A copper wire of negligible mass, `1 m` length and cross-sectional area `10^(-6) m^(2)` is kept on a smooth horizontal table with one end fixed. A ball of mass `1 kg` is attached to the other end. The wire and the ball are rotating with an angular velocity of `20 rad//s`. If the elongation in the wire is `10^(-3) m`.
a. Find the Young's modulus of the wire (in terms of `xx 10^(11) N//m^(2)`).
b. If for the same wire as stated above, the angular velocity is increased to `100 rad//s` and the wire breaks down, find the breaking stress (in terms of `xx 10^(10) N//m^(2)`).

Explanation

To solve the problem step by step, we will address both parts (a) and (b) of the question.

### Part (a): Finding Young's Modulus

1. **Identify the given values:**
- Length of the wire, \( L = 1 \, \text{m} \)
- Cross-sectional area, \( A = 10^{-6} \, \text{m}^2 \)
- Mass of the ball, \( m = 1 \, \text{kg} \)
- Angular velocity, \( \omega = 20 \, \text{rad/s} \)
- Elongation in the wire, \( \Delta L = 10^{-3} \, \text{m} \)

2. **Calculate the centrifugal force acting on the ball:**
The centrifugal force \( F \) acting on the ball when it is rotating is given by:
\[ F = m \omega^2 L \]
Substituting the values:
\[ F = 1 \times (20)^2 \times 1 = 400 \, \text{N} \]

3. **Use the formula for Young's modulus:**
The formula for Young's modulus \( Y \) is:
\[ Y = \frac{F L}{\Delta L A} \]
Substituting the values we have:
\[ Y = \frac{400 \times 1}{10^{-3} \times 10^{-6}} = \frac{400}{10^{-9}} = 400 \times 10^{9} \, \text{N/m}^2 \]

4. **Convert to the required format:**
We need to express this in terms of \( xx \times 10^{11} \):
\[ Y = 4 \times 10^{11} \, \text{N/m}^2 \]

### Answer for Part (a):
The Young's modulus of the wire is \( 4 \times 10^{11} \, \text{N/m}^2 \).

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### Part (b): Finding the Breaking Stress

1. **Identify the new angular velocity:**
- New angular velocity, \( \omega = 100 \, \text{rad/s} \)

2. **Calculate the new centrifugal force at breaking point:**
Using the same formula for force:
\[ F = m \omega^2 L \]
Substituting the new values:
\[ F = 1 \times (100)^2 \times 1 = 10000 \, \text{N} \]

3. **Calculate the breaking stress:**
The breaking stress \( \sigma \) is given by:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{10000}{10^{-6}} = 10^{10} \, \text{N/m}^2 \]

### Answer for Part (b):
The breaking stress of the wire is \( 1 \times 10^{10} \, \text{N/m}^2 \).

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