A `20kg `load is suspended by a wire of cross section `0.4mm^(2)`. The stress produced in `N//m^(2)` is

Options

A. `4.9xx10^(-6)`
B. `4.9xx10^(8)`
C. `49xx10^(8)`
D. `2.45xx10^(-6)`

Explanation

To find the stress produced in the wire when a 20 kg load is suspended, we can follow these steps:

### Step 1: Calculate the Force (F)
The force exerted by the load can be calculated using the formula:
\[ F = mg \]
where:
- \( m = 20 \, \text{kg} \) (mass of the load)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the force:
\[ F = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 2: Convert the Cross-Sectional Area (A)
The cross-sectional area is given as \( 0.4 \, \text{mm}^2 \). We need to convert this to square meters:
\[ A = 0.4 \, \text{mm}^2 = 0.4 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Stress (σ)
Stress is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]

Substituting the values we have:
\[ \sigma = \frac{196 \, \text{N}}{0.4 \times 10^{-6} \, \text{m}^2} \]

### Step 4: Perform the Calculation
Calculating the stress:
\[ \sigma = \frac{196}{0.4 \times 10^{-6}} = \frac{196}{0.4} \times 10^{6} \]
\[ \sigma = 490 \times 10^{6} \, \text{N/m}^2 \]
\[ \sigma = 4.9 \times 10^{8} \, \text{N/m}^2 \]

### Final Answer
The stress produced in the wire is:
\[ \sigma = 4.9 \times 10^{8} \, \text{N/m}^2 \]

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