Import Question JSON

Current Question (ID: 10274)

Question:

$\text{The value of } M\text{, as shown, for which the rod will be in equilibrium is:}$

Options:

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

Solution:

$\text{To achieve equilibrium, the moments on both sides of the fulcrum must be equal.}$ $\text{Moment = Mass × Distance}$ $\text{Left side moment: } 6 \text{ kg} \times 20 \text{ cm} = 120 \text{ kg} \cdot \text{cm}$ $\text{Right side moment: } M \times 30 \text{ cm}$ $\text{For equilibrium:}$ $120 = M \times 30$ $M = \frac{120}{30}$ $M = 4 \text{ kg}$ $\text{The value of } M \text{ is } 4 \text{ kg.}$

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
}