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Current Question (ID: 20567)

Question:

$\text{A circuit connected to an AC source of emf } \varepsilon = \varepsilon_0 \sin(100t) \text{ with } t \text{ in seconds, gives a phase difference of } \frac{\pi}{4} \text{ between the emf } \varepsilon \text{ and current } i. \text{ Which of the following circuits will exhibit this?}$

Options:

1. $\text{RL circuit with } R = 1 \text{ k} \Omega \text{ and } L = 10 \text{ mH}$
2. $\text{RL circuit with } R = 1 \text{ k} \Omega \text{ and } L = 1 \text{ mH}$
3. $\text{RC circuit with } R = 1 \text{ k} \Omega \text{ and } C = 1 \mu \text{F}$
4. $\text{RC circuit with } R = 1 \text{ k} \Omega \text{ and } C = 10 \mu \text{F}$

Solution:

$\text{Hint: } \tan \phi = \frac{X_L - X_C}{R}$ $X_C = R$ $\frac{1}{\omega C} = R$ $\frac{1}{100} = RC$ $R = 10^3 \ \Omega$ $C = 10^{-5} \ \text{F}$

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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
}