Import Question JSON

Current Question (ID: 20153)

Question:
$\text{A } 2 \, \mu\text{F capacitor is initially charged to a potential } V, \text{ storing an energy } E_1. \text{ The capacitor is then disconnected from the battery and connected in parallel with an identical uncharged capacitor. After the redistribution of charge, the new energy stored in the system is } E_2.$ $\text{What is the ratio } \frac{E_1}{E_2}?$
Options:
  • 1. $2$
  • 2. $4$
  • 3. $5$
  • 4. $6$
Solution:
$\text{Hint: } E = \frac{Q^2}{2C}$ $E_1 = \frac{Q^2}{2C}$ $E_2 = \frac{(Q/2)^2}{2C}$ $\therefore \frac{E_1}{E_2} = 4$ $\text{Option (2) is correct.}$

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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
}