A small hollow sphere which has a small hole in it is immersed in water to a depth of 40 cm, before any water is penetrated into it. If the surface tensionof water si `0.073 Nm^(-1)`, find the radius of the hole.

Explanation

To find the radius of the hole in the small hollow sphere immersed in water, we can use the relationship between pressure, surface tension, and the radius of the hole. Here’s a step-by-step solution:

### Step 1: Understand the Pressure at Depth
When the hollow sphere is immersed in water to a depth of 40 cm, the pressure at that depth can be calculated using the formula:
\[ P = \rho g h \]
where:
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the depth in meters (40 cm = 0.4 m).

### Step 2: Calculate the Pressure
Substituting the values into the pressure formula:
\[ P = (1000 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(0.4 \, \text{m}) \]
\[ P = 3920 \, \text{Pa} \]

### Step 3: Relate Surface Tension to Radius
The relationship between the surface tension (\( S \)), pressure (\( P \)), and radius (\( R \)) of the hole is given by:
\[ P = \frac{2S}{R} \]
Rearranging this formula to find the radius:
\[ R = \frac{2S}{P} \]

### Step 4: Substitute Surface Tension and Pressure
Now, substituting the values for surface tension (\( S = 0.073 \, \text{N/m} \)) and pressure (\( P = 3920 \, \text{Pa} \)):
\[ R = \frac{2 \times 0.073 \, \text{N/m}}{3920 \, \text{Pa}} \]

### Step 5: Calculate the Radius
Calculating the radius:
\[ R = \frac{0.146}{3920} \]
\[ R \approx 3.7 \times 10^{-5} \, \text{m} \]

### Final Answer
The radius of the hole is approximately \( 3.7 \times 10^{-5} \, \text{m} \).

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