A small hollow sphere having a small hole in it is immersed in water to a depth of 50cm, before any water penetrates into it. Calculate the radius of the hole, if the surface tension of water is `7.2 xx 10^(-2) Nm^(-1)`.

Explanation

To solve the problem of calculating the radius of the hole in a small hollow sphere immersed in water, we can follow these steps:

### Step 1: Understand the relationship between height, surface tension, and radius
The height of the water column that can be supported by the surface tension of the water is given by the formula:
\[ h = \frac{2 \sigma}{r \rho g} \]
where:
- \( h \) = height of the water column (in meters)
- \( \sigma \) = surface tension of water (in N/m)
- \( r \) = radius of the hole (in meters)
- \( \rho \) = density of water (in kg/m³)
- \( g \) = acceleration due to gravity (in m/s²)

### Step 2: Rearrange the formula to solve for the radius \( r \)
We can rearrange the formula to isolate \( r \):
\[ r = \frac{2 \sigma}{h \rho g} \]

### Step 3: Substitute the known values into the formula
We know the following values:
- Surface tension of water, \( \sigma = 7.2 \times 10^{-2} \, \text{N/m} \)
- Height \( h = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

Now, substituting these values into the rearranged formula:
\[ r = \frac{2 \times 7.2 \times 10^{-2}}{0.5 \times 10^3 \times 9.8} \]

### Step 4: Calculate the radius
Now, we perform the calculation:
1. Calculate the numerator:
\[ 2 \times 7.2 \times 10^{-2} = 0.144 \, \text{N/m} \]
2. Calculate the denominator:
\[ 0.5 \times 10^3 \times 9.8 = 490 \, \text{kg m/s}^2 \]
3. Now, divide the numerator by the denominator:
\[ r = \frac{0.144}{490} \approx 2.938 \times 10^{-5} \, \text{m} \]

### Conclusion
The radius of the hole in the hollow sphere is approximately:
\[ r \approx 2.938 \times 10^{-5} \, \text{m} \]

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