Maximum stress that can be applied to a wire which supports on elevator is `sigma`. Mass of elevator is `m ` and it is moved upwards with an acceleration of `g//2`. Minimum diameter of wire ( Neglecting weight of wire ) must be

Options

A. `sqrt(2mg)/(pisigma)`
B. `sqrt(3mg)/(2pisigma)`
C. `sqrt(5mg)/(2pisigma)`
D. `sqrt((6mg)/(pisigma))`

Explanation

To find the minimum diameter of the wire supporting an elevator moving upwards with an acceleration of \( \frac{g}{2} \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Elevator
The elevator has a mass \( m \) and is moving upwards with an acceleration \( a = \frac{g}{2} \). The forces acting on the elevator are:
- The weight of the elevator acting downwards: \( W = mg \)
- The tension \( T \) in the wire acting upwards.

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force acting on the elevator can be expressed as:
\[ T - mg = ma \]
Substituting \( a = \frac{g}{2} \) into the equation gives:
\[ T - mg = m \left(\frac{g}{2}\right) \]

### Step 3: Solve for Tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = mg + m \left(\frac{g}{2}\right) \]
\[ T = mg + \frac{mg}{2} = \frac{2mg}{2} + \frac{mg}{2} = \frac{3mg}{2} \]

### Step 4: Relate Tension to Stress
The maximum stress \( \sigma \) that can be applied to the wire is defined as:
\[ \sigma = \frac{T}{A} \]
where \( A \) is the cross-sectional area of the wire. For a wire with diameter \( d \), the area \( A \) can be expressed as:
\[ A = \frac{\pi d^2}{4} \]
Substituting this into the stress equation gives:
\[ \sigma = \frac{T}{\frac{\pi d^2}{4}} = \frac{4T}{\pi d^2} \]

### Step 5: Substitute for Tension \( T \)
Now substitute \( T = \frac{3mg}{2} \) into the stress equation:
\[ \sigma = \frac{4 \cdot \frac{3mg}{2}}{\pi d^2} \]
This simplifies to:
\[ \sigma = \frac{6mg}{\pi d^2} \]

### Step 6: Solve for Diameter \( d \)
Rearranging the equation to solve for \( d^2 \):
\[ d^2 = \frac{6mg}{\pi \sigma} \]
Taking the square root gives:
\[ d = \sqrt{\frac{6mg}{\pi \sigma}} \]

### Final Result
Thus, the minimum diameter \( d \) of the wire is:
\[ d = \sqrt{\frac{6mg}{\pi \sigma}} \]

---