Question

A 100m long wire having cross-sectional area \( 6.25 \times 10^{-4} \, {m}^2 \) and Young's modulus is \( 10^{10} \, {Nm}^{-2} \) is subjected to a load of 250N, then the elongation in the wire will be:

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

Hint:

The elongation in a wire is directly proportional to the applied force and the original length, and inversely proportional to the cross-sectional area and Young's modulus.

Answer 1

The Correct Option is D

Step 1: Formula for elongation. The elongation \( \Delta L \) in the wire can be calculated using the formula: \[ \Delta L = \frac{F L}{A Y} \] where: \( F = 250 \, {N} \) is the applied force, \( L = 100 \, {m} \) is the length of the wire, \( A = 6.25 \times 10^{-4} \, {m}^2 \) is the cross-sectional area of the wire, \( Y = 10^{10} \, {Nm}^{-2} \) is Young's modulus. 
Step 2: Substituting the values. \[ \Delta L = \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} = \frac{25000}{6.25 \times 10^{6}} = 4 \times 10^{-3} \, {m} \] Thus, the elongation in the wire is \( 4 \times 10^{-3} \, {m} \).