A lift is tied with thick iron wire and its mass is `1000kg`. If the maximum acceleration of the lift is `1.2ms^(2)` and the maximum stress of the wire is `1.4xx10^(8) Nm^(2)` what should be the minimum diameter of the wire?

Options

A. `10^(-2)m`
B. `10^(-4)m`
C. `10^(-6)m`
D. `0.5xx10^(-2)m`

Explanation

To solve the problem, we need to find the minimum diameter of the iron wire that can support a lift with a mass of 1000 kg, given the maximum acceleration and maximum stress. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the lift (m) = 1000 kg
- Maximum acceleration (a) = 1.2 m/s²
- Maximum stress (σ) = 1.4 × 10⁸ N/m²
- Acceleration due to gravity (g) = 9.8 m/s² (approximately)

### Step 2: Calculate the total force acting on the wire
The total force (F) acting on the wire when the lift is accelerating upwards can be calculated using the formula:
\[ F = m(g + a) \]
Substituting the known values:
\[ F = 1000 \, \text{kg} \times (9.8 \, \text{m/s}² + 1.2 \, \text{m/s}²) \]
\[ F = 1000 \, \text{kg} \times 11 \, \text{m/s}² \]
\[ F = 11000 \, \text{N} \]

### Step 3: Relate stress to force and area
Stress (σ) is defined as the force (F) per unit area (A):
\[ \sigma = \frac{F}{A} \]
The area (A) of the wire can be expressed in terms of its radius (r):
\[ A = \pi r^2 \]
Thus, we can rewrite the stress equation as:
\[ \sigma = \frac{F}{\pi r^2} \]

### Step 4: Rearrange the equation to find r
From the stress equation, we can solve for the radius (r):
\[ r^2 = \frac{F}{\pi \sigma} \]
Substituting the values of F and σ:
\[ r^2 = \frac{11000 \, \text{N}}{\pi \times 1.4 \times 10^8 \, \text{N/m}²} \]

### Step 5: Calculate r²
Calculating the denominator:
\[ \pi \times 1.4 \times 10^8 \approx 4.398 \times 10^8 \, \text{N/m}² \]
Now substituting this into the equation for r²:
\[ r^2 = \frac{11000}{4.398 \times 10^8} \]
\[ r^2 \approx 2.5 \times 10^{-5} \, \text{m}^2 \]

### Step 6: Calculate r
Taking the square root to find r:
\[ r \approx \sqrt{2.5 \times 10^{-5}} \]
\[ r \approx 0.005 \, \text{m} \]

### Step 7: Calculate the diameter
The diameter (d) is twice the radius:
\[ d = 2r \]
\[ d \approx 2 \times 0.005 \, \text{m} = 0.01 \, \text{m} \]
Thus, the minimum diameter of the wire is:
\[ d = 10 \, \text{mm} \]

### Final Answer
The minimum diameter of the wire should be **10 mm**.