Question

A steel wire of length 3 m and a copper wire of length 2.2 m are connected end to end. When the combination is stretched by a force, the net elongation is 1.05 mm. If the area of cross-section of each wire is 6 mm², then the load applied is: (Young’s moduli of steel and copper are respectively \(2 \times 10^{11} \, \text{Nm}^{-2}\) and \(1.1 \times 10^{11} \, \text{Nm}^{-2}\)).

• A. \(180 N\)
• B. \(90 N\)
• C. \(135 N\)
• D. \(120 N\)

Hint:

The total elongation in a series combination of wires is the sum of the elongations of individual wires under the same force.

Answer 1

The Correct Option is A

To determine the load applied to the combination of steel and copper wires, we need to use the concept of Young's modulus, which relates stress and strain in a material. 1. Given Data: - Length of steel wire, \( L_s = 3 \, \text{m} \) - Length of copper wire, \( L_c = 2.2 \, \text{m} \) - Total elongation, \( \Delta L = 1.05 \, \text{mm} = 1.05 \times 10^{-3} \, \text{m} \) - Cross-sectional area of each wire, \( A = 6 \, \text{mm}^2 = 6 \times 10^{-6} \, \text{m}^2 \) - Young’s modulus of steel, \( Y_s = 2 \times 10^{11} \, \text{Nm}^{-2} \) - Young’s modulus of copper, \( Y_c = 1.1 \times 10^{11} \, \text{Nm}^{-2} \) 2. Calculate the Elongation for Each Wire: - The total elongation is the sum of the elongations of the steel and copper wires: \[ \Delta L = \Delta L_s + \Delta L_c \] - Using Young's modulus formula \( \Delta L = \frac{F L}{A Y} \), we can express the elongations as: \[ \Delta L_s = \frac{F L_s}{A Y_s} \] \[ \Delta L_c = \frac{F L_c}{A Y_c} \] - Therefore: \[ \Delta L = \frac{F L_s}{A Y_s} + \frac{F L_c}{A Y_c} \] \[ 1.05 \times 10^{-3} = F \left( \frac{L_s}{A Y_s} + \frac{L_c}{A Y_c} \right) \] 3. Substitute the Given Values: \[ 1.05 \times 10^{-3} = F \left( \frac{3}{6 \times 10^{-6} \times 2 \times 10^{11}} + \frac{2.2}{6 \times 10^{-6} \times 1.1 \times 10^{11}} \right) \] \[ 1.05 \times 10^{-3} = F \left( \frac{3}{1.2 \times 10^{6}} + \frac{2.2}{6.6 \times 10^{5}} \right) \] \[ 1.05 \times 10^{-3} = F \left( 2.5 \times 10^{-6} + 3.333 \times 10^{-6} \right) \] \[ 1.05 \times 10^{-3} = F \left( 5.833 \times 10^{-6} \right) \] 4. Solve for the Force \( F \): \[ F = \frac{1.05 \times 10^{-3}}{5.833 \times 10^{-6}} \approx 180 \, \text{N} \] 5. Final Answer: - The load applied is: \[ \boxed{180 \, \text{N}} \] This corresponds to option (1).