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Current Question (ID: 10417)

Question:

$\text{A material has Poisson's ratio of 0.5. If a uniform rod made of it suffers a longitudinal strain of } 2 \times 10^{-3}, \text{ what is the percentage increase in volume?}$

Options:

1. $2\%$
2. $4\%$
3. $0\%$
4. $5\%$

Solution:

\text{To calculate the percentage increase in volume, we use the formula for volumetric strain:} \frac{\Delta V}{V} = \epsilon_L(1 - 2\nu) \text{Given Poisson's ratio } \nu = 0.5 \text{ and longitudinal strain } \epsilon_L = 2 \times 10^{-3} \text{Substituting the values:} \frac{\Delta V}{V} = (2 \times 10^{-3})(1 - 2 \times 0.5) \frac{\Delta V}{V} = (2 \times 10^{-3})(1 - 1) \frac{\Delta V}{V} = (2 \times 10^{-3}) \times 0 = 0 \text{The percentage increase in volume is } 0\% \text{The final answer is } 0

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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
}