A copper wire of negligible mass, length l and corss-section A is kept on a smooth horizontal table with one end fixed. A ball of mass m is attached to the other end. The wire and the ball are rotating with an angular velocity `omega`. If elongation in the wire is `Deltal`, obtain the expression for the Young's modulus.
`(mL^(2)omega)/(ADeltaL)`
`(m omega^(2)L^(2))/(A DeltaL)`
`(momega^(2)L)/(A DeltaL)`
`(m omegaL)/(A DeltaL)`

Options

A. `(mL^(2)omega)/(ADeltaL)`
B. `(m omega^(2)L^(2))/(A DeltaL)`
C. `(momega^(2)L)/(A DeltaL)`
D. `(m omegaL)/(A DeltaL)`

Answer

To derive the expression for Young's modulus for the given scenario, we will follow these steps:

### Step 1: Identify the forces acting on the ball
The ball of mass \( m \) is rotating in a circular path with radius \( L \). The centripetal force required to keep the ball in circular motion is given by:
\[ F_c = m \omega^2 L \]
where \( \omega \) is the angular velocity.

### Step 2: Relate the tension in the wire to the centripetal force
The tension \( T \) in the wire must provide the necessary centripetal force to keep the ball moving in a circle. Therefore, we can write:
\[ T = F_c = m \omega^2 L \]

### Step 3: Define stress and strain
Young's modulus \( Y \) is defined as the ratio of stress to strain.

- **Stress** is given by:
\[ \text{Stress} = \frac{T}{A} \]
where \( A \) is the cross-sectional area of the wire.

- **Strain** is defined as:
\[ \text{Strain} = \frac{\Delta L}{L} \]
where \( \Delta L \) is the elongation of the wire and \( L \) is the original length of the wire.

### Step 4: Substitute the expressions for stress and strain into the Young's modulus formula
Using the definitions of stress and strain, we can write Young's modulus as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{T}{A}}{\frac{\Delta L}{L}} = \frac{T \cdot L}{A \cdot \Delta L} \]

### Step 5: Substitute the expression for tension into the Young's modulus formula
Now, we substitute \( T = m \omega^2 L \) into the Young's modulus equation:
\[ Y = \frac{(m \omega^2 L) \cdot L}{A \cdot \Delta L} \]
This simplifies to:
\[ Y = \frac{m \omega^2 L^2}{A \Delta L} \]

### Conclusion
Thus, the expression for Young's modulus \( Y \) is:
\[ Y = \frac{m \omega^2 L^2}{A \Delta L} \]

### Final Answer
The correct option is:
\[ \frac{m \omega^2 L^2}{A \Delta L} \]

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