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Current Question (ID: 10400)

Question:

$\text{The area of cross-section of a wire of length 1.1 m is } 1 \text{ mm}^2. \text{ It is loaded with mass of 1 kg. If Young's modulus of copper is } 1.1 \times 10^{11} \text{ N/m}^2, \text{ then the increase in length will be: (If } g = 10 \text{ m/s}^2)$

Options:

1. $0.01 \text{ mm}$
2. $0.075 \text{ mm}$
3. $0.1 \text{ mm}$
4. $0.15 \text{ mm}$

Solution:

\text{Using the formula for extension: } l = mgL/(AY) \text{Where:} m = 1 \text{ kg} g = 10 \text{ m/s}^2 L = 1.1 \text{ m} A = 1 \text{ mm}^2 = 1 \times 10^{-6} \text{ m}^2 Y = 1.1 \times 10^{11} \text{ N/m}^2 l = (1 \times 10 \times 1.1)/(1.1 \times 10^{11} \times 10^{-6}) = 0.1 \text{ mm}

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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
}