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Current Question (ID: 10442)

Question:
When a man sits on a boat of length $2 \text{ m}$ and breadth $1 \text{ m}$ floating on a lake, the boat sinks by $2 \text{ cm}$. The mass of the man is:
Options:
  • 1. $25 \text{ kg}$
  • 2. $40 \text{ kg}$
  • 3. $60 \text{ kg}$
  • 4. $80 \text{ kg}$
Solution:
\text{Hint: Mass of man = volume of water displaced × density of water} \text{Step 1: Calculate the mass.} mg = \rho V_{\text{extra}} g m = \rho V_{\text{extra}} V_{\text{extra}} = \text{length} \times \text{breadth} \times \text{sinking depth} \times \text{conversion to meters} V_{\text{extra}} = 2\text{ m} \times 1\text{ m} \times 2\text{ cm} = 2 \times 1 \times 0.02\text{ m}^3 = 0.04\text{ m}^3 \rho = 1000\text{ kg/m}^3 m = 1000\text{ kg/m}^3 \times 0.04\text{ m}^3 = 40\text{ kg} \text{Therefore, the mass of the man is } 40\text{ kg}.

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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
}