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

Explanation

To solve the problem, we will use the relationship between stress, strain, and Young's modulus. Here are the steps to find the increase in length of the second wire:

### Step 1: Understand the relationship
The increase in length (extension) of a wire under a load is given by the formula:

\[ \Delta L = \frac{F L}{A Y} \]

where:
- \( \Delta L \) = increase in length (extension)
- \( F \) = force (weight suspended)
- \( L \) = original length of the wire
- \( A \) = cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

### Step 2: Determine the area of the wires
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:

\[ A = \frac{\pi}{4} d^2 \]

### Step 3: Set up the equations for both wires
Let:
- For the first wire, diameter \( d_1 \) and extension \( \Delta L_1 = 1 \, \text{cm} \)
- For the second wire, diameter \( d_2 = \frac{d_1}{2} \)

The area for the first wire is:

\[ A_1 = \frac{\pi}{4} d_1^2 \]

The area for the second wire is:

\[ A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4} \left(\frac{d_1}{2}\right)^2 = \frac{\pi}{4} \cdot \frac{d_1^2}{4} = \frac{\pi}{16} d_1^2 \]

### Step 4: Relate the extensions of the two wires
Using the formula for extension, we can write:

\[ \Delta L_1 = \frac{F L}{A_1 Y} \]

\[ \Delta L_2 = \frac{F L}{A_2 Y} \]

### Step 5: Find the ratio of extensions
Taking the ratio of the extensions:

\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{A_2} \]

Substituting the areas:

\[ \frac{\Delta L_2}{\Delta L_1} = \frac{\frac{\pi}{4} d_1^2}{\frac{\pi}{16} d_1^2} = \frac{\frac{1}{4}}{\frac{1}{16}} = 4 \]

### Step 6: Calculate the extension in the second wire
Since \( \Delta L_1 = 1 \, \text{cm} \):

\[ \Delta L_2 = 4 \Delta L_1 = 4 \times 1 \, \text{cm} = 4 \, \text{cm} \]

### Final Answer
The increase in length of the second wire will be **4 cm**.

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