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Current Question (ID: 19150)

Question:
$\text{In the arrangement shown in figure } a_1, a_2, a_3 \text{ and } a_4 \text{ are the accelerations of masses } m_1, m_2, m_3 \text{ and } m_4 \text{ respectively. Which of the following relation is true for this arrangement?}$
Options:
  • 1. $4a_1 + 2a_2 + a_3 + a_4 = 0$
  • 2. $a_1 + 4a_2 + 3a_3 + a_4 = 0$
  • 3. $a_1 + 4a_2 + 3a_3 + 2a_4 = 0$
  • 4. $2a_1 + 2a_2 + 3a_3 + a_4 = 0$
Solution:
$\text{Hint: } \Sigma \vec{T} \cdot \vec{a} = 0$

Import JSON File

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
}