A body of mass 1 kg is attached to one end of a wire and rotated in a horizontal circle of diameter 40 cm with a constant speed of 2m/s. What is the area of cross-section of the wire if the stress developed in the wire is `5 xx 10^(6) N//m^(2)` ?

Options

A. `2m m^(2)`
B. `3 m m^(2)`
C. `4m m^(2)`
D. `5m m^(2)`

Explanation

To solve the problem step by step, we will follow these calculations:

### Given Data:
- Mass of the body, \( m = 1 \, \text{kg} \)
- Diameter of the circle, \( d = 40 \, \text{cm} = 0.40 \, \text{m} \)
- Radius of the circle, \( r = \frac{d}{2} = \frac{0.40}{2} = 0.20 \, \text{m} \)
- Speed of rotation, \( v = 2 \, \text{m/s} \)
- Stress developed in the wire, \( \sigma = 5 \times 10^6 \, \text{N/m}^2 \)

### Step 1: Calculate the Centripetal Force
The centripetal force \( F \) required to keep the mass moving in a circle is given by the formula:
\[ F = \frac{m v^2}{r} \]
Substituting the known values:
\[ F = \frac{1 \times (2)^2}{0.20} = \frac{1 \times 4}{0.20} = \frac{4}{0.20} = 20 \, \text{N} \]

### Step 2: Relate Stress to Force and Area
Stress \( \sigma \) is defined as the force \( F \) per unit area \( A \):
\[ \sigma = \frac{F}{A} \]
Rearranging this equation to solve for the area \( A \):
\[ A = \frac{F}{\sigma} \]

### Step 3: Calculate the Area of Cross-section
Now substituting the values of force \( F \) and stress \( \sigma \):
\[ A = \frac{20 \, \text{N}}{5 \times 10^6 \, \text{N/m}^2} = \frac{20}{5 \times 10^6} = \frac{20}{5} \times 10^{-6} = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Convert Area to mm²
To convert the area from m² to mm²:
\[ A = 4 \times 10^{-6} \, \text{m}^2 = 4 \times 10^{-6} \times (1000 \, \text{mm})^2 = 4 \times 10^{-6} \times 10^6 \, \text{mm}^2 = 4 \, \text{mm}^2 \]

### Final Answer:
The area of cross-section of the wire is \( 4 \, \text{mm}^2 \).

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