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Current Question (ID: 20171)

Question:
$\text{A capacitor with capacitance } 5 \, \mu\text{F is charged to } 5 \, \mu\text{C. If the plates are pulled apart to reduce the capacitance to } 2 \, \mu\text{F, how much work is done?}$
Options:
  • 1. $3.75 \times 10^{-6} \, \text{J}$
  • 2. $6.25 \times 10^{-6} \, \text{J}$
  • 3. $2.55 \times 10^{-6} \, \text{J}$
  • 4. $2.16 \times 10^{-6} \, \text{J}$
Solution:
$\text{Hint: Work done } = U_f - U_i$ $\text{Work } = \frac{q^2}{2C_2} - \frac{q^2}{2C_1}$ $= \frac{(5 \times 10^{-6})^2}{2} \left( \frac{1}{2 \times 10^{-6}} - \frac{1}{1 \times 10^5} \right)$ $= \frac{15}{4} \times 10^{-6}$ $= 3.75 \times 10^{-6} \, \text{J}$

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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
}