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Current Question (ID: 10377)

Question:

$\text{A uniform wire is held at one end and stretched by applying force at the other end. If the product of the length of wire and area of cross-section of the wire remains unchanged, then Poisson's ratio of material of the wire will be:}$

Options:

1. $0$
2. $0.50$
3. $-0.5$
4. $\text{Infinity}$

Solution:

\text{The volume of the wire, } V = \pi r^2 l \text{Differentiating both sides, we get,} \Delta V = \pi (2r \Delta r)l + \pi r^2 \Delta l \text{As the volume of wire remains unchanged so, } \Delta V = 0 \text{Hence,} 0 = 2\pi rl \Delta r + \pi r^2 \Delta l \Rightarrow \frac{\Delta r}{r} \div \frac{\Delta l}{l} = -\frac{1}{2} \text{We know that,} \text{Poisson's ratio} = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{-\Delta r/r}{\Delta l/l} = \frac{1}{2} = 0.50

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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
}