Two block of mass m and M are connected means of a metal of a metal wire passing over a frictionless fixed pulley. The area of cross section of the wire is `6.67xx10^(-9)m^(2)` and its breaking stress is `2xx10^(9)Nm^(-2)`. If m = 1kg. Find the maximum value of M in kg for which the wire will not break. `(g=10m//s^(2))`.

Answer

To solve the problem, we need to find the maximum value of mass \( M \) such that the wire does not break when two blocks of mass \( m \) and \( M \) are connected by the wire over a frictionless pulley.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of block \( m = 1 \, \text{kg} \)
- Area of cross-section of the wire \( A = 6.67 \times 10^{-9} \, \text{m}^2 \)
- Breaking stress of the wire \( \sigma = 2 \times 10^9 \, \text{N/m}^2 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Maximum Tension in the Wire**:
The breaking stress is defined as the maximum tension per unit area that the wire can withstand:
\[ T_{\text{max}} = \sigma \times A \]
Substituting the given values:
\[ T_{\text{max}} = (2 \times 10^9 \, \text{N/m}^2) \times (6.67 \times 10^{-9} \, \text{m}^2) = 13.34 \, \text{N} \]

3. **Set Up the Equations of Motion**:
For the system of two blocks, the forces acting on them can be described as follows:
- For mass \( M \) (downward motion):
\[ M g - T = M a \quad \text{(1)} \]
- For mass \( m \) (upward motion):
\[ T - m g = m a \quad \text{(2)} \]

4. **Add the Two Equations**:
Adding equations (1) and (2):
\[ (M g - T) + (T - m g) = M a + m a \]
This simplifies to:
\[ (M - m) g = (M + m) a \]
Rearranging gives:
\[ a = \frac{(M - m) g}{M + m} \quad \text{(3)} \]

5. **Substitute for Tension \( T \)**:
From equation (2), we can express \( T \) in terms of \( a \):
\[ T = m g + m a \]
Substituting equation (3) into this:
\[ T = m g + m \left(\frac{(M - m) g}{M + m}\right) \]
Simplifying this:
\[ T = m g \left(1 + \frac{(M - m)}{M + m}\right) = m g \left(\frac{M + m + M - m}{M + m}\right) = m g \left(\frac{2M}{M + m}\right) \]

6. **Set \( T \) Equal to \( T_{\text{max}} \)**:
To find the maximum value of \( M \) before the wire breaks, set \( T = T_{\text{max}} \):
\[ m g \left(\frac{2M}{M + m}\right) = T_{\text{max}} \]
Substituting the known values:
\[ 1 \cdot 10 \left(\frac{2M}{M + 1}\right) = 13.34 \]
Simplifying:
\[ \frac{20M}{M + 1} = 13.34 \]

7. **Cross-Multiply and Solve for \( M \)**:
\[ 20M = 13.34(M + 1) \]
Expanding:
\[ 20M = 13.34M + 13.34 \]
Rearranging gives:
\[ 20M - 13.34M = 13.34 \]
\[ 6.66M = 13.34 \]
\[ M = \frac{13.34}{6.66} \approx 2 \, \text{kg} \]

### Conclusion:
The maximum value of \( M \) for which the wire will not break is approximately \( 2 \, \text{kg} \).