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Current Question (ID: 16578)

Question:
$\text{A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is } A \text{ metre}^2 \text{ and the separation is } t \text{ metre. The dielectric constants are } k_1 \text{ and } k_2 \text{ respectively. Its capacitance in farad will be:}$
Options:
  • 1. $\frac{\varepsilon_0 A}{t} \left( k_1 + k_2 \right)$
  • 2. $\frac{\varepsilon_0 A}{t} \frac{\left( k_1 + k_2 \right)}{2}$
  • 3. $\frac{2 \varepsilon_0 A}{t} \left( k_1 + k_2 \right)$
  • 4. $\frac{\varepsilon_0 A}{t} \frac{\left( k_1 - k_2 \right)}{2}$
Solution:
$\text{(2) The two capacitors are in parallel, so: } \mathcal{C} = \frac{\varepsilon_0 A}{2t} \left( k_1 + k_2 \right)$

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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
}