, the excess pressure due to surface tension must balance the hydrostatic pressure:
\[ \frac{4\sigma}{r} = \rho g h \]
- Substituting the known values:
\[ \frac{4 \times 7 \times 10^{-2}}{r} = 3924 \]

5. **Solving for Radius**:
- Rearranging the equation to solve for \( r \):
\[ r = \frac{4 \times 7 \times 10^{-2}}{3924} \]
- Calculating \( r \):
\[ r = \frac{0.28}{3924} \approx 7.13 \times 10^{-5} \, \text{m} \]

6. **Finding the Diameter**:
- The diameter \( d \) of the hole is twice the radius:
\[ d = 2r = 2 \times 7.13 \times 10^{-5} \approx 1.426 \times 10^{-4} \, \text{m} = 0.0001426 \, \text{m} \approx 0.1426 \, \text{mm} \]

### Final Answer:
The diameter of the hole is approximately **0.1426 mm**.