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Current Question (ID: 20187)

Question:
$\text{A parallel plate capacitor with a width of } 4 \text{ cm, length of } 8 \text{ cm and separation between the plates of } 4 \text{ mm is connected to a battery of } 20 \text{ V. A dielectric slab of dielectric constant } 5 \text{ having a length of } 1 \text{ cm, width of } 4 \text{ cm and thickness of } 4 \text{ mm is inserted between the plates of a parallel plate capacitor. The electrostatic energy of this system will be:}$ $\text{(where } \varepsilon_0 \text{ is the permittivity of free space)}$
Options:
  • 1. $120 \varepsilon_0 \text{ J}$
  • 2. $140 \varepsilon_0 \text{ J}$
  • 3. $240 \varepsilon_0 \text{ J}$
  • 4. $260 \varepsilon_0 \text{ J}$
Solution:
$\text{Hint: } E = \frac{1}{2} C_{\text{eff}} V^2$

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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
}