A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by `l`. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant. Find the ratio of the change in length of first wire versus second wire?

Options

A. `1/3`
B. `3/2`
C. `1/4`
D. `1/2`

Answer

To solve the problem, we need to find the ratio of the change in length of the first wire to the change in length of the second wire when the radius and applied force are halved for the second wire. We will use the formula for Young's modulus and the relationship between stress, strain, and the geometry of the wires.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) = applied force,
- \( A \) = cross-sectional area,
- \( \Delta L \) = change in length,
- \( L \) = original length.

2. **Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]

3. **Change in Length for the First Wire**:
For the first wire, let:
- \( F_1 = F \) (the original force),
- \( r_1 = r \) (the original radius),
- \( \Delta L_1 \) = change in length of the first wire.

The change in length can be expressed as:
\[ \Delta L_1 = \frac{F_1 \cdot L}{Y \cdot A_1} = \frac{F \cdot L}{Y \cdot \pi r^2} \]

4. **Change in Length for the Second Wire**:
For the second wire, we have:
- \( F_2 = \frac{F}{2} \) (the force is halved),
- \( r_2 = \frac{r}{2} \) (the radius is halved),
- \( A_2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} \).

The change in length for the second wire is:
\[ \Delta L_2 = \frac{F_2 \cdot L}{Y \cdot A_2} = \frac{\frac{F}{2} \cdot L}{Y \cdot \frac{\pi r^2}{4}} = \frac{F \cdot L}{2Y \cdot \frac{\pi r^2}{4}} = \frac{F \cdot L \cdot 4}{2Y \cdot \pi r^2} = \frac{2F \cdot L}{Y \cdot \pi r^2} \]

5. **Finding the Ratio of Changes in Length**:
Now, we find the ratio of the changes in length:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{F \cdot L}{Y \cdot \pi r^2}}{\frac{2F \cdot L}{Y \cdot \pi r^2}} = \frac{1}{2} \]

### Final Answer:
The ratio of the change in length of the first wire to the change in length of the second wire is:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{1}{2} \]