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Current Question (ID: 17315)

Question:

$\text{An } AC \text{ voltage is applied to a resistance } R \text{ and an inductor } L \text{ in series. If } R \text{ and the inductive reactance are both equal to } 3 \, \Omega, \text{ then the phase difference between the applied voltage and the current in the circuit will be:}$

Options:

1. $\pi/4$
2. $\pi/2$
3. $\text{zero}$
4. $\pi/6$

Solution:

$\tan \phi = \frac{X_L}{R} = \frac{L \omega}{R}$ $\tan \phi = \frac{3 \Omega}{3 \Omega}$ $\tan \phi = 1$ $\phi = \tan^{-1}(1)$ $\phi = 45^\circ$ $\phi = \frac{\pi}{4} \, \text{rad}$

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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
}