A 100 m long wire having cross sectional are `6.25 times10^(-4)M^(2)` and young` modulus is `10^(10)Nm^-2` is subjected to a load of 250N,then the elongation inthe wire will be

Options

A. `6.25 times10^(-3)`
B. `4times10^(-3)m `
C. `4 times 10^(-4)m`
D. `6.25 times 10^(-6)m`

Answer

To find the elongation in the wire, we can use the formula related to Young's modulus. The Young's modulus (Y) is defined as:

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

Where:
- \( F \) = force applied (in Newtons)
- \( A \) = cross-sectional area of the wire (in \( m^2 \))
- \( \Delta L \) = change in length (elongation) of the wire (in meters)
- \( L \) = original length of the wire (in meters)

Rearranging the formula to solve for \( \Delta L \):

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

Now, let's plug in the values given in the question:

- \( F = 250 \, N \)
- \( L = 100 \, m \)
- \( A = 6.25 \times 10^{-4} \, m^2 \)
- \( Y = 10^{10} \, N/m^2 \)

Now substituting the values into the formula:

\[ \Delta L = \frac{250 \, N \cdot 100 \, m}{6.25 \times 10^{-4} \, m^2 \cdot 10^{10} \, N/m^2} \]

Calculating the numerator:

\[ 250 \cdot 100 = 25000 \]

Now calculating the denominator:

\[ 6.25 \times 10^{-4} \cdot 10^{10} = 6.25 \times 10^{6} \]

Now substituting these values back into the equation for \( \Delta L \):

\[ \Delta L = \frac{25000}{6.25 \times 10^{6}} \]

Now simplifying:

\[ \Delta L = \frac{25000}{6.25} \times 10^{-6} \]

Calculating \( \frac{25000}{6.25} \):

\[ \frac{25000}{6.25} = 4000 \]

Thus,

\[ \Delta L = 4000 \times 10^{-6} \]

This can be written as:

\[ \Delta L = 4 \times 10^{-3} \, m \]

So, the elongation in the wire is:

\[ \Delta L = 4 \, mm \]

### Final Answer:
The elongation in the wire is \( 4 \, mm \).