Question

Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are $Y_1$ and $Y_2$. The combination behaves as a single wire then its Young's modulus is :

• A. $Y = \frac{Y_1 Y_2}{Y_1 + Y_2}$
• B. $Y = \frac{Y_1 Y_2}{2(Y_1 + Y_2)}$
• C. $Y = \frac{2 Y_1 Y_2}{Y_1 + Y_2}$
• D. $Y = \frac{2 Y_1 Y_2}{3(Y_1 + Y_2)}$

Hint:

Equivalent Young's modulus for equal length wires in series is the Harmonic Mean of their individual moduli. This is perfectly analogous to resistors in series if you consider the 'stiffness' of the wires.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When wires are joined end-to-end, they are in series. In a series combination, the tension (\(F\)) in each wire is the same. The total elongation is the sum of the individual elongations.
Step 2: Key Formula or Approach:
1. Young's Modulus: \(Y = \frac{FL}{A \Delta L} \implies \Delta L = \frac{FL}{AY}\).
2. Total elongation: \(\Delta L_{total} = \Delta L_1 + \Delta L_2\).
3. Equivalent wire parameters: Length = \(2L\), Area = \(A\).
Step 3: Detailed Explanation:
For the combination:
\[ \Delta L_{eq} = \frac{F(2L)}{AY_{eq}} \]
For the individual wires:
\[ \Delta L_1 = \frac{FL}{AY_1} \text{ and } \Delta L_2 = \frac{FL}{AY_2} \]
Equating the total extension:
\[ \frac{2FL}{AY_{eq}} = \frac{FL}{AY_1} + \frac{FL}{AY_2} \]
Dividing by \(\frac{FL}{A}\):
\[ \frac{2}{Y_{eq}} = \frac{1}{Y_1} + \frac{1}{Y_2} \implies \frac{2}{Y_{eq}} = \frac{Y_1 + Y_2}{Y_1 Y_2} \]
\[ Y_{eq} = \frac{2 Y_1 Y_2}{Y_1 + Y_2} \]
Step 4: Final Answer:
The equivalent Young's modulus of the series combination is \(\frac{2 Y_1 Y_2}{Y_1 + Y_2}\).