A steel wire of cross-sectional area 4 cm2 has elastic limit of `2.2 xx10^(8) N//m^(2)`. The maximum upward acceleration that can be given to a 1000 kg elevator supported by this steel wire if the stress is to exceed one-fourth of the elastic limit is [Take, g = 10 `m//s^(2) ` ]

Options

A. 10 `m//s^(2)`
B. `9 m//s^(2)`
C. 11`m //s^(2)`
D. `12 m//s^(2)`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Given Data
- Cross-sectional area of the steel wire, \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Elastic limit of the steel wire, \( \sigma_{\text{max}} = 2.2 \times 10^8 \, \text{N/m}^2 \)
- Mass of the elevator, \( m = 1000 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate One-Fourth of the Elastic Limit
We need to find the maximum stress that can be applied without exceeding one-fourth of the elastic limit:
\[ \sigma_{\text{allowed}} = \frac{1}{4} \sigma_{\text{max}} = \frac{1}{4} \times 2.2 \times 10^8 = 0.55 \times 10^8 \, \text{N/m}^2 \]

### Step 3: Calculate the Maximum Tension in the Wire
Using the formula for stress, we can relate stress to tension and area:
\[ \sigma = \frac{T}{A} \implies T = \sigma \times A \]
Substituting the values:
\[ T = 0.55 \times 10^8 \times 4 \times 10^{-4} = 0.55 \times 4 \times 10^4 = 2.2 \times 10^4 \, \text{N} \]

### Step 4: Apply Newton's Second Law
The tension in the wire must overcome the weight of the elevator and provide the necessary upward acceleration:
\[ T - mg = ma \]
Where \( T \) is the tension, \( mg \) is the weight of the elevator, and \( a \) is the upward acceleration. Rearranging gives:
\[ T = mg + ma \]
Factoring out \( m \):
\[ T = m(g + a) \]

### Step 5: Substitute Known Values
Substituting \( T \) from Step 3:
\[ 2.2 \times 10^4 = 1000(10 + a) \]

### Step 6: Solve for \( a \)
Now, we can solve for \( a \):
\[ 2.2 \times 10^4 = 10000 + 1000a \]
\[ 2.2 \times 10^4 - 10000 = 1000a \]
\[ 1.2 \times 10^4 = 1000a \]
\[ a = \frac{1.2 \times 10^4}{1000} = 12 \, \text{m/s}^2 \]

### Conclusion
The maximum upward acceleration that can be given to the elevator is:
\[ \boxed{12 \, \text{m/s}^2} \]