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)`

Options

A. a.`283xx10^(3)m`
B. b.`28.3xx10^(3)m`
C. c.`2.83xx10^(3)m`
D. d.`0.283xx10^(3)m`

Answer

To find the greatest length of an aluminum wire that can hang vertically without breaking, we can use the relationship between breaking stress, density, and length. Here is a step-by-step solution:

### Step 1: Understand the given values
- Breaking stress of aluminum, \( \sigma = 7.5 \times 10^7 \, \text{N/m}^2 \)
- Density of aluminum, \( \rho = 2.7 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) (standard value)

### Step 2: Relate mass, volume, and density
The mass \( m \) of the wire can be expressed in terms of its volume \( V \) and density \( \rho \):
\[ m = V \cdot \rho \]
Since the volume \( V \) of the wire can be expressed as the product of its cross-sectional area \( A \) and its length \( L \):
\[ V = A \cdot L \]
Thus, we can rewrite the mass as:
\[ m = A \cdot L \cdot \rho \]

### Step 3: Relate force to mass
The force \( F \) acting on the wire due to its weight is given by:
\[ F = m \cdot g = A \cdot L \cdot \rho \cdot g \]

### Step 4: Relate force to breaking stress
Breaking stress \( \sigma \) is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the expression for \( F \):
\[ \sigma = \frac{A \cdot L \cdot \rho \cdot g}{A} \]
This simplifies to:
\[ \sigma = L \cdot \rho \cdot g \]

### Step 5: Solve for the length \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{\sigma}{\rho \cdot g} \]

### Step 6: Substitute the known values
Now we can substitute the known values into the equation:
\[ L = \frac{7.5 \times 10^7 \, \text{N/m}^2}{(2.7 \times 10^3 \, \text{kg/m}^3) \cdot (9.81 \, \text{m/s}^2)} \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ \rho \cdot g = (2.7 \times 10^3) \cdot (9.81) \approx 26467.7 \, \text{kg/(m}^2\text{s}^2\text{)} \]

### Step 8: Calculate \( L \)
Now substitute this value back into the equation for \( L \):
\[ L = \frac{7.5 \times 10^7}{26467.7} \approx 2831.5 \, \text{m} \]

### Step 9: Round the answer
Rounding the answer gives:
\[ L \approx 2.83 \times 10^3 \, \text{m} \]

### Final Answer
The greatest length of aluminum wire that can hang vertically without breaking is approximately:
\[ \boxed{2.83 \times 10^3 \, \text{m}} \]