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NCERT Exemplar · Q13

Q.A copper and a steel wire of the same diameter are connected end to end. A deforming force FF is applied to this composite wire which causes a total elongation of 1cm. The two wires will have (Note: more than one of the given options may be correct.)

(a) the same stress.
(b) different stress.
(c) the same strain.
(d) different strain.
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When wires of the same diameter are connected end to end and stretched by the same force, the stress is identical in both, but the strain differs because Young’s modulus differs for copper and steel. The correct options are (A) and (D).

The key here is to understand what happens when two wires are joined end to end — a series arrangement. A single deforming force FF pulls on the composite wire. Since the wires are in series, the tension (and hence the force) is the same at every cross-section. Both wires have the same diameter, so their cross-sectional areas are equal. Stress is force per unit area: σ=F/A\sigma = F/A. Because FF and AA are the same for both wires, the stress in the copper wire equals the stress in the steel wire. That eliminates option (B) and confirms (A).

Now strain. Strain is the fractional change in length: ϵ=ΔL/L\epsilon = \Delta L / L. For a given stress, the strain depends on the material’s Young’s modulus YY through Hooke’s law: σ=Yϵ\sigma = Y \epsilon, so ϵ=σ/Y\epsilon = \sigma / Y. Since the stress is the same but copper and steel have different Young’s moduli (steel is stiffer, with a larger YY), the strains will be different. Steel, being stiffer, undergoes a smaller strain for the same stress; copper, being more flexible, undergoes a larger strain. Therefore the two wires have different strains, making (D) correct and (C) incorrect.

Let’s walk through it step by step.

  1. Series connection and force transmission. When wires are connected end to end and a force FF is applied at the free end, the tension is the same throughout the entire composite wire. There is no branching, so each wire experiences the same pulling force FF. This is a fundamental property of series connections in mechanics.

  2. Same diameter → same area. The problem states both wires have the same diameter. Therefore their cross-sectional areas AA are equal. Stress is defined as σ=F/A\sigma = F/A. With FF and AA identical, the stress in the copper wire equals the stress in the steel wire. So option (A) is correct, and (B) is wrong. …

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