Question

A steel wire of length 3.2 m (Y_s = 2.0 × 10¹¹ Nm^-2) and a copper wire of length 4.4 m (Y_c = 1.1 × 10¹¹ Nm^-2), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be:
\((Given: π = \frac{22}{7})\)

• A. 360
• B. 160
• C. 1080
• D. 154

Answer 1

The Correct Option is D

The net elongation of a wire system consisting of a steel wire and a copper wire occurs due to the applied load and depends on the Young's modulus and length of both wires. Let's solve this step by step.

Step 1: Determine the cross-sectional area of both wires.

Given the radius \( r = 1.4 \, \text{mm} = 1.4 \times 10^{-3} \, \text{m} \).

The cross-sectional area \( A \) is given by the formula: \(A = \pi r^2\).

Substitute the given values: \(A = \frac{22}{7} \times (1.4 \times 10^{-3})^2 \, \text{m}^2\).

Calculating this gives: \(A = \frac{22}{7} \times 1.96 \times 10^{-6} = 6.16 \times 10^{-6} \, \text{m}^2\).

Step 2: Calculate the elongation of each wire separately.

Elongation of steel wire (\( \Delta L_s \)):

The elongation \((\Delta L)\) due to a load is given by: \(\Delta L = \frac{F \cdot L}{A \cdot Y}\), where \( F \) is the force (load), \( L \) is the original length, \( A \) is the area, and \( Y \) is Young's modulus.

For the steel wire, \(\Delta L_s = \frac{F \cdot 3.2}{6.16 \times 10^{-6} \times 2.0 \times 10^{11}}\).

This simplifies to: \(\Delta L_s = \frac{F \cdot 3.2}{1.232 \times 10^6}\).

Elongation of copper wire (\( \Delta L_c \)):

For the copper wire, \(\Delta L_c = \frac{F \cdot 4.4}{6.16 \times 10^{-6} \times 1.1 \times 10^{11}}\).

This simplifies to: \(\Delta L_c = \frac{F \cdot 4.4}{6.776 \times 10^5}\).

Step 3: Calculate the total elongation and solve for the force \( F \).

The net elongation is given as \( 1.4 \, \text{mm} = 1.4 \times 10^{-3} \, \text{m} \).

Therefore: \(\Delta L_s + \Delta L_c = 1.4 \times 10^{-3}\).

Substituting the expressions for \( \Delta L_s \) and \( \Delta L_c \):

\(\frac{F \cdot 3.2}{1.232 \times 10^6} + \frac{F \cdot 4.4}{6.776 \times 10^5} = 1.4 \times 10^{-3}\).

This simplifies to: \(F \cdot ( \frac{3.2}{1.232 \times 10^6} + \frac{4.4}{6.776 \times 10^5}) = 1.4 \times 10^{-3}\).

Calculate the combined fraction: \(\frac{3.2}{1.232 \times 10^6} + \frac{4.4}{6.776 \times 10^5} = \frac{3.2 \times 6.776 + 4.4 \times 1.232}{1.232 \times 10^6 \times 6.776 \times 10^5} \approx 9.09 \times 10^{-9} \, \text{m/N}\).

Thus: \(F \cdot 9.09 \times 10^{-9} = 1.4 \times 10^{-3}\).

Solving for \( F \): \(F = \frac{1.4 \times 10^{-3}}{9.09 \times 10^{-9}} \approx 153.91 \, \text{N}\).

Therefore, the applied load is approximately 154 N.

Conclusion: The correct answer is 154 N.