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 diameter, as described by Young's modulus.

### Step 1: Understand the given data
- Mass (m) = 5 kg
- Length of the wire (L) = 30 m = 3000 cm (since 1 m = 100 cm)
- Initial diameter (d1) = 0.5 mm = 0.05 cm (since 1 mm = 0.1 cm)
- Extension produced (ΔL1) = 2.4 cm

### Step 2: Calculate the force applied
The force (F) applied due to the weight can be calculated using:
\[ F = m \cdot g \]
Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).
\[ F = 5 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]

### Step 3: Relate extension to diameter
From the theory of elasticity, the extension (ΔL) of a wire is inversely proportional to the square of its diameter (d):
\[ \Delta L \propto \frac{1}{d^2} \]
This means:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{d_1^2}{d_2^2} \]

### Step 4: Determine the new diameter
If the diameter is doubled:
- New diameter (d2) = 2 * d1 = 2 * 0.5 mm = 1 mm = 0.1 cm

### Step 5: Calculate the ratio of extensions
Now, we can use the ratio of extensions:
- Initial diameter (d1) = 0.05 cm
- New diameter (d2) = 0.1 cm

Using the formula:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{(0.05)^2}{(0.1)^2} = \frac{0.0025}{0.01} = \frac{1}{4} \]

### Step 6: Calculate the new extension
Now we can find the new extension (ΔL2):
\[ \Delta L_2 = \frac{1}{4} \Delta L_1 \]
Substituting ΔL1 = 2.4 cm:
\[ \Delta L_2 = \frac{1}{4} \cdot 2.4 \, \text{cm} = 0.6 \, \text{cm} \]

### Final Answer
The extension produced when the diameter is doubled is **0.6 cm**.

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