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Current Question (ID: 16598)

Question:
$\text{A parallel plate capacitor of capacitance } C \text{ is connected to a battery and is charged to a potential difference } V. \text{ Another capacitor of capacitance } 2C \text{ is connected to another battery and is charged to potential difference } 2V. \text{ The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is?}$
Options:
  • 1. $\text{zero}$
  • 2. $\frac{25CV^2}{6}$
  • 3. $\frac{3CV^2}{2}$
  • 4. $\frac{9CV^2}{2}$
Solution:
$\text{(3) Total charge } = (2C)(2V) + (C)(-V) = 3CV$ $\therefore \text{ Common potential } = \frac{3CV}{3C} = V$ $\therefore \text{ Energy } = \frac{1}{2}(3C)(V)^2 = \frac{3}{2}CV^2$

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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
}