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Current Question (ID: 20112)

Question:

$\text{The parallel combination of two air filled parallel plate capacitors of capacitance } C \text{ and } nC \text{ is connected to a battery of voltage, } V. \text{ When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant } K \text{ is placed between the two plates of the first capacitors. The new potential difference of the combined system is:}$

Options:

1. $\frac{V}{K+n}$
2. $\frac{nV}{K+n}$
3. $\frac{(n+1)V}{(K+n)}$
4. $V$

Solution:

$\text{Hint: } V_C = \frac{q_{\text{total}}}{C_{\text{eff}}}$ $\text{After full charging}$ $q = CV + nCV = (n+1)CV$ $\text{Due to insertion of dielectric}$ $q_1 = KCV_C, \quad q_2 = nCV_C$ $V_C = \frac{q_{\text{total}}}{C_{\text{eff}}} = \frac{(n+1)CV}{KC + nC} = \frac{(n+1)}{K+n}V$

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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
}