The breaking stress for a metal is `9xx10^(10)Nm^(-2)` The density of the metal is `9000kgm^(-3)` . If `g=10Nkg^(-1)` , the maximum length of the wire made of this metal which may be suspended without breaking is `10^(k)` m . Find the value of k.

Explanation

To solve the problem, we need to determine the maximum length of a wire made of a metal, given its breaking stress, density, and the acceleration due to gravity.

### Step-by-step Solution:

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

2. **Identify the Forces Acting on the Wire**:
When the wire is suspended, the force acting downwards due to gravity is the weight of the wire, which can be expressed as:
\[ F = mg \]
where \( m \) is the mass of the wire and \( g \) is the acceleration due to gravity.

3. **Express Mass in Terms of Density**:
The mass \( m \) of the wire can be expressed using its density \( \rho \) and volume \( V \):
\[ m = \rho V \]
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the wire. Therefore, we have:
\[ m = \rho A L \]

4. **Substituting Mass into the Force Equation**:
Substituting the expression for mass into the force equation gives:
\[ F = \rho A L g \]

5. **Relating Stress to Force and Area**:
Stress is defined as force per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the expression for \( F \):
\[ \sigma = \frac{\rho A L g}{A} = \rho L g \]

6. **Setting Up the Equation**:
Now, we can set the breaking stress equal to the expression we derived:
\[ 9 \times 10^{10} = \rho L g \]

7. **Substituting Known Values**:
We know:
- \( \rho = 9000 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{N/kg} \)
Substituting these values into the equation:
\[ 9 \times 10^{10} = 9000 \cdot L \cdot 10 \]

8. **Solving for Length \( L \)**:
Rearranging the equation to solve for \( L \):
\[ L = \frac{9 \times 10^{10}}{9000 \cdot 10} \]
Simplifying:
\[ L = \frac{9 \times 10^{10}}{90000} = \frac{9 \times 10^{10}}{9 \times 10^4} = 10^6 \, \text{m} \]

9. **Finding the Value of \( k \)**:
The maximum length \( L \) can be expressed as \( 10^k \, \text{m} \), where \( k = 6 \).

### Final Answer:
The value of \( k \) is \( 6 \).