When a certain weight is suspended from a long uniform wire, its length increases by `1 cm`. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increases in length will be

Options

A. 0.5 cm
B. 2 cm
C. 4 cm
D. 8 cm

Explanation

To solve the problem, we will use the relationship between the increase in length of a wire (ΔL) and its physical properties, specifically Young's modulus (Y), the force (F) applied, the original length (L), and the cross-sectional area (A) of the wire.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The first wire has an increase in length (ΔL1) of 1 cm when a certain weight (force F) is applied.
- The second wire has the same length and material but has a diameter that is half of the first wire.

2. **Recall the Formula for Deformation:**
The formula for the increase in length (ΔL) of a wire under a tensile force is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F \) = force applied,
- \( L \) = original length of the wire,
- \( A \) = cross-sectional area of the wire,
- \( Y \) = Young's modulus of the material.

3. **Determine the Cross-Sectional Area:**
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire. If the diameter is halved, the radius of the second wire (r2) is:
\[ r_2 = \frac{r_1}{2} \]
Therefore, the area of the second wire becomes:
\[ A_2 = \pi \left(\frac{r_1}{2}\right)^2 = \pi \frac{r_1^2}{4} = \frac{A_1}{4} \]

4. **Set Up the Ratio of Increases in Length:**
Since the same force, length, and Young's modulus are applied to both wires, we can set up the ratio of the increases in length:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{A_2} \]
Substituting the area of the second wire:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{\frac{A_1}{4}} = 4 \]

5. **Calculate the Increase in Length for the Second Wire:**
Since ΔL1 = 1 cm, we can find ΔL2:
\[ \Delta L_2 = 4 \cdot \Delta L_1 = 4 \cdot 1 \text{ cm} = 4 \text{ cm} \]

6. **Conclusion:**
The increase in length of the second wire (ΔL2) is 4 cm.

### Final Answer:
The increase in length will be **4 cm**.

---