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Current Question (ID: 17353)

Question:
$\text{The power factor of the given circuit is:}$
Options:
  • 1. $\frac{1}{2}$
  • 2. $\frac{1}{\sqrt{2}}$
  • 3. $\frac{\sqrt{3}}{2}$
  • 4. $0$
Solution:
$\text{Step 1: Find net voltage in the circuit}$ $V_N = \sqrt{V_R^2 + (V_L - V_C)^2}$ $= \sqrt{2^2 + (6 - 4)^2}$ $= 2\sqrt{2}$ $\text{Step 2: Express resistances in terms of current}$ $R = \frac{2}{I}$ $Z = \frac{V_N}{I} = \frac{2\sqrt{2}}{I}$ $\text{Step 3: Find power factor}$ $\cos \phi = \frac{R}{Z}$ $= \frac{1}{\sqrt{2}}$ $\text{NCERT Reference: Class XII, part 1, pg 247}$ $\text{Hint: Find net voltage in the circuit}$

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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
}