The extension of a wire by application of load is `0.3cm`. The extension in a wire of same material but of double the length and half the radius of cross section by the same load will be in `(cm)`

Options

A. `0.3`
B. `0.6`
C. `0.2`
D. `2.4`

Explanation

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

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

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

### Step 1: Understand the given parameters
We know:
- The extension of the first wire, \(\Delta L_1 = 0.3 \, \text{cm}\)
- The length of the second wire, \(L_2 = 2L_1\) (double the length)
- The radius of the second wire, \(r_2 = \frac{1}{2} r_1\) (half the radius)

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

\[ A = \pi r^2 \]

For the first wire:
\[ A_1 = \pi r_1^2 \]

For the second wire:
\[ A_2 = \pi \left(\frac{1}{2} r_1\right)^2 = \pi \frac{r_1^2}{4} = \frac{1}{4} A_1 \]

### Step 3: Set up the extension formula for both wires
Using the extension formula for both wires, we have:

For the first wire:
\[ \Delta L_1 = \frac{F \cdot L_1}{A_1 \cdot Y} \]

For the second wire:
\[ \Delta L_2 = \frac{F \cdot L_2}{A_2 \cdot Y} \]

### Step 4: Substitute the values
Substituting \(L_2 = 2L_1\) and \(A_2 = \frac{1}{4} A_1\) into the equation for \(\Delta L_2\):

\[ \Delta L_2 = \frac{F \cdot (2L_1)}{\left(\frac{1}{4} A_1\right) \cdot Y} \]

This simplifies to:

\[ \Delta L_2 = \frac{F \cdot 2L_1 \cdot 4}{A_1 \cdot Y} = \frac{8F \cdot L_1}{A_1 \cdot Y} \]

### Step 5: Relate \(\Delta L_2\) to \(\Delta L_1\)
From the first wire, we know:

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

Thus, we can express \(\Delta L_2\) in terms of \(\Delta L_1\):

\[ \Delta L_2 = 8 \Delta L_1 \]

### Step 6: Calculate the extension of the second wire
Now substituting \(\Delta L_1 = 0.3 \, \text{cm}\):

\[ \Delta L_2 = 8 \cdot 0.3 \, \text{cm} = 2.4 \, \text{cm} \]

### Final Answer
The extension in the second wire will be \(2.4 \, \text{cm}\).

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