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 approximately :

Options

A. 0.21 mm
B. 0.14 mm
C. 0.07 mm
D. 0.7 mm

Explanation

To solve the problem, we need to find the diameter of the hole in the hollow sphere that allows water to enter when the sphere is submerged up to a depth of 40 cm. We will use the relationship between the hydrostatic pressure and the excess pressure due to surface tension.

### Step-by-Step Solution:

1. **Understand the Concept**:
- When the hollow sphere is submerged in water, a drop forms at the hole due to surface tension. The water will enter the sphere when the hydrostatic pressure from the water column exceeds the excess pressure created by the surface tension of the drop.

2. **Identify the Given Values**:
- Depth of water (h) = 40 cm = 0.40 m
- Surface tension of water (T) = 7 x 10^(-2) N/m
- Density of water (ρ) = 1000 kg/m³ (approximate value)
- Acceleration due to gravity (g) = 10 m/s² (approximate value)

3. **Calculate Hydrostatic Pressure**:
- The hydrostatic pressure (P_h) at a depth h is given by:
\[ P_h = \rho g h \]
- Substituting the values:
\[ P_h = 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 0.40 \, \text{m} = 4000 \, \text{Pa} \]

4. **Relate Hydrostatic Pressure to Excess Pressure**:
- The excess pressure (P_e) due to surface tension for a drop is given by:
\[ P_e = \frac{2T}{R} \]
- Here, R is the radius of the hole.

5. **Set Up the Equation**:
- For water to start entering the sphere, the hydrostatic pressure must equal the excess pressure:
\[ P_h = P_e \]
- Therefore:
\[ 4000 = \frac{2 \times (7 \times 10^{-2})}{R} \]

6. **Solve for R**:
- Rearranging the equation to solve for R:
\[ R = \frac{2 \times (7 \times 10^{-2})}{4000} \]
- Calculating R:
\[ R = \frac{0.14}{4000} = 3.5 \times 10^{-5} \, \text{m} \]

7. **Calculate the Diameter**:
- The diameter (D) is twice the radius:
\[ D = 2R = 2 \times (3.5 \times 10^{-5}) = 7 \times 10^{-5} \, \text{m} \]
- Converting to millimeters:
\[ D = 7 \times 10^{-5} \, \text{m} = 0.07 \, \text{mm} \]

### Final Answer:
The diameter of the hole is approximately **0.07 mm**.