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Current Question (ID: 20143)

Question:

$\text{Two parallel plate capacitors, having capacitances of } C \text{ and } 3C, \text{ are connected in parallel and charged to a potential difference of } 18 \text{ V.}$ $\text{After the charging process, the battery is disconnected, and the space between the capacitor plates with capacitance } C \text{ is completely filled with a dielectric material of dielectric constant } 9. \text{ What will be the final potential difference across the combination of capacitors?}$

Options:

1. $3 \text{ V}$
2. $5 \text{ V}$
3. $9 \text{ V}$
4. $6 \text{ V}$

Solution:

$\text{Hint: } Q = CV$ $\text{Step 1: Find the initial charge on the capacitors.}$ $\text{The initial charge on the capacitors is given by:}$ $\Rightarrow Q = CV$ $\Rightarrow Q = 18C + (18 \times 3)C$ $\Rightarrow Q = 72C$ $\text{The total charge is conserved.}$ $\text{Step 2: Find the final potential difference across the combination of the capacitors.}$ $\text{The final potential difference across the combination of the capacitors is given by:}$ $\Rightarrow V_{\text{common}} = \frac{q_{\text{net}}}{C_{\text{net}}}$ $\Rightarrow V_{\text{common}} = \frac{(18 \times 4)C}{9C + 3C} = 6 \text{ V}$ $\text{Hence, option (4) is the correct answer.}$

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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
}