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Current Question (ID: 16588)

Question:

$\text{Five equal capacitors connected in series have a resultant capacitance of } 4 \, \mu\text{F.}$ $\text{The total energy stored in these when these are connected in parallel and charged to } 400 \, \text{V is:}$

Options:

1. $1 \, \text{J}$
2. $8 \, \text{J}$
3. $16 \, \text{J}$
4. $4 \, \text{J}$

Solution:

$\text{Hint: Energy stored, } E = \frac{1}{2} C_{\text{eq}} V^2.$ $\text{Step 1: Find the capacitance.}$ $\text{For series combination-}$ $C_{\text{eq}} = \frac{C}{5}$ $C = 20 \, \mu\text{F}$ $\text{For parallel,}$ $C_{\text{eq}} = nC$ $= 100 \, \mu\text{F}$ $\text{Step 2: Use the relation, to find the total energy stored.}$ $E = \frac{1}{2} C_{\text{eq}} V^2$ $= \frac{1}{2} \times 100 \times 10^{-6} \times (400)^2 = 8 \, \text{J}$

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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
}