Question

A $100\, m$ long wire having cross-sectional area $625 \times 10^{-4} \, m ^2$ and Young's modulus is $10^{10} Nm ^{-2}$ is subjected to a load of $250 \, N$, then the elongation in the wire will be:

• A. $6.25 \times 10^{-3} m$
• B. $4 \times 10^{-3} m$
• C. $4 \times 10^{-4} m$
• D. $6.25 \times 10^{-6} m$

Answer 1

The Correct Option is B

To find the elongation in the wire, we can use Hooke's law, which relates the elongation (or extension) of a material to the applied force, its Young's modulus, original length, and cross-sectional area. The formula for elongation \(\Delta L\) is given by:

\(\Delta L = \frac{F \times L}{A \times Y}\)

Where:

• \(F\) is the applied force,
• \(L\) is the original length of the wire,
• \(A\) is the cross-sectional area,
• \(Y\) is the Young's modulus of the material.

Given:

• \(F = 250 \, N\)
• \(L = 100 \, m\)
• \(A = 625 \times 10^{-4} \, m^2\)
• \(Y = 10^{10} \, Nm^{-2}\)

Substitute the given values into the formula:

\(\Delta L = \frac{250 \times 100}{625 \times 10^{-4} \times 10^{10}}\)

First, calculate the denominator:

\(625 \times 10^{-4} \times 10^{10} = 625 \times 10^{6} = 6.25 \times 10^{8}\)

Now, substitute into the equation:

\(\Delta L = \frac{250 \times 100}{6.25 \times 10^{8}}\)

Calculate the numerator:

\(250 \times 100 = 25000\)

Therefore:

\(\Delta L = \frac{25000}{6.25 \times 10^{8}}\)

\(\Delta L = 4 \times 10^{-3} \, m\)

Thus, the elongation in the wire is \(4 \times 10^{-3} \, m\).

Therefore, the correct answer is: \(4 \times 10^{-3} \, m\).