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Current Question (ID: 16571)

Question:

$\text{The capacitance of a parallel plate capacitor is } C. \text{ If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant } K \text{ is inserted between the plates, then the new capacitance will be:}$

Options:

1. $\frac{KC}{2(K+1)}$
2. $\frac{2KC}{K+1}$
3. $\frac{5KC}{4K+1}$
4. $\frac{4KC}{3K+1}$

Solution:

$C_1 = \frac{k \varepsilon_0 A}{d} \times 4$ $C_2 = \frac{k \varepsilon_0 A}{3d} \times 4$ $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ $C_{eq} = \frac{4KC}{3K+1}$

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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
}