9. An iron bar and a brass bar of equal length and cross-sectional area, are joined at
one end. A tensile force is applied to the free ends as shown.
If the Young’s modulus of iron is twice that of brass, then which is the following is
correct?
a. The extension of brass is twice that of iron
b. The extension of iron is twice that of brass
c. The stress of brass is twice that is iron
d. The stress of iron is twice that of brass

Comparison of Extension and Stress in Iron and Brass Bars

Given Data

Two bars — one made of iron and the other of brass — have:

• Equal length (L)
• Equal cross-sectional area (A)
• Are joined end to end (in series)
• A tensile force (F) applied at the free ends

Young’s modulus relation:

Y_Fe = 2 Y_Br

Formula for Extension under Axial Load

ΔL = (F × L) / (A × Y)

For Each Bar

Extension in Iron:

ΔL_Fe = (F × L) / (A × Y_Fe)

Extension in Brass:

ΔL_Br = (F × L) / (A × Y_Br)

Given Y_Fe = 2Y_Br, therefore:

ΔL_Br = (F × L) / [A × (Y_Fe/2)] = 2 × (F × L) / (A × Y_Fe) = 2ΔL_Fe

Hence, the extension of brass is twice that of iron.

Stress and Strain

Stress for each bar:

σ = F / A

Since both bars carry the same force and have the same area:

σ_Fe = σ_Br

Strain:

ε = ΔL / L

Because L is the same for both, the strain ratio equals the extension ratio:

ε_Br = 2 ε_Fe

Evaluation of Options

• Option A: “The extension of brass is twice that of iron” — Correct.
• Option B: “The extension of iron is twice that of brass” — Incorrect.
• Option C: “The stress of brass is twice that of iron” — Incorrect.
• Option D: “The stress of iron is twice that of brass” — Incorrect.

Key Points to Remember

• For bars in series: force is the same.
• If cross-sectional areas are equal, stresses are equal.
• Extension ΔL ∝ 1/Y (when F, L, and A are constant).
• Lower Young’s modulus → greater extension and greater strain under the same stress.
• In this problem, brass (with half the modulus of iron) has double the extension and strain.

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