Question

A copper wire of length \(2.4\,\text{m}\) and an aluminum wire of length \(0.7\,\text{m}\), both having diameter \(2\,\text{mm}\), are connected end to end. When stretched by a load, the obtained elongation is found to be \(0.6\,\text{mm}\). The applied load is \((Y_{\text{Cu}}=1.2\times10^{11}\,\text{N/m}^2,\ Y_{\text{Al}}=0.7\times10^{11}\,\text{N/m}^2)\):

• 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, the same force acts through each wire and total elongation is the sum of individual elongations.

Answer 1

The Correct Option is C

Step 1: Model each wire as a spring.
For an elastic wire, $F = k\Delta l$ where the stiffness $k = \frac{AY}{L}$.
When two wires are joined end to end (in series), the same force acts through both.

Step 2: Calculate the common cross-sectional area.
Both wires have diameter $d = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}$, so radius $r = 10^{-3}\,\text{m}$:
\[ A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6}\,\text{m}^2 \]

Step 3: Calculate the stiffness of each wire.
For copper:
\[ k_1 = \frac{AY_1}{L_1} = \frac{\pi \times 10^{-6} \times 1.2 \times 10^{11}}{2.4} = \pi \times 10^{-6} \times 0.5 \times 10^{11} = 5\pi \times 10^4\,\text{N/m} \] For aluminium:
\[ k_2 = \frac{AY_2}{L_2} = \frac{\pi \times 10^{-6} \times 0.7 \times 10^{11}}{0.7} = \pi \times 10^{-6} \times 10^{11} = \pi \times 10^5\,\text{N/m} \]

Step 4: Find the effective stiffness for series combination.
For springs in series: $\frac{1}{k_\text{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$:
\[ \frac{1}{k_\text{eff}} = \frac{1}{5\pi \times 10^4} + \frac{1}{\pi \times 10^5} = \frac{1}{\pi \times 10^4}\left(\frac{1}{5} + \frac{1}{10}\right) = \frac{3}{10\pi \times 10^4} \] \[ k_\text{eff} = \frac{10\pi \times 10^4}{3}\,\text{N/m} \]

Step 5: Use $F = k_\text{eff} \Delta l$ to find the load.
Total elongation $\Delta l = 0.6\,\text{mm} = 6 \times 10^{-4}\,\text{m}$:
\[ F = k_\text{eff} \times \Delta l = \frac{10\pi \times 10^4}{3} \times 6 \times 10^{-4} = \frac{10\pi \times 6 \times 10^0}{3} = 20\pi\,\text{N} \]

Step 6: State the answer.
\[ \boxed{20\pi\,\text{N}} \]