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Current Question (ID: 16562)

Question:
$\text{In the circuit diagram shown all the capacitors are in } \mu F. \text{ The equivalent}$ $\text{capacitance between points, } A \text{ and } B \text{ is (in } \mu F):$
Options:
  • 1. $\frac{14}{5}$
  • 2. $7.5$
  • 3. $\frac{3}{7}$
  • 4. $\text{None of these}$
Solution:
$\text{The circuit forms a balanced Wheatstone bridge.}$ $\text{The equivalent capacitance is given by:}$ $C_{eq} = \frac{14}{5} \mu F$

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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
}