A steel rod has a radius of cross-sectional area is R= 10 mm and length L = 90 cm ls fixed at one end, a force `F = 3.14 * 10^4 N` stretches it along its length, the stress in the rod is equal to (take `pi` = 3.14)

Options

A. `10^6 N/m^2`
B. `3.14* 10^-5 N/m^2`
C. `10^4 N/m^2`
D. `3.14* 10^-4 N/m^2`

Answer

To find the stress in the steel rod, we will follow these steps:

### Step 1: Convert the radius from mm to meters
Given:
- Radius \( R = 10 \, \text{mm} \)

To convert mm to meters:
\[ R = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} = 0.01 \, \text{m} \]

### Step 2: Convert the length from cm to meters
Given:
- Length \( L = 90 \, \text{cm} \)

To convert cm to meters:
\[ L = 90 \, \text{cm} = 90 \times 10^{-2} \, \text{m} = 0.90 \, \text{m} \]

### Step 3: Calculate the cross-sectional area of the rod
The formula for the cross-sectional area \( A \) of a circle is:
\[ A = \pi R^2 \]
Substituting \( \pi = 3.14 \) and \( R = 0.01 \, \text{m} \):
\[ A = 3.14 \times (0.01)^2 = 3.14 \times 0.0001 = 3.14 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Calculate the stress in the rod
Stress \( \sigma \) is defined as the force \( F \) applied per unit area \( A \):
\[ \sigma = \frac{F}{A} \]
Given:
- Force \( F = 3.14 \times 10^4 \, \text{N} \)

Substituting the values:
\[ \sigma = \frac{3.14 \times 10^4}{3.14 \times 10^{-4}} \]

### Step 5: Simplify the expression
\[ \sigma = \frac{3.14}{3.14} \times \frac{10^4}{10^{-4}} = 1 \times 10^{4 + 4} = 1 \times 10^8 \, \text{N/m}^2 \]
Thus, the stress in the rod is:
\[ \sigma = 10^8 \, \text{N/m}^2 \]

### Final Answer
The stress in the rod is \( 10^8 \, \text{N/m}^2 \).

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