Find the greatest length of copper wire, that can hang without breaking. Breaking stress `=7.2 xx 10^(7) N//m^(2)`. Density of copper `7.2 g//c c. g=10 m//s^(2)`.

Options

A. a.100 m
B. b.1000 m
C. c.150 m
D. d.1500 m

Answer

To find the greatest length of copper wire that can hang without breaking, we will use the given breaking stress and density of copper. Here’s a step-by-step solution:

### Step 1: Understand the relationship between breaking stress, force, and area
Breaking stress (σ) is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]

### Step 2: Relate force to weight
The force acting on the wire due to its weight can be expressed as:
\[ F = mg \]
where \(m\) is the mass of the wire and \(g\) is the acceleration due to gravity.

### Step 3: Express mass in terms of volume and density
The mass \(m\) of the wire can be expressed as:
\[ m = \text{Volume} \times \text{Density} = V \cdot \rho \]
where \(V\) is the volume of the wire and \(\rho\) is the density of copper.

### Step 4: Relate volume to area and length
The volume \(V\) of the wire can be expressed in terms of its cross-sectional area \(A\) and length \(L\):
\[ V = A \cdot L \]
Thus, we can rewrite the mass as:
\[ m = A \cdot L \cdot \rho \]

### Step 5: Substitute mass into the force equation
Substituting the expression for mass into the force equation gives:
\[ F = A \cdot L \cdot \rho \cdot g \]

### Step 6: Substitute force into the breaking stress equation
Now substituting \(F\) back into the breaking stress equation:
\[ \sigma = \frac{A \cdot L \cdot \rho \cdot g}{A} \]
This simplifies to:
\[ \sigma = L \cdot \rho \cdot g \]

### Step 7: Solve for length \(L\)
Rearranging the equation to solve for \(L\):
\[ L = \frac{\sigma}{\rho \cdot g} \]

### Step 8: Substitute the known values
Now, substitute the given values:
- Breaking stress \(\sigma = 7.2 \times 10^7 \, \text{N/m}^2\)
- Density of copper \(\rho = 7.2 \, \text{g/cm}^3 = 7.2 \times 10^3 \, \text{kg/m}^3\)
- Acceleration due to gravity \(g = 10 \, \text{m/s}^2\)

Substituting these values:
\[ L = \frac{7.2 \times 10^7}{7.2 \times 10^3 \cdot 10} \]

### Step 9: Simplify the expression
\[ L = \frac{7.2 \times 10^7}{7.2 \times 10^4} = \frac{10^7}{10^4} = 10^3 = 1000 \, \text{m} \]

### Conclusion
The greatest length of copper wire that can hang without breaking is:
\[ \boxed{1000 \, \text{m}} \]