A wire is loaded by 6 kg at its one end, the increase in length is 12 mm. If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be

Options

A. 6 mm
B. 3 mm
C. 24 mm
D. 48 mm

Explanation

To solve the problem, we will use the formula for the extension of a wire under a load, which is given by:

\[ \Delta L = \frac{F \cdot L_0}{A \cdot Y} \]

Where:
- \(\Delta L\) = increase in length (extension)
- \(F\) = force applied (weight of the load)
- \(L_0\) = original length of the wire
- \(A\) = cross-sectional area of the wire
- \(Y\) = Young's modulus of the material

### Step 1: Understand the initial conditions
Given:
- Load \(F = 6 \, \text{kg}\) (which can be converted to Newtons by multiplying by \(g \approx 9.81 \, \text{m/s}^2\))
- Increase in length \(\Delta L = 12 \, \text{mm} = 0.012 \, \text{m}\)
- The radius of the wire is \(r\).

### Step 2: Calculate the initial cross-sectional area
The cross-sectional area \(A\) of the wire is given by:

\[ A = \pi r^2 \]

### Step 3: Determine the new conditions after doubling the radius
If the radius is doubled, the new radius \(r' = 2r\). The new cross-sectional area \(A'\) becomes:

\[ A' = \pi (2r)^2 = \pi (4r^2) = 4A \]

### Step 4: Analyze the effect on the increase in length
Since the force \(F\), original length \(L_0\), and Young's modulus \(Y\) remain unchanged, we can see how the increase in length will change with the new area.

Using the extension formula again for the new area:

\[ \Delta L' = \frac{F \cdot L_0}{A' \cdot Y} \]

Substituting \(A' = 4A\):

\[ \Delta L' = \frac{F \cdot L_0}{4A \cdot Y} \]

### Step 5: Relate the new extension to the old extension
From the original extension formula, we have:

\[ \Delta L = \frac{F \cdot L_0}{A \cdot Y} \]

Now, substituting this into the equation for \(\Delta L'\):

\[ \Delta L' = \frac{1}{4} \cdot \Delta L \]

### Step 6: Calculate the new increase in length
Since \(\Delta L = 12 \, \text{mm}\):

\[ \Delta L' = \frac{1}{4} \cdot 12 \, \text{mm} = 3 \, \text{mm} \]

### Final Answer
The increase in length when the radius of the wire is doubled is **3 mm**.

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