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Current Question (ID: 16592)

Question:

$\text{Two condensers of capacity } 0.3 \, \mu\text{F} \text{ and } 0.6 \, \mu\text{F} \text{ are connected in series.}$ $\text{The combination is connected across a potential of } 6 \, \text{V.}$ $\text{The ratio of energies stored by the condensers will be:}$

Options:

1. $\frac{1}{2}$
2. $2$
3. $\frac{1}{4}$
4. $4$

Solution:

$\text{When capacitors are connected in series, the charge } Q \text{ on each is the same.}$ $\text{Energy stored in a capacitor is given by } E = \frac{1}{2} C V^2.$ $\text{For capacitors in series:}$ $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ $\Rightarrow C_{eq} = \frac{0.3 \times 0.6}{0.3 + 0.6} = 0.2 \, \mu\text{F}$ $\text{Energy ratio: } \frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{0.6}{0.3} = 2.$

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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
}