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Current Question (ID: 17331)

Question:

$\text{In a circuit, } L, C \text{ and } R \text{ are connected in series with an alternating voltage source of frequency } f. \text{ The current leads the voltage by } 45^\circ. \text{ The value of } C \text{ will be:}$

Options:

1. $\frac{1}{2\pi f \left(2\pi f L + R\right)}$
2. $\frac{1}{\pi f \left(2\pi f L + R\right)}$
3. $\frac{1}{2\pi f \left(2\pi f L - R\right)}$
4. $\frac{1}{\pi f \left(2\pi f L - R\right)}$

Solution:

$\text{(1) } \tan \phi = \frac{X_C - X_L}{R}$ $\Rightarrow \tan 45^\circ = \frac{\frac{1}{2\pi f C} - 2\pi f L}{R}$ $\Rightarrow C = \frac{1}{2\pi f \left(2\pi f L + R\right)}$

Import JSON File

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
}