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Current Question (ID: 16553)

Question:
$\text{The equivalent capacitance between points } a \text{ and } b \text{ in the network shown below is:}$
Options:
  • 1. $5 \text{ C}$
  • 2. $4 \text{ C}$
  • 3. $3 \text{ C}$
  • 4. $2 \text{ C}$
Solution:
$\text{The capacitors are arranged in a combination of series and parallel.}$ $\text{The two capacitors in series at the top and bottom each have an equivalent capacitance of } \frac{C}{2}. $ $\text{These are in parallel with the middle capacitor, giving:}$ $C_{eq} = \frac{C}{2} + C + \frac{C}{2} = 2C. $

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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
}