Import Question JSON

Current Question (ID: 10154)

Question:

$\text{The value of } M, \text{ as shown, for which the rod will be in equilibrium is:}$ $\text{[A rod is supported at a fulcrum point, with a 6 kg mass at 20 cm from the fulcrum on the left side, and mass } M \text{ at 30 cm from the fulcrum on the right side]}$

Options:

1. $1 \text{ kg}$
2. $2 \text{ kg}$
3. $4 \text{ kg}$
4. $6 \text{ kg}$

Solution:

$\text{For equilibrium, balance torques:}$ $6 \text{ kg} \times 20 \text{ cm} = M \times 30 \text{ cm}$ $120 = 30M$ $M = \frac{120}{30} = 4 \text{ kg}$ $\text{The final answer is 4 kg.}$

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
}