Question

A copper wire of length 2.4 m and an aluminum wire of length 0.7 m, both having diameter 2 mm, are connected end to end. When stretched by a load, the obtained elongation is found to be 0.6 mm. The applied load is
(Given: Young’s modulus of copper \( Y_C = 1.2 \times 10^{11} \, \text{Nm}^{-2} \), Young’s modulus of aluminum \( Y_A = 0.7 \times 10^{11} \, \text{Nm}^{-2} \))

• A. \( 12\pi \, \text{N} \)
• B. \( 24\pi \, \text{N} \)
• C. \( 20\pi \, \text{N} \)
• D. \( 80\pi \, \text{N} \)

Hint:

For wires connected in series, total elongation is the sum of individual elongations. Use \( \Delta L = \frac{FL}{AY} \) for each segment.

Answer 1

The Correct Option is C

The total elongation is the sum of elongation in copper and aluminum: \[ \Delta L = \Delta L_{\text{Cu}} + \Delta L_{\text{Al}} \] Using the formula for elongation under load: \[ \Delta L = \frac{F L}{A Y} \] Let \( F \) be the applied force, \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \), so: \[ A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \] Now calculate elongations: For copper: \[ \Delta L_{\text{Cu}} = \frac{F \times 2.4}{\pi \times 10^{-6} \times 1.2 \times 10^{11}} = \frac{2.4F}{1.2\pi \times 10^5} \] For aluminum: \[ \Delta L_{\text{Al}} = \frac{F \times 0.7}{\pi \times 10^{-6} \times 0.7 \times 10^{11}} = \frac{0.7F}{0.7\pi \times 10^5} = \frac{F}{\pi \times 10^5} \] Total elongation: \[ \frac{2.4F}{1.2\pi \times 10^5} + \frac{F}{\pi \times 10^5} = \frac{0.6}{1000} = 6 \times 10^{-4} \] Simplify: \[ \frac{2F}{\pi \times 10^5} + \frac{F}{\pi \times 10^5} = \frac{3F}{\pi \times 10^5} \] \[ \frac{3F}{\pi \times 10^5} = 6 \times 10^{-4} \Rightarrow F = \frac{6 \times 10^{-4} \times \pi \times 10^5}{3} \] \[ F = 20\pi \, \text{N} \]