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 given metal that can be suspended without breaking, using the provided breaking stress, density, and gravitational acceleration.

### Step-by-Step Solution:

1. **Understand the Concept of Breaking Stress**:
Breaking stress (\( \sigma \)) is defined as the maximum stress that a material can withstand before failure. It is given by the formula:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force applied and \( A \) is the cross-sectional area.

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

3. **Express Mass in Terms of Density and Volume**:
The mass \( m \) of the wire can be expressed in terms of its density (\( \rho \)) and volume (\( V \)):
\[ m = \rho \cdot V \]
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( L \) is the length of the wire.

4. **Substituting Volume into the Mass Equation**:
Substituting the expression for volume into the mass equation gives:
\[ m = \rho \cdot (A \cdot L) \]

5. **Substituting Mass into the Force Equation**:
Now substituting this expression for \( m \) into the force equation:
\[ F = \rho \cdot (A \cdot L) \cdot g \]

6. **Substituting Force into the Breaking Stress Equation**:
Now we substitute \( F \) into the breaking stress formula:
\[ \sigma = \frac{\rho \cdot (A \cdot L) \cdot g}{A} \]
The area \( A \) cancels out:
\[ \sigma = \rho \cdot L \cdot g \]

7. **Rearranging to Find Length**:
Rearranging the equation to solve for \( L \):
\[ L = \frac{\sigma}{\rho \cdot g} \]

8. **Substituting the Given Values**:
Now we substitute the given values into the equation:
- Breaking stress \( \sigma = 9 \times 10^{10} \, \text{N/m}^2 \)
- Density \( \rho = 9000 \, \text{kg/m}^3 \)
- Gravitational acceleration \( g = 10 \, \text{N/kg} \)

Thus,
\[ L = \frac{9 \times 10^{10}}{9000 \times 10} \]

9. **Calculating the Length**:
Simplifying the expression:
\[ L = \frac{9 \times 10^{10}}{9 \times 10^{3}} = 10^{6} \, \text{m} \]

10. **Expressing Length in Terms of \( 10^k \)**:
According to the question, the maximum length is expressed as \( 10^k \, \text{m} \). Here, we have:
\[ L = 10^{6} \, \text{m} \]
Therefore, comparing gives us \( k = 6 \).

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