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Current Question (ID: 16561)

Question:
$\text{Two capacitors of capacity } 2 \, \mu\text{F} \text{ and } 3 \, \mu\text{F} \text{ are charged to the same potential difference of } 6 \, \text{V.}$ $\text{Now they are connected with opposite polarity as shown.}$ $\text{After closing switches } S_1 \text{ and } S_2, \text{ their final potential difference becomes:}$
Options:
  • 1. $\text{zero}$
  • 2. $\frac{4}{3} \, \text{V}$
  • 3. $3 \, \text{V}$
  • 4. $\frac{6}{5} \, \text{V}$
Solution:
$\text{Hint: The final potential of each capacitor is equal.}$ $\text{Step 1: Find the initial charges on the capacitors.}$ $\text{Initially, } Q = CV$ $q = 18 - 12 = 6 \, \mu\text{C}$ $\text{Step 2: Find the final charges on the capacitors.}$ $\frac{Q_1}{3} = \frac{6 - Q_1}{2}$ $Q_1 = \frac{18}{5} \, \mu\text{C}$ $\text{Step 3: Find the final potential difference across the capacitors.}$ $V = \frac{Q_1}{3}$ $\Rightarrow V = \frac{6}{5} \, \text{V}$ $\text{Hence, option (4) is the correct answer.}$

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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
}