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Current Question (ID: 20157)

Question:
$\text{A square loop of resistance } 16 \, \Omega \text{ is connected with a battery of } 9 \, \text{V}$ $\text{and an internal resistance of } 1 \, \Omega. \text{ In steady state, find the energy}$ $\text{stored in a capacitor of capacity } C = 4 \, \mu \text{F} \text{ as shown. (at steady}$ $\text{state current divides symmetrically)}$
Options:
  • 1. $51.84 \, \mu \text{J}$
  • 2. $12.96 \, \mu \text{J}$
  • 3. $25.92 \, \mu \text{J}$
  • 4. $103.68 \, \mu \text{J}$
Solution:
$\text{Hint: } U = \frac{1}{2} C V^2$ $\text{Equivalent circuit}$ $i = \frac{9}{4 + 1} = 1.8 \, \text{A}$ $\Rightarrow \frac{i}{2} = 0.9 \, \text{A}$ $(V_P - V_Q) = 0.9 \times 6 - 0.9 \times 2$ $V_C = 3.6 \, \text{V}$ $U = \frac{1}{2} C V^2 = \frac{1}{2} \times 4 \times 3.6 \times 3.6 \, \mu \text{J}$ $= 25.92 \, \mu \text{J}$

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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
}