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 determine the maximum value of mass \( M \) such that the wire does not break when two blocks of mass \( m \) and \( M \) are connected over a frictionless pulley.

### Step-by-Step Solution:

1. **Understanding Breaking Stress**:
The breaking stress of the wire is given as \( 2 \times 10^9 \, \text{N/m}^2 \). Breaking stress is the maximum stress that the wire can withstand before breaking.

2. **Calculating Maximum Tension**:
The maximum tension \( T_{\text{max}} \) in the wire can be calculated using the formula:
\[ T_{\text{max}} = \text{Breaking Stress} \times \text{Area of Cross Section} \]
Given:
- Area of cross section \( A = 6.67 \times 10^{-9} \, \text{m}^2 \)
- Breaking stress \( \sigma = 2 \times 10^9 \, \text{N/m}^2 \)

Substituting the 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. **Setting Up the Equations of Motion**:
For the block of mass \( M \):
\[ M g - T = M a \quad \text{(1)} \]
For the block of mass \( m \) (where \( m = 1 \, \text{kg} \)):
\[ T - m g = m a \quad \text{(2)} \]
Here, \( g = 10 \, \text{m/s}^2 \).

4. **Substituting Values in Equation (2)**:
Substituting \( m = 1 \, \text{kg} \):
\[ T - 1 \cdot 10 = 1 \cdot a \]
Simplifying:
\[ T - 10 = a \quad \text{(3)} \]

5. **Substituting Equation (3) into Equation (1)**:
From equation (3), we can express \( a \):
\[ a = T - 10 \]
Substitute \( a \) in equation (1):
\[ M g - T = M (T - 10) \]
Expanding:
\[ M g - T = M T - 10M \]
Rearranging gives:
\[ M g + 10M = T + M T \]
Factoring out \( T \):
\[ T(1 + M) = M g + 10M \]
Thus:
\[ T = \frac{M g + 10M}{1 + M} \]

6. **Setting \( T \) to Maximum Tension**:
We know \( T \) cannot exceed \( T_{\text{max}} = 13.34 \, \text{N} \):
\[ \frac{M g + 10M}{1 + M} \leq 13.34 \]
Substituting \( g = 10 \):
\[ \frac{M(10 + 10)}{1 + M} \leq 13.34 \]
Simplifying:
\[ \frac{20M}{1 + M} \leq 13.34 \]

7. **Cross Multiplying**:
\[ 20M \leq 13.34(1 + M) \]
Expanding:
\[ 20M \leq 13.34 + 13.34M \]
Rearranging gives:
\[ 20M - 13.34M \leq 13.34 \]
\[ 6.66M \leq 13.34 \]
Thus:
\[ M \leq \frac{13.34}{6.66} \approx 2 \, \text{kg} \]

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