A force is applied to a steel wire `A`,rigidly clamped at one end.As a result elongation in the wire is `0.2mm`.If same force is applied to another steel wire `B` of double the length and a diameter `2.4` times that of the wire `A`,the elevation in the wire `B` will be (wires having uniform circular cross sections

Options

A. `2.77times10^(-2)mm`
B. `6.9times10^(-2)mm`
C. `6.06times10^(-2)mm`
D. `3.0times10^(-2)mm`

Answer

To solve the problem, we will use the formula for elongation (ΔL) of a wire under tension, which is given by:

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

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

### Step 1: Identify the parameters for wire A
Let:
- \( L_1 \) = length of wire A
- \( A_1 \) = cross-sectional area of wire A
- \( \Delta L_1 = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)

The elongation for wire A can be expressed as:
\[ \Delta L_1 = \frac{F \cdot L_1}{A_1 \cdot Y} \]

### Step 2: Identify the parameters for wire B
For wire B:
- The length \( L_2 = 2L_1 \) (double the length of wire A)
- The diameter \( d_2 = 2.4 d_1 \) (where \( d_1 \) is the diameter of wire A)
- The cross-sectional area \( A_2 \) can be calculated as:
\[ A_2 = \pi \left(\frac{d_2}{2}\right)^2 = \pi \left(\frac{2.4 d_1}{2}\right)^2 = \pi \left(1.2 d_1\right)^2 = 1.44 \pi \left(\frac{d_1}{2}\right)^2 = 1.44 A_1 \]

### Step 3: Write the elongation for wire B
The elongation for wire B can be expressed as:
\[ \Delta L_2 = \frac{F \cdot L_2}{A_2 \cdot Y} \]

Substituting \( L_2 \) and \( A_2 \):
\[ \Delta L_2 = \frac{F \cdot (2L_1)}{(1.44 A_1) \cdot Y} \]

### Step 4: Relate elongation of wire B to wire A
Now, we can relate the elongations:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{F \cdot (2L_1)}{(1.44 A_1) \cdot Y} \cdot \frac{A_1 \cdot Y}{F \cdot L_1} \]

This simplifies to:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{2L_1}{1.44 A_1} \cdot \frac{A_1}{L_1} = \frac{2}{1.44} \]

### Step 5: Calculate the elongation of wire B
Now substituting \( \Delta L_1 = 0.2 \, \text{mm} \):
\[ \Delta L_2 = \Delta L_1 \cdot \frac{2}{1.44} = 0.2 \cdot \frac{2}{1.44} \]
Calculating this gives:
\[ \Delta L_2 = 0.2 \cdot 1.3889 \approx 0.27778 \, \text{mm} \]

### Step 6: Final answer
Thus, the elongation in wire B is approximately:
\[ \Delta L_2 \approx 0.06944 \, \text{mm} = 6.944 \times 10^{-2} \, \text{mm} \]