A weight of 1.0 kg is suspended form the lower end of a wire of cross- section ` 10 mm^(2)`. What is the magnitude and direction of stress produced in it ? Take g=`9.8 ms^(-2)`.

Answer

To solve the problem, we need to calculate the stress produced in the wire when a weight of 1.0 kg is suspended from it. Here are the steps to find the magnitude and direction of the stress:

### Step 1: Calculate the Weight of the Suspended Mass
The weight (W) of the mass can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 1.0 \, \text{kg} \) (mass)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ W = 1.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

### Step 2: Understand the Concept of Stress
Stress (\( \sigma \)) is defined as the force (F) applied per unit area (A). The formula for stress is:
\[ \sigma = \frac{F}{A} \]
In this case, the force is the weight of the mass, which acts downward.

### Step 3: Convert the Cross-Sectional Area to SI Units
The cross-sectional area is given as \( 10 \, \text{mm}^2 \). We need to convert this to square meters (m²):
\[ A = 10 \, \text{mm}^2 = 10 \times 10^{-6} \, \text{m}^2 = 1.0 \times 10^{-5} \, \text{m}^2 \]

### Step 4: Calculate the Stress
Now we can calculate the stress using the weight calculated in Step 1 and the area from Step 3:
\[ \sigma = \frac{W}{A} = \frac{9.8 \, \text{N}}{1.0 \times 10^{-5} \, \text{m}^2} \]
\[ \sigma = 9.8 \times 10^{5} \, \text{N/m}^2 \]

### Step 5: Determine the Direction of Stress
The direction of stress is determined by the direction of the force applied. Since the weight is acting downward, the tension in the wire will act upward. Therefore, the stress in the wire will also act in the upward direction.

### Final Answer
- **Magnitude of Stress**: \( 9.8 \times 10^{5} \, \text{N/m}^2 \)
- **Direction of Stress**: Upward

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