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Current Question (ID: 20586)

Question:
$\text{In a series } LR \text{ circuit } X_L = R \text{ and power factor of the circuit is } P_1. \text{ When capacitor with capacitance } C \text{ such that } X_L = X_C \text{ is put in series, the power factor becomes } P_2. \text{ The ratio } \frac{P_1}{P_2} \text{ is:}$
Options:
  • 1. $\frac{1}{2}$
  • 2. $\frac{1}{\sqrt{2}}$
  • 3. $\frac{\sqrt{3}}{\sqrt{2}}$
  • 4. $2:1$
Solution:
$\text{Hint: } P = \frac{R}{Z}$

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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
}