Maximum pressure difference between inner and outer surface of a thin walled circular tube of radius r and thickness `Deltar(Deltar lt lt r)` so that tube won't break is (Breaking stress of the material is `sigma_(0))`

Options

A. `(sigma_(0)r)/(Deltar)`
B. `(Deltarsigma_(0))/(r)`
C. `(2sigma_(0)r)/(Delta)`
D. `sigma_(0)`

Explanation

To solve the problem of finding the maximum pressure difference between the inner and outer surfaces of a thin-walled circular tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a thin-walled circular tube with an inner radius \( r \) and thickness \( \Delta r \) (where \( \Delta r \ll r \)).
- The breaking stress of the material is given as \( \sigma_0 \).

2. **Identify Forces Acting on the Tube**:
- At any cross-section of the tube, there is tension \( T \) acting on the material due to the pressure difference between the inner and outer surfaces.

3. **Relating Tension to Pressure Difference**:
- The net tension acting towards the center of the tube can be expressed in terms of the pressure difference \( \Delta P \) and the area \( A \) of the cross-section.
- The force due to the pressure difference can be given as:
\[ F = \Delta P \cdot A \]

4. **Area of the Cross-section**:
- The area \( A \) for the thin-walled tube can be approximated as:
\[ A \approx 2 \pi r \Delta r \]

5. **Using the Breaking Stress**:
- The breaking stress \( \sigma_0 \) is defined as the maximum tension per unit area that the material can withstand:
\[ \sigma_0 = \frac{T}{A} \]
- Rearranging gives:
\[ T = \sigma_0 \cdot A \]

6. **Substituting for Area**:
- Substitute the expression for area into the tension equation:
\[ T = \sigma_0 \cdot (2 \pi r \Delta r) \]

7. **Relating Tension to Pressure Difference**:
- The net force due to the pressure difference can also be expressed in terms of tension:
\[ \Delta P \cdot A = T \]
- Substituting for \( A \):
\[ \Delta P \cdot (2 \pi r \Delta r) = T \]

8. **Combining the Equations**:
- Now, substituting the expression for \( T \):
\[ \Delta P \cdot (2 \pi r \Delta r) = \sigma_0 \cdot (2 \pi r \Delta r) \]

9. **Solving for Pressure Difference**:
- Canceling \( 2 \pi r \Delta r \) from both sides (assuming \( \Delta r \neq 0 \)):
\[ \Delta P = \sigma_0 \]

### Final Result:
Thus, the maximum pressure difference \( \Delta P \) that the tube can withstand without breaking is:
\[ \Delta P = \sigma_0 \]