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Current Question (ID: 20166)

Question:
$\text{The equivalent capacitance between } A \text{ and } B \text{ in the circuit given below is:}$ $\begin{array}{c} \text{A} \quad \begin{array}{c} 6 \, \mu\text{F} \\ 5 \, \mu\text{F} \end{array} \quad \begin{array}{c} 2 \, \mu\text{F} \\ 5 \, \mu\text{F} \end{array} \quad \begin{array}{c} 4 \, \mu\text{F} \\ 2 \, \mu\text{F} \end{array} \quad \text{B} \end{array}$
Options:
  • 1. $3.6 \, \mu\text{F}$
  • 2. $2.4 \, \mu\text{F}$
  • 3. $4.9 \, \mu\text{F}$
  • 4. $5.4 \, \mu\text{F}$
Solution:
$\text{Hint: } C_{||} = C_1 + C_2 \text{ and } \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}$

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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
}