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Current Question (ID: 16550)

Question:

$\text{Two capacitors of capacitance } 6 \, \mu\text{F and } 3 \, \mu\text{F are connected in series with a battery of } 30 \, \text{V.}$ $\text{The charge on } 3 \, \mu\text{F capacitor is:}$

Options:

1. $3 \, \mu\text{C}$
2. $1.5 \, \mu\text{C}$
3. $60 \, \mu\text{C}$
4. $900 \, \mu\text{C}$

Solution:

$\text{Hint: The charge on each of the capacitors connected in the series is the same and it is equal to the charge on its equivalent capacitor.}$ $\text{Step 1: Calculation of equivalent capacitance.}$ $\frac{1}{C_{\text{eq}}} = \frac{1}{6} + \frac{1}{3}$ $C_{\text{eq}} = 2 \, \mu\text{F}$ $\text{Step 2: Calculate the charge.}$ $Q = C_{\text{eq}} V$ $= 2 \times 10^{-6} \times 30$ $= 60 \, \mu\text{C}.$

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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
}