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Current Question (ID: 10422)

Question:
$\text{The work done per unit volume to stretch the length of a wire by 1\% with a constant cross-sectional area will be:}$ $(Y = 9 \times 10^{11} \text{ N/m}^2)$
Options:
  • 1. $9 \times 10^{11} \text{ J}$
  • 2. $4.5 \times 10^{7} \text{ J}$
  • 3. $9 \times 10^{7} \text{ J}$
  • 4. $4.5 \times 10^{11} \text{ J}$
Solution:
\text{The work done per unit volume to stretch the wire is given by the formula } U_v = \frac{1}{2}Y\text{(strain)}^2 \text{Given } Y = 9 \times 10^{11} \text{ N/m}^2 \text{ and strain = 1\% = 0.01} \text{Substituting the values:} U_v = \frac{1}{2} \times (9 \times 10^{11}) \times (0.01)^2 = \frac{1}{2} \times 9 \times 10^{11} \times 10^{-4} = 4.5 \times 10^{7} \text{ J/m}^3 \text{The correct option is 2.}

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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
}