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Current Question (ID: 16574)

Question:

$\text{A parallel plate capacitor has capacitance } C. \text{ If it is equally filled with parallel layers of materials of dielectric constants } K_1 \text{ and } K_2, \text{ its capacity becomes } C_1. \text{ The ratio of } C_1 \text{ to } C \text{ is:}$

Options:

1. $K_1 + K_2$
2. $\frac{K_1 K_2}{K_1 - K_2}$
3. $\frac{K_1 + K_2}{K_1 K_2}$
4. $\frac{2 K_1 K_2}{K_1 + K_2}$

Solution:

$C_A = \frac{K_1 \varepsilon_0 A}{d/2}, \ C_B = \frac{K_2 \varepsilon_0 A}{d/2}$ $\therefore C_{eq} = \frac{C_A C_B}{C_A + C_B} = \left( \frac{2 K_1 K_2}{K_1 + K_2} \right) \frac{\varepsilon_0 A}{d}$

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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
}