Question

A force is applied to a steel wire ' $A$ ', rigidly clamped at one end As a result elongation in the wire is $02\, mm$ If same force is applied to another steel wire ' $B$ ' of double the length and a diameter $24$ times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)

• A. $2.77 \times 10^{-2} mm$
• B. $6.9 \times 10^{-2} mm$
• C. $6.06 \times 10^{-2} mm$
• D. $3.0 \times 10^{-2} mm$

Answer 1

The Correct Option is B

To determine the elongation in wire 'B', we apply the principle of elasticity. The elongation of a material is given by the formula:

\(\delta L = \frac{F \cdot L}{A \cdot E}\)

Where:

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

Given:

• Elongation in wire 'A': \(0.2 \, \text{mm} = \delta L_A\)
• Length of wire 'B': \(2L\) (double the length of wire 'A')
• Diameter of wire 'B': \(24d\) (24 times the diameter of wire 'A')

Consequently, the cross-sectional area for wire 'A' is:

\(A_A = \frac{\pi d^2}{4}\)

For wire 'B', the diameter is \(24d\), hence its area will be:

\(A_B = \frac{\pi (24d)^2}{4} = 576 \cdot \frac{\pi d^2}{4} = 576 \cdot A_A\)

Since the force \(F\) and material (Young's modulus, \(E\)) remain the same, we compare the elongations:

\(\frac{\delta L_B}{\delta L_A} = \frac{F(2L)}{576A_A \cdot E} \bigg/ \frac{FL}{A_A \cdot E} = \frac{2L}{576L} = \frac{1}{288}\)

The elongation in wire 'B' is:

\(\delta L_B = \delta L_A \times \frac{1}{288} = 0.2 \, \text{mm} \times \frac{1}{288} = 0.000694 \, \text{mm} \approx 6.9 \times 10^{-2} \, \text{mm}\)

Hence, the correct option is:
$6.9 \times 10^{-2} \, \text{mm}$