A steel wire of 1.5 m long and of radius 1 mm is attached with a load 3 kg at one end the other end of the wire is fixed it is whirled in a vertical circle with a frequency 2 Hz. Find the elongation of the wire when the weight is at the lowest position `Y=2xx10^(11)N//m^(2)` and `(g=10m//s^(2))`

Options

A. `1.77xx10^(-3)m`
B. `7.17xx10^(-3)m`
C. `3.17xx10^(-7)m`
D. `1.37xx10^(-7)m`

Answer

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

### Step 1: Identify the Given Values
- Length of the wire, \( L = 1.5 \, \text{m} \)
- Radius of the wire, \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Mass of the load, \( m = 3 \, \text{kg} \)
- Frequency of rotation, \( f = 2 \, \text{Hz} \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Forces at the Lowest Point
At the lowest position, the forces acting on the wire are:
1. Gravitational force, \( F_g = mg \)
2. Centrifugal force, \( F_c = m \omega^2 L \)

Where \( \omega \) (angular velocity) can be calculated using the frequency:
\[ \omega = 2\pi f = 2\pi \times 2 = 4\pi \, \text{rad/s} \]

Now, we can calculate the gravitational force:
\[ F_g = mg = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \]

Next, calculate the centrifugal force:
\[ F_c = m \omega^2 L = 3 \, \text{kg} \times (4\pi)^2 \times 1.5 \, \text{m} \]
Calculating \( (4\pi)^2 \):
\[ (4\pi)^2 = 16\pi^2 \approx 157.9137 \]
Thus,
\[ F_c = 3 \times 157.9137 \times 1.5 \approx 710.61 \, \text{N} \]

### Step 3: Total Force Acting on the Wire
The total force \( F \) acting on the wire at the lowest point is:
\[ F = F_g + F_c = 30 \, \text{N} + 710.61 \, \text{N} \approx 740.61 \, \text{N} \]

### Step 4: Calculate the Elongation of the Wire
The elongation \( \Delta L \) of the wire can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]
Where \( A \) is the cross-sectional area of the wire:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

Now substituting the values into the elongation formula:
\[ \Delta L = \frac{740.61 \, \text{N} \cdot 1.5 \, \text{m}}{2 \times 10^{11} \, \text{N/m}^2 \cdot \pi \times 10^{-6} \, \text{m}^2} \]

Calculating the denominator:
\[ 2 \times 10^{11} \cdot \pi \times 10^{-6} \approx 6.2832 \times 10^{5} \, \text{N} \]

Now substituting back:
\[ \Delta L = \frac{740.61 \cdot 1.5}{6.2832 \times 10^{5}} \approx \frac{1110.915}{6.2832 \times 10^{5}} \approx 1.77 \times 10^{-3} \, \text{m} \]

### Final Answer
The elongation of the wire is approximately:
\[ \Delta L \approx 1.77 \times 10^{-3} \, \text{m} \text{ or } 1.77 \, \text{mm} \]

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