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Current Question (ID: 10382)

Question:

A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of $\text{1 cm}$. The two wires will have:

Options:

1. (a) the same stress (b) different stress
2. (a) (d)
3. (b), (c)
4. (c), (d)

Solution:

Hint: The tension or compression in the wires will be the same. Step 1: Find the stress in both the wires. Consider the diagram where a deforming force $F$ is applied to the combination. $\text{For steel wire, } Y_{\text{steel}} = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{strain}_{\text{steel}}}$ $\text{For copper wire, } Y_{\text{copper}} = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{strain}_{\text{copper}}}$ $\text{where } F \text{ is tension in each wire and } A \text{ is the cross-section area of each wire.}$ $\text{As } F \text{ and } A \text{ are the same for both the wires,}$ $\text{Therefore, stress will be the same for both the wires.}$ $\text{Step 2: Find the strain in the two wires.}$ $\text{The strain in the wire is given by;}$ $\text{Strain} = \frac{\text{Stress}}{Y}$ $\Rightarrow \text{Strain}_{\text{steel}} = \frac{\text{Stress}}{Y_{\text{steel}}}$ $\Rightarrow \text{Strain}_{\text{copper}} = \frac{\text{Stress}}{Y_{\text{copper}}}$ $\text{As } Y_{\text{steel}} \neq Y_{\text{copper}}$ $\text{Therefore, the two wires will have different strains.}$ H$\text{ence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}