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Current Question (ID: 20122)

Question:

$A \ 10 \, \mu\text{F} \text{ capacitor is fully charged to a potential difference of } 50 \, \text{V.}$ $\text{After disconnecting the voltage source, it is connected in parallel with an initially uncharged capacitor.}$ $\text{The potential difference across both capacitors becomes } 20 \, \text{V.}$ $\text{The capacitance of the second capacitor is:}$

Options:

1. $10 \, \mu\text{F}$
2. $15 \, \mu\text{F}$
3. $20 \, \mu\text{F}$
4. $30 \, \mu\text{F}$

Solution:

$\text{Hint: Charge is conserved.}$ $\text{Step: Find the capacitance of the second capacitor.}$ $\text{Initially, the capacitor is fully charged, then the charge on the capacitor is given by;} \ Q = CV = 10 \times 10^{-6} \times 50$ $\Rightarrow Q = 500 \times 10^{-6} = 500 \, \mu\text{C}$ $\text{The final charge on the capacitor is given by;} \ Q = CV = 10 \times 10^{-6} \times 20$ $\Rightarrow Q = 200 \times 10^{-6} = 200 \, \mu\text{C}$ $\text{From charge conservation, the charge on the unknown capacitor is given by;} \ C = 500 \, \mu\text{C} - 200 \, \mu\text{C} = 300 \, \mu\text{C}$ $\text{The capacitance of the second capacitor is given by;} \ C = \frac{Q}{V} = \frac{300 \, \mu\text{C}}{20} = 15 \, \mu\text{F}$ $\text{Hence, option (2) is the correct answer.}$

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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
}