To double the length of a iron wire having `0.5 cm^(2)` area of cross-section, the required force will be `(Y = 10^(12) "dyne/cm"^(2))`

Options

A. `1.0 xx 10^(-7) N`
B. `1.0 xx 10^(7) N`
C. `0.5 xx 10^(-7)N`
D. `0.5 xx 10^(12)` dyne

Answer

To find the force required to double the length of an iron wire with a cross-sectional area of \(0.5 \, \text{cm}^2\) and a Young's modulus \(Y = 10^{12} \, \text{dyne/cm}^2\), we can follow these steps:

### Step 1: Understand the problem
We need to double the length of the wire. If the original length is \(L\), the new length will be \(L_2 = 2L\). The change in length (\(\Delta L\)) can be expressed as:
\[ \Delta L = L_2 - L = 2L - L = L \]

### Step 2: Use the formula for Young's modulus
Young's modulus (\(Y\)) is defined as:
\[ Y = \frac{F \cdot L}{\Delta L \cdot A} \]
Where:
- \(F\) is the force applied,
- \(L\) is the original length,
- \(\Delta L\) is the change in length,
- \(A\) is the cross-sectional area.

### Step 3: Rearrange the formula to solve for force \(F\)
Rearranging the formula gives:
\[ F = Y \cdot \frac{\Delta L \cdot A}{L} \]

### Step 4: Substitute the known values
We know:
- \(Y = 10^{12} \, \text{dyne/cm}^2\)
- \(\Delta L = L\)
- \(A = 0.5 \, \text{cm}^2\)

Substituting these values into the equation:
\[ F = 10^{12} \cdot \frac{L \cdot 0.5}{L} \]

### Step 5: Simplify the expression
The \(L\) in the numerator and denominator cancels out:
\[ F = 10^{12} \cdot 0.5 \]
\[ F = 0.5 \times 10^{12} \, \text{dyne} \]

### Step 6: Final result
Thus, the required force to double the length of the iron wire is:
\[ F = 5 \times 10^{11} \, \text{dyne} \]