When a weight of `5 kg` is suspended from a copper wire of length `30 m` and diameter `0.5 mm`, the length of the wire increases by `2.4 cm`. If the diameter is doubled, the extension produced is

Options

A. `1.2cm`
B. `0.6`
C. `0.3cm`
D. `0.15cm`

Explanation

To solve the problem step by step, we will use the relationship between the extension of a wire and its physical properties, specifically Young's modulus.

### Step 1: Understand the initial conditions
We have a copper wire with:
- Length (L) = 30 m = 3000 cm (since we will convert everything to cm for consistency)
- Diameter (d) = 0.5 mm = 0.05 cm
- Weight (W) = 5 kg

### Step 2: Calculate the initial area of the wire
The area (A) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi \left(\frac{d}{2}\right)^2 \]
Substituting the diameter:
\[ A = \pi \left(\frac{0.05}{2}\right)^2 = \pi \left(0.025\right)^2 = \pi \times 0.000625 \approx 0.0019635 \, \text{cm}^2 \]

### Step 3: Calculate the force applied
The force (F) applied due to the weight is given by:
\[ F = W \times g \]
Where \( g \approx 9.81 \, \text{m/s}^2 \). Thus:
\[ F = 5 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 49.05 \, \text{N} \]

### Step 4: Use the formula for extension
The extension (\( \Delta L \)) of the wire is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Where \( Y \) is Young's modulus for copper (a constant).

### Step 5: Analyze the effect of doubling the diameter
When the diameter is doubled, the new diameter \( d' \) becomes:
\[ d' = 2 \times 0.05 \, \text{cm} = 0.1 \, \text{cm} \]
The new area \( A' \) becomes:
\[ A' = \pi \left(\frac{d'}{2}\right)^2 = \pi \left(\frac{0.1}{2}\right)^2 = \pi \left(0.05\right)^2 = \pi \times 0.0025 \approx 0.007854 \, \text{cm}^2 \]

### Step 6: Calculate the new extension
Since the force and length remain constant, we can relate the extensions:
\[ \Delta L' = \frac{F \cdot L}{A' \cdot Y} \]
Using the relationship between the areas:
\[ \Delta L' = \Delta L \cdot \frac{A}{A'} \]
Substituting the values:
\[ \Delta L' = 2.4 \, \text{cm} \cdot \frac{0.0019635}{0.007854} \]
Calculating this gives:
\[ \Delta L' = 2.4 \, \text{cm} \cdot \frac{0.0019635}{0.007854} \approx 2.4 \, \text{cm} \cdot 0.25 = 0.6 \, \text{cm} \]

### Final Answer
The extension produced when the diameter is doubled is \( 0.6 \, \text{cm} \).

---