There is a small hole in a hollow sphere. When it is taken to a depth of 20cm under water, water enters in it. What will be the radius of the hole in `mum` if the surface tension of water is 0.07N/m . (Take `g = 10m//s^(2)`)

Explanation

To solve the problem, we need to find the radius of the hole in the hollow sphere when it is submerged at a depth of 20 cm under water, given the surface tension of water is 0.07 N/m.

### Step-by-step Solution:

1. **Understand the Problem**:
- We have a hollow sphere with a small hole.
- The sphere is submerged at a depth of 20 cm in water.
- Water enters the hole due to the pressure difference created by the water column above it.

2. **Identify Given Values**:
- Depth \( h = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Surface tension \( \gamma = 0.07 \, \text{N/m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

3. **Calculate the Pressure at Depth**:
- The pressure exerted by the water column at depth \( h \) is given by:
\[ P = \rho g h \]
- Here, \( \rho \) (density of water) is approximately \( 1000 \, \text{kg/m}^3 \).
- Substituting the values:
\[ P = 1000 \times 10 \times 0.2 = 2000 \, \text{Pa} \]

4. **Relate Pressure to Surface Tension**:
- The pressure difference due to surface tension can be expressed as:
\[ \Delta P = \frac{2\gamma}{r} \]
- Where \( r \) is the radius of the hole.

5. **Set the Pressures Equal**:
- Since the pressure due to the water column is equal to the pressure difference due to surface tension:
\[ 2000 = \frac{2 \times 0.07}{r} \]

6. **Solve for \( r \)**:
- Rearranging the equation gives:
\[ r = \frac{2 \times 0.07}{2000} \]
- Calculating \( r \):
\[ r = \frac{0.14}{2000} = 0.00007 \, \text{m} = 70 \, \mu m \]

### Final Answer:
The radius of the hole is \( 70 \, \mu m \).