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 find the diameter of the hole in the hollow sphere, we can use the relationship between excess pressure and the height of water column supported by surface tension. Here’s a step-by-step solution:

### Step 1: Understand the relationship between pressure and surface tension
The excess pressure (ΔP) inside the hollow sphere due to surface tension can be expressed as:
\[ \Delta P = \frac{2T}{R} \]
where \( T \) is the surface tension and \( R \) is the radius of the hole.

### Step 2: Relate excess pressure to the height of the water column
The pressure at a depth \( h \) in a fluid is given by:
\[ P = \rho g h \]
where \( \rho \) is the density of the fluid (water), \( g \) is the acceleration due to gravity, and \( h \) is the depth of the water column.

### Step 3: Set the two pressure equations equal
At the depth where water enters the hollow sphere, the excess pressure due to surface tension is equal to the hydrostatic pressure:
\[ \frac{2T}{R} = \rho g h \]

### Step 4: Solve for the radius \( R \)
Rearranging the equation to solve for \( R \):
\[ R = \frac{2T}{\rho g h} \]

### Step 5: Substitute the known values
Given:
- Surface tension \( T = 7 \times 10^{-2} \, \text{N/m} \)
- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
- Depth \( h = 40 \, \text{cm} = 0.4 \, \text{m} \)

Substituting these values into the equation:
\[ R = \frac{2 \times (7 \times 10^{-2})}{1000 \times 9.8 \times 0.4} \]

### Step 6: Calculate \( R \)
Calculating the denominator:
\[ 1000 \times 9.8 \times 0.4 = 3920 \]
Now substituting:
\[ R = \frac{0.14}{3920} \approx 3.57 \times 10^{-5} \, \text{m} \]

### Step 7: Find the diameter
The diameter \( d \) of the hole is twice the radius:
\[ d = 2R = 2 \times 3.57 \times 10^{-5} \approx 7.14 \times 10^{-5} \, \text{m} = 0.0000714 \, \text{m} = 0.0714 \, \text{mm} \]

### Step 8: Convert to mm
To express this in mm:
\[ d \approx 1/14 \, \text{mm} \]

### Final Answer
The diameter of the hole is approximately \( \frac{1}{14} \, \text{mm} \).

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