When a certain weight is suspended from a long uniform wire, its length increases by 1 cm. If the same weight is suspended from the other wire of the same material and length but having a diameter half of the first one then the increase 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 concept of elasticity and the formula for the elongation of a wire under a load. The elongation (increase in length) of a wire is given by the formula:

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

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

### Step 1: Understand the relationship between diameter and area
The cross-sectional area \(A\) of a wire with diameter \(d\) is given by:

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

If the diameter is halved, the new diameter \(d' = \frac{d}{2}\). The new area \(A'\) will be:

\[ A' = \frac{\pi (d')^2}{4} = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi \frac{d^2}{4}}{4} = \frac{\pi d^2}{16} = \frac{A}{4} \]

### Step 2: Relate the elongation of the two wires
Let’s denote:
- The increase in length of the first wire as \(\Delta L_1 = 1 \text{ cm}\)
- The increase in length of the second wire as \(\Delta L_2\)

Using the elongation formula for both wires:

For the first wire:
\[ \Delta L_1 = \frac{F L}{A Y} \]

For the second wire:
\[ \Delta L_2 = \frac{F L}{A' Y} \]

### Step 3: Substitute the area of the second wire
Since \(A' = \frac{A}{4}\), we can substitute this into the equation for \(\Delta L_2\):

\[ \Delta L_2 = \frac{F L}{\left(\frac{A}{4}\right) Y} = \frac{4 F L}{A Y} \]

### Step 4: Relate \(\Delta L_2\) to \(\Delta L_1\)
Now, we can relate \(\Delta L_2\) to \(\Delta L_1\):

\[ \Delta L_2 = 4 \Delta L_1 \]

Since we know that \(\Delta L_1 = 1 \text{ cm}\):

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

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

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