Q129 / 180 · COMEDK 2022 · Physics
A copper and a steel wire of same diameter are connected end to end. A deforming force $F_1$ is applied to the wire which causes an elongation of $1 \mathrm{~cm}$. The two wires will have
- Athe same strain
- Bdifferent stress
- Correct answerCdifferent strain
- DNone of these
Solution
Let the copper wire have Young's modulus $Y_C$ and the steel wire have Young's modulus $Y_S$. Both wires have the same diameter, so they have the same cross-sectional area $A$.
When connected end to end and subjected to a deforming force $F_1$, the tension in both wires is the same, equal to $F_1$.
The stress in each wire is given by $\sigma = \dfrac{F_1}{A}$. Since $F_1$ and $A$ are the same for both wires, the stress is the same in both wires.
The strain in each wire is given by $\epsilon = \dfrac{\sigma}{Y}$. Since the Young's moduli $Y_C$ and $Y_S$ are different for copper and steel, the strains $\epsilon_C = \dfrac{F_1}{A Y_C}$ and $\epsilon_S = \dfrac{F_1}{A Y_S}$ will be different.
Therefore, the two wires will have different strains.
Answer: different strain
When connected end to end and subjected to a deforming force $F_1$, the tension in both wires is the same, equal to $F_1$.
The stress in each wire is given by $\sigma = \dfrac{F_1}{A}$. Since $F_1$ and $A$ are the same for both wires, the stress is the same in both wires.
The strain in each wire is given by $\epsilon = \dfrac{\sigma}{Y}$. Since the Young's moduli $Y_C$ and $Y_S$ are different for copper and steel, the strains $\epsilon_C = \dfrac{F_1}{A Y_C}$ and $\epsilon_S = \dfrac{F_1}{A Y_S}$ will be different.
Therefore, the two wires will have different strains.
Answer: different strain
Hence, the correct answer is option: 'C'.