There is a small hole in a hollow sphere . The water enters in it when it is taken to depth of 40 cm under water. The surface tension of water is 0.07 N/m. The diameter of hole is-

Options

A. 7 mm
B. 0.07 mm
C. 0.0007 mm
D. 0.7 m

Answer

To solve the problem of determining the diameter of the hole in the hollow sphere, we can follow these steps:

### Step 1: Understand the Forces Involved
When the hollow sphere is submerged at a depth of 40 cm, the water enters through the hole due to the pressure exerted by the water column above it. The force due to the surface tension of the water at the edge of the hole opposes this entry.

### Step 2: Identify the Relevant Formulas
1. **Force due to Surface Tension (F_s)**:
\[ F_s = T \cdot L \]
where \( T \) is the surface tension and \( L \) is the circumference of the hole. For a hole of radius \( R \), \( L = 2\pi R \), thus:
\[ F_s = 2\pi R T \]

2. **Pressure Force (F_p)**:
The pressure force due to the water column is given by:
\[ F_p = P \cdot A \]
where \( P \) is the pressure and \( A \) is the area of the hole. The pressure at a depth \( h \) is given by:
\[ P = \rho g h \]
The area \( A \) of the hole is:
\[ A = \pi R^2 \]
Therefore:
\[ F_p = \rho g h \cdot \pi R^2 \]

### Step 3: Set the Forces Equal
For the water to enter the sphere, the pressure force must overcome the force due to surface tension:
\[ F_p = F_s \]
Substituting the expressions for \( F_p \) and \( F_s \):
\[ \rho g h \cdot \pi R^2 = 2\pi R T \]

### Step 4: Simplify the Equation
Cancel \( \pi \) from both sides:
\[ \rho g h \cdot R^2 = 2 R T \]
Now, divide both sides by \( R \) (assuming \( R \neq 0 \)):
\[ \rho g h \cdot R = 2 T \]

### Step 5: Solve for Radius \( R \)
Rearranging gives:
\[ R = \frac{2T}{\rho g h} \]

### Step 6: Substitute Known Values
- Surface tension \( T = 0.07 \, \text{N/m} \)
- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- Depth \( h = 40 \, \text{cm} = 0.4 \, \text{m} \)

Substituting these values into the equation:
\[ R = \frac{2 \cdot 0.07}{1000 \cdot 10 \cdot 0.4} \]
Calculating the denominator:
\[ 1000 \cdot 10 \cdot 0.4 = 4000 \]
Thus:
\[ R = \frac{0.14}{4000} = 3.5 \times 10^{-5} \, \text{m} \]

### Step 7: Calculate the Diameter
The diameter \( D \) is:
\[ D = 2R = 2 \cdot 3.5 \times 10^{-5} = 7 \times 10^{-5} \, \text{m} = 0.07 \, \text{mm} \]

### Final Answer
The diameter of the hole is \( 0.07 \, \text{mm} \).

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