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

Given:

Two wires joined end to end.

Length of each wire = L
Radius of each wire = r
Young’s modulus of first wire = Y₁
Young’s modulus of second wire = Y₂

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Step 1: Elongation of each wire

Cross-sectional area, A = πr²

For a tensile force F:

ΔL₁ = FL / (A Y₁)

ΔL₂ = FL / (A Y₂)

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Step 2: Total elongation

ΔL = ΔL₁ + ΔL₂

ΔL = FL/(A Y₁) + FL/(A Y₂)

ΔL = FL (Y₁ + Y₂) / (A Y₁Y₂)

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Step 3: Define effective Young’s modulus

Total length = 2L

Y = F(2L) / (A ΔL)

Substitute ΔL:

Y = F(2L) / [A × FL(Y₁ + Y₂) / (A Y₁Y₂)]

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Step 4: Simplify

Y = 2Y₁Y₂ / (Y₁ + Y₂)

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Final Answer:

Effective Young’s modulus,
Y = 2Y₁Y₂ / (Y₁ + Y₂)