The breaking stress of aluminium is `7.5 xx 10^7 Nm^(-2)` Find the greatest length of aluminum wire that can hang vertically without breaking Density of aluminium is `2.7 xx 10^3 kg m^(-3)`

Explanation

To find the greatest length of aluminum wire that can hang vertically without breaking, we can follow these steps:

### Step 1: Understand the relationship between breaking stress, force, and area.
Breaking stress (σ) 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 of the wire.

### Step 2: Relate the force to the weight of the wire.
The force acting on the wire due to gravity is equal to its weight, which can be expressed as:
\[ F = mg \]
where \(m\) is the mass of the wire and \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).

### Step 3: Express the mass of the wire in terms of its volume and density.
The mass of the wire can be calculated using its volume and density:
\[ m = V \cdot \rho \]
The volume \(V\) of the wire can be expressed as:
\[ V = A \cdot L \]
where \(L\) is the length of the wire.

### Step 4: Substitute the expressions for mass and volume into the force equation.
Substituting for \(m\) in the force equation gives:
\[ F = (A \cdot L) \cdot \rho \cdot g \]

### Step 5: Substitute the expression for force into the breaking stress equation.
Now substituting \(F\) 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 6: Solve for the length \(L\).
Rearranging the equation to solve for \(L\):
\[ L = \frac{\sigma}{\rho \cdot g} \]

### Step 7: Substitute the known values.
Now we can substitute the values for breaking stress, density, and acceleration due to gravity:
- Breaking stress, \(\sigma = 7.5 \times 10^7 \, \text{N/m}^2\)
- Density of aluminum, \(\rho = 2.7 \times 10^3 \, \text{kg/m}^3\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

Substituting these values into the equation:
\[ L = \frac{7.5 \times 10^7}{(2.7 \times 10^3) \cdot (9.8)} \]

### Step 8: Calculate the value of \(L\).
Calculating the denominator:
\[ (2.7 \times 10^3) \cdot (9.8) = 26406 \, \text{kg m/s}^2 \]
Now substituting back:
\[ L = \frac{7.5 \times 10^7}{26406} \approx 2846.4 \, \text{m} \]

### Final Answer:
The greatest length of aluminum wire that can hang vertically without breaking is approximately \(2846.4 \, \text{m}\).

---